DUET Last 10 Years 2011-2021 Question Papers With Answer Key || Delhi University Entrance Test conducted by the NTA
DUMTech Microwave Electronics
Topic: ME MTECH S2
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- A family of arcs is obtained in the Smith chart by varying normalized reactance in a range of, [Question ID = 4165]
- 0 to 1 [Option ID = 16654]
- -8 to +8 [Option ID = 16655]
- 0 to +8 [Option ID = 16656]
- -1 to 0 [Option ID = 16657]
- -8 to +8 [Option ID = 16655]
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- A 75 O quarter wave transformer is connected to a 125 O TV receiver. The impedance at the input of this transformer will be, [Question ID = 4166]
- 25 O [Option ID = 16658]
- 45 O [Option ID = 16659]
- 75 O [Option ID = 16660]
- 100 O [Option ID = 16661]
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- 45 O [Option ID = 16659]
- If a coaxial line has dimensions a × b cm, the conductivity of the conductors is s and the permittivity of the filled medium is e. Conductance per unit length of this coaxial transmission line is given as, [Question ID = 4167]
- 2ps
ln(b/a) [Option ID = 16662] - 2pe
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ln(b/a) [Option ID = 16663] - 1
2psln(b/a) [Option ID = 16664] - 1
2peln(b/a) [Option ID = 16665]
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- 2ps
ln(b/a) [Option ID = 16662]
- 2ps
- A lossless transmission line used in a TV Receiver has a capacitance of 50 pF/m and an inductance of 20 nH/m. The characteristic impedance of a 10 meter long line will be then, [Question ID = 4168]
- 50 O [Option ID = 16666]
- 20 O [Option ID = 16667]
- 40 O [Option ID = 16668]
- 100 O [Option ID = 16669]
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- 20 O [Option ID = 16667]
- A one-half wavelength lossless line of 50 O is terminated to a load impedance of 50+j50 O, its input impedance is, [Question ID = 4169]
- 50+j50 O [Option ID = 16670]
- 50-j50 O [Option ID = 16671]
- 25+j25 O [Option ID = 16672]
- 25-j25 O [Option ID = 16673]
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- 50+j50 O [Option ID = 16670]
- ZL
If a line of characteristic impedance Zo is connected to a load of impedance ZL, then the normalized load impedance
Zo at a distance d is equal to, [Question ID = 4170]- 1-G(d)
1 + G (d) [Option ID = 16674] - G(d)
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1- G(d) [Option ID = 16675] - G (d)
1 + G (d) [Option ID = 16676] - 1+ G (d)
1-G(d) [Option ID = 16677]
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- 1+ G (d)
1-G(d) [Option ID = 16677]
- 1-G(d)
- A copper rectangular cavity resonator is structured by 3x4x1 cm³. The resonant frequency for TM110 mode, is [Question ID = 4171]
- 6.25 GHz [Option ID = 16678]
- 5.81 GHz [Option ID = 16679]
- 8.65 GHz [Option ID = 16680]
- 2.04 GHz [Option ID = 16681]
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- 6.25 GHz [Option ID = 16678]
- A transmission line of characteristics impedance Zo is terminated in a load having VSWR=2. The normalized load impedance Zn of the transmission line is [Question ID = 4172]
- Zo [Option ID = 16682]
- 1 [Option ID = 16683]
- 0 [Option ID = 16684]
- 2 [Option ID = 16685]
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- 2 [Option ID = 16685]
- In a Square coaxial transmission line shown in Fig., if the space between the conductors is filled with a material having a relative permittivity of 4.0 and the capacitance per unit length is 6.3 e F/m, the characteristic impedance of this line will be,
2 mm
2 mm
4 mm [Question ID = 4173]- 30 O [Option ID = 16686]
- 50 O [Option ID = 16687]
- 40 O [Option ID = 16688]
- 20 O [Option ID = 16689]
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- 30 O [Option ID = 16686]
- A parabolic antenna is operating at 5 GHz and has an antenna power gain of 216. The diameter of this antenna is [Question ID = 4174]
- 8 cm [Option ID = 16690]
- 16 cm [Option ID = 16691]
- 25 cm [Option ID = 16692]
- 36 cm [Option ID = 16693]
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- 36 cm [Option ID = 16693]
- If a signal of 30 MHz is transmitted through a coaxial cable, which has a capacitance of 40 pF/m and an inductance of 900 nH/m. The propagation velocity for a 1m long cable is given by, [Question ID = 4175]
- 2.36×108 m/s [Option ID = 16694]
- 1.66×108 m/s [Option ID = 16695]
- 2.66×108 m/s [Option ID = 16696]
- 1.33×108 m/s [Option ID = 16697]
- 1.66×108 m/s [Option ID = 16695]
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- A perpendicularly polarized wave is incident at an angle of ?i = 15° and it is propagating from medium 1 to medium 2. Medium 1 is characterized by er1 = 9, µr1 = 1, s1= 0 and medium 2 is a free space. If the electric field intensity E1 = 3.0 mV/m, the value of the magnetic field intensity Hi, will be, [Question ID = 4176]
- 23.87 µA/m [Option ID = 16698]
- 48.53 µA/m [Option ID = 16699]
- 15.46 µA/m [Option ID = 16700]
- 10 µA/m [Option ID = 16701]
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- 23.87 µA/m [Option ID = 16698]
- If capacitor plates of area 'A' are placed 'd' distance apart and the capacitor is filled with a material having dielectric constant e, then the current through this capacitor is expressed as, [Question ID = 4177]
- e dV
Ad dt [Option ID = 16702] - Ad dV
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e dt [Option ID = 16703] - eA dV
d dt [Option ID = 16704] - d dV
eA dt [Option ID = 16705]
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- eA dV
d dt [Option ID = 16704]
- e dV
- For a 4 GHz wave propagating in a free Space, the propagation constant of this wave will be, [Question ID = 4178]
- j 3.14 [Option ID = 16706]
- j 5.53 [Option ID = 16707]
- j 2.09 [Option ID = 16708]
- j 10.8 [Option ID = 16709]
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- j 2.09 [Option ID = 16708]
- A transmission line of characteristic impedance Zo terminated by a load R= (1/2)Zo as shown in Fig., is connected to a matched source by a switch for a short time t to produce a pulse on the line.
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V
Zo
A
B
S--- Content provided by FirstRanker.com ---
Zo
Zo/2
If the propagation time along the line is T, where T > t, the values of the reflection coefficient at point A and point B are, respectively [Question ID = 4179]- 1, -1/3 [Option ID = 16710]
- 1, -1 [Option ID = 16711]
- 2, -1 [Option ID = 16712]
- 1/2, -1/2 [Option ID = 16713]
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- 1, -1/3 [Option ID = 16710]
- ? /4 transformer is connected in between generator and load, which have impedence ratio ZL / ZG =0.16 and ZG = 50 O. The characteristic impedance of a ? /4 transformer is [Question ID = 4180]
- 8 O [Option ID = 16714]
- 20 O [Option ID = 16715]
- 312 O [Option ID = 16716]
- 25 O [Option ID = 16717]
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- 20 O [Option ID = 16715]
- The input power needed to generate an output power of 3 W from an amplifier with a power gain of 30 dB is [Question ID = 4181]
- 1 W [Option ID = 16718]
- 100 mW [Option ID = 16719]
- 1 mW [Option ID = 16720]
- 3 mW [Option ID = 16721]
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- 3 mW [Option ID = 16721]
- The VSWR of a coaxial line which has a reflection coefficient of 0.6e-j<60°, is [Question ID = 4182]
- 4 [Option ID = 16722]
- 1 [Option ID = 16723]
- 1.6 [Option ID = 16724]
- 2 [Option ID = 16725]
- 4 [Option ID = 16722]
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- When EM waves travel along a metallic waveguide in which there is a discontinuity due to some lossy material filled in a small length and the other end is terminated with a matched load. The discontinuity will cause [Question ID = 4183]
- Progressive waves throughout the line. [Option ID = 16726]
- Progressive waves on the generator side and standing waves on the matched load side. [Option ID = 16727]
- Standing waves on the generator side and progressive waves on the matched load side. [Option ID = 16728]
- Standing waves thoughout the line. [Option ID = 16729]
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- Standing waves on the generator side and progressive waves on the matched load side. [Option ID = 16728]
- The power output of an input power of 10 mW in the following system, is
Pin = 10 mW
25 dB--- Content provided by FirstRanker.com ---
-35 dB
Pout [Question ID = 4184]- 5 dBm [Option ID = 16730]
- 10 dB [Option ID = 16731]
- 0 dBm [Option ID = 16732]
- 10 dBm [Option ID = 16733]
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- 0 dBm [Option ID = 16732]
- At microwave frequency, hybrid(H), admittance (Y) and impedance (Z) parameters can not be measured because, [Question ID = 4185]
- Short and open circuit conditions are realizable over broad frequency range. [Option ID = 16734]
- Short and open circuit conditions are unrealizable over broad frequency range. [Option ID = 16735]
- Only short circuit condition is realizable over broad frequency range. [Option ID = 16736]
- Only open circuit condition Is realizable over broad frequency range. [Option ID = 16737]
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- Short and open circuit conditions are unrealizable over broad frequency range. [Option ID = 16735]
- Noise temperature (T) and gain(G) are two important parameters of satellite antenna. Out of the following, the antenna having largest G/T ratio is [Question ID = 4186]
- Cassegrain antenna [Option ID = 16738]
- Pyramidal antenna [Option ID = 16739]
- Parabolic reflector [Option ID = 16740]
- Dipole antenna [Option ID = 16741]
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- Parabolic reflector [Option ID = 16740]
- A generator of 50 O internal impedance and operating frequency of 1 GHz feeds a 75 O load via a coaxial line of characteristics impedance 50 O. The reflection coefficient on the feed line is, [Question ID = 4187]
- 0.2 [Option ID = 16742]
- 0.67 [Option ID = 16743]
- 1.5 [Option ID = 16744]
- 1.05 [Option ID = 16745]
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- 0.2 [Option ID = 16742]
- A lossless transmission line is excited by a signal of voltage 5?0° V at 1.2 MHz. If the line is terminated by ZL (= Zo) at a distance 1km, the electrical length at the end point of the line is [Question ID = 4188]
- 4p [Option ID = 16746]
- 6p [Option ID = 16747]
- 8p [Option ID = 16748]
- 2p [Option ID = 16749]
- 8p [Option ID = 16748]
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- The modulated microwave signal with 1 kHz square wave modulating signal is detected by [Question ID = 4189]
- Tunable detector [Option ID = 16750]
- VSWR meter [Option ID = 16751]
- Spectrum analyzer [Option ID = 16752]
- Slotted line [Option ID = 16753]
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- Tunable detector [Option ID = 16750]
- A parabolic dish has a gain of 40 dB at 3GHz. The diameter of this dish is [Question ID = 4190]
- 4.08 m [Option ID = 16754]
- 8.02 m [Option ID = 16755]
- 3.04 m [Option ID = 16756]
- 1.25 m [Option ID = 16757]
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- 4.08 m [Option ID = 16754]
- A satellite operates with 12 GHz at a distance of 36000 km has effective isotropic radiated power (EIRP) of 20 dBW.The flux density at receiving antenna of each station is [Question ID = 4191]
- 2.21 × 10-14 W/m² [Option ID = 16758]
- 3.6 × 10-14 W/m² [Option ID = 16759]
- 0.614 × 10-14 W/m² [Option ID = 16760]
- 2.0 × 10-14 W/m² [Option ID = 16761]
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- 0.614 × 10-14 W/m² [Option ID = 16760]
- A broadside array operating at 100cm wavelength consist of 4 half-wave dipoles spaced 50cm apart. Each element carries radio frequency current In the same phase and of magnitude 0.5 A. The radiated power will be [Question ID = 4192]
- 34p² W [Option ID = 16762]
- 16p² W [Option ID = 16763]
- 20p² W [Option ID = 16764]
- 50p² W [Option ID = 16765]
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- 20p² W [Option ID = 16764]
- Charge needed within a unit sphere centred at the origin for producing a potential field, V = - 6r5/e for the distance r = 1 meter, is [Question ID = 4193]
- 30 p Coulomb [Option ID = 16766]
- 240 p Coulomb [Option ID = 16767]
- 120 p Coulomb [Option ID = 16768]
- 180 p Coulomb [Option ID = 16769]
- 120 p Coulomb [Option ID = 16768]
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- A material has conductivity of 10-2 mho/m and a relative permittivity of 4. The frequency for which conduction current in the medium is equal to displacement current is [Question ID = 4194]
- 90 MHz [Option ID = 16770]
- 180 MHz [Option ID = 16771]
- 27 MHz [Option ID = 16772]
- 45 MHz [Option ID = 16773]
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- 45 MHz [Option ID = 16773]
- For a electricity short dipole of length 80 cm operating at 30 MHz, the loss resistance (Rloss) is 1.5 O. Its radiation efficiency is [Question ID = 4195]
- 92.4% [Option ID = 16774]
- 94.9% [Option ID = 16775]
- 102.5% [Option ID = 16776]
- 86.7% [Option ID = 16777]
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- 94.9% [Option ID = 16775]
- A Hertzien dipole of length ?/50 has an efficiency of 6.5%. The total quality factor for this dipole is [Question ID = 4196]
- 1.1 [Option ID = 16778]
- 0.20 [Option ID = 16779]
- 3.7 [Option ID = 16780]
- 5.5 [Option ID = 16781]
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- 0.20 [Option ID = 16779]
- The value of integral ?0p sin³ a da is given by [Question ID = 4197]
- 8/3 [Option ID = 16782]
- 4/3 [Option ID = 16783]
- 1/2 [Option ID = 16784]
- 2/3 [Option ID = 16785]
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- 4/3 [Option ID = 16783]
- p/2
The region specified by {(?,f,z): 3 = ? = 5, p/6 = f = p/3 ,3 = z = 4.5} in cylindrical coordinates has volume of [Question ID = 4198]- 4.712 [Option ID = 16786]
- 9.424 [Option ID = 16787]
- 2.725 [Option ID = 16788]
- 5.725 [Option ID = 16789]
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- 4.712 [Option ID = 16786]
- dx
With the initial condition x(1) = 0.5 the solution of the differential equation, t dt + x = t is [Question ID = 4199]- x = t/2 [Option ID = 16790]
- x = t - 1/2 [Option ID = 16791]
- x = t2 - 1/2 [Option ID = 16792]
- x= t2/2 [Option ID = 16793]
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- x = t/2 [Option ID = 16790]
- The Newton - Raphson method is used to solve the equation f(x) = x3 - 5x2 + 6x - 8 = 0. Taking the initial guess as x = 5, the solution obtained at the end of the first iteration is [Question ID = 4200]
- 2.2903 [Option ID = 16794]
- 4.515 [Option ID = 16795]
- 4.2903 [Option ID = 16796]
- 2.515 [Option ID = 16797]
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- 4.2903 [Option ID = 16796]
- The inverse Laplace transform of the function
s+5
(s + 1)(s+3) is equal to [Question ID = 4201]- 2e-t + e-3t [Option ID = 16798]
- e-t + 2e-3t [Option ID = 16799]
- e-t- 2e-3t [Option ID = 16800]
- 2e-t-e-3t [Option ID = 16801]
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- 2e-t-e-3t [Option ID = 16801]
- A 4 GHz carrier is amplitude modulated by a low-pass message signal is to be ideally sampled. The minimum frequency of the sampling impulse train should be [Question ID = 4202]
- 4 MHz [Option ID = 16802]
- 8 GHz [Option ID = 16803]
- 4 GHz [Option ID = 16804]
- 8 MHz [Option ID = 16805]
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- 8 MHz [Option ID = 16805]
- ?1?
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A message signal given by mt = ?3? cos?1t amplitude-modulated with a carrier of frequency ?c to generate
s(t) = [1+m(t)]cos?ct. What is the power efficiency achieved by this modulation scheme? [Question ID = 4203]- 20% [Option ID = 16806]
- 11.11% [Option ID = 16807]
- 8.33% [Option ID = 16808]
- 25% [Option ID = 16809]
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- 20% [Option ID = 16806]
- The number of quantization levels with 8-bits required to reduce the quantization noise by a factor of 4 would be [Question ID = 4204]
- 1024 [Option ID = 16810]
- 64 [Option ID = 16811]
- 256 [Option ID = 16812]
- 512 [Option ID = 16813]
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- 512 [Option ID = 16813]
- An ideal band - pass channel 500 Hz - 2000 Hz is deployed for communication. A modem is designed to transmit bits at the rate of 4800 bits/s using 16 - QAM. The roll off factor of a pulse with a raised cosine spectrum that utilizes the entire frequency band is [Question ID = 4205]
- 0.20 [Option ID = 16814]
- 0.25 [Option ID = 16815]
- 0.30 [Option ID = 16816]
- 0.15 [Option ID = 16817]
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- 0.25 [Option ID = 16815]
- sin(pt)
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The Nyquist rate for the signal s(t) = ---------------- is given by
pt [Question ID = 4206]- 1200 Hz [Option ID = 16818]
- 400 Hz [Option ID= 16819]
- 600 Hz [Option ID = 16820]
- 1400 Hz [Option ID = 16821]
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- 1200 Hz [Option ID = 16818]
- The Vestigial Side Band (VSB) modulation is preferred in TV systems because It reduces the bandwidth requirement to half it avoids phase distortion at low frequency [Question ID = 4207]
- 1 only [Option ID = 16822]
- 2 only [Option ID = 16823]
- Neither 1 nor 2 [Option ID = 16824]
- Both 1 and 2 [Option ID = 16825]
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- Neither 1 nor 2 [Option ID = 16824]
- A communication channel distributed by Gaussian noise has a bandwidth of 6 kHz and S/N ratio of 15. The maximum transmission rate that such a channel can support is [Question ID = 4208]
- 48 kbits/s [Option ID = 16826]
- 24 kbits/sec [Option ID = 16827]
- 2.4 kbits/s [Option ID = 16828]
- 32 kbits/s [Option ID = 16829]
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- 24 kbits/sec [Option ID = 16827]
- In the given circuit, the value of Vx is
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5A
0.5 V
10 O
+
V 20 O--- Content provided by FirstRanker.com ---
8O
0.25 V [Question ID = 4209]- 12 V [Option ID = 16830]
- 10 V [Option ID = 16831]
- 0.8 V [Option ID = 16832]
- 8 V [Option ID = 16833]
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- 8V [Option ID = 16833]
- In the circuit shown, the current I flowing through the 50 O resistor will be zero if the value of Capacitor C (in µF) is
5? sin(5000t)
50 O--- Content provided by FirstRanker.com ---
I
1 mH
1 mH
1 mH
C [Question ID = 4210]- 15 µF [Option ID = 16834]
- 20 µF [Option ID = 16835]
- 10 µF [Option ID = 16836]
- 22.5 µF [Option ID = 16837]
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- 20 µF [Option ID = 16835]
- In the circuit shown, the initial voltages across the capacitors C1 and C2 are 1V and 3V, respectively. The switch is closed at time t = 0. The total energy dissipated (in Joules) in the resistor R until steady state is reached, is
+
C1 = 3F
R = 10 O
+--- Content provided by FirstRanker.com ---
C2 = 1F [Question ID = 4211]- 1.5 J [Option ID = 16838]
- 1.0 J [Option ID = 16839]
- 0.5 J [Option ID = 16840]
- 2 J [Option ID = 16841]
- 1.5 J [Option ID = 16838]
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- The driving point impedance of the following network is given by Z(s) = 2+ ---------------------
Z(s)
L
C--- Content provided by FirstRanker.com ---
R
0.2s
+0.1s + 2 The component values are [Question ID = 4212]- L = 0.1 H, R = 0.5 O, C = 5 F [Option ID = 16842]
- L = 5 H, R = 0.5 O, C = 0.1 F [Option ID = 16843]
- L = 0.1 H, R = 2 O, C = 5 F [Option ID = 16844]
- L = 5H, R = 2 O, C = 0.1 F [Option ID = 16845]
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- L = 0.1 H, R = 2 O, C = 5 F [Option ID = 16844]
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