Download GATE GATE ECE Question Paper With Solutions

Download GATE (Graduate Aptitude Test in Engineering) GATE ECE Question Paper With Solutions

This post was last modified on 18 December 2019

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GATE Previous Last 10 Years 2010-2020 Question Papers With Solutions And Answer Keys

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GATE_SOLUTION

The strategies that the company uses to sell its products include house to house marketing.
The boat arrived at dawn
As the positions of book R & S are fixed. The books P, Q and T can be arranged in 3! = 6 ways
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When he did not come home, she pictured him lying dead on the roadside somewhere.
Let t be the time taken by the machines when they work simultaneously.

1/t = 1/4 + 1/2
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1/t = 3/4

t = 4/3
Given is the % of illiterates

So % of literates will be
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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 x 0.4 + 0.5 × 0.6 = 0.46T

And total literate in 2011 will be
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0.5 x 0.6 + 0.5 × 0.6 = 0.6T

Increase = 0.6T-0.46T = 0.14T

% increase = (0.14T / 0.46T) ×100 = 30.43
Lohit Seema Rahul Mathew
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Doctor Dancer Teacher Engineer
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.

	F	M
2001	40%	50%
2011	60%	60%

	F	M
2001	40%	60%
2011	50%	50%

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P Q

Start time 8 AM 8 AM

Working (210/360) × 12 = 7 hrs (210/360) × 12 = 8 hrs

Breaks 15 minutes each (2 breaks) = 30 minutes 20 minute break (1 break) = 20 minutes

? paid working hours = 7 hrs + 8 hrs - 30 minutes – 20 minutes = 14 hrs 10 minutes

Paid = 14 × 200 + (10/60) × 200

Paid = 2833.33

Budget left = 3000 - 2833.33 = 166.67

	P	Q
Start time	8 AM	8 AM
Working	(210/360) × 12 = 7 hrs	(210/360) × 12 = 8 hrs
Breaks	15 minutes each (2 breaks) = 30 minutes	20 minute break (1 break) = 20 minutes

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As it is given that R is sharing an office with T. So only option (D) is correct.

Electronics Engineering

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

e^z at z=0 ? e⁰ ? No derivative
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ln z at z = 0 ? ln(0) = 8 does not exists

1/(1-z) at z = 1 ? 1/0 = 8 does not exists

But cos z exists for all values of z so it is analytic over the entire complex plane.
As no supply is connected hence fermi level will be constant.
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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).

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By reciprocity theorem,

I = 1A
Let output of NAND gate is M and output of NOR gate is N

M=EN AND D
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And N=EN NOR D

N=(EN+D)'

When EN = 0

M = 1 and N = 0

So both PMOS and NMOS will be OFF
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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'
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? Option (A) is correct.
Since it is a upper triangular matrix eigen values will bee 2, 1, 3, 2

distinct eigen values are three

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x (dy/dx) = y

When n = -1

x (dy/dx) = y

dy/y = dx/x

ln(y) = -ln(x) + ln(c)
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ln(xy) = ln(c)

xy = c

This represents rectangular hyperbola.

Now for n = +1

x (dy/dx) = y
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y dy = -x dx

y²/2 = -x²/2 + C

x² + y² = 2c

This represents family of circles.

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let H(z) = (z-a)(z-b) / (z-c)(z-d)

H(1/z) = ((1/z)-a)((1/z)-b) / ((1/z)-c)((1/z)-d)

H(1/z) = (z-a)(z-b) / (z-c)(z-d) * cd/ab

H(z) H(1/z) = 1

zeros are a,b, 1/a, 1/b
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given zero is a = -1/2 + j/2

as h(n) is real valued signal another zero must be complex conjugate of this

b = -1/2 - j/2

Now z3 = 1/a

z3 = 1-j
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as h(n) is real valued signal another zero must be complex conjugate of this

Z4 = 1 + j

z1 = -1/2 + j/2, z2 = -1/2 - j/2, Z3 = 1- j, Z4 = 1+ j
? sin(x)/x dy dx
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y=0 to x, x=0 to p

By changing order of integration

x=p, y=x to p

? sin(x)/x dx dy

? sin(x) dx
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[-cosx] = 2
R_rad = 80p² (dl/?)²

R_rad ? f²

?R/R = 2 ?f/f
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?R/R = 2 × 1%

?R/R = 2%
y(s) is unit step response

y(s) = G(s) * 1/s = (3-s) / s(s+1)(s+3)
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y(s) = A/s + B/(s+1) + C/(s+3)

y(s) = 1/s - 2/(s+1) + 1/(s+3)

y(t) = u(t) – 2e^-t u(t) + e^-3t u(t)
B
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E

C
P

N
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P

C

N

P

B
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N

E
If we consider a total cylinder then by gauss law ? D.ds = Qenclosed

But Qenclosed = Q * H
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And we are considering only 1/4 th of the cylinder

D = Q.H / 4

E = Q.H / 4e₀
By rearranging the circuit,
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Truth table:

A B F

0 0 1

0 1 0

1 0 0

1 1 1

So it is XNOR gate.
When Vs is +ve

Diode will be reserve biased
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V_L = (R₂ / (R₁ + R₂)) * V_s

V_L = (50 / (50+50)) * 8

V_L = 4V ...(i)

When Vs is -ve

Diode will be forward biased
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V_L = V_S = -10V ...(ii)

From (i) and (ii)

Average value = (4+(-10)) / 2

Average value = -3

A	B	F
0	0	1
0	1	0
1	0	0
1	1	1

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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)
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3E[X] + 3E[Y] = 33

E[X] + E[Y] = 11
We know that

NML = V_IL - V_OL
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NMH = V_OH - V_IH
V.D = ?_v - This is Gauss law

VXE = -?B/?t - This is faraday law of electromagnetic induction

VxB=0 - This is Gauss law in magnetostatics which states magnetic monopole does not exists.
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VxH= J+ ?D/?t - This is modified form of ampere's circuital law.
at F = 10 Hz we have one pole

At F = 10² Hz we can see two more poles are added as slope is decreased by 40 dB/decade

At F = 10³ Hz we have a zero
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At F = 10⁴ Hz we have two zero's

At F = 10⁵ Hz we have two pole's

At F = 10⁶ we have one pole

Total poles Np = 6

And total zeros Nz = 3
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x(t) = cos(2p f_ct + k m(t))

? Q(t) = 2pf_ct + k m(t)

And f_i = (1/2p) (dQ(t)/dt)

f_i = (1/2p) (d/dt) [2pf_ct+km(t)]
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f_i = f_c + k/(2p) (dm(t)/dt)

f_{i max} = f_c + k/(2p) (dm(t)/dt)_max

f_{i max} = 50 kHz+5× ((1-(-1)) / ((7-6)×10^-3))

f_{i max} = 50 kHz + 10 kHz

f_{i max} = 60 kHz
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f_{i min} = f_c + k/(2p) (dm(t)/dt)_min

f_{i min} = 50kHz+5×((-1-1) / ((9-7)×10^-3))

= 50 kHz - 5kHz

f_{i min} = 45 kHz

f_min / f_max = 45 / 60 = 0.75
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D₁ = Q₁ Q₂'

D₂ = Q₁

Sol.

Present State Excitation Next state

Q1 Q2 D1 D2 Q1+ Q2+

0 0 1 0 1 0

1 0 1 1 1 1

1 1 0 1 0 1

0 1 0 0 0 0

0 0 1 0

As three states are there
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Frequency of output = Frequency of Q2 = 12 kHz / 3 = 4 kHz
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)

0001 0000111
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00111100110

0010 1100001
Ans. 0.367

Probability density function (Pdf) = d/dx (CDF)
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CDF = 1-e^-x , x=0

Pdf = e^-x, x=0

0 , x<0

Pr(z>2 | z>1) = Pr(z>2) / Pr(z>1)

= (?₂⁸ e^-x dx) / (?₁⁸ e^-x dx)
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= (e^-2) / (e^-1) = e^-1 ˜ 0.367
Pr(Z>2|Z>1)=0.367
DC value and phase shift does not affect time period of a signal.
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So it is equivalent to find time period of

x(t)=2cos(t)+sin(2p/3 t)+cos(p/2 t + 4)

?₁ = 1, T₁ = 2p/?₁ = 2p ˜ 6.28 second

?₂ = 2p/3, T₂ = 2p/?₂ = 3 second

?₃ = p/2, T₃ = 2p/?₃ = 4 second
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Now overall T = LCM (T₁, T₂, T₃)

= LCM (2, 3, 4)

overall T = 12 seconds
accumulation mode
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Flat band condition

Inversion mode
(1/(2pi)) ? dz / z² =0

For poles:
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Consider z² = 0 => z=0,0
V-V/3-0.6=0
In figure (i)

V =3-0.6

?V =2.4V
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V-V/3-0.6

2.4-V/3-0.6

?V=1.8V

In figure (ii)

Present State	Excitation	Next state
0	0	1	0	1	0
1	0	1	1	1	1
1	1	0	1	0	1
0	1	0	0	0	0
0	0	1	0

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Every MOS transistor has same V_G = 3V

? V₁ = V₂ = V_{out 2} = V_G - V_T

= 3-0.6

? V_{out 2} = 2.4 V

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Transfer Function

C(S)/R(S)=(S+1)/(S²+1+(S/S²+1))

C(S)/R(S)=((S+1)/(S²+S+1))

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This post was last modified on 18 December 2019

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