Download GATE (Graduate Aptitude Test in Engineering) GATE ECE Question Paper With Solutions
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
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2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
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2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
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2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
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2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
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2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
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2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
FirstRanker.com - FirstRanker's Choice
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GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
25
? ?
0 1 0
A 0 0 1 and C 1 0 0
?1 ?2 ?3
??
??
? = =
??
??
??
44. Same current will flow through both NMOS & PMOS
? ID1 = ID2
( ) ( )
2 2 p
n
GSN TN GSP TP
Np
? cox
? cox
? V ? V ? V ? V
2 L 2 L
?? ? ? ? ?
?=
? ? ? ?
? ? ? ?
( ) ( )
22
NP
100 ? 1.5 ? 0.7 400 1.5 ? 0.9
LL
?? ? ? ? ?
? ? = ?
? ? ? ?
? ? ? ?
( )
( )
N
p
/L
94
/ L 16 10
?
? = ?
?
= 0.225
dd
GSN GSP
V
V V 1.5V
2
??
= = =
??
??
45.
22
10,
2
1
2
1, 0
1
0
22
c
V m n
f
ab
For T m n
VV
fc
aa
? ? ? ?
=+
? ? ? ?
? ? ? ?
? = =
??
= = =
??
??
For T ? 11, m = 1, n = 1
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
25
? ?
0 1 0
A 0 0 1 and C 1 0 0
?1 ?2 ?3
??
??
? = =
??
??
??
44. Same current will flow through both NMOS & PMOS
? ID1 = ID2
( ) ( )
2 2 p
n
GSN TN GSP TP
Np
? cox
? cox
? V ? V ? V ? V
2 L 2 L
?? ? ? ? ?
?=
? ? ? ?
? ? ? ?
( ) ( )
22
NP
100 ? 1.5 ? 0.7 400 1.5 ? 0.9
LL
?? ? ? ? ?
? ? = ?
? ? ? ?
? ? ? ?
( )
( )
N
p
/L
94
/ L 16 10
?
? = ?
?
= 0.225
dd
GSN GSP
V
V V 1.5V
2
??
= = =
??
??
45.
22
10,
2
1
2
1, 0
1
0
22
c
V m n
f
ab
For T m n
VV
fc
aa
? ? ? ?
=+
? ? ? ?
? ? ? ?
? = =
??
= = =
??
??
For T ? 11, m = 1, n = 1
26
22 2
1
2
22
22
22
22
2 2 2
22
2
2
11
2
b
1
2
/2 1
2 11
2
11
11
22 11
1
2
4
3
1
3
1
3
3
3 1.732
=+
=
=
+
= ? =
+
+
=
+
? = +
?=
?=
?=
?=
==
c
c
c
V
f
a
f
Given
f
Va
V
ab
aa
ab
ab
ab
b
ab
b a b
ba
b
a
b
a
a
b
width
hight
46.
and y(t) = z(t) + p(t)
?Ryy ( ?) = R zz ( ?) + R pp( ?)+ R pz( ?)+ R zp( ?)
now x(t) & z(t) are uncorrelated.
?Rpz( ?) = R zp( ?) = 0
?R yy( ?) = R zz( ?) + R pp( ?)
So the power spectral relation can be given by Fourier transform of the above relation.
?S yy(f) = S zz(f) + S pp(f)
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GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
25
? ?
0 1 0
A 0 0 1 and C 1 0 0
?1 ?2 ?3
??
??
? = =
??
??
??
44. Same current will flow through both NMOS & PMOS
? ID1 = ID2
( ) ( )
2 2 p
n
GSN TN GSP TP
Np
? cox
? cox
? V ? V ? V ? V
2 L 2 L
?? ? ? ? ?
?=
? ? ? ?
? ? ? ?
( ) ( )
22
NP
100 ? 1.5 ? 0.7 400 1.5 ? 0.9
LL
?? ? ? ? ?
? ? = ?
? ? ? ?
? ? ? ?
( )
( )
N
p
/L
94
/ L 16 10
?
? = ?
?
= 0.225
dd
GSN GSP
V
V V 1.5V
2
??
= = =
??
??
45.
22
10,
2
1
2
1, 0
1
0
22
c
V m n
f
ab
For T m n
VV
fc
aa
? ? ? ?
=+
? ? ? ?
? ? ? ?
? = =
??
= = =
??
??
For T ? 11, m = 1, n = 1
26
22 2
1
2
22
22
22
22
2 2 2
22
2
2
11
2
b
1
2
/2 1
2 11
2
11
11
22 11
1
2
4
3
1
3
1
3
3
3 1.732
=+
=
=
+
= ? =
+
+
=
+
? = +
?=
?=
?=
?=
==
c
c
c
V
f
a
f
Given
f
Va
V
ab
aa
ab
ab
ab
b
ab
b a b
ba
b
a
b
a
a
b
width
hight
46.
and y(t) = z(t) + p(t)
?Ryy ( ?) = R zz ( ?) + R pp( ?)+ R pz( ?)+ R zp( ?)
now x(t) & z(t) are uncorrelated.
?Rpz( ?) = R zp( ?) = 0
?R yy( ?) = R zz( ?) + R pp( ?)
So the power spectral relation can be given by Fourier transform of the above relation.
?S yy(f) = S zz(f) + S pp(f)
27
now power of y(t)=
( )
( ) ( )
( ) ( ) ( )
2
yy
zz pp
pp xx
s f df
P s f df S f df
now S f H w S f
?
?
??
??
? = +
=?
?
??
&
51
10 10 1
42
17.5
P
P watt
? = ? + ? ?
?=
47. For the minimization of the energy in the error signal there are different approaches like,
Prony?s method, Pade approximation. As g(n) has three samples.
Consider them as g(-1) , g(0) , g(1) we can minimise E(h,g) by making h(n) = g(n) using
rectangular window and Parseval?s there of OTFT.
Based on which 10g(-1) + g(1) = 10(-3) + 3
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
25
? ?
0 1 0
A 0 0 1 and C 1 0 0
?1 ?2 ?3
??
??
? = =
??
??
??
44. Same current will flow through both NMOS & PMOS
? ID1 = ID2
( ) ( )
2 2 p
n
GSN TN GSP TP
Np
? cox
? cox
? V ? V ? V ? V
2 L 2 L
?? ? ? ? ?
?=
? ? ? ?
? ? ? ?
( ) ( )
22
NP
100 ? 1.5 ? 0.7 400 1.5 ? 0.9
LL
?? ? ? ? ?
? ? = ?
? ? ? ?
? ? ? ?
( )
( )
N
p
/L
94
/ L 16 10
?
? = ?
?
= 0.225
dd
GSN GSP
V
V V 1.5V
2
??
= = =
??
??
45.
22
10,
2
1
2
1, 0
1
0
22
c
V m n
f
ab
For T m n
VV
fc
aa
? ? ? ?
=+
? ? ? ?
? ? ? ?
? = =
??
= = =
??
??
For T ? 11, m = 1, n = 1
26
22 2
1
2
22
22
22
22
2 2 2
22
2
2
11
2
b
1
2
/2 1
2 11
2
11
11
22 11
1
2
4
3
1
3
1
3
3
3 1.732
=+
=
=
+
= ? =
+
+
=
+
? = +
?=
?=
?=
?=
==
c
c
c
V
f
a
f
Given
f
Va
V
ab
aa
ab
ab
ab
b
ab
b a b
ba
b
a
b
a
a
b
width
hight
46.
and y(t) = z(t) + p(t)
?Ryy ( ?) = R zz ( ?) + R pp( ?)+ R pz( ?)+ R zp( ?)
now x(t) & z(t) are uncorrelated.
?Rpz( ?) = R zp( ?) = 0
?R yy( ?) = R zz( ?) + R pp( ?)
So the power spectral relation can be given by Fourier transform of the above relation.
?S yy(f) = S zz(f) + S pp(f)
27
now power of y(t)=
( )
( ) ( )
( ) ( ) ( )
2
yy
zz pp
pp xx
s f df
P s f df S f df
now S f H w S f
?
?
??
??
? = +
=?
?
??
&
51
10 10 1
42
17.5
P
P watt
? = ? + ? ?
?=
47. For the minimization of the energy in the error signal there are different approaches like,
Prony?s method, Pade approximation. As g(n) has three samples.
Consider them as g(-1) , g(0) , g(1) we can minimise E(h,g) by making h(n) = g(n) using
rectangular window and Parseval?s there of OTFT.
Based on which 10g(-1) + g(1) = 10(-3) + 3
28
= -27
48. I r= 0.75 I 5
? Forward current = I D = - 0.75 I S
? I S(e
vo/nvt
? 1) = ? 0.75 I s
Now Take n = 1
?e
vo/vT
= 0.25
? V D = V T l n (0.25)
? V R = ? V T l n (0.25)
23
19
1.38 10 300
1.386
1.6 10
?
?
??
= ? ? ?
?
? V R = 35.87 mv
49. Given differential equation is of Cauchy ? Euler differential equation type.
So let x = e
z
? z = ln x
The differential equation can be written as,
D (D ?1) ? 3D + 3 = 0
? D
2
? 4 D + 3 = 0
? D = 1, 3
3
12
3
12
? = +
? = +
zz
y C e C e
y C x C x
Now y (1) = 1
? C 1 + C 2 = 1 ?(i)
And y(2) = 14
? 2C 1 + 8C 2 = 14 ?(ii)
From (i) and (ii)
C 1 = ?1, C 2 = 2
? y = ?x + 2x
3
? y(1.5) = ?1.5 + 2(1.5)
3
? y(1.5) = 5.25
50. We know that,
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
25
? ?
0 1 0
A 0 0 1 and C 1 0 0
?1 ?2 ?3
??
??
? = =
??
??
??
44. Same current will flow through both NMOS & PMOS
? ID1 = ID2
( ) ( )
2 2 p
n
GSN TN GSP TP
Np
? cox
? cox
? V ? V ? V ? V
2 L 2 L
?? ? ? ? ?
?=
? ? ? ?
? ? ? ?
( ) ( )
22
NP
100 ? 1.5 ? 0.7 400 1.5 ? 0.9
LL
?? ? ? ? ?
? ? = ?
? ? ? ?
? ? ? ?
( )
( )
N
p
/L
94
/ L 16 10
?
? = ?
?
= 0.225
dd
GSN GSP
V
V V 1.5V
2
??
= = =
??
??
45.
22
10,
2
1
2
1, 0
1
0
22
c
V m n
f
ab
For T m n
VV
fc
aa
? ? ? ?
=+
? ? ? ?
? ? ? ?
? = =
??
= = =
??
??
For T ? 11, m = 1, n = 1
26
22 2
1
2
22
22
22
22
2 2 2
22
2
2
11
2
b
1
2
/2 1
2 11
2
11
11
22 11
1
2
4
3
1
3
1
3
3
3 1.732
=+
=
=
+
= ? =
+
+
=
+
? = +
?=
?=
?=
?=
==
c
c
c
V
f
a
f
Given
f
Va
V
ab
aa
ab
ab
ab
b
ab
b a b
ba
b
a
b
a
a
b
width
hight
46.
and y(t) = z(t) + p(t)
?Ryy ( ?) = R zz ( ?) + R pp( ?)+ R pz( ?)+ R zp( ?)
now x(t) & z(t) are uncorrelated.
?Rpz( ?) = R zp( ?) = 0
?R yy( ?) = R zz( ?) + R pp( ?)
So the power spectral relation can be given by Fourier transform of the above relation.
?S yy(f) = S zz(f) + S pp(f)
27
now power of y(t)=
( )
( ) ( )
( ) ( ) ( )
2
yy
zz pp
pp xx
s f df
P s f df S f df
now S f H w S f
?
?
??
??
? = +
=?
?
??
&
51
10 10 1
42
17.5
P
P watt
? = ? + ? ?
?=
47. For the minimization of the energy in the error signal there are different approaches like,
Prony?s method, Pade approximation. As g(n) has three samples.
Consider them as g(-1) , g(0) , g(1) we can minimise E(h,g) by making h(n) = g(n) using
rectangular window and Parseval?s there of OTFT.
Based on which 10g(-1) + g(1) = 10(-3) + 3
28
= -27
48. I r= 0.75 I 5
? Forward current = I D = - 0.75 I S
? I S(e
vo/nvt
? 1) = ? 0.75 I s
Now Take n = 1
?e
vo/vT
= 0.25
? V D = V T l n (0.25)
? V R = ? V T l n (0.25)
23
19
1.38 10 300
1.386
1.6 10
?
?
??
= ? ? ?
?
? V R = 35.87 mv
49. Given differential equation is of Cauchy ? Euler differential equation type.
So let x = e
z
? z = ln x
The differential equation can be written as,
D (D ?1) ? 3D + 3 = 0
? D
2
? 4 D + 3 = 0
? D = 1, 3
3
12
3
12
? = +
? = +
zz
y C e C e
y C x C x
Now y (1) = 1
? C 1 + C 2 = 1 ?(i)
And y(2) = 14
? 2C 1 + 8C 2 = 14 ?(ii)
From (i) and (ii)
C 1 = ?1, C 2 = 2
? y = ?x + 2x
3
? y(1.5) = ?1.5 + 2(1.5)
3
? y(1.5) = 5.25
50. We know that,
29
( )
( )
C
C
dV t
I t C
dt
=
And capacitor will be charged by the following equation
V C(t) = V S(1 ? e
?t/?
)
( )
( )
?t/
CS
d
I t C ? V 1 ? e
dt
?
??
=
??
( )
( )
( ) ?t/R t ?C
S
C
V
I t e
Rt
?=
Given R(t) = R O
t
1?
T
??
??
??
Now R O = 1 and C=1
? T = 3R OC = 3
( )
t
R t 1 ?
3
??
?=
??
??
( )
??
??
??
=?
??
??
??
?t
t
1?
3
C
1
& I t e
t
1?
3
At
T3
t sec
22
==
I C(t) = 2
3
e
?
= 0.099
I C(t) ? 0.1 mA
51. V S = 10 V
Voltage across capacitor will be
V C(t) = 10(1 ? e
?t/RC
)
R C = 500 ? 10 ? 10
?6
= 5 ? 10
?3
sec
At t = 2 ms = 2 ? 10
?3
sec
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
25
? ?
0 1 0
A 0 0 1 and C 1 0 0
?1 ?2 ?3
??
??
? = =
??
??
??
44. Same current will flow through both NMOS & PMOS
? ID1 = ID2
( ) ( )
2 2 p
n
GSN TN GSP TP
Np
? cox
? cox
? V ? V ? V ? V
2 L 2 L
?? ? ? ? ?
?=
? ? ? ?
? ? ? ?
( ) ( )
22
NP
100 ? 1.5 ? 0.7 400 1.5 ? 0.9
LL
?? ? ? ? ?
? ? = ?
? ? ? ?
? ? ? ?
( )
( )
N
p
/L
94
/ L 16 10
?
? = ?
?
= 0.225
dd
GSN GSP
V
V V 1.5V
2
??
= = =
??
??
45.
22
10,
2
1
2
1, 0
1
0
22
c
V m n
f
ab
For T m n
VV
fc
aa
? ? ? ?
=+
? ? ? ?
? ? ? ?
? = =
??
= = =
??
??
For T ? 11, m = 1, n = 1
26
22 2
1
2
22
22
22
22
2 2 2
22
2
2
11
2
b
1
2
/2 1
2 11
2
11
11
22 11
1
2
4
3
1
3
1
3
3
3 1.732
=+
=
=
+
= ? =
+
+
=
+
? = +
?=
?=
?=
?=
==
c
c
c
V
f
a
f
Given
f
Va
V
ab
aa
ab
ab
ab
b
ab
b a b
ba
b
a
b
a
a
b
width
hight
46.
and y(t) = z(t) + p(t)
?Ryy ( ?) = R zz ( ?) + R pp( ?)+ R pz( ?)+ R zp( ?)
now x(t) & z(t) are uncorrelated.
?Rpz( ?) = R zp( ?) = 0
?R yy( ?) = R zz( ?) + R pp( ?)
So the power spectral relation can be given by Fourier transform of the above relation.
?S yy(f) = S zz(f) + S pp(f)
27
now power of y(t)=
( )
( ) ( )
( ) ( ) ( )
2
yy
zz pp
pp xx
s f df
P s f df S f df
now S f H w S f
?
?
??
??
? = +
=?
?
??
&
51
10 10 1
42
17.5
P
P watt
? = ? + ? ?
?=
47. For the minimization of the energy in the error signal there are different approaches like,
Prony?s method, Pade approximation. As g(n) has three samples.
Consider them as g(-1) , g(0) , g(1) we can minimise E(h,g) by making h(n) = g(n) using
rectangular window and Parseval?s there of OTFT.
Based on which 10g(-1) + g(1) = 10(-3) + 3
28
= -27
48. I r= 0.75 I 5
? Forward current = I D = - 0.75 I S
? I S(e
vo/nvt
? 1) = ? 0.75 I s
Now Take n = 1
?e
vo/vT
= 0.25
? V D = V T l n (0.25)
? V R = ? V T l n (0.25)
23
19
1.38 10 300
1.386
1.6 10
?
?
??
= ? ? ?
?
? V R = 35.87 mv
49. Given differential equation is of Cauchy ? Euler differential equation type.
So let x = e
z
? z = ln x
The differential equation can be written as,
D (D ?1) ? 3D + 3 = 0
? D
2
? 4 D + 3 = 0
? D = 1, 3
3
12
3
12
? = +
? = +
zz
y C e C e
y C x C x
Now y (1) = 1
? C 1 + C 2 = 1 ?(i)
And y(2) = 14
? 2C 1 + 8C 2 = 14 ?(ii)
From (i) and (ii)
C 1 = ?1, C 2 = 2
? y = ?x + 2x
3
? y(1.5) = ?1.5 + 2(1.5)
3
? y(1.5) = 5.25
50. We know that,
29
( )
( )
C
C
dV t
I t C
dt
=
And capacitor will be charged by the following equation
V C(t) = V S(1 ? e
?t/?
)
( )
( )
?t/
CS
d
I t C ? V 1 ? e
dt
?
??
=
??
( )
( )
( ) ?t/R t ?C
S
C
V
I t e
Rt
?=
Given R(t) = R O
t
1?
T
??
??
??
Now R O = 1 and C=1
? T = 3R OC = 3
( )
t
R t 1 ?
3
??
?=
??
??
( )
??
??
??
=?
??
??
??
?t
t
1?
3
C
1
& I t e
t
1?
3
At
T3
t sec
22
==
I C(t) = 2
3
e
?
= 0.099
I C(t) ? 0.1 mA
51. V S = 10 V
Voltage across capacitor will be
V C(t) = 10(1 ? e
?t/RC
)
R C = 500 ? 10 ? 10
?6
= 5 ? 10
?3
sec
At t = 2 ms = 2 ? 10
?3
sec
30
V C (2 ms) =
?2
5
10 1 ? e
??
??
??
??
V C (2 ms) = 3.3 V
For
T
2
to T diode will be off so capacitor will not charge further
? V C ( 3 msec) = 3.3V
52. By greens theorem
NM
xdy ? ydx ? dxdy
xy
?? ??
=
??
??
??
? ? ?
( ) ( ) xdy ? ydx 1 1 dxdy =+
? ? ?
2 dxdy
??
dxdy
??
= area of the region
( )
2
1
23
2
??
?
?? = ? +
??
??
6
2
? ??
+
??
??
( ) xdy ? ydx 12 ? = + ?
?
53.
Overall
( )
( )
C
2
K
Gs
s s 3s 2
=
++
32
( ) 3 2 0 q s s s s k ? = + + + =
3
2
1
0
s 1 2
s 3 k
6 ? k
s
3
sk
Auxiliary equation is 3s
2
+ k = 0
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
25
? ?
0 1 0
A 0 0 1 and C 1 0 0
?1 ?2 ?3
??
??
? = =
??
??
??
44. Same current will flow through both NMOS & PMOS
? ID1 = ID2
( ) ( )
2 2 p
n
GSN TN GSP TP
Np
? cox
? cox
? V ? V ? V ? V
2 L 2 L
?? ? ? ? ?
?=
? ? ? ?
? ? ? ?
( ) ( )
22
NP
100 ? 1.5 ? 0.7 400 1.5 ? 0.9
LL
?? ? ? ? ?
? ? = ?
? ? ? ?
? ? ? ?
( )
( )
N
p
/L
94
/ L 16 10
?
? = ?
?
= 0.225
dd
GSN GSP
V
V V 1.5V
2
??
= = =
??
??
45.
22
10,
2
1
2
1, 0
1
0
22
c
V m n
f
ab
For T m n
VV
fc
aa
? ? ? ?
=+
? ? ? ?
? ? ? ?
? = =
??
= = =
??
??
For T ? 11, m = 1, n = 1
26
22 2
1
2
22
22
22
22
2 2 2
22
2
2
11
2
b
1
2
/2 1
2 11
2
11
11
22 11
1
2
4
3
1
3
1
3
3
3 1.732
=+
=
=
+
= ? =
+
+
=
+
? = +
?=
?=
?=
?=
==
c
c
c
V
f
a
f
Given
f
Va
V
ab
aa
ab
ab
ab
b
ab
b a b
ba
b
a
b
a
a
b
width
hight
46.
and y(t) = z(t) + p(t)
?Ryy ( ?) = R zz ( ?) + R pp( ?)+ R pz( ?)+ R zp( ?)
now x(t) & z(t) are uncorrelated.
?Rpz( ?) = R zp( ?) = 0
?R yy( ?) = R zz( ?) + R pp( ?)
So the power spectral relation can be given by Fourier transform of the above relation.
?S yy(f) = S zz(f) + S pp(f)
27
now power of y(t)=
( )
( ) ( )
( ) ( ) ( )
2
yy
zz pp
pp xx
s f df
P s f df S f df
now S f H w S f
?
?
??
??
? = +
=?
?
??
&
51
10 10 1
42
17.5
P
P watt
? = ? + ? ?
?=
47. For the minimization of the energy in the error signal there are different approaches like,
Prony?s method, Pade approximation. As g(n) has three samples.
Consider them as g(-1) , g(0) , g(1) we can minimise E(h,g) by making h(n) = g(n) using
rectangular window and Parseval?s there of OTFT.
Based on which 10g(-1) + g(1) = 10(-3) + 3
28
= -27
48. I r= 0.75 I 5
? Forward current = I D = - 0.75 I S
? I S(e
vo/nvt
? 1) = ? 0.75 I s
Now Take n = 1
?e
vo/vT
= 0.25
? V D = V T l n (0.25)
? V R = ? V T l n (0.25)
23
19
1.38 10 300
1.386
1.6 10
?
?
??
= ? ? ?
?
? V R = 35.87 mv
49. Given differential equation is of Cauchy ? Euler differential equation type.
So let x = e
z
? z = ln x
The differential equation can be written as,
D (D ?1) ? 3D + 3 = 0
? D
2
? 4 D + 3 = 0
? D = 1, 3
3
12
3
12
? = +
? = +
zz
y C e C e
y C x C x
Now y (1) = 1
? C 1 + C 2 = 1 ?(i)
And y(2) = 14
? 2C 1 + 8C 2 = 14 ?(ii)
From (i) and (ii)
C 1 = ?1, C 2 = 2
? y = ?x + 2x
3
? y(1.5) = ?1.5 + 2(1.5)
3
? y(1.5) = 5.25
50. We know that,
29
( )
( )
C
C
dV t
I t C
dt
=
And capacitor will be charged by the following equation
V C(t) = V S(1 ? e
?t/?
)
( )
( )
?t/
CS
d
I t C ? V 1 ? e
dt
?
??
=
??
( )
( )
( ) ?t/R t ?C
S
C
V
I t e
Rt
?=
Given R(t) = R O
t
1?
T
??
??
??
Now R O = 1 and C=1
? T = 3R OC = 3
( )
t
R t 1 ?
3
??
?=
??
??
( )
??
??
??
=?
??
??
??
?t
t
1?
3
C
1
& I t e
t
1?
3
At
T3
t sec
22
==
I C(t) = 2
3
e
?
= 0.099
I C(t) ? 0.1 mA
51. V S = 10 V
Voltage across capacitor will be
V C(t) = 10(1 ? e
?t/RC
)
R C = 500 ? 10 ? 10
?6
= 5 ? 10
?3
sec
At t = 2 ms = 2 ? 10
?3
sec
30
V C (2 ms) =
?2
5
10 1 ? e
??
??
??
??
V C (2 ms) = 3.3 V
For
T
2
to T diode will be off so capacitor will not charge further
? V C ( 3 msec) = 3.3V
52. By greens theorem
NM
xdy ? ydx ? dxdy
xy
?? ??
=
??
??
??
? ? ?
( ) ( ) xdy ? ydx 1 1 dxdy =+
? ? ?
2 dxdy
??
dxdy
??
= area of the region
( )
2
1
23
2
??
?
?? = ? +
??
??
6
2
? ??
+
??
??
( ) xdy ? ydx 12 ? = + ?
?
53.
Overall
( )
( )
C
2
K
Gs
s s 3s 2
=
++
32
( ) 3 2 0 q s s s s k ? = + + + =
3
2
1
0
s 1 2
s 3 k
6 ? k
s
3
sk
Auxiliary equation is 3s
2
+ k = 0
31
And for roots on imaginary axis s
1
row = 0
6 ? k
0
3
?=
? k = 6
54.
m(t) has frequency range 5 kHz to 15 kHz
Now it is amplitude modulated
f(t) = A (1 + m(t)) cos2? f ct where f c = 600 kHz
? AM signal will have highest frequency = f c + f m (max)
= 600 + 15 = 615 kHz
And AM signal will have lowest frequency = f c ? f m (max)
= 600 ? 15 = 585 kHz
It is a band pass signal so we use bandpass sampling
s
2fH
f 1.2
k
=?
H
HL
f
K
f ? f
=
=
615
615 ? 585
K = 20.5
We select K = 20
s
2 615
f 1.2
20
?
? = ?
? f s = 73.8 kHz
Now L = 256
And 2
n
= L = 256
? n = 8
Bitrate = R b = nf s
?R b = 8 ? 73.8 ? 103
?R b = 0.59 Mbps
FirstRanker.com - FirstRanker's Choice
2
GATE_SOLUTION
GA
1. The strategies that the company uses to sell its products include house to house marketing.
2. The boat arrived at down
3. As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
4. When he did not come home, she pictured him lying dead on the roadside somewhere.
5. Let t be the time taken by the machines when they work simultaneously.
?
1 1 1
t 4 2
=+
?
13
t4
=
?
4
t
3
=
6. Given is the % of illiterates
So % of literates will be
F M
2001 40% 50%
2011 60% 60%
And population distribution is
F M
2001 40% 60%
2011 50% 50%
Let total population in both the years as T.
So total literate in 2001 will be
0.4 ? 0.4 + 0.5 ? 0.6 = 0.46T
And total literate in 2011 will be
0.5 ? 0.6 + 0.5 ? 0.6 = 0.6T
? Increase = 0.6T ? 0.46T = 0.14T
0.14T
% increase 100 30.43
0.46T
? = ? =
7. Lohit Seema Rahul Mathew
Doctor Dancer Teacher Engineer
3
8. As first line says Indian history was written by British historians was extremely well
documented and researched, but not always impartial.
So option (C) can be interfered from given passage.
9. P Q
Start time 8 AM 8 AM
Working
210
12 7 hrs
360
?=
210
12 8 hrs
360
?=
Breaks 15 minutes each 20 minute break
(2 breaks) (1 break)
= 30 minutes = 20 minutes
? paid working hours = 7 hrs + 8 hrs ? 30 minutes ? 20 minutes
= 14 hrs 10 minutes
? Paid = 14 ? 200 +
10
200
60
?
? Paid = 2833.33
? Budget left = 3000 ? 2833.33 = 166.67
10. As it is given that R is sharing an office with T. So only option (D) is correct.
Electronics Engineering
1. A function F(z) is said to be analytic at a point z = a then F(z) has a derivative at z = a and derivative
exists at each neighbouring point of z = a in domain D.
1
z
e at z 0 e No derivative
?
= ? ? ? ? ? ?
ln z at z = 0 ? ln(0) = ?? ? does not exists
1
1 ? z
at z = 1 ?
1
0
=? ? does not exists
But cos z exists for all values of z so it is analytic over the entire complex plane.
2. As no supply is connected hence fermi level will be constant.
In P type semiconductor Fermi level should be closer to EV.
In N type semiconductor Fermi level should be closer to EC.
In P
++
type semiconductor due to large doping Fermi level enters into valance band.
Hence answer is (B).
3. By reciprocity theorem,
4
1I
55
=
? I = 1A
4. let output of NAND gate is M and output of NOR gate is N
N
M E ? D ?=
And
N
N E D =+
N
N E ? D ?=
When EN = 0
M = 1 and N = 0
So both PMOS and NMOS will be OFF
So F will be at high impedance
When EN = 1
M D & N D ==
So this CMOS will act as not gate
? F will be D
? Option (A) is correct.
5. Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2
? distinct eigen values are three
6.
n
dy x
?
dx y
??
=
??
??
When n = ?1
dy x
?
dx y
=
dy dx
?
yx
?=
?lny = ? ln(x) + ln(c)
? ln(xy) = ln(c)
?xy = c
This represents rectangular hyperbola.
5
Now for n = +1
dy x
?
dx y
=
?ydy = ?x dx
22
y ?x
c
22
? = +
? x
2
+ y
2
= 2c
This represents family of circles.
7.
( )
( ) ( )
( ) ( )
z ? a z ? b
let H z
z ? c z ? d
=
11
? a ? b
1 zz
H
11 z
? c ? d
zz
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
11
z ? z ?
1 ab
H
11 z
z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
( )
( ) ( )
( ) ( )
11
z ? a z ? b z ? z ?
1 ab
H z ? H
11 z
z ? c z ? d z ? z ?
cd
? ? ? ?
? ? ? ?
??
? ? ? ?
?=
??
? ? ? ?
??
? ? ? ?
? ? ? ?
? zeros are
11
a, b, ,
ab
given zero is
11
aj
22
=+
as h(n) is real valued signal another zero must be complex conjugate of this
11
b ? j
22
?=
Now
3
11
z
11
a
j
22
==
+
6
2
1j
=
+
( ) 2 1 ? j
2
=
z 3 = 1 ? j
as h(n) is real valued signal another zero must be complex conjugate of this
z 4 = 1 + j
1 2 3 4
1 1 1 1
z j z ? j z 1 ? j z 1 j
2 2 2 2
? = + = = = +
8.
By changing order of integration
yx x
x 0 y 0
sinx
dy dx
x
= =?
==
??
??
??
??
??
x0
sinx
x dx
x
?
=
?
?
x0
xsinx dx
?
=
?
?
? ?
0
?cosx 2
?
?=
9. R rad =
2
2
dl
80
??
?
??
?
??
??
?
??
??
2
2
dlf
80
C
?R rad? l
2
f
2
7
Now frequency is constant
?R rad? f
2
Rl
2
Rl
??
?=
= 2 ? 1%
R
2%
R
?
?=
10. y(s) is unit step response
( ) ( )
1
y s G s
s
? = ?
( ) ( )
3 ? s
s s 1 s 3
=
++
A B C
s s 1 s 3
= + +
++
( )
1 2 1
y s ?
s s 1 s 3
? = +
++
? y(t) = u(t) ? 2e
?t
u(t) + e
?3t
u(t)
11.
12.
8
If we consider a total cylinder then by gauss law
enclosed
D ? ds Q =
?
But Q enclosed = Q ? H
And we are considering only
1
4
th of the cylinder
Q ? H
D
4
?=
0
Q ? H
E
4
?=
?
13. By rearranging the circuit,
Truth table:
A B F
0 0 1
0 1 0
1 0 0
1 1 1
So it is XNOR gate.
9
14. When V S is +ve
Diode will be reserve biased
2
LS
12
R
VV
RR
=
+
L
50
V8
50 50
? = ?
+
? V L = 4V ?(i)
When V S is ?ve
Diode will be forward biased
? VL = VS = ?10V ?(ii)
From (i) and (ii)
( ) 4 ?10
Average value ?3
2
+
==
? Average value = ?3
15. We know that
E[AX + BY] = AE[X] + BE[Y]
? E[2X + Y] = 2E[X] + E[Y] = 0 ?(i)
And E[X + 2Y] = E[X] + 2E[Y] = 33 ?(ii)
Adding (i) and (ii)
3E[X] + 3E[Y] = 33
? E[X] + E[Y] = 11
10
16. We know that
L I O
LL
NM V ? V =
H OH
IH
NM V ? V =
Now,
0 TP DD Tn
IL
2V ? V ? V kV
V
1k
+
=
+
( ) ( )
22
OL in TP in DD TP in TP
V V ? V V ? V ? V k V ? V = + +
( ) ( )
22
OH in Tn in Tn in DD TP
1
V V ? V V ? V V ? V ? V
k
= + +
( )
DD TP O TP
IH
V V k 2V V
V
1k
+ + +
=
+
Where
( )
( )
w / L n
k
w / L P
=
? as W P ? ? NM L ? and NM H ?
17.
? = ?
V
?D
This is Gauss law
B
E?
t
?
? ? =
?
This is faraday law of electromagnetic induction
B0 ? ? =
This is Gauss law in magnetostatics which states magnetic monopole does not exists.
D
HJ
t
?
? ? = +
?
This is modified form of ampere?s circuital law.
18. at F = 10 Hz we have one pole
At F = 10
2
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10
3
Hz we have a zero
At F = 10
4
Hz we have two zero?s
At F = 10
5
Hz we have two pole?s
11
At F = 10
6
we have one pole
? Total poles N P = 6
And total zeros N Z = 3
19. x(t) = cos(2? fct + km(t))
? Q(t) = 2?fct + km(t)
And ( )
?
=
??
1
fi Q(t)
2t
( )
1
2 ct km t
2t
?
= ? + ??
??
??
f
( )
?
=+
??
k
fi fc m t
2t
( )
???
? = +
??
??
??
max
max
k
fi fc m t
2t
( )
( )
? = + ?
?
max
?3
1 ? ?1
fi 50 kHz 5
7 ? 6 10
?fi max = 50 kHz + 10 kHz
?fi max= 60 kHz
And ( ) ( )
???
=+
??
??
??
min
min
k
fi fc m t
2t
( )
?3
?1 ? 1
50kHz 5
9 ? 7 10
+?
?
= 50 kHz ? 5kHz
fi min = 45 kHz
? = =
min
max
f 45
0.75
f 60
20.
12 1
D Q ? Q =
21
DQ =
12
Present State Excitation Next state
Q 1 Q 2 D 1 D 2
1
Q
+
2
Q
+
0 0 1 0 1 0
1 0 0 1 0 1
0 1 0 0 0 0
As three states are there
Frequency of output = Frequency of Q 2 =
12 kHz
3
= 4 kHz
21. As it is given that it is linear hamming code addition of two codes will produce another code.
(Here we are talking about mod 2 addition)
0 0 0 1 0 0 0 0 1 1 1
0 0 1 1 1 1 0 0 1 1 0
0 0 1 0 1 1 0 0 0 0 1
?
?
?
22. Ans. 0367
Sol. Probability density function (Pdf) = ( )
d
CDF
dx
?
? ?
?=
?
?
?
?
?x
e , x 0
0 , x 0
Now
( )
( ) ( )
( )
? ? ? ??
??
? ? =
?
Pr z 2 z 1
Pr z 2 z 1
Pr z 1
( )
( )
Pr z 2
Pr z 1
?
=
?
?x
2
?x
1
e dx
e dx
?
?
=
?
?
( )
( )
? ?2
? ?1
?1 e ? e
?1 e ? e
?
?
=
13
=
?2
?1
e
e
1
e
=
( )
Pr Z 2 Z 1 0.367 ? ? ? =
23. DC value and phase shift does not affect time period of a signal.
So it is equivalent to find time period of
( ) ( )
2
x t 2cos t 3sin 4cos t
32
?? ? ? ? ?
= ? + +
? ? ? ?
? ? ? ?
11
1
2
T 2 second
?
? ? = ? = =
?
22
2
22
T 3 second
3
??
? = = =
?
33
3
2
T 4 second
2
??
? = = =
?
Now overall T = LCM (T 1, T 2, T 3)
= LCM (2, 3, 4)
? overall T = 12 seconds
24.
25.
( )
=
+
?
?
2
2
2
z1
z1
1
dz
2i
z
For poles :
Consider
2
z 0 z 0,0 = ? =
14
Now f(z) = (z
2
+ 1)
2
( )
( )
( )
( )
?
=
?
n?1
n
C
fz
2i
dz f a
n ? 1 !
z ? a
( )
( ) ( ) ( )
??
?
= = =
??
?
??
2?1
1 2 i
f a f' a f' 0
2 i 2 ? 1 !
Now f?(z) = 2(z
2
+ 1) (2z)
f?(0) = 2(0 + 1) (0) = 0
? So answer is zero.
26. Let output of MUX is M
So M = AQ AQ +
? M = AQ
And D = MQ
= MQ +
D = A Q Q ?+
Present State Input Next State
Q A Q
+
= D
0 0 1
0 1 1
1 0 1
1 1 0
State Diagram:-
27. Given V TN = 0.6V, V SB = 0 and ? = 0
In figure (i)
15
In figure (ii)
Ever MOS transistor has same V G = 3V
? V 1 = V 2 = Vout 2 = VG ? VT
= 3 ? 0.6
? Vout 2 = 2.4 V
28.
2
2
22
s1
s1
s
s 1 s s 1
1
s
+
+
=
+ + +
+
16
( )
( )
2
2
2
2
s1
s s s 1
TF
s1
1
s s s 1
+
++
=
+
+
++
2
32
s1
TF
s 2s s 1
+
?=
+ + +
29.
( )
o
P ?1 N Vth +?
( ) ? ? ( )
2
o
vth 1
1 1 1
P N Vth 1 dx 2 ? Vth ? 1 1 ? Vth
4 4 4
+
? + = = =
?
P 1(1 + N < Vth)
( ) ? ? ( )
Vth?1
1
?2
1 1 1
P N Vth ? 1 dx Vth ? 1 2 Vth 1
4 4 4
? = = + = +
?
P e = P(0)P o (N > Vth + 1) + P(1) P 1(N < Vth ? 1)
( ) ( )
e
11
P 0.2 1 ? Vth 0.8 Vth 1
44
= ? + ? +
=0.05 ? 0.5Vth + 0.2Vth + 0.2
P e = 0.25 + 0.15Vth
For Vth = 0 ? Pe = 0.25
For Vth = 1 ? Pe = 0.4
For Vth = ?1 ? Pe = 0.1
? Minimum probability of error = 0.1
30. Ans. 0.231
Sol. 1 ? e
?
?
x
= 0.5
e
?
?
x
= 0.5
now ? = 3 ? 10
4
cm
?1
( )
4
?ln 0.5
x
3 10
?=
?
31.
( )
2
D P GSP TP
P
1
I ? cox V ? V
2L
? ??
=
??
??
17
= ( )
2
?6
1
30 10 10 2 ? 1
2
? ? ? ?
I D = 150 ?A
Now,
? ??
=
??
??
m D n
N
g 2I ? cox
L
= ? ? ? ? ?
?6 ?6
m
g 2 150 10 60 10 5
? gm = 300 ? 10
?6
s
Now A v = ?gm (r ds || r ds)
( ) ( ) ( )
?6 6 6
?300 10 6 10 6 10 = ? ? ?
= ?300 ? 10
?6
? 3 ? 10
6
? A V = ?900
32. Given that
h(0) = 1, h(1) = a, h(2) = b and h(n) = 0 otherwise
2
( ) 1
jw jw j w
H e ae be
??
? = + +
Now y(n) = 0 for all n
Now ( )
?? ? ? ? ?
? ? ? ?
? ? ? ?
=+
? j n j n
22
12
x n C e C e
If we consider
? ??
??
??
? j n
2
1
Ce as input then
? ?? ?
+ ??
??
??
??
= + +
??
??
? j2 ?
j
2
2
1
Output C 1 ae be
?
?
??
?? = + +
??
??
j
j
2
1
Output C 1 ae be ?(i)
If we consider
? ??
??
??
jn
2
2
Ce as input then
? j2
?j
2
2
2
Output C 1 ae be
? ?? ?
??
??
??
??
= + +
??
??
18
?j
?j
2
2
C 1 ae be
?
?
??
?? = + +
??
??
?(ii)
Both output (i) and (ii) will be zero if
a = 0, b = 1
33.
( )
? ??
=
??
??
2
n ox
D gs T
?c
I ? ? V ? V
2L
( )
?7
2 300 3.45 10 10
5 ? 0.7
21
?? ??
= ? ?
??
??
? I D = 25.5 mA
34. Current through FET having 3
L
? ??
=
??
??
will be I 1
( )
( )
2
1
1
/L
I 1mA
/L
?
? = ?
?
1
3
I mA
2
?=
Now,
( )
( )
out 1
3
/ L 4
II
/
?
=?
?
40 3
mA
10 2
=?
?I out = 6mA
35. Quantum Efficiency
e
p
R
R
?=
R e = Corresponding Electron Rate (electrons/sec)
Rp = Incident Photon Rate (Photons/sec)
p
in P
ep
in
I
PI
R , R , R
q h P
= = =
?
Now
P/q
in/h
I
P
?
?=
19
So
P/q
P
in/h in
I
Ih hR
P qP q
?
? ?
? = = =
q q q
R
h hc hc
? ? ? ??
? = = = ? ?
??
?
??
q = 1.6 ? 10
?19
c, h = 6.63 ? 10
?34
Js, C = 3 ? 10
8
m/s
R
1.24
??
=
36.
Performing star to delta conversion
20
Where
1
R
Z2
jWCR
1
3
??
??
=
??
?? +
??
eq 1
R
ZZ
jWCR
1
3
??
??
?=
??
??
+
??
eq
2R
Z
jWCR
3
1
3
??
??
?=
??
??
+
??
Now R = 1kW, C = 1?F and W=1000 rad/sec
?Zeq = 0.66 ? 0.2178j
eq
V
I
Z
?=
( ) 2sin 1000t
0.66 ? 0.2178j
=
?1
22
21
? sin 1000t ? tan
3
0.66 0.2178
?? ??
=
?? ??
?? ??
+
= 3.16 sin(1000t + 18.43?)
? I ? 3 sin(1000t) + cos (1000t)
37.
I Zmax = 60 mA
L
20
I 20mA
1000
==
21
As I Zmin not given,
I Zmin = 0 mA
Now I S = I Z + I L
?I Smin = I Zmin + I L
= 0 + 20 mA
?I Smin = 20 mA
Now
SZ
S
V ? V
I
200
=
S
V ? 20
20mA
200
?=
? V S = 24V
Now I Smax = I Zmax + I L
= 60 + 20
I Smax = 80 mA
SZ
S
V ? V
I
200
?=
S
V ? 20
80mA
200
?=
? V S = 36 V
38.
Sol.
I
Ha
2
=?
??
For wire ? 1
1
I
H
2r
=
?
For wire ? 2
2
2I
H
2 3r
=
?
Magnetic field will be circular and can be find out by right hand rule
Both fields will add at middle region
? at dotted line
22
H = H 1 + H 2
5I
H
6r
?=
?
Now B = ?oH
=
?
o
? 5I
B
6r
39.
Sol.
g
d
V
d
?
=
?
Now,
( )
22
0
22
0
dk
d d 1 1
? ? 2
d d d c
2c ?
?
?
= = ? ? = ? ?
???
??
22
0
d
d
c?
??
=
?
??
22
0 88
g
22
0
1
c?
V 2 10 2 10
c?
??
?
= = ? ? = ?
?
??
22
0
2
?
3
?
? ? ? =
p
22
0
c 3c
Now, V
1
k2
?2
c3
? ? ? ?
= = = = =
?
?
??
88
p
3
V 3 10 4.5 10 m / s
2
= ? ? = ?
8
p
V 4.5 10 m / s =?
40. f(?1) = 0
So only option (B) and (C) are possible
Let?s try option (B)
f(x) = 2 x 1 +
23
( )
( )
( )
2 x 1 for x 1 0
fx
?2 x 1 for x 1 0
+ + ? ?
?
?=
?
+ + ?
?
?
( )
( )
( )
2 x 1 for x ?1
fx
?2 x 1 for x ?1
+? ?
?
?=
?
+?
?
?
( )
2 for x ?1
f' x
?2 for x ?1
? ?
?=
?
?
?
( ) f' x 2 ??
? option (B) is correct.
41.
( )
( )
( )
Cs
Gs
Rs
=
? C(s) = G(s) ? R(s)
( )
2
1
s s 2s 1
=
++
( )
( )
2
1
Cs
s s 1
?=
+
( )
( )
( )
2
A B C
Cs
s s 1
s1
? = + +
+
+
? A(s + 1)
2
+ Bs(s + 1) + Cs = 1
? As
2
+ 2As + A + Bs
2
+ Bs + Cs = 1
? A + B = 0
? 2A + B + C = 0
? A = 1
So B = ?1
And C = ?1
( )
( )
2
1 ?1 ?1
Cs
s s 1
s1
? = + +
+
+
? C(t) = (1 ? e
?t
? te
?t
) u(t)
At t ? ? stedy state will occur
24
? C(?) = 1
Now we are asked to find time at which 94% of the steady state value reached.
? C(t) = 1 ? e
?t
? te
?t
= 0.94
? e
?t
+ te
?t
= 0.06
? e
?t
(1 + t) = 0.06
Now from the given options try all option you will get t = 4.50 sec.
42.
( ) ( )
N?1
kn
N
n0
X k x n W
=
=
?
We are obtaining X(1) correctly
? k = 1
( ) ( ) ( ) ( ) ( ) ( ) ( )
1 2 3 4 5
6 6 6 6 6
x 1 x 0 x 1 W x 2 W x 3 W x 4 W x 5 W ? = + + + + +
We know that
N
k
K
2
N N
W ?W
+
=
30
66
W ?W ?1 ? = =
41
66
W ?W =
52
66
W ?W =
? comparing with given graph
a 1 = 1, a 2 = W 6,
2
36
aW =
43.
( )
22
1
Hs
s 3s 2s 1
=
+ + +
? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? = +
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
11
22
33
x 0 1 0 x 0
x 0 0 1 x 0 u
x ?1 ?2 ?3 x 1
? ? ? ? ? ? ? ?
??
??
=+
??
??
??
1
2
3
x
& y 1 0 0 x 0 u
x
25
? ?
0 1 0
A 0 0 1 and C 1 0 0
?1 ?2 ?3
??
??
? = =
??
??
??
44. Same current will flow through both NMOS & PMOS
? ID1 = ID2
( ) ( )
2 2 p
n
GSN TN GSP TP
Np
? cox
? cox
? V ? V ? V ? V
2 L 2 L
?? ? ? ? ?
?=
? ? ? ?
? ? ? ?
( ) ( )
22
NP
100 ? 1.5 ? 0.7 400 1.5 ? 0.9
LL
?? ? ? ? ?
? ? = ?
? ? ? ?
? ? ? ?
( )
( )
N
p
/L
94
/ L 16 10
?
? = ?
?
= 0.225
dd
GSN GSP
V
V V 1.5V
2
??
= = =
??
??
45.
22
10,
2
1
2
1, 0
1
0
22
c
V m n
f
ab
For T m n
VV
fc
aa
? ? ? ?
=+
? ? ? ?
? ? ? ?
? = =
??
= = =
??
??
For T ? 11, m = 1, n = 1
26
22 2
1
2
22
22
22
22
2 2 2
22
2
2
11
2
b
1
2
/2 1
2 11
2
11
11
22 11
1
2
4
3
1
3
1
3
3
3 1.732
=+
=
=
+
= ? =
+
+
=
+
? = +
?=
?=
?=
?=
==
c
c
c
V
f
a
f
Given
f
Va
V
ab
aa
ab
ab
ab
b
ab
b a b
ba
b
a
b
a
a
b
width
hight
46.
and y(t) = z(t) + p(t)
?Ryy ( ?) = R zz ( ?) + R pp( ?)+ R pz( ?)+ R zp( ?)
now x(t) & z(t) are uncorrelated.
?Rpz( ?) = R zp( ?) = 0
?R yy( ?) = R zz( ?) + R pp( ?)
So the power spectral relation can be given by Fourier transform of the above relation.
?S yy(f) = S zz(f) + S pp(f)
27
now power of y(t)=
( )
( ) ( )
( ) ( ) ( )
2
yy
zz pp
pp xx
s f df
P s f df S f df
now S f H w S f
?
?
??
??
? = +
=?
?
??
&
51
10 10 1
42
17.5
P
P watt
? = ? + ? ?
?=
47. For the minimization of the energy in the error signal there are different approaches like,
Prony?s method, Pade approximation. As g(n) has three samples.
Consider them as g(-1) , g(0) , g(1) we can minimise E(h,g) by making h(n) = g(n) using
rectangular window and Parseval?s there of OTFT.
Based on which 10g(-1) + g(1) = 10(-3) + 3
28
= -27
48. I r= 0.75 I 5
? Forward current = I D = - 0.75 I S
? I S(e
vo/nvt
? 1) = ? 0.75 I s
Now Take n = 1
?e
vo/vT
= 0.25
? V D = V T l n (0.25)
? V R = ? V T l n (0.25)
23
19
1.38 10 300
1.386
1.6 10
?
?
??
= ? ? ?
?
? V R = 35.87 mv
49. Given differential equation is of Cauchy ? Euler differential equation type.
So let x = e
z
? z = ln x
The differential equation can be written as,
D (D ?1) ? 3D + 3 = 0
? D
2
? 4 D + 3 = 0
? D = 1, 3
3
12
3
12
? = +
? = +
zz
y C e C e
y C x C x
Now y (1) = 1
? C 1 + C 2 = 1 ?(i)
And y(2) = 14
? 2C 1 + 8C 2 = 14 ?(ii)
From (i) and (ii)
C 1 = ?1, C 2 = 2
? y = ?x + 2x
3
? y(1.5) = ?1.5 + 2(1.5)
3
? y(1.5) = 5.25
50. We know that,
29
( )
( )
C
C
dV t
I t C
dt
=
And capacitor will be charged by the following equation
V C(t) = V S(1 ? e
?t/?
)
( )
( )
?t/
CS
d
I t C ? V 1 ? e
dt
?
??
=
??
( )
( )
( ) ?t/R t ?C
S
C
V
I t e
Rt
?=
Given R(t) = R O
t
1?
T
??
??
??
Now R O = 1 and C=1
? T = 3R OC = 3
( )
t
R t 1 ?
3
??
?=
??
??
( )
??
??
??
=?
??
??
??
?t
t
1?
3
C
1
& I t e
t
1?
3
At
T3
t sec
22
==
I C(t) = 2
3
e
?
= 0.099
I C(t) ? 0.1 mA
51. V S = 10 V
Voltage across capacitor will be
V C(t) = 10(1 ? e
?t/RC
)
R C = 500 ? 10 ? 10
?6
= 5 ? 10
?3
sec
At t = 2 ms = 2 ? 10
?3
sec
30
V C (2 ms) =
?2
5
10 1 ? e
??
??
??
??
V C (2 ms) = 3.3 V
For
T
2
to T diode will be off so capacitor will not charge further
? V C ( 3 msec) = 3.3V
52. By greens theorem
NM
xdy ? ydx ? dxdy
xy
?? ??
=
??
??
??
? ? ?
( ) ( ) xdy ? ydx 1 1 dxdy =+
? ? ?
2 dxdy
??
dxdy
??
= area of the region
( )
2
1
23
2
??
?
?? = ? +
??
??
6
2
? ??
+
??
??
( ) xdy ? ydx 12 ? = + ?
?
53.
Overall
( )
( )
C
2
K
Gs
s s 3s 2
=
++
32
( ) 3 2 0 q s s s s k ? = + + + =
3
2
1
0
s 1 2
s 3 k
6 ? k
s
3
sk
Auxiliary equation is 3s
2
+ k = 0
31
And for roots on imaginary axis s
1
row = 0
6 ? k
0
3
?=
? k = 6
54.
m(t) has frequency range 5 kHz to 15 kHz
Now it is amplitude modulated
f(t) = A (1 + m(t)) cos2? f ct where f c = 600 kHz
? AM signal will have highest frequency = f c + f m (max)
= 600 + 15 = 615 kHz
And AM signal will have lowest frequency = f c ? f m (max)
= 600 ? 15 = 585 kHz
It is a band pass signal so we use bandpass sampling
s
2fH
f 1.2
k
=?
H
HL
f
K
f ? f
=
=
615
615 ? 585
K = 20.5
We select K = 20
s
2 615
f 1.2
20
?
? = ?
? f s = 73.8 kHz
Now L = 256
And 2
n
= L = 256
? n = 8
Bitrate = R b = nf s
?R b = 8 ? 73.8 ? 103
?R b = 0.59 Mbps
32
55.
0 is represented by p(t)
And 1 is represented by q(t)
And ? 1(t) and ? 2(t) are orthogonal signal set
(i) p(t) = ? 1(t) and q(t) = ? ? 1(t)
So signal space diagram will be,
? dmin 1 = 2
(ii) p(t) = ? 1(t) and q(t) = ( ) ?
2
Et
So signal space diagram will be
2
dmin E 1 ? = +
Now bit error probability is same in both cases
? dmin 1 = dmin 2
1 E 2 +=
? E = 3
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This post was last modified on 18 December 2019