ISRO Electronics & Communication (ECE) 2017 Question Paper & Solutions

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All ISRO Electronics & Communication (ECE) papers

80 questions.

Question 1

Topic: Analog Circuits

If the transformer and diodes in the following circuit are ideal, Find out the value of Capacitor (Co) provides 5% ripple voltage across RL. (Vin = 100sin(2\pi × 1000 × t), Np = 10, Ns1 = 5, Ns2 = 5, RL = 1kΩ)

  • A. 4.7 μF
  • B. 8.7 μF
  • C. 10 μF
  • D. 12 μF

Answer: C. 10 μF

Explanation: For a Full Wave Rectifier with a C-filter, the ripple factor is given by $\gamma = \frac{1}{4\sqrt{3}fCR_{L}}$. Alternatively, using peak values, $V_{r(rms)} = \frac{V_{max}}{4\sqrt{3}fCR_L}$. Given $f = 1000$ Hz, $R_{L} = 1000 \Omega$, and $\gamma = 5\%$. The capacitance is calculated as $C = \frac{50}{2 \times 100 \times 2.5 \times 1000} = 10 \mu F$ based on the voltage drop parameters.

Question 2

Topic: Microwave Engineering

Relationship between doppler frequency shifts of two radars A and B having 0.1 foot and 0.05 foot wavelengths, approaching the target at 1000 feet per second and 2000 feet per second rate respectively, will be

  • A. Doppler frequency shift of radar A will be one-fourth of doppler frequency shift of radar B
  • B. Doppler frequency shift of radar A will be one-half of doppler frequency shift of radar B
  • C. Doppler frequency shift of radar A will be double of doppler frequency shift of radar B
  • D. Doppler frequency shifts of radar A and radar B will be same

Answer: A. Doppler frequency shift of radar A will be one-fourth of doppler frequency shift of radar B

Explanation: Doppler shift for Radar A: $f_{D,A} = \frac{2v_{A}}{\lambda_{A}} = \frac{2 \times 1000}{0.1} = 2 \times 10^4$ Hz. Doppler shift for Radar B: $f_{D,B} = \frac{2v_{B}}{\lambda_{B}} = \frac{2 \times 2000}{0.05} = 8 \times 10^4$ Hz. Therefore, the Doppler shift of Radar A is one-fourth of the Doppler shift of Radar B.

Question 3

Topic: Analog Circuits

A sinusoidal input which can be reproduced in an OP-AMP without any distortion having slew rate of $10\pi~V/\mu s$ and 5V peak output amplitude, has the maximum frequency of

  • A. 1 kHz
  • B. 1 MHz
  • C. 31.42 kHz
  • D. 31.42 MHz

Answer: B. 1 MHz

Explanation: The maximum frequency avoiding slew rate distortion is given by $f_{m} = \frac{SR}{2\pi A_m}$. Given $SR = 10\pi~V/\mu s = 10\pi \times 10^6$ V/s, and $A_m = 5$ V. Thus, $f_{m} = \frac{10\pi \times 10^6}{2\pi \times 5} = 10^6$ Hz = 1 MHz.

Question 4

Topic: Electronic Devices

Which of the following is NOT a characteristic of Schottky Diode ?

  • A. Thermoinic emission of carriers across Schottky barrier
  • B. Current conduction in Schottky diodes is by majority carriers
  • C. Switching speed of Schottky diodes is less compared to p-n junction diodes
  • D. Schottky diode comprises of Metal-Semiconductor junction

Answer: C. Switching speed of Schottky diodes is less compared to p-n junction diodes

Explanation: A Schottky diode is formed by a metal-semiconductor junction. Because conduction is primarily by majority carriers, there is negligible minority charge carrier storage time. This makes its switching speed much higher, not lower, than that of a standard p-n junction diode.

Question 5

Topic: Material Science

The temperature below which certain materials are antiferromagnetic and above which they are paramagnetic is called

  • A. Weiss temperature
  • B. Curie temperature
  • C. Neel temperature
  • D. None of the above

Answer: C. Neel temperature

Explanation: The Neel temperature ($T_N$) is the critical temperature above which an antiferromagnetic material becomes paramagnetic. Below this temperature, the material's magnetic susceptibility decreases as temperature drops.

Question 6

Topic: Material Science

In metals, the thermal conductivity K and electrical conductivity $\sigma$ are related as $\frac{K}{\sigma T}=L$. L is known as

  • A. Lattice constant
  • B. Lorenz number
  • C. Lanevin function
  • D. Larmor number

Answer: B. Lorenz number

Explanation: The Wiedemann-Franz law states that the ratio of the electronic contribution of the thermal conductivity ($K$) to the electrical conductivity ($\sigma$) of a metal is proportional to the temperature ($T$). The proportionality constant $L$ is known as the Lorenz number.

Question 7

Topic: Material Science

In a specimen of ferromagnetic material with saturation magnetization as 8000 Gauss, as the flux density is increased from 0 to 2.5 T, $\mu_{r}$ will

  • A. Increase
  • B. Decrease
  • C. First decrease then increase
  • D. First increase then decrease

Answer: D. First increase then decrease

Explanation: From the B-H curve of a ferromagnetic material, the relative permeability $\mu_{r}$ is proportional to the slope ($B/H$). Initially, as H increases, the slope steepens ($\mu_r$ increases) until it reaches the knee of the curve, after which the material begins to saturate and the slope flattens ($\mu_r$ decreases).

Question 8

Topic: Microwave Engineering

The cavity magnetron uses strapping to

  • A. Prevent mode jumping
  • B. Prevent cathode back-heating
  • C. Ensure bunching
  • D. Improve the phase focussing effect

Answer: A. Prevent mode jumping

Explanation: Strapping in a cavity magnetron involves connecting alternate anode poles with conductive rings. This helps to maintain the dominant $\pi$-mode of oscillation and suppresses adjacent, unwanted modes, thereby preventing mode jumping.

Question 9

Topic: Microwave Engineering

The TWT is sometimes preferred to the magnetron as a radar transmitter output tube because it is

  • A. Capable of a larger duty cycle
  • B. A more efficient amplifier
  • C. More broadband
  • D. Less noisy

Answer: C. More broadband

Explanation: Traveling Wave Tubes (TWTs) are non-resonant microwave amplifiers. Because they do not rely on resonant cavities for operation like klystrons or magnetrons, they are capable of operating over a much broader bandwidth.

Question 10

Topic: Microwave Engineering

One of the reason why conventional vacuum valve tubes not used at microwave frequencies is that their

  • A. Noise increase
  • B. It has less transit time
  • C. Shunt capacitive reactance become too large
  • D. Series induction reactance become too small

Answer: A. Noise increase

Explanation: At microwave frequencies, conventional vacuum tubes suffer from inter-electrode capacitance effects and electron transit-time effects. Additionally, the noise generated within the tube tends to increase significantly with frequency, degrading performance.

Question 11

Topic: Network Theory

Y-parameter of a two port network is $\begin{bmatrix}5 & 2 \\ 1 & 3\end{bmatrix}$. A $1~\Omega$ resistor is connected in parallel to the network as shown. Find out the Y parameter of the whole network.

  • A. $\begin{bmatrix}6 & 1 \\ 0 & 4\end{bmatrix}$
  • B. $\begin{bmatrix}3 & 1 \\ 1 & 2\end{bmatrix}$
  • C. $\begin{bmatrix}4 & 1 \\ 0 & 2\end{bmatrix}$
  • D. $\begin{bmatrix}6 & 3 \\ 2 & 4\end{bmatrix}$

Answer: A. $\begin{bmatrix}6 & 1 \\ 0 & 4\end{bmatrix}$

Explanation: When two-port networks are connected in parallel, their Y-parameter matrices add up. The Y-parameter matrix for a series resistor $R=1~\Omega$ connected between input and output ports is $\begin{bmatrix}1/R & -1/R \\ -1/R & 1/R\end{bmatrix} = \begin{bmatrix}1 & -1 \\ -1 & 1\end{bmatrix}$. Adding this to the original matrix gives $\begin{bmatrix}5+1 & 2-1 \\ 1-1 & 3+1\end{bmatrix} = \begin{bmatrix}6 & 1 \\ 0 & 4\end{bmatrix}$.

Question 12

Topic: Network Theory

For the circuit shown, the impedance $Z_{1}$ looking into port 1 is $K_{1}\frac{s+2}{s+5}$. Find the impedance $Z_{2}$ looking into the output port, where $K_{1}$ and $K_{2}$ are constants.

  • A. $K_{2}\frac{s}{s+5}$
  • B. $\frac{s+5}{s K_{2}}$
  • C. $K_{2}\frac{s}{s+6}$
  • D. $\frac{s+6}{s K_{2}}$

Answer: A. $K_{2}\frac{s}{s+5}$

Explanation: $Z_{1}$ is the equivalent impedance from port 1, which equals $R_1 || (R_2 + Ls) = \frac{R_1(s + R_2/L)}{s + (R_1+R_2)/L}$. Matching with the given expression, $R_2/L = 2$ and $(R_1+R_2)/L = 5$. $Z_{2}$ from the other side is $(R_1+R_2) || Ls = \frac{s L(R_1+R_2)}{sL + R_1 + R_2} = \frac{s(R_1+R_2)}{s + 5}$. This takes the form $K_{2}\frac{s}{s+5}$.

Question 13

Topic: Communication Systems

The R=1/3 convolution encoder defined by transfer functions $H1(z) = 1 + z^{-1}$, $H2(z) = 1 + z^{-2}$, $H3(z) = 1 + z^{-1} + z^{-2}$ is

  • A. Recursive and K = 3
  • B. Systematic and K = 2
  • C. Non-recursive and K = 3
  • D. Non-recursive and K = 2

Answer: C. Non-recursive and K = 3

Explanation: Since all transfer functions are finite polynomials in $z^{-1}$, there is no feedback loop, meaning it is a Non-Recursive encoder. The highest degree of $z^{-1}$ is 2, indicating 2 memory elements ($m=2$). The constraint length $K$ is $m + 1 = 3$.

Question 14

Topic: Signals and Systems

The match filter response $h(t)$ for a given input signal $S(t)$ sampled at $t=T$ is

  • A. $h(t) = S(T-t)$
  • B. $h(t) = S(t-T)$
  • C. $h(t) = S(-t)$
  • D. $h(t) = S(t)$

Answer: A. $h(t) = S(T-t)$

Explanation: By definition, the impulse response of a matched filter designed to maximize the signal-to-noise ratio at sampling time $t=T$ is the time-reversed and shifted version of the input signal: $h(t) = S(T-t)$.

Question 15

Topic: Digital Circuits

Consider a logic circuit with 4 inputs $I_{0}, I_{1}, I_{2}, I_{3}$ producing outputs A and B such that $A = \overline{I}_{3}\overline{I}_{2}I_{1} + I_{3}$ and $B = \overline{I}_{3}I_{2} + I_{3}$. The circuit acts as a

  • A. 4:1 Multiplexer
  • B. De-Multiplexer
  • C. BCD circuit
  • D. 4 to 2 Priority Encoder

Answer: D. 4 to 2 Priority Encoder

Explanation: The given boolean expressions denote the logic for a 4-to-2 Priority Encoder. The output $B$ handles the MSB which is high if $I_3$ or $I_2$ is high. Output $A$ is high if $I_3$ is high, or if $I_3, I_2$ are low and $I_1$ is high. This follows the standard priority encoding scheme where higher indexed inputs take precedence.

Question 16

Topic: Digital Circuits

For a logic circuit with inputs A, B, and C, if the output is $Z = Y_3 + Y_2$ where intermediate signals are $Y_3 = A \cdot B \cdot (\overline{A \cdot \overline{C}})$ and $Y_2 = A \cdot B \cdot C$, the simplified expression for Z is

  • A. $\overline{A\overline{C}}AB$
  • B. $ABC$
  • C. $ABC + ACB$
  • D. $ABC + CB$

Answer: B. $ABC$

Explanation: We can simplify the boolean expression step by step. First, simplify $Y_3$: $Y_3 = AB(\overline{A\overline{C}})$. Applying De Morgan's law: $Y_3 = AB(\overline{A} + C) = AB\overline{A} + ABC$. Since $A\overline{A} = 0$, $Y_3 = ABC$. We are given $Y_2 = ABC$. The final output is $Z = Y_3 + Y_2 = ABC + ABC = ABC$.

Question 17

Topic: Digital Circuits

Consider a shift register where the frequency of input signal x is $f_s$ and the shift register is clocked at the positive edge of $2f_s$. The time offset between the output A (after the 1st stage) and output B (after the 3rd stage) is

  • A. $1/(2f_s)$
  • B. $1/f_s$
  • C. $3/(2f_s)$
  • D. $1/(4f_s)$

Answer: B. $1/f_s$

Explanation: The clock frequency is $2f_s$, which means the time period of one clock cycle is $T_{clk} = 1/(2f_s)$. Data reaches output A after 1 clock cycle ($T_{clk}$) and reaches output B after 3 clock cycles ($3T_{clk}$). The time offset between A and B is $3T_{clk} - T_{clk} = 2T_{clk} = 2 \times 1/(2f_s) = 1/f_s$.

Question 18

Topic: Microwave Engineering

An air filled rectangular waveguide R1 is operating at the frequency 2 GHz and another air filled rectangular waveguide R2 is operating at 4 GHz. The guide wavelengths of these waveguides at their respective frequencies are equal. If the cut-off frequency of waveguide R1 is 1 GHz, what is the cut-off frequency of the waveguide R2 in GHz?

  • A. $\sqrt{10}$
  • B. $\sqrt{11}$
  • C. $\sqrt{12}$
  • D. $\sqrt{13}$

Answer: D. $\sqrt{13}$

Explanation: The guide wavelength is given by $\lambda_g = \frac{\lambda}{\sqrt{1 - (f_c/f)^2}}$. For R1: $f_1 = 2\text{ GHz}$, $f_{c1} = 1\text{ GHz}$, so $\lambda_1 = C/2$. The guide wavelength is $\lambda_{g1} = \frac{C/2}{\sqrt{1 - (1/2)^2}} = \frac{C}{2\sqrt{3/4}} = \frac{C}{\sqrt{3}}$. For R2: $f_2 = 4\text{ GHz}$, so $\lambda_2 = C/4$. Equating $\lambda_{g1} = \lambda_{g2}$, we get $\frac{C}{\sqrt{3}} = \frac{C/4}{\sqrt{1 - (f_{c2}/4)^2}}$. Solving this yields $1 - (f_{c2}/4)^2 = 3/16$, leading to $(f_{c2})^2 = 13$, so $f_{c2} = \sqrt{13}\text{ GHz}$.

Question 19

Topic: Electromagnetics

An electromagnetic wave propagates through a lossless insulator with a velocity $1.5 \times 10^{10}\text{ cm/s}$. Calculate the relative permittivity ($\epsilon_r$) and relative permeability ($\mu_r$) of the insulator if its intrinsic impedance is $90\pi\text{ ohms}$.

  • A. $\epsilon_r = 2.66, \mu_r = 1.5$
  • B. $\epsilon_r = 1.5, \mu_r = 2.66$
  • C. $\epsilon_r = 1.2, \mu_r = 2.0$
  • D. $\epsilon_r = 2.0, \mu_r = 1.2$

Answer: A. $\epsilon_r = 2.66, \mu_r = 1.5$

Explanation: The velocity in a lossless medium is $v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$. Given $v = 1.5 \times 10^8\text{ m/s}$ and $c = 3 \times 10^8\text{ m/s}$, we get $\sqrt{\mu_r \epsilon_r} = 2$. The intrinsic impedance is $\eta = 120\pi \sqrt{\frac{\mu_r}{\epsilon_r}} = 90\pi$. This simplifies to $\sqrt{\frac{\mu_r}{\epsilon_r}} = \frac{3}{4}$, or $\frac{\mu_r}{\epsilon_r} = \frac{9}{16}$. Solving the two equations $\mu_r \epsilon_r = 4$ and $\mu_r = \frac{9}{16} \epsilon_r$, we get $\epsilon_r^2 = 64/9 \Rightarrow \epsilon_r = 2.66$ and $\mu_r = 1.5$.

Question 20

Topic: Microwave Engineering

A square waveguide carries $TE_{11}$ mode whose axial magnetic field is given by $H_z = H_0 \cos(\frac{\pi x}{\sqrt{8}}) \cos(\frac{\pi y}{\sqrt{8}})\text{ A/m}$ where waveguide dimensions are in cm. What is the cut-off frequency of the mode?

  • A. 5.5 GHz
  • B. 6.5 GHz
  • C. 7.5 GHz
  • D. 8.5 GHz

Answer: C. 7.5 GHz

Explanation: Comparing the given $H_z$ expression with the standard form $H_z = H_0 \cos(\frac{m\pi x}{a}) \cos(\frac{n\pi y}{b})$, we identify $m=1$, $n=1$, and the dimensions $a = b = \sqrt{8}\text{ cm} = \sqrt{8} \times 10^{-2}\text{ m}$. The cut-off frequency is $f_c = \frac{c}{2} \sqrt{(\frac{m}{a})^2 + (\frac{n}{b})^2}$. Substituting the values, $f_c = \frac{3 \times 10^8}{2} \sqrt{\frac{1}{8 \times 10^{-4}} + \frac{1}{8 \times 10^{-4}}} = \frac{3 \times 10^8}{2} \sqrt{\frac{2}{8 \times 10^{-4}}} = 7.5\text{ GHz}$.

Question 21

Topic: Analog Circuits

A ramp signal ($V_t$: 0 to 5V with period 100 $\mu$s) is compared with a Soft-Start Signal provided by an N-channel MOSFET ($Q_1$). If $Q_1$ has a low threshold voltage of 0.7 V, negligible ON resistance, and is driven by an RC network (R = 100 k$\Omega$, C = 1 nF) connected to a 5V source, what is the duty cycle of the output signal of the comparator after 100 $\mu$s?

  • A. 17.4%
  • B. 34.8%
  • C. 0%
  • D. 50%

Answer: B. 34.8%

Explanation: The MOSFET gate voltage is governed by the RC charging equation $V_G = 5(1 - e^{-t/\tau})$, where $\tau = RC = 100\text{ }\mu\text{s}$. The MOSFET turns ON when $V_G > 0.7\text{ V}$, which occurs at $t = 15\text{ }\mu\text{s}$. After $100\text{ }\mu\text{s}$, the MOSFET is fully ON, fixing the non-inverting terminal of the comparator at a steady-state voltage of 1.75 V. The comparator gives a positive logic output as long as the ramp voltage $V_t < 1.75\text{ V}$. Since the ramp goes from 0 to 5 V in $100\text{ }\mu\text{s}$, it reaches 1.75 V at $t = 35\text{ }\mu\text{s}$. Therefore, the signal is HIGH for 35 out of $100\text{ }\mu\text{s}$, yielding a duty cycle of $35\%$ (closest to option b, 34.8%).

Question 22

Topic: Power Electronics

A high speed digital Subsystem requires three voltages $V_1$, $V_2$ and $V_3$ with 1:2:1 power ratings respectively. The power supply is designed with a distributed power conversion scheme. The primary converter has efficiency $\eta_1$ and feeds $V_1$ directly, while also feeding two secondary converters (with efficiencies $\eta_2$ and $\eta_3$) that supply $V_2$ and $V_3$. What is the overall power conversion efficiency?

  • A. $\frac{4 \times \eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + 2 \times \eta_3 + \eta_2}$
  • B. $\frac{\eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + 2 \times \eta_3 + \eta_2}$
  • C. $\frac{4 \times \eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + \eta_2 + \eta_3}$
  • D. $\frac{2 \times \eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + \eta_2 + \eta_3}$

Answer: A. $\frac{4 \times \eta_1 \times \eta_2 \times \eta_3}{\eta_2 \eta_3 + 2 \times \eta_3 + \eta_2}$

Explanation: Let the output power values be $P_1 = P$, $P_2 = 2P$, and $P_3 = P$. The total useful output power is $4P$. The input power to the primary converter is $P_{in} = \frac{P_1 + P_2/\eta_2 + P_3/\eta_3}{\eta_1} = \frac{P(1 + 2/\eta_2 + 1/\eta_3)}{\eta_1}$. The overall efficiency is $\eta_{overall} = \frac{\text{Total Output Power}}{P_{in}} = \frac{4P}{\frac{P(1 + 2/\eta_2 + 1/\eta_3)}{\eta_1}}$. Simplifying this expression gives $\frac{4\eta_1\eta_2\eta_3}{\eta_2\eta_3 + 2\eta_3 + \eta_2}$.

Question 23

Topic: Analog Circuits

For a silicon NPN transistor circuit with its base grounded, emitter connected to $V_{EE} = -8\text{ V}$ through $R_E = 2.4\text{ k}\Omega$, and collector connected to $V_{CC} = 10\text{ V}$ through $R_C = 1.9\text{ k}\Omega$, the value of base current $I_B$ is (Assume $\beta = 100$ and $V_{BE} = 0.7\text{ V}$)

  • A. $26.47\text{ }\mu\text{A}$
  • B. $52.94\text{ }\mu\text{A}$
  • C. $13.235\text{ }\mu\text{A}$
  • D. $30.11\text{ }\mu\text{A}$

Answer: D. $30.11\text{ }\mu\text{A}$

Explanation: Applying KVL to the base-emitter loop: $V_B - V_{BE} - I_E R_E - V_{EE} = 0$. Since the base is grounded ($V_B = 0\text{ V}$), we have $0 - 0.7 - I_E(2.4\text{ k}\Omega) - (-8) = 0$. Solving for $I_E$, we get $I_E = \frac{7.3\text{ V}}{2.4\text{ k}\Omega} \approx 3.04\text{ mA}$. The base current is $I_B = \frac{I_E}{1 + \beta} = \frac{3.04\text{ mA}}{101} = 30.11\text{ }\mu\text{A}$.

Question 24

Topic: Computer Organization

Which one of the following statement is not true for static random access memory (SRAM)

  • A. Static RAM stores data in the form of charge
  • B. They have low capacity, but offer high speed
  • C. It doesn't require periodic refreshing
  • D. They are made up of six CMOS transistor

Answer: A. Static RAM stores data in the form of charge

Explanation: Static RAM (SRAM) uses bistable latching circuitry (made up of transistors, typically 6 per cell) to store each bit, not capacitors. Dynamic RAM (DRAM), on the other hand, stores data in the form of charge on a capacitor and therefore requires periodic refreshing. Thus, the statement that SRAM stores data in the form of charge is incorrect.

Question 25

Topic: Signals and Systems

Which of the following statement is not true

  • A. Autocorrelation function and energy spectral density forms a Fourier transform pair
  • B. Autocorrelation function of a real valued energy signal is a real valued odd function
  • C. The value of autocorrelation function of a power signal at the origin is equal to the average power of the signal
  • D. Autocorrelation function is the inverse Fourier transform of power spectral density

Answer: B. Autocorrelation function of a real valued energy signal is a real valued odd function

Explanation: The autocorrelation function $R_{xx}(\tau)$ of a real-valued signal is an even function, meaning $R_{xx}(\tau) = R_{xx}(-\tau)$. Therefore, the statement that it is a real-valued odd function is incorrect.

Question 26

Topic: Electrical Machines

The Eddy current loss is proportional to the

  • A. Frequency
  • B. Square of the frequency
  • C. Cube of the frequency
  • D. Square root of the frequency

Answer: B. Square of the frequency

Explanation: According to the classical eddy current loss formula, the eddy current loss in a magnetic core is directly proportional to the square of the frequency of the alternating magnetic field ($P_e \propto f²$).

Question 27

Topic: Analog Circuits

In a given op-amp circuit, an input current $I_{in}$ flows into a node connecting the non-inverting terminal and a resistor $R_s$ to ground. The output $V_0$ is fed back to the inverting terminal via a voltage divider consisting of $R_A$ and $R_B$. Find the output voltage $V_0$.

  • A. $(R_A/R_B) \times I_{in} \times R_s$
  • B. $I_{in} \times R_s \times (1 + R_A/R_B)$
  • C. $(R_B/R_A) \times I_{in} \times R_s$
  • D. $I_{in} \times R_s \times (1 + R_B/R_A)$

Answer: B. $I_{in} \times R_s \times (1 + R_A/R_B)$

Explanation: The voltage at the non-inverting terminal is $V_p = I_{in} \times R_s$. By the virtual short concept, the inverting terminal voltage $V_n$ equals $V_p$. The feedback network is a simple voltage divider, meaning $V_n = V_0 \frac{R_B}{R_A + R_B}$. Equating the two, we get $I_{in} \times R_s = V_0 \frac{R_B}{R_A + R_B}$. Solving for $V_0$ yields $V_0 = I_{in} \times R_s \times (1 + \frac{R_A}{R_B})$.

Question 28

Topic: Electronic Devices

Match the following regions of a noise spectrum (Relative Amplitude vs Frequency) to their dominant noise types:
A: Low frequency region
B: Spikes at mains frequency
C: Spikes at high switching frequencies
D: Flat wideband noise floor

  • A. A: Thermal Noise, B: Power line pick up, C: Power supply switching noise, D: 1/f noise
  • B. A: Power line pick up, B: Thermal Noise, C: 1/f noise, D: Power supply switching noise
  • C. A: 1/f noise, B: Power line pick up, C: Power supply switching noise, D: Thermal Noise
  • D. A: Power supply switching noise, B: 1/f noise, C: Power line pick up, D: Thermal Noise

Answer: C. A: 1/f noise, B: Power line pick up, C: Power supply switching noise, D: Thermal Noise

Explanation: In a typical noise spectrum: Region A at very low frequencies corresponds to 1/f noise (flicker noise). Region B consists of narrow spikes typically at 50/60 Hz and harmonics, caused by power line pick up. Region C contains higher frequency spikes caused by power supply switching circuits. Region D is the wideband, flat noise floor known as Thermal (Johnson) noise.

Question 29

Topic: Signals and Systems

Consider a system with $x(t)$ as input and $y(t)$ as output. The signal $x(t)$ is split into two branches: the first branch passes through a filter A and is multiplied by $j$, and the second branch is direct. The sum of these two branches passes through a filter B to produce $y(t)$. To obtain an output $Y(f)$ which contains only the positive frequency components of $X(f)$ shifted towards the origin by $f_c$, which combination of filters A and B is appropriate?

  • A. A is a signum function $\text{sgn}(f)$ and B is an ideal low-pass filter
  • B. A is a signum function $\text{sgn}(f)$ and B is an ideal high-pass filter
  • C. A is an ideal low-pass filter and B is a signum function $\text{sgn}(f)$
  • D. A is an ideal high-pass filter and B is a signum function $\text{sgn}(f)$

Answer: A. A is a signum function $\text{sgn}(f)$ and B is an ideal low-pass filter

Explanation: For the first branch, passing the signal through a signum function $\text{sgn}(f)$ and multiplying by $j$ effectively acts as a Hilbert transform. Adding this to the original signal mathematically cancels out the negative frequency components while doubling the positive frequency components. Afterwards, a Low-Pass Filter (LPF) applied as block B can isolate the desired down-converted baseband signal.

Question 30

Topic: Electromagnetics

The electric field of a linearly polarized electromagnetic wave is given by $E_i = \hat{a}_x E_0(x,y) e^{-jkz}$ is incident upon a linearly polarized antenna whose electric field polarization is expressed as $E_a = (\hat{a}_x + \hat{a}_y) E(r, \theta, \phi)$. Find the polarization loss factor.

  • A. 1/2
  • B. 3/2
  • C. 2/3
  • D. 1/4

Answer: A. 1/2

Explanation: The Polarization Loss Factor (PLF) is calculated using the normalized dot product of the incident wave and antenna polarization vectors: $PLF = \frac{|\mathbf{E}_{inc} \cdot \mathbf{E}_{ant}|^2}{|\mathbf{E}_{inc}|^2 |\mathbf{E}_{ant}|^2}$. Here, $\mathbf{E}_{inc}$ is in the $\hat{a}_x$ direction and $\mathbf{E}_{ant}$ is in the $(\hat{a}_x + \hat{a}_y)$ direction. Taking the dot product gives $1$. The squared magnitudes are $|\hat{a}_x|^2 = 1$ and $|\hat{a}_x + \hat{a}_y|^2 = 1^2 + 1^2 = 2$. Therefore, $PLF = \frac{1^2}{1 \times 2} = \frac{1}{2}$.

Question 31

Topic: Microwave Engineering

A lossless T-junction two way power divider has a source impedance, input transmission line impedance and output port load impedance of 50Ω. Find the output characterization impedances so that the input power is divided in a 2:1 ratio.

  • A. $Z_1 = 150\Omega$, $Z_2 = 75\Omega$
  • B. $Z_1 = 50\Omega$, $Z_2 = 100\Omega$
  • C. $Z_1 = 60\Omega$, $Z_2 = 120\Omega$
  • D. $Z_1 = 30\Omega$, $Z_2 = 60\Omega$

Answer: A. $Z_1 = 150\Omega$, $Z_2 = 75\Omega$

Explanation: Since the divider is lossless, $P_1^+ = P_2^- + P_3^-$. The division ratio is $\alpha = P_2^-/P_3^- = Z_{03}/Z_{02} = 2/1$. For matching at port 1, $Z_{01} = \frac{Z_{02}Z_{03}}{Z_{02}+Z_{03}}$. Thus, $Z_{02} = Z_{01}(1 + 1/\alpha) = 50(1 + 1/2) = 75 \Omega$ and $Z_{03} = Z_{01}(1 + \alpha) = 50(1 + 2) = 150 \Omega$.

Question 32

Topic: Digital Circuits

Following circuit implements a (Circuit with inputs $A_0$, $A_1$, $I_0$, an inverter on $I_0$, two AND gates, and an OR gate giving output Y)

  • A. De-Multiplexer
  • B. Multiplexer
  • C. $Y = I_0(A_0 + A_1)$
  • D. $Y = I_0(\overline{A}_1 + A_0)$

Answer: B. Multiplexer

Explanation: The output of the logic circuit can be written as $Y = A_0I_0 + A_1\overline{I}_0$. This is the standard boolean expression for a $2 \times 1$ Multiplexer where $I_0$ acts as the select line choosing between inputs $A_0$ and $A_1$.

Question 33

Topic: Digital Circuits

The frequency of the output Y is (Circuit showing a 32-bit register M and a clock frequency F synchronized for XOR and register)

  • A. $\frac{MF}{2^{32}}$
  • B. $\frac{2MF}{2^{32}}$
  • C. $\frac{F}{2^{32}}$
  • D. $\frac{2^{32}F}{M}$

Answer: A. $\frac{MF}{2^{32}}$

Explanation: The circuit represents a phase accumulator typically used in a Direct Digital Synthesizer (DDS). If M is the 32-bit input tuning word and F is the synchronized clock frequency, the output frequency is given by $\frac{M \cdot F}{2^{32}}$.

Question 34

Topic: Antennas and Wave Propagation

A gain-standard horn is known to have a gain G = 10. It is being used to measure the gain of a large directional antenna by the comparison method. When the antenna being measured is connected to the receiver it is found to be necessary to insert an attenuator adjust to attenuate by 23 dB in order to have the same receiver output that was observed with the horn connected. What is the gain of the large antenna?

  • A. 13 db
  • B. 23 db
  • C. 33 dB
  • D. 230 dB

Answer: C. 33 dB

Explanation: Let the gain of the unknown antenna be $G_1$ (in dB) and receiver loss be $L$. The gain of the horn antenna is $G = 10$ dB. The output of the receiver when the horn is used equals the output when the unknown gain antenna is used plus the loss of the attenuator. Therefore, $L + G = L + G_1 - 23$. Substituting $G = 10$, we get $10 = G_1 - 23$, which yields $G_1 = 33$ dB.

Question 35

Topic: Antennas and Wave Propagation

A paraboloidal-reflector antenna is designed for operation at 3 GHz. Its largest aperture dimension is 20 feet. It is desired to build a scale model of this antenna with the largest apeture dimension scaled to 18 inches. At what frequency must this model be operated in order to have the same pattern as the full-size antenna?

  • A. 10 GHz
  • B. 20 GHz
  • C. 40 GHz
  • D. 4 GHz

Answer: C. 40 GHz

Explanation: For the scale model to have the same radiation pattern, the electrical dimensions must be equal, so $D_1/\lambda_1 = D_2/\lambda_2$. Replacing $\lambda$ with $c/f$, this becomes $D_1 f_1 = D_2 f_2$. The full-size dimension $D_1 = 20 \text{ feet} = 240 \text{ inches}$, and $f_1 = 3$ GHz. The model dimension $D_2 = 18 \text{ inches}$. Thus, $f_2 = \frac{240 \times 3}{18} = 40$ GHz.

Question 36

Topic: Antennas and Wave Propagation

An antenna has a radiation resistance of 72Ω, a loss resistance of 8Ω and power gain of 16. Calculate its directivity.

  • A. 15.8
  • B. 16.8
  • C. 17.8
  • D. 18.7

Answer: C. 17.8

Explanation: The radiation efficiency $\eta_r = \frac{R_{rad}}{R_{rad} + R_{loss}} = \frac{72}{72 + 8} = \frac{72}{80} = 0.9$. The relationship between power gain ($G_p$) and directive gain ($G_d$) is $G_p = \eta_r \times G_d$. Therefore, directivity $G_d = \frac{16}{0.9} = \frac{160}{9} \approx 17.8$.

Question 37

Topic: Electromagnetics

The current density at the surface of a thick metal plate is $100\text{ A/m}^2$. What is the skin depth if the current density at a depth of 0.0059 cm is $0.272\text{ A/m}^2$? (Assume $\ln(10) \approx 2.3$, $\ln(2.72) \approx 1$)

  • A. 5 $\mu$m
  • B. 10 $\mu$m
  • C. 15 $\mu$m
  • D. 20 $\mu$m

Answer: B. 10 $\mu$m

Explanation: Current density decays exponentially: $J = J_s e^{-\alpha z}$. Given $J_s = 100$, $J = 0.272$, and $z = 0.0059 \times 10^{-2}$ m. Then $0.272 = 100 e^{-\alpha z}$, leading to $e^{-\alpha z} = 0.00272$. Taking natural log, $-\alpha z = \ln(0.00272) = \ln(2.72) - \ln(1000) = 1 - 3(2.3) = -5.9$. Solving for $\alpha$, we get $\alpha = \frac{5.9}{0.0059 \times 10^{-2}} = 10^5$. Skin depth $\delta = 1/\alpha = 10^{-5}$ m = 10 $\mu$m.

Question 38

Topic: Engineering Mathematics

A random variable z, has a probability density function $f(z)$ where $f(z)=e^{-z}$ for $0 \le z < \infty$, the probability of $0 \le z \le 2$ will be approximately.

  • A. 0.368
  • B. 0.135
  • C. 0.393
  • D. 0.865

Answer: D. 0.865

Explanation: The probability is the integral of the PDF over the interval. $P(0 \le z \le 2) = \int_{0}^{2} e^{-z} dz$. Evaluating the integral gives $[-e^{-z}]_0^2 = -[e^{-2} - e^0] = 1 - e^{-2}$. Since $e^{-2} \approx 0.135$, the probability is $1 - 0.135 = 0.865$.

Question 39

Topic: Engineering Mathematics

Evaluate $\lim_{x \to 2} \frac{\sqrt{x²+5}-3}{x²-2x}$

  • A. 1/3
  • B. $\infty$
  • C. -3
  • D. 0

Answer: A. 1/3

Explanation: This limit is of the form $0/0$. Applying L'Hospital's rule, we differentiate the numerator and denominator: $\lim_{x \to 2} \frac{\frac{d}{dx}(\sqrt{x²+5}-3)}{\frac{d}{dx}(x²-2x)} = \lim_{x \to 2} \frac{\frac{2x}{2\sqrt{x²+5}}}{2x-2}$. Substituting $x=2$ yields $\frac{2/\sqrt{9}}{2} = \frac{2/3}{2} = 1/3$.

Question 40

Topic: Signals and Systems

What is the R.M.S of following waveform if the average value is zero? (A sawtooth-like waveform with peak positive amplitude 4V, and negative peak A occurring at $t=0$, completing a cycle at $T$)

  • A. $\sqrt{\frac{17}{3}}\text{ V}$
  • B. $\sqrt{\frac{8}{3}}\text{ V}$
  • C. $\sqrt{\frac{5}{3}}\text{ V}$
  • D. $\sqrt{\frac{1}{3}}\text{ V}$

Answer: C. $\sqrt{\frac{5}{3}}\text{ V}$

Explanation: For the average value to be zero, the area above the time axis must equal the area below. Based on the geometry provided in the solution, solving for the negative peak gives $A = -1$ V. The signal can be described as $x(t) = \frac{8}{T}t - 1$ for $0 \le t \le T/2$ and $x(t) = -1$ for $T/2 < t < T$. The RMS value is $\sqrt{\frac{1}{T} \int_{0}^{T} x²(t) dt}$, which evaluates to $\sqrt{5/3}$ V.

Question 41

Topic: Radar and Navigation

A radar system uses TWTA as high power RF source for transmiting 300 W Peak Power. The efficiency of transmitter during pulse is 75% and the transmit duty is 25%. If the DC power required during pulse off period is 20W. The average power dissipation in TWTA is

  • A. 81.25 W
  • B. 125 W
  • C. 50 W
  • D. 40 W

Answer: D. 40 W

Explanation: During the pulse ON period, transmitted power is 300 W and efficiency is 75%, so total input DC power is $P_{del} = \frac{300}{0.75} = 400$ W. The power dissipated during ON time is $400 - 300 = 100$ W. During pulse OFF period, the DC power (and thus dissipation) is 20 W. The average power dissipation is calculated over the duty cycle: $P_{avg} = (100 \times 0.25) + (20 \times 0.75) = 25 + 15 = 40$ W.

Question 42

Topic: Electronic Devices

Which of the following is NOT true about opto-couplers?

  • A. It is a solid state device to isolate two parts of a circuit
  • B. It can act as an input device or output device but not both
  • C. Combines a Light Emitting Diode and a Photo Transistor in a single package
  • D. it prevents electrical noise or voltage transients of one circuit from integrating with other circuit

Answer: B. It can act as an input device or output device but not both

Explanation: An optocoupler transfers electrical signals between two isolated circuits using light. It typically combines an LED and a phototransistor in one package, isolating noise and transients. The claim that it can only act as an input or output but not both in complex configurations is incorrect, marking (b) as the false statement.

Question 43

Topic: Electronic Devices

Which of the noise types is dominant in Metal Semiconductor Field Effect Transistor (MESFET) when compared to Bipolar Junction Transistor (BJT)?

  • A. Thermal Noise
  • B. Shot Noise
  • C. Flicker Noise
  • D. All of the above

Answer: C. Flicker Noise

Explanation: 1/f noise, also known as flicker noise, is much higher in field-effect transistors (like MOSFETs and MESFETs) than in BJTs due to surface traps and generation-recombination centers. Thus, flicker noise is dominant in MESFETs compared to BJTs.

Question 44

Topic: Radar and Navigation

Which factor determines the range resolution of a radar?

  • A. Size of the antenna
  • B. Bandwidth of the transmitted pulse
  • C. Power radiated form the antenna
  • D. Centre frequency of the radar

Answer: B. Bandwidth of the transmitted pulse

Explanation: The theoretical range resolution is given by $S_r = \frac{C_0 \cdot \tau}{2}$, where $\tau$ is the pulse width. Since pulse width is inversely proportional to bandwidth, the bandwidth of the transmitted pulse is the primary factor that determines the range resolution.

Question 45

Topic: Communication Systems

A signal having frequency component from DC-2 kHz is to be Pulse code Modulated with a 6 bit Encoder. Minimum carrier bandwidth required is?

  • A. 4 kHz
  • B. 128 kHz
  • C. 16 kHz
  • D. 12 kHz

Answer: D. 12 kHz

Explanation: The maximum frequency component of the signal is $f_{max} = 2$ kHz. The minimum sampling frequency is the Nyquist rate, $f_s = 2f_{max} = 4$ kHz. With a 6-bit encoder ($n=6$), the bit rate is $R_b = n \times f_s = 6 \times 4\text{ kHz} = 24$ kbps. The minimum baseband carrier bandwidth required for transmission is $R_b / 2 = 12$ kHz.

Question 46

Topic: Analog Circuits

Which of the following is NOT true for Schmitt Trigger ?

  • A. Schmitt trigger can be used as Sine-to-Square Wave converter
  • B. Schmitt trigger uses OP-AMP in Open Loop Mode.
  • C. Hysteresis exists in Schmit Trigger
  • D. None of the above

Answer: B. Schmitt trigger uses OP-AMP in Open Loop Mode.

Explanation: A Schmitt trigger operates using positive feedback, meaning it uses an Op-Amp in a closed-loop mode, not open-loop. Therefore, statement (b) is incorrect. It does exhibit hysteresis and can be used to convert a sine wave into a square wave.

Question 47

Topic: Digital Circuits

Two signals $f_a$ and $f_b$ are given as input to EX-OR to measure phase difference. The average output voltage will be

  • A. A triangular wave passing through the origin with a positive peak at $\pi/2$ and negative peak at $-\pi/2$
  • B. A sawtooth wave passing through the origin
  • C. A triangular wave strictly in the positive voltage plane bounded between 0 and V
  • D. A sawtooth wave strictly in the positive voltage plane

Answer: A. A triangular wave passing through the origin with a positive peak at $\pi/2$ and negative peak at $-\pi/2$

Explanation: When two signals with the same duty cycle are applied to an EX-OR phase detector, the average output voltage $V_{avg}$ varies as a triangular function of the phase difference $\theta_e$. It reaches a positive maximum at a phase difference of $\pi/2$ and a negative maximum at $-\pi/2$.

Question 48

Topic: Signals and Systems

Consider the signal $X(t) = \begin{cases} 2\cos(t) + \cos(2t) & t < 0 \\ 2\sin(t) + \sin(2t) & t \ge 0 \end{cases}$. The signal $X(t)$ is:

  • A. Periodic with period $= 2\pi$
  • B. Periodic with period $= \pi$
  • C. Non-periodic
  • D. Periodic with period $= \pi/2$

Answer: C. Non-periodic

Explanation: For a signal to be strictly periodic, it must repeat its exact pattern from $-\infty$ to $+\infty$. Since this signal has two different definitions for $t < 0$ and $t \ge 0$, it does not satisfy the periodicity condition across all time, making it non-periodic.

Question 49

Topic: Signals and Systems

The system $y(t) = x(2t) + 3$ is

  • A. Linear and time invariant
  • B. Causal and linear
  • C. Non-linear and time variant
  • D. Linear and memoryless

Answer: C. Non-linear and time variant

Explanation: A system is time-invariant if a delay in the input produces the same delay in the output. For $y(t) = x(2t) + 3$, delaying the input yields $x(2t - t_0) + 3$, but delaying the output yields $x(2(t - t_0)) + 3 = x(2t - 2t_0) + 3$, making it time-variant. Furthermore, the addition of the constant term $+3$ means it violates the superposition principle (specifically the zero-in, zero-out property), making it non-linear.

Question 50

Topic: Signals and Systems

Consider the system $R[m] = \sum_{n=0}^{N-1} y[n]x[n-m]$ where $y[n]$ and $x[n]$ are real periodic signals with period N. The above output can be obtained using

  • A. $IFFT\{FFT[Y] \times conj(FFT[X])\}$
  • B. $IFFT\{FFT[Y] \times FFT[X]\}$
  • C. $IFFT\{FFT[Y] \times (FFT[-X])\}$
  • D. $IFFT\{FFT[-Y] \times (FFT[X])\}$

Answer: A. $IFFT\{FFT[Y] \times conj(FFT[X])\}$

Explanation: The given expression defines the circular cross-correlation of two periodic signals $y[n]$ and $x[n]$. In the frequency domain, circular cross-correlation corresponds to the multiplication of the DFT of the first signal by the complex conjugate of the DFT of the second signal. Thus, $R(m) = IFFT\{FFT[Y] \times conj(FFT[X])\}$.

Question 51

Topic: Signals and Systems

Consider the system defined by $\frac{d^2y}{dt²} + (a+b)\frac{dy}{dt} + ab = x(t)$; $a>0$, $b>0$. can be realized using which impulse response function.

  • A. $h(t)=(e^{-at}+e^{-bt})u(t)$
  • B. $h(t)=(e^{-at} \times e^{-bt})u(t)$
  • C. $h(t)=(e^{at}+e^{-bt})u(t)$
  • D. $h(t)=e^{-(a+b)t}u(t)$

Answer: A. $h(t)=(e^{-at}+e^{-bt})u(t)$

Explanation: Taking the Laplace transform of the given differential equation (as interpreted in the source solution): $s² Y(s) + (a+b)s Y(s) + ab = X(s)$. This rearranges to $Y(s) = \frac{X(s) - ab}{s² + (a+b)s}$. For the impulse response, $X(s) = 1$, giving $H(s) = \frac{1-ab}{s[s+(a+b)]}$. Applying partial fractions yields $h(t) = \frac{1-ab}{a+b}[1 - e^{-(a+b)t}]u(t)$, noting the source solution indicates a discrepancy with the exact provided options.

Question 52

Topic: Signals and Systems

A continuous time signal has frequency content at f = 10 MHz, 50 MHz and 70 MHz. The signal is sampled at sampling frequency of 56 MHz. The frequency content of output will be

  • A. 10 MHz
  • B. 10 MHz and 6 MHz
  • C. 10 MHz, 6 MHz and 14 MHz
  • D. 46 MHz

Answer: C. 10 MHz, 6 MHz and 14 MHz

Explanation: When a signal is sampled at $f_s$, the aliased frequencies fold down into the baseband according to $|f_{in} - kf_s|$. For $f_1 = 10$ MHz, the frequency is natively 10 MHz. For $f_2 = 50$ MHz, the baseband alias is $|56 - 50| = 6$ MHz. For $f_3 = 70$ MHz, the baseband alias is $|56 - 70| = 14$ MHz. Thus, the output frequency content will be 10 MHz, 6 MHz, and 14 MHz.

Question 53

Topic: Control Systems

The steady state response for an input $X(s) = K/s$ to a system whose transfer function is $H(s) = 1/((s+5)(s+2))$ in time domain is

  • A. $K/10$
  • B. $e^{-5t}u(t)$
  • C. $e^{-2t}u(t)$
  • D. $(e^{-5t}+e^{-2t}) u(t)$

Answer: A. $K/10$

Explanation: The output response in the s-domain is $Y(s) = H(s) \times X(s) = \frac{K}{s(s+2)(s+5)}$. To find the steady-state response in the time domain, we apply the Final Value Theorem: $\lim_{t \to \infty} y(t) = \lim_{s \to 0} s Y(s)$. Therefore, $Y_{ss} = \lim_{s \to 0} s \left( \frac{K}{s(s+2)(s+5)} \right) = \frac{K}{(0+2)(0+5)} = \frac{K}{10}$.

Question 54

Topic: Control Systems

Consider a closed loop stable phase locked loop system having an open loop transfer function $G(s) = K(K_1 + K_2/s)(1/s)$. The system is capable of producing zero steady state error E(t) for

  • A. Phase step only
  • B. Constant velocity and phase step
  • C. Acceleration
  • D. Jerk

Answer: B. Constant velocity and phase step

Explanation: The open-loop transfer function is $G(s) = \frac{K(K_2 + sK_1)}{s²}$, which indicates two poles at the origin (Type 2 system). A Type 2 system produces zero steady-state error for step inputs (phase step) and ramp inputs (constant velocity). It produces a finite steady-state error for parabolic inputs (acceleration).

Question 55

Topic: Microwave Engineering

The modes in a reflex klystron

  • A. Result from excessive oscillating frequencies of the cavity
  • B. Correspond to different oscillating frequencies of the cavity
  • C. Are caused by spurious frequency modulation
  • D. All give much the same frequency but different transit times

Answer: D. All give much the same frequency but different transit times

Explanation: Different modes of oscillation in a reflex klystron correspond to the electron beam transit times equal to $(n + 3/4)$ cycles. While these modes produce slightly varying output powers and tuning ranges, they all basically sustain oscillation near the fundamental resonant frequency of the single cavity.

Question 56

Topic: Material Science

A dielectric is subjected to alternating electric field. The dielectric losses are proportional to

  • A. Real part of the dielectric constant
  • B. Imaginary part of the dielectric constant
  • C. Both Real and Imaginary part of the dielectric constant.
  • D. None of the above

Answer: B. Imaginary part of the dielectric constant

Explanation: The dielectric losses in a material subjected to an alternating electric field are governed by the relation $W(t) = \frac{1}{2} \omega \epsilon_0 \epsilon_r'' E_0^2$. Here, $\epsilon_r''$ represents the imaginary part of the complex relative permittivity (dielectric constant), meaning the loss is strictly proportional to this imaginary component.

Question 57

Topic: Digital Circuits

The 4:1 Multiplexer implemented with data inputs $I_0 = 0, I_1 = \overline{A}, I_2 = A, I_3 = 1$ and select lines $S_1 = B, S_0 = C$ implements

  • A. $\Sigma(1, 6, 3, 7)$
  • B. $\Sigma(1, 2, 5, 7)$
  • C. $\Sigma(2, 3, 4, 5)$
  • D. $\Sigma(1, 3, 4, 7)$

Answer: A. $\Sigma(1, 6, 3, 7)$

Explanation: The output of a 4-to-1 MUX is $Y = \overline{S}_1 \overline{S}_0 I_0 + \overline{S}_1 S_0 I_1 + S_1 \overline{S}_0 I_2 + S_1 S_0 I_3$. Substituting the inputs: $Y = \overline{B}\overline{C}(0) + \overline{B}C(\overline{A}) + B\overline{C}(A) + BC(1)$. Expanding this gives $Y = \overline{A}\overline{B}C + AB\overline{C} + BC(\overline{A} + A) = \overline{A}\overline{B}C + AB\overline{C} + \overline{A}BC + ABC$. These correspond to minterms 1, 6, 3, and 7, meaning $Y = \Sigma(1, 3, 6, 7)$.

Question 58

Topic: Digital Circuits

In a digital circuit the set-up time violation can be fixed by

  • A. Increasing the clock frequency
  • B. Increasing the delay of data path logic
  • C. Slowing the clock frequency
  • D. None of the above

Answer: C. Slowing the clock frequency

Explanation: A setup time violation occurs when the combinational data path takes too long to stabilize before the next active clock edge. Slowing down the clock frequency increases the clock period, providing more time for the data to propagate and satisfy the setup requirements.

Question 59

Topic: Digital Circuits

Number of J-K flip flops in modulo 16 binary up-counter are

  • A. 16
  • B. 4
  • C. 8
  • D. 2

Answer: B. 4

Explanation: For a binary counter, the maximum modulo $M$ achievable with $n$ flip-flops is $M = 2^n$. For a modulo-16 counter, $16 = 2^n$, which yields $n = 4$. Thus, 4 J-K flip-flops are required.

Question 60

Topic: Antennas and Wave Propagation

The solid angle subtended by the sun as viewed from the earth is $\Omega = 4 \times 10^{-5}$ steradian. A microwave antenna designed to be used for studying the microwave radiation from the sun has a very narrow beam whose equivalent solid angle is approximately equal to that subtended by the sun. What is the approximate directivity, D?

  • A. $10^5$
  • B. $\pi \times 10^5$
  • C. $\pi \times 10^6$
  • D. $10^6$

Answer: B. $\pi \times 10^5$

Explanation: The directivity $D$ of an antenna is given by the relation $D = \frac{4\pi}{\Omega_A}$. Given that the solid angle $\Omega_A = 4 \times 10^{-5}$ steradians, substituting this into the equation yields $D = \frac{4\pi}{4 \times 10^{-5}} = \pi \times 10^5$.

Question 61

Topic: Engineering Mathematics

Find the area of the region bounded by the curves $y=x²$, $y=1/x$ and $x=1/2$ (see in graphical representative figure)

  • A. $\ln 2 - 7/24$
  • B. $\ln 2 + 7/24$
  • C. $\ln 3 - 5/24$
  • D. $\ln 3 + 5/24$

Answer: A. $\ln 2 - 7/24$

Explanation: To find the area bounded by $y = x²$, $y = 1/x$, and $x = 1/2$, we integrate the difference between the upper curve and the lower curve. Based on the provided solution, the integration simplifies to evaluating the bounds which yields $\ln 2 - 7/24$.

Question 62

Topic: Engineering Mathematics

Find the Eigen values in the system $\begin{bmatrix}4 & 1 \\ 3 & 2\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \lambda \begin{bmatrix}x \\ y\end{bmatrix}$

  • A. 1 and 5
  • B. 4 and 3
  • C. 1 and 2
  • D. 2 and 4

Answer: A. 1 and 5

Explanation: The characteristic equation is $|A - \lambda I| = 0$. This gives $(4-\lambda)(2-\lambda) - 3 = 0$. Expanding this, we get $\lambda² - 6\lambda + 8 - 3 = 0$, which simplifies to $\lambda² - 6\lambda + 5 = 0$. Factoring gives $(\lambda-5)(\lambda-1)=0$, so the eigenvalues are $\lambda = 1, 5$.

Question 63

Topic: Control Systems

Find the transient response of a pair of complex poles as given below (Poles at $-a \pm j\omega_0$)

  • A. An exponentially decaying sine wave starting at the origin
  • B. An underdamped step response with a steady state value
  • C. A growing sinusoidal wave
  • D. A purely sinusoidal wave

Answer: A. An exponentially decaying sine wave starting at the origin

Explanation: A system with a pair of complex conjugate poles at $s = -a \pm j\omega_0$ has a transfer function $T(s) = \frac{1}{(s+a)^2 + \omega_0^2}$. The inverse Laplace transform of this function (impulse response) is $y(t) = e^{-at} \sin(\omega_0 t) u(t)$, which represents an exponentially decaying sine wave that starts at zero.

Question 64

Topic: Electronic Devices

A Sensistor exhibits the following characteristics

  • A. Either increase or decrease in resistance value with increase in temperature based on the material type used in the thermistor
  • B. Always increase in resistance value with increase in temperature
  • C. Always decrease in resistance value with increase in temperature.
  • D. No change in resistance value with temperature.

Answer: B. Always increase in resistance value with increase in temperature

Explanation: A sensistor is a heavily doped semiconductor resistor whose resistance changes with temperature. It has a positive temperature coefficient, meaning its resistance always increases exponentially with an increase in temperature.

Question 65

Topic: Electromagnetics

When an electromagnetic wave is incident on an object having surface roughness comparable to the wavelength, then

  • A. Specular reflection occurs
  • B. Absorption occurs
  • C. Diffused scattering occurs
  • D. None of the above

Answer: C. Diffused scattering occurs

Explanation: If the surface roughness of an object is comparable to the wavelength of the incident electromagnetic wave, the wave is scattered in multiple directions rather than reflecting perfectly. This phenomenon is known as diffused scattering.

Question 66

Topic: Analog Circuits

For the following circuit, determine the output voltage $v_o$ in terms of input voltages $V_1$ and $V_2$ assuming $A_1$ and $A_2$ are ideal op-amps (A1 is a non-inverting amplifier with gain $1+1/10$, feeding into A2 which acts as a subtractor/amplifier)

  • A. $11 V_2 - V_1$
  • B. $V_2 - 11V_1$
  • C. $11(V_2 - V_1)$
  • D. None of the above

Answer: C. $11(V_2 - V_1)$

Explanation: For Op-amp $A_1$, it is a non-inverting amplifier: $V_{o1} = (1 + 1/10)V_1 = \frac{11}{10}V_1$. For Op-amp $A_2$, it applies a gain to both $V_{o1}$ and $V_2$. Using superposition: $V_o = -10V_{o1} + 11V_2 = -10(\frac{11}{10}V_1) + 11V_2 = 11(V_2 - V_1)$.

Question 67

Topic: Electronic Devices

Dominant mechanism for motion of charge carriers in forward and reverse biased silicon p-n junctions are

  • A. Drift in forward bias, diffusion in reverse bias
  • B. Diffusion in forward bias, drift in reverse bias
  • C. Diffusion in both forward and reverse bias
  • D. Drift in both forward and reverse bias

Answer: B. Diffusion in forward bias, drift in reverse bias

Explanation: In a forward-biased p-n diode, the dominant mechanism for the motion of charge carriers is the diffusion of majority carriers across the junction. In reverse bias, the small current that flows is due to the drift of minority charge carriers swept across the junction by the strong electric field in the depletion region.

Question 68

Topic: Antennas and Wave Propagation

In a telecommunication trans-receive system, the transmitting antenna with antenna aperture of 1m is fed with 1W of power at 10 GHz. The receive antenna with antenna aperture of 0.5m located at 1km away receives x mW of power. If the transmitting frequency changes to 20 GHz, what will happen to receive power?

  • A. Increase by 3 dB
  • B. Increase by 6 dB
  • C. Decrease by 3 dB
  • D. Decrease by 6 dB

Answer: B. Increase by 6 dB

Explanation: The received power $P_r$ is proportional to the transmitting antenna gain $G_t$. For a parabolic dish, $G_t$ is proportional to the square of the frequency ($f²$). Thus, $P_r \propto f²$. When the frequency doubles from 10 GHz to 20 GHz, the received power increases by a factor of 4. In decibels, this is an increase of $10 \log_{10}(4) = 6\text{ dB}$.

Question 69

Topic: Antennas and Wave Propagation

Find the maximum directivity of an antenna whose power density is given by $W_{rad} = \hat{a}_r \frac{A_0 \sin \theta}{r²}\text{ W/m}^2$ where $A_0$ is peak value of power density. $\theta$ is the usual spherical coordinate and $\hat{a}_r$ is the radial unit vector.

  • A. $4/\pi$
  • B. $2/\pi$
  • C. $3/\pi$
  • D. $5/\pi$

Answer: A. $4/\pi$

Explanation: The radiation intensity is $U = r² W_{rad} = A_0 \sin \theta$. The total radiated power is $P_{rad} = \iint U d\Omega = \int_0^{2\pi} \int_0^\pi A_0 \sin² \theta d\theta d\phi = 2\pi A_0 (\pi/2) = \pi² A_0$. The maximum directivity is $D_{max} = \frac{4\pi U_{max}}{P_{rad}} = \frac{4\pi A_0}{\pi² A_0} = \frac{4}{\pi}$.

Question 70

Topic: Power Electronics

A DC series motor is driven by a chopper circuit. The supply voltage is 220 V and the duty cycle is 25%. Determine the DC voltage applied to the motor

  • A. 165 V
  • B. 55 V
  • C. 220 V
  • D. 110 V

Answer: B. 55 V

Explanation: For a step-down chopper, the average output DC voltage is given by $V_{o(av)} = \alpha V_s$, where $\alpha$ is the duty cycle and $V_s$ is the supply voltage. Given $\alpha = 0.25$ and $V_s = 220\text{ V}$, the voltage applied to the motor is $V_{o(av)} = 0.25 \times 220 = 55\text{ V}$.

Question 71

Topic: Power Electronics

A single-phase full-wave AC phase controller feeds power to a resistive load of 100$\Omega$ from a 220 V, 50 Hz supply. What will be the R.M.S. output voltage at delay angles $\alpha_1=\alpha_2=\alpha=\pi/2$ of both transistor?

  • A. $\frac{220}{\sqrt{2}}\text{ V}$
  • B. $\sqrt{2 \times 220}\text{ V}$
  • C. $\sqrt{2} \times 110\text{ V}$
  • D. $\sqrt{2} \times 220\text{ V}$

Answer: A. $\frac{220}{\sqrt{2}}\text{ V}$

Explanation: For a single-phase full-wave AC phase controller, the RMS output voltage is $V_{o(rms)} = \frac{V_m}{\sqrt{2}} \sqrt{\frac{1}{\pi} [(\pi - \alpha) + \frac{\sin 2\alpha}{2}]}$. Given $V_s = 220\text{ V}$ RMS (so $V_m = 220\sqrt{2}$) and $\alpha = \pi/2$. Substituting these values: $V_{o(rms)} = 220 \sqrt{\frac{1}{\pi} (\pi - \pi/2)} = 220 \sqrt{1/2} = \frac{220}{\sqrt{2}}\text{ V}$.

Question 72

Topic: Power Electronics

A half-wave rectifier is used to charge a 12 V battery through a resistance 'R'. The input transformer is fed by 34 V AC with turns ratio 2:1. Calculate the conduction period of the diode.

  • A. $136^\circ$
  • B. $120^\circ$
  • C. $173^\circ$
  • D. $137^\circ$

Answer: B. $120^\circ$

Explanation: The RMS secondary voltage is $V_s = 34 / 2 = 17\text{ V}$. The diode conducts when the supply voltage exceeds the battery voltage: $V_s \sqrt{2} \sin \theta > 12\text{ V}$. This gives $17\sqrt{2} \sin \theta = 12$, leading to $\theta = \sin^{-1}(\frac{12}{17\sqrt{2}}) \approx 29.94^\circ$. The conduction period is $\pi - 2\theta = 180^\circ - 2(30^\circ) = 120^\circ$.

Question 73

Topic: Analog Circuits

Following is the small signal high frequency equivalent circuit of a common source amplifier. $V_o/V_i$ will be of the form ($K, z_1, a_0, a_1, a_2, a_3$ are constants containing circuit elements)

  • A. $K(s-z_1)/(a_0+a_1s+a_2s^2)$
  • B. $K \cdot s/(a_0+a_1s+a_2s^2+a_3s^3)$
  • C. $K(s-z_1)/(a_0+a_1s)$
  • D. $K/(a_0+a_1s)$

Answer: A. $K(s-z_1)/(a_0+a_1s+a_2s^2)$

Explanation: The high-frequency equivalent circuit of a Common Source amplifier includes the gate-drain capacitance ($C_{gd}$) creating a Right Half Plane (RHP) zero. It also has two poles caused by the input and output node capacitances. Thus, the transfer function has one zero and two poles, taking the form $\frac{K(s-z_1)}{a_0+a_1s+a_2s^2}$.

Question 74

Topic: Analog Circuits

Phase of the transfer function of the following circuit is (Circuit with a capacitor C in series and a resistor R in parallel/shunt to ground)

  • A. $\tan^{-1}(1/\omega RC)$
  • B. $\tan^{-1}(\omega RC)$
  • C. $\tan^{-1}(RC/\omega)$
  • D. $\tan^{-1}(\omega/RC)$

Answer: A. $\tan^{-1}(1/\omega RC)$

Explanation: The circuit acts as a high-pass filter. The transfer function is $V_o/V_i = \frac{R}{R + 1/(j\omega C)} = \frac{j\omega RC}{1 + j\omega RC}$. Alternatively written as $\frac{R}{R - j/\omega C}$, the phase angle is $\tan^{-1}(\frac{1}{\omega RC})$.

Question 75

Topic: Measurements and Instrumentation

The error in measurement of a dc voltmeter with input signal: 1.5 V, voltage range: 2V, accuracy: $\pm(25$ ppm of reading + 5 ppm of range) is

  • A. $\pm 50\text{ }\mu\text{V}$
  • B. $\pm 30\text{ }\mu\text{V}$
  • C. $\pm 47.5\text{ }\mu\text{V}$
  • D. $\pm 10\text{ }\mu\text{V}$

Answer: C. $\pm 47.5\text{ }\mu\text{V}$

Explanation: The error is computed by summing the reading error and the range error components. Error $= \pm [ (1.5 \times 25 \times 10^{-6}) + (2 \times 5 \times 10^{-6}) ] = \pm (37.5 \times 10^{-6} + 10 \times 10^{-6}) = \pm 47.5\text{ }\mu\text{V}$.

Question 76

Topic: Control Systems

Characteristic equation of $H(s)$ is given as $3s^4+2s^3+5s^2+s+2=0$ is

  • A. unstable
  • B. stable
  • C. all poles in right half plane
  • D. unstable with only one pole in right half plane

Answer: A. unstable

Explanation: Applying the Routh-Hurwitz criterion to the polynomial $3s^4+2s^3+5s^2+s+2=0$, the first column of the Routh array evaluates to 3, 2, 3.5, -0.443, and 2. Since there are two sign changes (from positive to negative and back to positive), there are two poles in the right half of the s-plane, making the system unstable.

Question 77

Topic: Control Systems

Consider the feedback system $G(s) = \frac{K(s+4)}{s(s+1)}$, $H(s) = \frac{1}{s+2}$. The value of gain for which system is marginally stable is

  • A. $K = 4$
  • B. $K = 6$
  • C. $K = 10$
  • D. $K = 2$

Answer: B. $K = 6$

Explanation: The characteristic equation is $1 + G(s)H(s) = 0$, which yields $s(s+1)(s+2) + K(s+4) = 0 \Rightarrow s³ + 3s^2 + (K+2)s + 4K = 0$. Using the Routh array, marginal stability occurs when a row of zeros is formed, which happens when the product of the inner terms equals the product of the outer terms. Thus, $3(K+2) = 4K$, giving $3K + 6 = 4K$, so $K = 6$.

Question 78

Topic: Control Systems

Consider the Bode plots (magnitude and phase) of two different open loop transfer functions of two unity feedback systems. The open loop transfer functions have poles in right half plane. The closed loop system formed from these open loop systems. Which of the following holds true?

  • A. Closed loop system with I is stable and with II is unstable
  • B. Closed loop systems using I and II both are unstable
  • C. Closed loop system with I is unstable and II is stable
  • D. Closed loop system with I and II are stable

Answer: A. Closed loop system with I is stable and with II is unstable

Explanation: For system I, both the Gain Margin (GM = 4 dB) and Phase Margin (PM = $20^\circ$) are positive, indicating a stable system. For system II, both the Gain Margin (GM = -4 dB) and Phase Margin (PM = $-20^\circ$) are negative, indicating an unstable system. Therefore, system I is stable and system II is unstable.

Question 79

Topic: Digital Circuits

The circuit is formed as shown below. The output S and C implement (Logic diagram with inputs A, B, two NOT gates, two AND gates, and an OR gate)

  • A. Two bit adder with sum and carry respectively
  • B. Two bit subtractor with sum and borrow respectively
  • C. $S = \overline{A}B + A\overline{B}; C = \overline{A}B$
  • D. None of the above

Answer: C. $S = \overline{A}B + A\overline{B}; C = \overline{A}B$

Explanation: The logic circuit evaluates to $S = A\overline{B} + \overline{A}B$ (which is the XOR operation for the difference) and $C = \overline{A}B$ (which is the borrow operation). This logic specifically implements a half subtractor, and the exact boolean expressions for it correspond to option (c).

Question 80

Topic: Digital Circuits

What is the value of the register formed from D flip-flops using $Q_0$, $Q_1$ and $Q_2$ as output ($Q_0 Q_1 Q_2$) after 14 cycles? (A 3-bit shift register with the inverted output of the last stage fed back to the input of the first stage)

  • A. 110
  • B. 000
  • C. 001
  • D. 011

Answer: A. 110

Explanation: The circuit represents a 3-bit Johnson counter, which has a modulo of $2n = 6$ states. Assuming it starts at 000, the sequence repeats every 6 cycles. After 12 clock cycles, the counter returns to the initial state 000. After the 13th cycle, the state is 100. After the 14th cycle, the state shifts to 110.