ISRO Electronics & Communication (ECE) 2012 Question Paper & Solutions

Attempt the ISRO Electronics & Communication (ECE) 2012 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.

All ISRO Electronics & Communication (ECE) papers

80 questions.

Question 1

Topic: Electronic Devices

$1~cm^{3}$ of pure Germanium at $20^{\circ}C$ contains about $4.2\times10^{22}$ atoms, $2.5\times10^{13}$ free electrons and $2.5\times10^{13}$ holes. 0.0001% of Arsenic doping donates an extra $10^{17}$ free electrons in the same volume. The approximate number of holes in one $cm^{3}$ in the doped semiconductor under equilibrium condition is

  • A. $6.25\times10^{9}$
  • B. $2.5\times10^{9}$
  • C. $10.5\times10^{9}$
  • D. $10^{17}$

Answer: A. $6.25\times10^{9}$

Explanation: In pure Germanium, $n_{i}=n=p=2.5\times10^{13}$. When it is doped with arsenic, it donates an extra $10^{17}$ free electrons. So, number of electrons now are $n=(2.5\times10^{13}+10^{17})$. $n_{i}$ (intrinsic concentration) will remain same. Using mass action law, $n\cdot p=n_{i}^{2} \Rightarrow (2.5\times10^{13}+10^{17})\times p=(2.5\times10^{13})^{2} \Rightarrow p=6.25\times10^{9}$.

Question 2

Topic: Electronic Devices

Measurement of Hall coefficient enables the determination of

  • A. Temperature coefficient and thermal conductivity
  • B. Mobility and concentration of charge carriers
  • C. Fermi level and forbidden energy gap
  • D. None of the above

Answer: B. Mobility and concentration of charge carriers

Explanation: Hall experiment is used to determine whether a semiconductor is n-type or p-type and the carrier concentration. Also by simultaneously measuring the conductivity $\sigma$, the mobility $\mu$ can be calculated.

Question 3

Topic: Electronic Devices

Silicon is not suitable for fabrication of light emitting diodes because it is

  • A. An indirect band gap semiconductor
  • B. A direct band gap semiconductor
  • C. A wide band gap semiconductor
  • D. A narrow band gap semiconductor

Answer: A. An indirect band gap semiconductor

Explanation: Since silicon is an indirect band gap semiconductor so electron cannot fall directly to the valence band but must undergo a momentum change as well as change in energy. So, energy is released as heat along with light, so not used.

Question 4

Topic: Electronic Devices

The band gap of elements arranged in ascending order is

  • A. Diamond, Ge, Si
  • B. Si, Ge, Diamond
  • C. Ge, Si, Diamond
  • D. Diamond, Si, Ge

Answer: C. Ge, Si, Diamond

Explanation: Band gap of elements at 300 K: $Si \rightarrow 1.1~eV$, $Ge \rightarrow 0.67~eV$, Diamond (C) $\rightarrow 5.5~eV$. The ascending order is Ge, Si, Diamond.

Question 5

Topic: Electronic Devices

The following statements are made for NMOS and PMOS:
1. The carrier mobility in NMOS is higher
2. PMOS require less area than NMOS
3. NMOS circuits are smaller than PMOS
4. PMOS are faster in switching.
Of these, the true statements are:

  • A. 2 and 4 only
  • B. 2 and 3 only
  • C. 1 and 3 only
  • D. 1, 2 and 3 only

Answer: C. 1 and 3 only

Explanation: In n-MOS, conduction is mainly due to electrons and in p-MOS conduction is due to holes. $\mu_{n~si}=3800~cm^{2}/Vsec$, $\mu_{p~si}=1800~cm^{2}/Vsec$. So, higher the mobility, faster is the switching and n-MOS requires far lesser area than p-MOS.

Question 6

Topic: Electronic Devices

The following statements are made for FETs:
1. In n-channel depletion mode MOS in the active region, the control voltage $V_{GS}$ is negative.
2. NMOS in depletion mode is cut off for $V_{GS}=0$
3. NMOS in enhance mode is cut off for $V_{GS}=0$
4. There is no path between source and drain in the enhancement mode
Of these, the true statements are

  • A. 1 and 2 only
  • B. 1 and 3 only
  • C. 3 and 4 only
  • D. 1, 3 and 4 only

Answer: D. 1, 3 and 4 only

Explanation: In depletion mode, the gate voltage is applied such that the channel length reduces, so in n-channel depletion mode MOSFET, the gate voltage $V_{GS}$ is negative. In depletion mode, the channel between source and drain is present, even at $V_{GS}=0$, $I_{D}\ne 0$ or the MOSFET is not cut-off when $V_{GS}=0$. In enhancement mode, the gate voltage is applied such that to create a channel between the source and drain, as no path exists between source and drain initially. So, n-MOS in enhancement mode is cut-off when $V_{GS}=0$.

Question 7

Topic: Electronic Devices

Special handling precautions should be taken when working with MOS devices. The statement which is not true is

  • A. All test equipment should be grounded.
  • B. MOS devices should have their all leads shorted together during shipment and storage.
  • C. Never remove or insert MOS devices with the power on.
  • D. MOS devices do not require grounding straps used for CMOS.

Answer: D. MOS devices do not require grounding straps used for CMOS.

Explanation: All CMOS devices should be placed on ground bench surface and operators should ground themselves prior to handling devices, since a worker can be statically charged with respect to the bench surface. Wrist straps with skin are strongly recommended.

Question 8

Topic: Analog Circuits

If the input to the circuit of Figure is a sine wave, then the output will be

  • A. Half wave rectified sine wave
  • B. A full wave rectified sine wave
  • C. A square wave
  • D. A triangular wave

Answer: C. A square wave

Explanation: Since, the gain of op-amp is always very high, so $V_{out}=+V_{sat}$ when $V_{non-inverting}>V_{inverting}$ and $V_{out}=-V_{sat}$ when $V_{inverting}>V_{non-inverting}$. So, for positive half cycle, $V_{inverting}>V_{non-inverting} \Rightarrow V_{out}=-V_{sat}$. For negative half cycle, $V_{inverting}<V_{non-inverting} \Rightarrow V_{out}=+V_{sat}$. The output wave is a square wave.

Question 9

Topic: Analog Circuits

A high gain Op-Amp has $R_{f}=10^{5}~\Omega$ and $R_{i}=10^{3}~\Omega$. The correct statement is

  • A. The non-inverting gain and inverting gain are nearly same.
  • B. The inverting gain is very much higher than non-inverting gain
  • C. The non-inverting gain is very much higher than inverting gain
  • D. The inverting and non-inverting gains are unrelated

Answer: A. The non-inverting gain and inverting gain are nearly same.

Explanation: $A_{noninverting}=(1+\frac{R_{F}}{R_{i}})=(1+\frac{10^{5}}{10^{3}})=101$. $A_{inverting}=-\frac{R_{F}}{R_{i}}=-\frac{10^{5}}{10^{3}}=-100$. The magnitude of non-inverting and inverting gains are nearly same.

Question 10

Topic: Analog Circuits

For various types of oscillators, the correct statement is

  • A. LC oscillators are more stable than crystal oscillators
  • B. Crystal oscillators have highest Q
  • C. Phase-shift oscillators have the widest range of frequency
  • D. Wien bridge oscillator is used where a single frequency oscillator is required

Answer: B. Crystal oscillators have highest Q

Explanation: Advantage of crystal oscillator: (i) It is one of the most stable oscillator. (ii) High Q-factor of the range of $10^{4}$ to $10^{6}$ compared to $10^{2}$ for LC oscillator. (iii) No ageing effect. Phase shift oscillator is used in audio range frequency.

Question 11

Topic: Analog Circuits

For Butterworth and Chebyshev filters, the correct statement is

  • A. Butterworth response has a sharp cut-off
  • B. Chebyshev response has a flat response in the pass band
  • C. Butterworth response has ripples in the pass band
  • D. Chebyshev response has ripples in the pass band

Answer: D. Chebyshev response has ripples in the pass band

Explanation: Difference between Butterworth and Chebyshev filter: The magnitude response ($|H(\omega)|$) of Butterworth filter decreases with increase in frequency from 0 to $\infty$, while the magnitude response of the Chebyshev filter fluctuates or show ripples in the pass band and stop-band depending on the type of the filter.

Question 12

Topic: Analog Circuits

In a feedback series regulator circuit, the output voltage is regulated by controlling the

  • A. Magnitude of the I/P voltage
  • B. Gain of the feedback transistor
  • C. Voltage drop across the series pass transistor
  • D. Reference voltage

Answer: D. Reference voltage

Explanation: In regulator circuit, output does not change due to the change in the load voltage or the input. But the output can change due to change of reference voltage.

Question 13

Topic: Analog Circuits

In a full-wave rectifier circuit with centre tap transformer, if voltage between one end of secondary winding and centre tap is 300 V peak, then PIV (peak inverse voltage) is

  • A. 300 V
  • B. 150 V
  • C. 600 V
  • D. 900 V

Answer: C. 600 V

Explanation: $PIV=2V_{m}$ (for centre-tap full wave rectifier) $=2\times300=600~V$.

Question 14

Topic: Network Theory

The rms value of the periodic waveform as shown in Figure is

  • A. $\sqrt{3/2}A$
  • B. $\sqrt{2/3}A$
  • C. $\sqrt{1/3}A$
  • D. $\sqrt{2}A$

Answer: B. $\sqrt{2/3}A$

Explanation: $V_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}V^{2}dt}$. $V_{rms}=\sqrt{\frac{1}{T}\left[\int_{0}^{T/2}\left(\frac{2A}{T}t\right)^{2}dt+\int_{T/2}^{T}(-A)^{2}dt\right]} = \sqrt{\frac{1}{T}\left[\frac{4A^{2}}{T^{2}}\left(\frac{t^{3}}{3}\right)_{0}^{T/2}+A^{2}[t]_{T/2}^{T}\right]} = \sqrt{\frac{4A^{2}}{3T^{3}}\cdot\frac{T^{3}}{8}+\frac{A^{2}}{T}\cdot\frac{T}{2}} = A\sqrt{\frac{1}{6}+\frac{1}{2}}=A\sqrt{\frac{2}{3}}$.

Question 15

Topic: Network Theory

The approximate equivalent resistance between terminals A and B for the following infinite ladder network comprising of 1 $\Omega$ and 2 $\Omega$ resistors is

  • A. 1 $\Omega$
  • B. 2 $\Omega$
  • C. 4 $\Omega$
  • D. 0.5 $\Omega$

Answer: B. 2 $\Omega$

Explanation: Let the equivalent resistance be $R_{in}$. Since the ladder is infinite, adding one more stage will not change the equivalent resistance. So, $R_{in}=1+\frac{2R_{in}}{2+R_{in}} = \frac{2+R_{in}+2R_{in}}{2+R_{in}} = \frac{3R_{in}+2}{2+R_{in}}$. $R_{in}^{2}+2R_{in}=3R_{in}+2 \Rightarrow R_{in}^{2}-R_{in}-2=0$. Therefore, $(R_{in}-2)(R_{in}+1)=0 \Rightarrow R_{in}=2~\Omega$.

Question 16

Topic: Network Theory

If the step response to the input step amplitude of 1 V is given by $V_{o}(t)=(1-e^{-t/RC})$, the network can be represented by

  • A. $V_{i}$ $V_{0}$
  • B. $V_{i}$ CR $V_{0}$
  • C. $V_{i}$ RV $V_{0}$
  • D. $V_{i}$ RC WE $V_{0}$

Answer: A. $V_{i}$ $V_{0}$

Explanation: $V_{o}(t)=(1-e^{-t/RC})$, $V_{i}(t)=1~V$. Taking the Laplace T/F of equation (i) $V_{O}(s)=\frac{1}{s}-\frac{1}{s+\frac{1}{RC}} = \frac{1}{s(RCs+1)}$. $V_{i}(s)=\frac{1}{s}$ (Taking the Laplace of equation (ii)). $\frac{V_{o}(s)}{V_{i}(s)}=\frac{1}{RCs+1}$. It can be represented as an RC low pass network with R in series and C in parallel.

Question 17

Topic: Network Theory

For a parallel RC circuit as shown in Figure, $R=20~\Omega$ and $C=0.9~pF$, which is to be matched to $50~\Omega$ over a bandwidth, The circuit has to be matched over 6 GHz to 18 GHz. What is the very best $\Gamma$ that can be achieved

  • A. 0
  • B. 0.05
  • C. 0.1
  • D. 1

Answer: C. 0.1

Explanation: Detailed explanation not provided in the source.

Question 18

Topic: Analog Circuits

In this circuit, V is a sinusoid input with frequency f, current i contains

  • A. Only f
  • B. All harmonics of f
  • C. Only even harmonics of f
  • D. Only odd harmonics of f

Answer: B. All harmonics of f

Explanation: Here, the two diodes are in parallel and connected in opposite directions. So, for positive half cycle, diode $D_{1}$ will allow the voltage to flow. For negative half cycle, diode $D_{2}$ will allow the voltage. Thus, current i contains all harmonics of f.

Question 19

Topic: Communication Systems

The effect of phase error of local oscillator on spectral components in coherent detection of SSB Amplitude modulated signal is

  • A. The frequency of demodulated signals varies with fixed amplitude
  • B. Amplitude of demodulated signals varies with fixed frequency
  • C. Both amplitude and frequency vary
  • D. The recovered signal has phase offset with fixed amplitude

Answer: A. The frequency of demodulated signals varies with fixed amplitude

Explanation: Detailed explanation not provided in the source.

Question 20

Topic: Communication Systems

The maximum transmission power efficiency of DSB-SC Amplitude modulation is

  • A. 25%
  • B. 33.33%
  • C. 50%
  • D. 100%

Answer: D. 100%

Explanation: DSB-SC is an amplitude modulation but without carrier wave, $\eta_{power~efficiency}=\frac{Carrier~power}{Total~transmitted~power} = 0$. Also $\eta_{transmission} + power~efficiency = 1$. $\eta_{transmission~efficiency} = 1 = 100\%$.

Question 21

Topic: Communication Systems

For identical total transmitted power, the ratio of $S/N$ for wide band FM to $S/N$ of DSB-AM (with $m_{a}=1$) is (Given FM = 15 kHz and frequency deviation $=\pm75~kHz$)

  • A. 112.5
  • B. 300
  • C. 75
  • D. 450

Answer: A. 112.5

Explanation: $(FOM)_{wide~bandFM}=\frac{3}{2}\beta^{2}=\frac{(S/N)_{o}}{(S/N)_{i}}$. $(FOM)_{DSBAM}=\frac{1}{3}=\frac{(S/N)_{o}}{(S/N)_{i}}$. Since the total transmitted power is same for both wide band FM and DSB-AM, then $(\frac{S}{N})_{input}$ for both will also be same. $\beta=\frac{\Delta f}{f_{m}}=\frac{75}{15}=5$. $\frac{(S/N)_{wide~band~FM}}{(S/N)_{DSB-AM}}=\frac{3/2\beta^{2}}{1/3} = \frac{9}{2}\times25=112.5$.

Question 22

Topic: Communication Systems

A signal $m(t)=3cos(6000\pi t)+4cos(7000\pi t) + 5cos (10000\pi t)$ to be sampled. What is the minimum sampling rate for band pass sampling for truth full representation?

  • A. 2000 Hz
  • B. 3000 Hz
  • C. 5000 Hz
  • D. 10000 Hz

Answer: D. 10000 Hz

Explanation: $m(t)=3~cos(6000\pi t)+4~cos(7000\pi t) + 5~cos(10000\pi t)$. Minimum sampling rate $= 2~max(f_{m1}, f_{m2}, f_{m3}) = 2\times(5000) = 10000~Hz$.

Question 23

Topic: Digital Circuits

Minimum number of bits required to represent maximum value of given analogue signal with 0.1% accuracy is

  • A. 8 bits
  • B. 9 bits
  • C. 10 bits
  • D. 12 bits

Answer: C. 10 bits

Explanation: Accuracy = 0.1% = 0.001 = $\frac{1}{10^{3}} \approx \frac{1}{2^{n}}$. So, $n \approx 10~bits$.

Question 24

Topic: Communication Systems

An Ideal power limited communication channel with additive white Gaussian noise is having 4 kHz band width and Signal to Noise ratio of 255. The channel capacity is

  • A. 8 kilo bits/sec
  • B. 9.63 kilo bits/sec
  • C. 16 kilo bits/sec
  • D. 32 kilo bits/sec

Answer: D. 32 kilo bits/sec

Explanation: Channel capacity is given as $C=B~log_{2}(1+\frac{S}{N})$. $C = 4\times10^{3}log_{2}(1+255) = 4\times10^{3}log_{2}256 = 4\times10^{3}log_{2}2^{8} = 32~kilo~bits/sec$.

Question 25

Topic: Communication Systems

An Ideal power limited communication channel with additive white Gaussian noise with 1 MHz band width and Signal to Noise ratio of 15 is transmitting the information at theoretical maximum rate. If the Signal to Noise ratio is reduced to 7, how much bandwidth is required to maintain same rate

  • A. $15/7$ MHz
  • B. $4/3$ MHz
  • C. 2 MHz
  • D. None of these

Answer: B. $4/3$ MHz

Explanation: $C_{1}=B_{1}log_{2}(1+\frac{S}{N}) = 10^{6}log_{2}(1+15) = 4\times10^{6}~bits/sec$. $C_{2}=B_{2}log_{2}(1+\frac{S}{N}) = B_{2}log_{2}(1+7) = 3B_{2}~bits/sec$. To maintain the same capacity $C_{1}=C_{2} \Rightarrow 4\times10^{6}=3B_{2} \Rightarrow B_{2}=\frac{4}{3}~MHz$.

Question 26

Topic: Communication Systems

Following constellation diagram represents

  • A. 16 PSK
  • B. 16 QAM
  • C. 16 APSK
  • D. All of above

Answer: B. 16 QAM

Explanation: Detailed explanation not provided in the source.

Question 27

Topic: Communication Systems

The information rate to be transmitted is 900 bits/sec, FEC is $3/4$, the symbol rate of QPSK modulation scheme is

  • A. 225 symbols/sec
  • B. 450 symbols/sec
  • C. 600 symbols/sec
  • D. 675 symbols/sec

Answer: C. 600 symbols/sec

Explanation: Information rate $= 900~bits/sec$. Forward error correction is $3/4$. So, total number of bits/sec $= 900\times\frac{4}{3} = 1200~bits/sec$. Since it is QPSK modulation, so symbol rate is $=\frac{R_{b}}{2} = \frac{1200}{2} = 600~bits/sec$.

Question 28

Topic: Electromagnetics

Electric displacement density D at any point on spherical surface of radius R with a charge Q at the centre in a medium with dielectric constant $\epsilon$ is

  • A. $Q/(4\pi\epsilon~R^{2})$
  • B. $Q/(4\pi R^{2})$
  • C. $(Q/4\pi\epsilon R)^{2}$
  • D. $Q/(4\pi~R)^{2}$

Answer: B. $Q/(4\pi R^{2})$

Explanation: Electric field intensity, E at any spherical surface of radius R is $E = \frac{Q}{4\pi\epsilon R^{2}}$. Also, $D=\epsilon\vec{E}$, so $D=\frac{Q}{4\pi R^{2}}$.

Question 29

Topic: Electromagnetics

Which of the following does not represent Maxwell's Equation?

  • A. $\oint E\cdot ds=-\int\frac{\partial B}{\partial t}\cdot da$
  • B. $\oint D\cdot da=\int\rho\cdot dV$
  • C. $\nabla\times H=B+J$
  • D. $\oint B\cdot da=0$

Answer: C. $\nabla\times H=B+J$

Explanation: Maxwell's equations: $\oint\vec{D}\cdot\vec{ds}=\int\rho_{v}dv$, $\oint\vec{E}\cdot\vec{ds}=-\int\frac{\partial\vec{B}}{\partial t}\cdot ds$, $\oint\vec{B}\cdot\vec{ds}=0$, $\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$, $\nabla\cdot\vec{B}=0$, $\nabla\cdot\vec{D}=\rho_{v}$, $\nabla\times\vec{H}=\vec{J_{C}}+\vec{J_{D}}$ or $\oint\vec{H}\cdot\vec{dl}=\oint\vec{J_{C}}\cdot ds$. So incorrect option is (c).

Question 30

Topic: Electromagnetics

For a homogeneous and isotropic medium with a charge density $\rho$ and dielectric constant $\epsilon$ , the Poisson's equation for electrical potential V is

  • A. $\nabla^{2}V^{\prime}=\frac{\rho}{\epsilon}$
  • B. $\nabla^{2}V^{\prime}=-\frac{\rho}{\epsilon}$
  • C. $\nabla\cdot V=\frac{\rho}{\epsilon}$
  • D. $\nabla\cdot V=-\frac{\rho}{\epsilon}$

Answer: B. $\nabla^{2}V^{\prime}=-\frac{\rho}{\epsilon}$

Explanation: From Gauss law $\nabla\cdot\vec{D}=\rho_{v}$, $\vec{D}=\epsilon\vec{E}$. So, $\nabla\cdot(\epsilon\vec{E})=\rho_{v}$. Since medium is homogeneous and isotropic, so $\nabla\cdot\vec{E}=\frac{\rho_{v}}{\epsilon}$. Also, $\vec{E}=-\frac{\partial V}{\partial r}=-\nabla V$. Put in equation (i) $\nabla\cdot(-\nabla V)=\frac{\rho_{v}}{\epsilon} \Rightarrow \nabla^{2}V=-\frac{\rho_{v}}{\epsilon}$.

Question 31

Topic: Electromagnetics

For a plane travelling EM wave, the correct equation for characteristic impedance Z for the medium with permittivity of $\epsilon$ and permeability of $\mu$ is

  • A. $Z=\sqrt{\frac{\mu}{\epsilon}}$
  • B. $Z=\sqrt{\frac{\epsilon}{\mu}}$
  • C. $Z=\sqrt{(\mu\times\epsilon)}$
  • D. $Z=\frac{1}{\sqrt{\mu\times\epsilon}}$

Answer: A. $Z=\sqrt{\frac{\mu}{\epsilon}}$

Explanation: $Z=\sqrt{\frac{j\omega\mu}{\sigma+j\omega\epsilon}} \rightarrow$ Intrinsic impedance in conducting medium. Now suppose $\sigma=0$ (for lossless or perfectly dielectric medium) $Z=\sqrt{\frac{\mu}{\epsilon}}$.

Question 32

Topic: Electromagnetics

A plane wave propagating in the dielectric medium has an electric field given as $E_{x}=E_{0}cos(2.6\times10^{10}t-100~z)$. The phase velocity of plane wave is

  • A. $10^{6}m/sec$
  • B. $2.6\times10^{8}m/sec$
  • C. $3.0\times10^{8}m/sec$
  • D. $4.0\times10^{8}m/sec$

Answer: B. $2.6\times10^{8}m/sec$

Explanation: $E_{x}=E_{0}cos(2.6\times10^{10}t-100z)$. Comparing it with standard equation $E_{x}=E_{o}cos(\omega t-\beta z)$, $\omega=2.6\times10^{10}rad/sec.$, $\beta=100~rad/m$. Phase velocity, $v_{p}=\frac{\omega}{\beta}=\frac{2.6\times10^{10}}{100} = 2.6\times10^{8}m/sec$.

Question 33

Topic: Electromagnetics

A plane wave propagating in the dielectric medium has an electric field given as $E_{x}=E_{0}cos(3\times10^{10}t-100~z)$. The dielectric constant of the medium is

  • A. 1.33
  • B. 1.00
  • C. 1.01
  • D. 2.15

Answer: B. 1.00

Explanation: $E_{x}=E_{0}cos(3\times10^{10}t-100z)$. $\omega=3\times10^{10}$, $\beta=100$. $v_{\rho}=\frac{\omega}{\beta}=\frac{1}{\sqrt{\mu~\epsilon}}=\frac{c}{\sqrt{\epsilon}}$. $3\times10^{8}=\frac{3\times10^{8}}{\sqrt{\epsilon}} \Rightarrow \sqrt{\epsilon}=1 \Rightarrow \epsilon=1$.

Question 34

Topic: Electromagnetics

A transmission line of $50~\Omega$ characteristic impedance terminated by a load impedance of $50-j50$. The magnitude of reflection coefficient at the load is

  • A. $\sqrt{5/3}$
  • B. $\sqrt{3/5}$
  • C. $3\sqrt{5}$
  • D. None of above

Answer: D. None of above

Explanation: $Z_{L}=50-j50$, $Z_{o}=50$. $\Gamma=\frac{Z_{L}-Z_{0}}{Z_{L}+Z_{0}}=\frac{50-j50-50}{50-j50+50} = \frac{-j50}{100-j50}=\frac{-j}{2-j}$. $|\Gamma|=\frac{1}{\sqrt{5}}$.

Question 35

Topic: Electromagnetics

The outer conductor diameter and the inner conductor diameter dimensions of a coaxial cable are made twice to its original value. (Coaxial line is filled with dielectric material having dielectric constant $\epsilon_{r}=4.0,$). The correct statement is

  • A. impedance remain same to it's original value
  • B. impedance reduces to half of it's original value
  • C. impedance increase and becomes twice of original value
  • D. None of above

Answer: A. impedance remain same to it's original value

Explanation: $Z_{o}=\frac{138}{\sqrt{\epsilon_{r}}}ln\frac{D}{d}$, where, D = Diameter of outer conductor, d = Diameter of inner conductor. Now, when the outer conductor and inner conductor dimensions are doubled, the ratio $D/d$ remains same, so there is no change to the original value of impedance.

Question 36

Topic: Electromagnetics

A non dispersive media has

  • A. Group velocity > phase velocity
  • B. Group velocity < phase velocity
  • C. Group velocity = phase velocity
  • D. None of above

Answer: C. Group velocity = phase velocity

Explanation: In non-dispersive media, the speed of wave is independent of frequency. Here the carrier wave propagates at the phase speed, the modulation envelope propagates at the group speed. Thus, the non-dispersive media, both the phase velocity and group velocity are same.

Question 37

Topic: Electromagnetics

A plane wave is characterized by $\overline{E}=(0.5\hat{x}+\hat{y}e^{j\pi/2})e^{j\omega t-jkz}$. The wave is

  • A. Linearly Polarized
  • B. Circularly Polarized
  • C. Unpolarized
  • D. Elliptically Polarized

Answer: D. Elliptically Polarized

Explanation: Here $\vec{E}=(0.5\hat{x}+\hat{y}e^{j\pi/2})e^{j\omega t-jkz}$. $E_{x}\ne E_{y}$ but $\phi=\frac{\pi}{2}$. So, it is elliptically polarized wave.

Question 38

Topic: Electromagnetics

If a right hand circularly polarized wave is incident on a perfect conductor. The reflected wave is

  • A. Right circularly polarized wave
  • B. Left circularly polarized wave
  • C. Linear polarized wave
  • D. None of above

Answer: B. Left circularly polarized wave

Explanation: If a right hand circularly polarized wave is incident on a perfect conductor, then the reflected wave is left circularly polarized wave.

Question 39

Topic: Electromagnetics

In which of the following cases the divergence of electric field is zero?

  • A. When electric field flows uniform through the tube
  • B. When electric field is released from the tube
  • C. When electric field is within the closed tube
  • D. None of above

Answer: C. When electric field is within the closed tube

Explanation: Divergence of electric field is zero, when the number of electric fields emerging from the tube is equal to incoming field lines. So, the option (c) is correct because all electric field lines are emerging but reside inside the tube.

Question 40

Topic: Electromagnetics

Skin depth at 2 GHz for a gold conductor with $\sigma=4.55\times10^{7}S/m$ is $1.5~\mu m$. Skin depth (in $\mu m$) at 8 GHz and 18 GHz is

  • A. 3.00, 4.50
  • B. 0.75, 0.50
  • C. 0.375, 0.055
  • D. 1.50, 1.00

Answer: B. 0.75, 0.50

Explanation: Here, $\delta=\frac{1}{\sqrt{\pi f\mu\sigma}}$. So, $\delta_{1}=1.5~\mu m$ for $f_{1}=2~GHz$. $\sigma=4.55\times10^{7}S/m$, $\mu=fixed$. $\delta\propto\frac{1}{\sqrt{f}}$. $f_{2}=8GHz$, $f_{3}=18~GHz$. $\frac{\delta_{1}}{\delta_{2}}=\sqrt{\frac{f_{2}}{f_{1}}} \Rightarrow \frac{1.5\times10^{-6}}{\delta_{2}}=\sqrt{\frac{8}{2}} \Rightarrow \delta_{2}=0.75~\mu m$. $\frac{\delta_{1}}{\delta_{3}}=\sqrt{\frac{f_{3}}{f_{1}}} \Rightarrow \frac{1.5\times10^{-6}}{\delta_{3}}=\sqrt{\frac{18}{2}} \Rightarrow \delta_{3}=0.50~\mu m$.

Question 41

Topic: Electromagnetics

The S matrix of a lossless line having length $\lambda/4$ is

  • A. $\begin{bmatrix}1&0\\ 0&1\end{bmatrix}$
  • B. $\begin{bmatrix}1&-1\\ -1&1\end{bmatrix}$
  • C. $\begin{bmatrix}0&j\\ j&0\end{bmatrix}$
  • D. None of above

Answer: D. None of above

Explanation: $S=\begin{bmatrix}0&e^{-j\beta l}\\ e^{-j\beta l}&0\end{bmatrix}$. $l=\frac{\lambda}{4}$. $\beta l=\frac{2\pi}{\lambda}\times\frac{\lambda}{4}=\frac{\pi}{2}$. $e^{-j\beta l}=e^{-j\frac{\pi}{2}}=-j$. $S=\begin{bmatrix}0&-j\\ -j&0\end{bmatrix}$.

Question 42

Topic: Electromagnetics

In the following Smith Chart, the movement from y along a constant VSWR circle to y1 needs addition of

  • A. Capacitance in series with y
  • B. Capacitance in shunt with y
  • C. TRL in series with y
  • D. Stub with y

Answer: B. Capacitance in shunt with y

Explanation: Detailed explanation based on the provided text is implicit in Smith Chart properties, moving from y to y1 requires addition of capacitance in shunt with y.

Question 43

Topic: Electromagnetics

If the RF transmission is terminated in its Characteristic impedance $Z_{C}$, which of the following statements is wrong?

  • A. The input impedance of transmission line becomes $Z_{c}$.
  • B. The transmission line acts as an infinite long transmission line.
  • C. The VSWR becomes one
  • D. The VSWR becomes infinite

Answer: D. The VSWR becomes infinite

Explanation: If the RF transmission is terminated in its characteristic impedance $Z_{0}$ then $Z_{L}=Z_{0}$. Reflection coefficient $\Gamma=\frac{Z_{L}-Z_{o}}{Z_{L}+Z_{o}}=0$. $VSWR=\frac{1+\Gamma}{1-\Gamma}=1$. Therefore, VSWR becoming infinite is the wrong statement.

Question 44

Topic: Electromagnetics

The loss-less, quarter wavelength long, RF transmission of characteristic impedance $50~\Omega$ is terminated in pure resistive load impedance $100~\Omega$, the input impedance is

  • A. $5~\Omega$
  • B. $10~\Omega$
  • C. $25~\Omega$
  • D. $500~\Omega$

Answer: C. $25~\Omega$

Explanation: $Z_{o}=50~\Omega$, $Z_{L}=100~\Omega$. Since the wave is quarter wavelength long line, the input impedance is given as $Z_{in}=\frac{Z_{0}^{2}}{Z_{L}}=\frac{2500}{100} = 25~\Omega$.

Question 45

Topic: Electromagnetics

A lossless matching circuit is shown in the Figure. The values satisfying the matching condition are

  • A. $X_{S1}=-j25.1$, $X_{p}=+j100$, $X_{S2}=-j50$
  • B. $X_{S1}=+j25.1$, $X_{p}=-j100$, $X_{S2}=-j50$
  • C. $X_{S1}=-j25.1$, $X_{p}=-j100$, $X_{S2}=-j50$
  • D. $X_{S1}=+j25.1$, $X_{p}=+j100$, $X_{S2}=+j100$

Answer: B. $X_{S1}=+j25.1$, $X_{p}=-j100$, $X_{S2}=-j50$

Explanation: Detailed explanation not provided in the source.

Question 46

Topic: Electromagnetics

The true statement for wave propagation in rectangular wave guides filled with air is

  • A. The phase velocity is equal to velocity of light in air
  • B. The phase velocity is always more than the velocity of light in air
  • C. The phase velocity is always less than the velocity of light in air
  • D. The phase velocity is undefined.

Answer: B. The phase velocity is always more than the velocity of light in air

Explanation: Let the velocity of light in air is c, $c=\frac{1}{\sqrt{\mu_{0}\epsilon_{0}}}$. Let $v_{p}$ is the phase velocity of light which is propagated down the guide and is given by $v_{p}=\frac{c}{\cos \theta}=\frac{c}{\sqrt{1-(\frac{f_{c}}{f_{0}})^{2}}}$. Since $\cos\theta \le 1$, so $v_{p} \ge c$. Also from the equation, when the wave can propagate in waveguide when $f_{c}>f_{0}$, so since $\frac{f_{c}}{f_{0}}<1$, $v_{p}>c$. Phase velocity is always greater than velocity of light in air.

Question 47

Topic: Electromagnetics

As the frequency approaches to cut-off frequency of wave guide, the correct statements is

  • A. The phase velocity of waves tends to zero
  • B. The phase velocity of waves tends to velocity of light
  • C. The phase velocity of waves tends to infinite
  • D. None of these

Answer: C. The phase velocity of waves tends to infinite

Explanation: Phase velocity, $v_{p}=\frac{c}{\sqrt{1-(\frac{f_{c}}{f_{0}})^{2}}}$. As the wave frequency $f_{o}$ approaches to cut-off frequency $f_{c}$, then $\frac{f_{c}}{f_{0}}$ reaches to 1. Therefore $1-(\frac{f_{c}}{f_{0}})^{2} \approx 0$. So, $v_{p}=\frac{c}{0} \rightarrow \infty$.

Question 48

Topic: Electromagnetics

The EM wave which does not propagate through a rectangular waveguide with $a=30$ mm and $b=15$ mm is

  • A. EM wave of frequency 4.5 GHz
  • B. EM wave of frequency 5.5 GHz
  • C. EM wave of frequency 50 GHz
  • D. EM wave of frequency 300 GHz

Answer: A. EM wave of frequency 4.5 GHz

Explanation: Critical frequency for the dominant mode $TE_{10}$ is $f_{c}=\frac{v_{0}}{2a}=\frac{3\times10^{8}}{2\times30\times10^{-3}}=5~GHz$. So, the waves having the frequency above cut-off frequency will only pass. So, the EM wave of frequency 4.5 GHz will not pass.

Question 49

Topic: Communication Systems

The minimum input signal that gives output signal to noise ratio of 10 dB in a system that has an input impedance equal to $50~\Omega$, a noise figure of 4 dB and bandwidth of 100 kHz is

  • A. $0.32~\mu V$
  • B. $1.41~\mu V$
  • C. $0.245~\mu V$
  • D. None of above

Answer: C. $0.245~\mu V$

Explanation: $R_{in}=50~\Omega$, N.F. = 4 dB, $(\frac{S}{N})_{o}=10dB$, B.W. = 100 kHz. $(\frac{S}{N})_{i}=10~log_{10}N.F + 10~log_{10}(\frac{S}{N})_{o}$. $10~log_{10}S-10~log_{10}N=4+10$, where $N=KTB$. $10~log_{10}N=10~log_{10}KT+10~log_{10}B = -174~dBm+10log_{10}10^{5} = -174+50=-124$. $10~log S=-124+14 = -110 \Rightarrow S=10^{-11}$. $\frac{V^{2}}{R_{in}}=10^{-11} \Rightarrow V=0.245~\mu V$.

Question 50

Topic: Communication Systems

In the given circuit, containing $A_{m}cos(\omega_{m}t)$, $A_{c}cos(\omega_{c}t)$ In-phase Power Divider, and $A_{m}sin(\omega_{m}t)$ with a $-\pi/2$ phase shift, $V_{out}$ is

  • A. Upper Sideband $(\omega_{L}+\omega_{m})$
  • B. Lower Sideband $(\omega_{L}-\omega_{m})$
  • C. Both Upper and Lower Sideband $(\omega_{L}\pm\omega_{m})$
  • D. None of above

Answer: B. Lower Sideband $(\omega_{L}-\omega_{m})$

Explanation: This is SSB-SC modulation with suppressed carrier using phase discriminator method. When the two signals apart from in-phase, then it is SSB-modulation with lower sideband $(\omega_{L}-\omega_{m})$.

Question 51

Topic: Communication Systems

The RF amplifier A is having 3 dB NF and 5 dB of gain, amplifier B is having 3.5 dB NF and 15 dB. For overall noise figure of the cascaded system, the correct statement is

  • A. A as first stage and B as second stage offers minimum overall NF
  • B. B as first stage and A as second stage minimum overall NF
  • C. Both the combination will offer same overall NF
  • D. Overall NF will be less than dB

Answer: A. A as first stage and B as second stage offers minimum overall NF

Explanation: RF amplifier A: $F_{A}=3~dB=10^{0.3}=2$, $G_{A}=5~dB=3.162$. RF amplifier B: $F_{B}=3.5~dB=10^{0.35}=2.2387$, $G_{B}=15~dB=31.6277$. When RF amplifier A is used as Ist stage, $F=F_{A}+\frac{F_{B}-1}{G_{A}} = 2+\frac{1.2387}{3.162}=2.3917$. When RF amplifier B is used as Ist stage, $F=F_{B}+\frac{F_{A}-1}{G_{B}} = 2.2387+\frac{1}{31.6277}=2.270$. Clearly, when RF amplifier A is at first stage, and B as second stage, then minimum N.F. is there. (Note: calculations in source have slight typos but lead to the same comparative conclusion).

Question 52

Topic: Communication Systems

For an RF Circuit with RF = 2.5 GHz, LO = 2.64 GHz, and IF = 140 MHz, the correct statement is

  • A. IF spectrum is inverted and image is at 2.780 GHz
  • B. IF spectrum is not inverted and image is at 2.780 GHz
  • C. IF spectrum is not inverted and image is at 2.360 GHz
  • D. IF spectrum is not inverted and image is at 2.220 GHz

Answer: B. IF spectrum is not inverted and image is at 2.780 GHz

Explanation: $f_{si}$ (image frequency) $= f_{s}+2 I.F = 2.5~GHz+2\times140~MHz = 2.780~GHz$. $I.F. = f_{LO}-f_{s} = (2.64-2.5)GHz = 140~MHz$ and also $f_{LO}>f_{s}$ therefore I.F. spectrum is not inverted.

Question 53

Topic: Electromagnetics

The gain of a parabolic dish antenna of 1m diameter and 80% efficiency at 9.5 GHz is

  • A. 39 dB
  • B. 78 dB
  • C. 40 dB
  • D. 45 dB

Answer: A. 39 dB

Explanation: $Gain=\eta\frac{4\pi A_{e}}{\lambda^{2}}$. $c=f\lambda \Rightarrow \lambda=\frac{3\times10^{8}}{9.5\times10^{9}}=0.0315$. $A_{e}=\frac{\pi D^{2}}{4}=\frac{\pi}{4}$. $Gain=\frac{0.80\times4\pi\times\pi}{4\times(0.0315)^{2}}=7917.62$. Gain in $dB=10log_{10}(7917.62)=39~dB$.

Question 54

Topic: Electromagnetics

In a uniform linear array, four isotropic radiating elements are spaced $\lambda/4$ apart. The progressive phase shift between the elements required for forming the main beam at $60^{\circ}$ of the end fire is

  • A. $-\pi$
  • B. $-\pi/2$
  • C. $-\pi/4$
  • D. $-\pi/8$

Answer: C. $-\pi/4$

Explanation: We have, $\varphi=\beta d~cos~\theta+\alpha$. For forming the main beam at $60^{\circ}$ ($0 = \beta d\cos 60^{\circ} + \alpha$), at $\theta=60^{\circ}$ and $d=\lambda/4$, $\varphi=\frac{2\pi}{\lambda}\times\frac{\lambda}{4}[\cos 60^{\circ}-1]$ (if relative to end-fire maximum calculation difference, source states $\alpha=-\beta d$, then $\varphi = \beta d (\cos\theta - 1)$ resulting in $-\pi/4$ radians).

Question 55

Topic: Electromagnetics

The radiation resistance of a thin line centre fed $\lambda/2$ antenna with sinusoid current distribution is

  • A. $73~\Omega$
  • B. $73+j42.5~\Omega$
  • C. $50~\Omega$
  • D. $75~\Omega$

Answer: A. $73~\Omega$

Explanation: Detailed explanation not provided in the source.

Question 56

Topic: Electromagnetics

To increase gain of a parabolic reflector antenna by 6 dBi, the correct statement is (Assuming efficiency 70%)

  • A. Diameter is to be increased to 6 folds
  • B. Diameter is to be increased to 4 folds
  • C. Diameter is to be increased to 2 folds
  • D. None of above

Answer: C. Diameter is to be increased to 2 folds

Explanation: $G=\eta\frac{4\pi A_{e}}{\lambda^{2}}=\eta\frac{4\pi\cdot\pi D^{2}}{4} = X~dB$ (suppose). If the diameter is increased to 2 folds i.e. $D'=2D$ then gain in dB will increase by $10~log_{10}4 = 6.02~dBi$.

Question 57

Topic: Electromagnetics

The waveguide Magic Tee can be used as

  • A. An Isolator
  • B. T/R switch in trans-receiver
  • C. Adjustable phase shifter
  • D. All of above

Answer: D. All of above

Explanation: The Magic-Tee is a four port rectangular waveguide used to provide perfect isolation between the ports. It can be used as isolator, T/R switch in trans-receiver, adjustable phase-shifter.

Question 58

Topic: Electromagnetics

The microwave device working on the principle of Faraday rotation is called as

  • A. Rotary joint
  • B. Circulator
  • C. Isolator
  • D. None of above

Answer: B. Circulator

Explanation: A circulator is a passive non-reciprocal three or four port device works on the principle of Faraday rotation.

Question 59

Topic: Electromagnetics

The phase difference between adjacent anode poles of Magnetron is generally

  • A. zero radians
  • B. $\pi/4$ radians
  • C. $\pi/2$ radians
  • D. $\pi$ radians

Answer: D. $\pi$ radians

Explanation: Reentrant oscillations are possible only if the total phase shift around the structure is an integer of $2\pi$ radians. i.e. $\phi_{n}=\frac{2\pi m}{N}$, N = number of reentrant cavities. Magnetron oscillations are ordinarily operated in $\pi$-mode.

Question 60

Topic: Electronic Devices

Which of the following devices cannot be used as microwave oscillator

  • A. IMPATT diode
  • B. PN Junction diode
  • C. TRPATT diode
  • D. BARRITT diode

Answer: B. PN Junction diode

Explanation: P-N junction diode works as a rectifier.

Question 61

Topic: Control Systems

For a Phase Locked Loop system, the open loop transfer function is $G(s)=\frac{1000}{s(\frac{s}{1192}+1)}$. The phase margin is $50^{\circ}$ and the crossover frequency is approximately $1000~rad/s$. A time delay is introduced in the phase detector reducing the phase margin to $40^{\circ}$. The maximum permissible time delay (in seconds) is

  • A. $10^{-4}$
  • B. 0.174
  • C. $1.192\times10^{-3}$
  • D. $0.174\times10^{-3}$

Answer: C. $1.192\times10^{-3}$

Explanation: Detailed step-by-step mathematical derivation for this specific option is obfuscated in the source text, but the maximum permissible phase delay formula applies the change in phase margin over the crossover frequency $\omega_c$.

Question 62

Topic: Analog Circuits

In this circuit, the relationship between $V_{i}$ and $V_{0}$ is

  • A. $V_{o}=V_{i}e^{V_{o}/V_{\tau}}$
  • B. $V_{0}=V_{T}ln\frac{V_{i}}{R_{s}I_{S}}$
  • C. $V_{o}=V_{i}-I_{C}R_{S}$
  • D. None of above

Answer: D. None of above

Explanation: $I_{1}=\frac{V_{i}-0}{R_{s}}=I_{c}$. $I_{C}=I_{s}e^{\frac{V_{BE}}{V_{T}}}$. $V_{BE}=V_{T}ln\frac{I_{C}}{I_{S}}$. Also $V_{o}=-V_{BE} = -V_{T}ln\frac{I_{1}}{I_{S}} = -V_{T}ln\frac{V_{i}}{I_{S}R_{s}}$.

Question 63

Topic: Analog Circuits

Output impedance of a source follower is

  • A. $1/g_{m}+r_{d}$
  • B. $r_{d}/(1+g_{m}\cdot r_{d})$
  • C. $r_{d}$
  • D. None of these

Answer: B. $r_{d}/(1+g_{m}\cdot r_{d})$

Explanation: Out impedance $=\frac{r_{d}}{1+\mu}$. $R_{o}=\frac{r_{d}}{1+g_{m}r_{d}}$.

Question 64

Topic: Control Systems

For a negative feedback system shown in Figure, the equivalent transfer function is

  • A. $\frac{s^{3}+5s^{2}+6s}{s^{3}+7s^{2}+12s+3}$
  • B. $\frac{s^{3}+5s^{2}+6s}{s^{3}+5s^{2}+4s-3}$
  • C. $\frac{s^{2}+5s+4}{s^{3}+7s^{2}+12s+3}$
  • D. $\frac{s^{2}+5s+4}{s^{3}+5s^{2}+4s-3}$

Answer: C. $\frac{s^{2}+5s+4}{s^{3}+7s^{2}+12s+3}$

Explanation: $\frac{C(s)}{R(s)}=\frac{G(s)}{1+G(s)H(s)} = \frac{\frac{s+1}{s(s+2)}}{1+\frac{s+1}{s(s+2)}\cdot\frac{s+3}{s+4}} = \frac{(s+1)(s+4)}{s(s^{2}+6s+8)+(s^{2}+4s+3)} = \frac{s^{2}+5s+4}{s^{3}+6s^{2}+8s+s^{2}+4s+3} = \frac{s^{2}+5s+4}{s^{3}+7s^{2}+12s+3}$.

Question 65

Topic: Digital Circuits

If $(146)_{x}+(313)_{x-2}=(246)_{8},$ then the value of base x is

  • A. 5
  • B. 6
  • C. 7
  • D. 9

Answer: C. 7

Explanation: $(146)_{x}+(313)_{x-2}=(246)_{8}$. Converting into common base 10 as $x^{2}+4x+6+3(x-2)^{2}+1(x-2)+3=2\times8^{2}+4\times8+6 \Rightarrow 4x^{2}-7x-147=0 \Rightarrow x=7$.

Question 66

Topic: Digital Circuits

The truth table for implementing a boolean variable F is given by ... where d represents don't care states. The minimized expression for F is

  • A. $B\cdot\overline{C}+A\cdot\overline{C}+A\cdot B$
  • B. $A\cdot B+\overline{C}$
  • C. $\overline{A}\cdot B\cdot\overline{C}+A\cdot\overline{B}\cdot\overline{C}+A\cdot B\cdot C$
  • D. None of above

Answer: B. $A\cdot B+\overline{C}$

Explanation: Solving via K-map using the given truth table minterms and don't care conditions results in $F = \overline{C}+AB$.

Question 67

Topic: Digital Circuits

The state transition diagram for a sequence generator is shown in Figure. It is designed using D F/Fs and combinational logic blocks $L_{1}$, $L_{2}$ and $L_{3}$ and is initialized at $(Q_{2}=1, Q_{1}=0, Q_{0}=1)$. The minimized expressions for $O_{0}$, $O_{1}$ and $O_{2}$ are

  • A. $O_{2}=Q_{2}; O_{1}=Q_{1}; Q_{0}=Q_{0}$
  • B. $O_{2}=Q_{1}\cdot\overline{Q}_{0};O_{1}=\overline{Q}_{0}\cdot Q_{1}\cdot\overline{Q}_{2}+Q_{0}\cdot\overline{Q}_{1}\cdot Q_{2}; O_{0}=\overline{Q}_{0}\cdot Q_{1}\cdot Q_{2}$
  • C. $O_{2}=Q_{1};O_{1}=Q_{0}+\overline{Q}_{2};O_{0}=Q_{1}\cdot Q_{2}$
  • D. None of above

Answer: C. $O_{2}=Q_{1};O_{1}=Q_{0}+\overline{Q}_{2};O_{0}=Q_{1}\cdot Q_{2}$

Explanation: From the state transition diagram, writing the present state and next state equations and solving the K-maps for inputs $D_2, D_1, D_0$ gives $D_2 = Q_1$, $D_1 = \overline{Q}_2 + Q_0$, and $D_0 = Q_2 Q_1$.

Question 68

Topic: Digital Circuits

The number of comparators needed in a 8-bit flash type A to D converter is

  • A. 8
  • B. 16
  • C. 255
  • D. 256

Answer: C. 255

Explanation: The number of comparators in n-bit flash type ADC is $=2^{n}-1$. For $n=8$, number of comparators $=2^{8}-1 = 255$.

Question 69

Topic: Microprocessors

The register in the 8085 microprocessor that is used to keep track of the memory address of next op-code to be run in the program is the

  • A. Stack pointer
  • B. Program counter
  • C. Accumulator
  • D. None of above

Answer: B. Program counter

Explanation: Program counter is a 16-bit register which stores the address of the next instruction to be executed in the memory.

Question 70

Topic: Microprocessors

The Complement Accumulator (CMA) instruction of 8085 processor on execution affects

  • A. Zero Flag
  • B. Sign Flag
  • C. Carry Flag
  • D. None of these

Answer: D. None of these

Explanation: CMA instruction does not affect any flag but it just complements the contents of the accumulator.

Question 71

Topic: Microprocessors

"Cycle Stealing" in microprocessor parlance refers to

  • A. Special type of DMA access by an external device
  • B. RESET operation at power ON by microprocessor
  • C. Interrupt Acknowledge cycle
  • D. None of above

Answer: A. Special type of DMA access by an external device

Explanation: Cycle stealing is a method of accessing RAM without interfacing with CPU. It is similar to DMA for allowing I/O controllers to read or write without CPU intervention.

Question 72

Topic: Computer Organization

Principle of "locality" is used in context of

  • A. Addressing lowest Memory address by microprocessor
  • B. Addressing Highest Memory address by microprocessor
  • C. Accessing Cache memory locations
  • D. None of these

Answer: C. Accessing Cache memory locations

Explanation: Principle of locality is a phenomenon describing the same value, or related storage, being frequently accessed for performance optimization such as cache.

Question 73

Topic: Electromagnetics

If A, B and C are vectors and $A=1\cdot u_{x}+2\cdot u_{y}+3\cdot u_{z},$ $B=1\cdot u_{x}+1\cdot u_{y}+1\cdot u_{z}$ and $C=3\cdot u_{x}+2\cdot u_{y}+1\cdot u_{z}$ then $(A\times B)\cdot C$ is

  • A. 2
  • B. 1
  • C. 0
  • D. -1

Answer: C. 0

Explanation: $\vec{A}=1\hat{u}_{x}+2\hat{u}_{y}+3\hat{u}_{z}$, $\vec{B}=1\hat{u}_{x}+1\hat{u}_{y}+1\hat{u}_{z}$, $\vec{C}=3\hat{u}_{x}+2\hat{u}_{y}+1\hat{u}_{z}$. $(\vec{A}\times\vec{B}) = -\hat{u}_{x}+2\hat{u}_{y}-\hat{u}_{z}$. $(\vec{A}\times\vec{B})\cdot\vec{C} = (-\hat{u}_{x}+2\hat{u}_{y}-\hat{u}_{z})\cdot(3\hat{u}_{x}+2\hat{u}_{y}+1\hat{u}_{z}) = 0$.

Question 74

Topic: Mathematics

The system of equations $x+y+z=6$, $2x+y+z=7$, $x+2y+z=8$ has

  • A. A unique solution
  • B. No solution
  • C. An infinite number of solutions
  • D. None of these

Answer: A. A unique solution

Explanation: Forming the augmented matrix $[A:B] = \begin{bmatrix}1&1&1&6\\ 2&1&1&7\\ 1&2&1&8\end{bmatrix}$. Applying row operations $R_{2}\leftarrow R_{2}-2R_{1}$ and $R_{3}\leftarrow R_{3}-R_{1}$ gives $\begin{bmatrix}1&1&1&6\\ 0&-1&-1&-5\\ 0&1&0&2\end{bmatrix}$. Here Rank of A = Rank of AB = Number of variables. Thus, it is a unique solution.

Question 75

Topic: Mathematics

If $x=e^{y+e^{y+e^{y+e^{y+...}}}}$ then $\frac{dy}{dx}$ is

  • A. $(1-x)$
  • B. $(1-x)/x$
  • C. $1/x$
  • D. $x/(1-x)$

Answer: B. $(1-x)/x$

Explanation: $x=e^{y+e^{y+e^{y+...}}}$ It can also be written as $x=e^{y+x}$. Taking log of both sides $log x = y+x$. Differentiating both sides $\frac{1}{x}=\frac{dy}{dx}+1 \Rightarrow \frac{dy}{dx}=\frac{1-x}{x}$.

Question 76

Topic: Mathematics

The area bounded by the curves $y^{2}=x$, $y=x$ is given by

  • A. $2/3$
  • B. $1/2$
  • C. $1/6$
  • D. $1/3$

Answer: C. $1/6$

Explanation: $A=\int_{0}^{1}\int_{x}^{\sqrt{x}}dy dx=\int_{0}^{1}[y]_{x}^{\sqrt{x}}dx=\int_{0}^{1}[\sqrt{x}-x]dx=[\frac{2}{3}x^{3/2}-\frac{x^{2}}{2}]_{0}^{1}=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$.

Question 77

Topic: Mathematics

The eigen values of the matrix $\begin{bmatrix}cos~\alpha & sin~\alpha \\ sin~\alpha & cos~\alpha\end{bmatrix}$ are

  • A. $\pm cos~\alpha$
  • B. $\pm sin~\alpha$
  • C. $tan~\alpha$ & $cot~\alpha$
  • D. $cos~\alpha \pm sin~\alpha$

Answer: D. $cos~\alpha \pm sin~\alpha$

Explanation: Let $A=\begin{bmatrix}cos~\alpha & sin~\alpha \\ sin~\alpha & cos~\alpha\end{bmatrix}$. Eigen values are given as $|sI-A|=0 \Rightarrow \begin{vmatrix}s-cos~\alpha & -sin~\alpha \\ -sin~\alpha & s-cos~\alpha\end{vmatrix}=0 \Rightarrow (s-cos~\alpha)^{2}-sin^{2}\alpha=0 \Rightarrow (s-cos~\alpha)=\pm sin~\alpha \Rightarrow s=cos~\alpha\pm sin~\alpha$.

Question 78

Topic: Signals and Systems

The Laplace Transform of $x(t)$ formed by convolution operator is $x(t)=\{t\cdot u(t)\}\otimes\{cos~2\pi t\cdot u(t)\}$

  • A. $1/[s(s^{2}+4\pi^{2})]$
  • B. $2\pi/[s(s^{2}+4\pi^{2})]$
  • C. $1/[s^{2}(s^{2}+4\pi^{2})]$
  • D. $s^{3}/(s^{2}+4\pi^{2})$

Answer: A. $1/[s(s^{2}+4\pi^{2})]$

Explanation: $x(t)=\{t\cdot u(t)\}\otimes\{cos~2\pi t~u(t)\}$. In the Laplace domain, convolution becomes multiplication: $X(s)=\frac{1}{s^{2}}\cdot\frac{s}{s^{2}+(2\pi)^{2}}=\frac{1}{s(s^{2}+4\pi^{2})}$.

Question 79

Topic: Mathematics

A random variable X has $\overline{X}=0$ and $\sigma_{X}^{2}=1$. Form a new random variable $Y=2X+1$. The values of $\overline{Y}$ and $\sigma_{Y}^{2}$ are

  • A. 0 and 1
  • B. 1 and 2
  • C. 1 and 4
  • D. None of above

Answer: C. 1 and 4

Explanation: Let $Y=2X+1$. $\overline{Y}=E[Y]=E[2X+1]=2E[X]+1 = 2(0)+1=1$. $\sigma_{y}^{2}=V[Y]=V[2X+1]=2^{2}V[X]=4(1)=4$.

Question 80

Topic: Mathematics

Person X can solve 80% of the ISRO question paper and Person Y can solve 60%. The probability that at least one of them will solve a problem from the question paper, selected at random is

  • A. 0.48
  • B. 0.70
  • C. 0.88
  • D. 0.92

Answer: D. 0.92

Explanation: $P(X)=0.80$, $P(Y)=0.60$. P(at least one can solve) $=P(X)+P(Y)-P(X\cap Y) = 0.80+0.60-(0.80\times0.60) = 1.40-0.48 = 0.92$.