ISRO Electronics & Communication (ECE) 2014 Question Paper & Solutions

Attempt the ISRO Electronics & Communication (ECE) 2014 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: Network Theory

In the circuit shown below, switch $S_{1}$ and $S_{2}$ are in open and close position respectively for long time. At $t=t_{0}$ switch $S_{1}$ is closed and switch $S_{2}$ is opened. What would be the current through $R_{1}$ immediately after the transition of switches?

  • A. 0 mA
  • B. 1 mA
  • C. 0.5 mA
  • D. 2 mA

Answer: B. 1 mA

Explanation: At $t=t_{0}^{-}$, steady state is reached and the capacitor acts as an open circuit. $V_c(t_{0}^{-}) = V_c(t_{0}^{+}) = -5\text{ V}$.
At $t=t_{0}^{+}$, $S_{1}$ is closed and $S_{2}$ is opened. Using nodal analysis at node A (between the 10 k$\Omega$ resistors and $R_1$):
$\frac{V_A-10}{10} + \frac{V_A}{10} + \frac{V_A+5}{5} = 0$
$\frac{2V_A}{5} = 0 \Rightarrow V_A = 0\text{ V}$
So, $V_{5\text{ k}\Omega} = I_{5\text{ k}\Omega}(5\text{ k}\Omega) = 5\text{ V}$
$I_{5\text{ k}\Omega} = 1\text{ mA}$.

Question 2

Topic: Digital Electronics

A digital board has a unipolar square clock of 250 MHz. If the clock on the board at all places should have all the harmonic components which have more than 10% of DC value, the board has to be designed for at least

  • A. 250 MHz
  • B. 750 MHz
  • C. 1250 MHz
  • D. 2500 MHz

Answer: C. 1250 MHz

Explanation: Based on the official answer key, the correct option is (c). Detailed mathematical derivation is not provided in the reference, but representing the significant harmonics of a square wave often requires accounting for up to the 5th harmonic (i.e., $5 \times 250\text{ MHz} = 1250\text{ MHz}$) to accurately reconstruct the sharp edges of the clock.

Question 3

Topic: Network Theory

A switch is connected in between a 12 V battery and an uncharged capacitor and a 1 k$\Omega$ resistor. At the time instant when the switch is closed, the voltage across the capacitor is

  • A. 6 V
  • B. 12 V
  • C. 0 V
  • D. 24 V

Answer: C. 0 V

Explanation: A capacitor doesn't allow a sudden change in voltage. Voltage (just before switching) = Voltage (just after switching) = 0 V.

Question 4

Topic: Digital Electronics

Which type of timing violation will occur, If a digital IC is operated at clock frequency which is higher than its specified maximum clock frequency?

  • A. Hold violation
  • B. Setup violation
  • C. Propagation delay
  • D. All of above

Answer: B. Setup violation

Explanation: Based on the official answer key, the correct option is (b). Operating at a higher clock frequency reduces the clock period, making it difficult for the data to arrive and stabilize before the next active clock edge, resulting in a setup time violation.

Question 5

Topic: Electromagnetics

If the maximum effective aperture of a radiating small loop of constant current is $\frac{3}{(32\pi)}\text{ m}^{2}$, the wavelength of the radiated electromagnetic wave is

  • A. 5000 mm
  • B. 100 mm
  • C. 1000 mm
  • D. 500 mm

Answer: D. 500 mm

Explanation: Using $A_e = \frac{\lambda²}{4\pi}G$.
Also, $G = 1.5$ for a small loop antenna.
$\lambda = \sqrt{\frac{4\pi A_e}{G}} = \sqrt{\frac{4\pi \times 3}{32\pi \times 1.5}} = \sqrt{\frac{1}{4}} = 0.5\text{ m} = 500\text{ mm}$.

Question 6

Topic: Radar and Navigation

A radar system uses magnetron as high power RF source. Its transmitter emits 300 kW RF power at 10% duty factor. If the efficiency of radar transmitter during pulse is 60% and the power required during pulse off period is 1 kW. The average power dissipation in radar is,

  • A. 50.9 kW
  • B. 301 kW
  • C. 20.9 kW
  • D. 150.5 kW

Answer: C. 20.9 kW

Explanation: Based on the official answer key, the correct option is (c). Detailed mathematical derivation is not provided in the reference, but the computation involves calculating the power dissipated during the ON time due to inefficiency and combining it with the power consumed during the OFF time.

Question 7

Topic: Communication Systems

Match the following:

A. Gaussian distribution
B. Rayleigh distribution
C. Poisson distribution
D. Uniform distribution

1. Calls on a telephone channel
2. Random number
3. Thermal noise
4. Fading channel in wireless communication

  • A. A-3, B-1, C-4, D-2
  • B. A-3, B-4, C-2, D-1
  • C. A-1, B-4, C-3, D-2
  • D. A-3, B-4, C-1, D-2

Answer: D. A-3, B-4, C-1, D-2

Explanation: Rayleigh distribution is used in mobile computing for flat fading channel.
Poisson distribution is applied when events can't be counted in whole numbers, occurrences are independent, and the average frequency of occurrence is known (e.g., calls on a telephone channel).
Gaussian distribution is used for most of the natural events like thermal noise.

Question 8

Topic: Optical Instrumentation

In a fiber-optic cable, which phenomenon occurs for signal to propagate along the inner core

  • A. Modulation
  • B. Reflection
  • C. Refraction
  • D. All of the above

Answer: B. Reflection

Explanation: An optical fiber has high refractive index for core ($n_1$) and low refractive index for cladding ($n_2$). Thus the propagation of wave inside the fibre is through total internal reflection.

Question 9

Topic: Microprocessors

Which of the following computation will take minimum number of clock cycles, if it is implemented on 8051 micro-controller?
Following is variable declaration in c code:
unsigned int a, b;

  • A. b = a * a
  • B. b = a * 3
  • C. b = a / 3
  • D. b = a / 4

Answer: D. b = a / 4

Explanation: Based on the official answer key, the correct option is (d). Dividing by powers of 2 (like 4) is often optimized as bitwise right-shift operations, which generally execute faster and take fewer clock cycles than multiplication or division by non-power-of-2 values.

Question 10

Topic: Electromagnetics

Using a quarter wave transformer of characteristic impedance 1 $\Omega$, a load impedance of 100 $\Omega$ will be transformed to an input impedance of value

  • A. 1 $\Omega$
  • B. 100 $\Omega$
  • C. 0.01 $\Omega$
  • D. 0.1 $\Omega$

Answer: C. 0.01 $\Omega$

Explanation: $Z_{in}(L=\frac{\lambda}{4}) = \frac{Z_0^2}{Z_L} = \frac{1^2}{100} = 0.01\text{ }\Omega$.

Question 11

Topic: Analog Circuits

What is the output signal level of following circuit?
(The circuit is a non-inverting operational amplifier with $Z_1 = 10\text{ k}\Omega$ connected to ground, and feedback impedance $Z_2$ consisting of a $1\text{ }\mu\text{F}$ capacitor in parallel with a $40\text{ k}\Omega$ resistor. The input is a $1\text{ V}_{rms}$, $200\text{ kHz}$ sine wave connected to the non-inverting terminal.)

  • A. ~5 $V_{rms}$
  • B. ~1 $V_{rms}$
  • C. ~4 $V_{rms}$
  • D. ~0.1 $V_{rms}$

Answer: B. ~1 $V_{rms}$

Explanation: Voltage gain $(A_v) = 1 + \frac{Z_2}{Z_1}$
where, $Z_2 = (\frac{1}{j400\pi\times 10^3 \times 10^{-6}} \parallel 40\text{ k}\Omega)$
$Z_1 = 10\text{ k}\Omega$
So, $Z_2 = \frac{40\times 10^3}{1+j16000\pi}$
$|Z_2| \simeq 0.8\text{ }\Omega$
$\frac{Z_2}{Z_1} = 0.8 \times 10^{-4}$
$A_v = 1 + \frac{|Z_2|}{Z_1} = 1.00008$
$V_o \simeq 1\text{ V}_{rms}$.

Question 12

Topic: Communication Systems

The effect of channel coding in communication channels is

  • A. Reduced errors
  • B. Increased bandwidth
  • C. Reduced terminal size
  • D. All of above

Answer: D. All of above

Explanation: Channel coding in communication channel is used to reduce the errors while source coding is used to reduce the number of bits for transmission. Note: According to the official ISRO answer key, the answer is option (d).

Question 13

Topic: Electromagnetics

For a point charge having electric flux density, $D=\frac{Q}{4\pi r^{2}}a_{r}$, where $a_{r}$ is the unit vector in radial direction; volume charge density $\rho_{v}$ is

  • A. 1
  • B. 0
  • C. Infinite
  • D. $\frac{Q}{4\pi}$

Answer: B. 0

Explanation: Using the divergence theorem in spherical coordinates:
$\nabla\cdot D = \frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}D_{r}) + \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta D_{\theta}) + \frac{1}{r\sin\theta}\frac{\partial D_{\phi}}{\partial\phi}$
$\nabla\cdot D = \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{Q}{4\pi r^{2}}\right) = 0$ (everywhere except at $r=0$). Since volume charge density $\rho_{v} = \nabla\cdot D$, $\rho_{v}$ evaluates to 0.

Question 14

Topic: Digital Signal Processing

What is Discrete Time Fourier Transform (DTFT) of following signal?
$x(n)=a^{n}$ where $n\ge 0$, $-1<a<1$

  • A. $X(\omega)=\frac{1}{(1-ae^{-j\omega})}$
  • B. $X(\omega)=\frac{1}{(1+ae^{-j\omega})}$
  • C. $X(\omega)=\frac{a}{(1-ae^{-j\omega})}$
  • D. $X(\omega)=\frac{a}{(1+ae^{-j\omega})}$

Answer: A. $X(\omega)=\frac{1}{(1-ae^{-j\omega})}$

Explanation: $x(n) = a^{n}u(n)$; $-1<a<1$
$X(\omega) = \sum_{n=0}^{\infty}a^{n}e^{-j\omega n} = \sum_{n=0}^{\infty}(ae^{-j\omega})^{n} = \frac{1}{1-ae^{-j\omega}}$

Question 15

Topic: Electromagnetics

In the near field, the radiated waves take the form of

  • A. Spherical wave
  • B. Plane wave
  • C. Cylindrical wave
  • D. Conical wave

Answer: A. Spherical wave

Explanation: In the near field of a radiating source, the radiated waves take the form of spherical waves. These evolve into cylindrical waves in the intermediate region and finally appear as plane waves in the far field.

Question 16

Topic: Analog Circuits

The zener diode shown in following figure is temperature compensated and current gain $\beta$ of transistor is very high. If current through 100 ohm resistor is 55 mA at $25^{\circ}C$, what is the approximate current through it at $65^{\circ}C$?

  • A. 55 mA
  • B. 54 mA
  • C. 56 mA
  • D. 100 mA

Answer: C. 56 mA

Explanation: $V_{base} = 6.2\text{ V}$, $I(25^{\circ}C) = 55\text{ mA}$.
$V_{BE}$ decreases by $2.5\text{ mV}$ with every $1^{\circ}C$ rise in temperature.
So, $V_{BE}(65^{\circ}C) = V_{BE}(25^{\circ}C) - 2.5 \times 10^{-3} \times 40 = 0.7 - 0.1 = 0.6\text{ V}$.
Voltage across $100\text{ }\Omega$ resistor $V_{100\text{ }\Omega} = 6.2 - 0.6 = 5.6\text{ V}$.
Current $I = \frac{5.6}{100} = 56\text{ mA}$.

Question 17

Topic: Communication Systems

We wish to transmit a digital voice at 32 Kbps through a communication channel having bandwidth of 3000 Hz. The received signal to noise power ratio is 30 dB. State which of the following statement is true?

  • A. To transmit data error-free through this channel, channel equalizer circuit should be used at the receiver.
  • B. To transmit data error-free through this channel, the voice signal should be compressed by 2.
  • C. To transmit data error-free through this channel, a channel code with coding gain of 3 dB should be used.
  • D. It is not possible to transmit data error-free through this channel.

Answer: D. It is not possible to transmit data error-free through this channel.

Explanation: $R = 32\text{ kbps}$, $B = 3000\text{ Hz}$, $SNR = 30\text{ dB} = 1000$.
Using Shannon's channel capacity: $C = B\log_2(1+SNR) = 3000\log_2(1001) \approx 3000 \times 9.967 \approx 29.9\text{ kbps}$.
As $C < R$, it is not possible to transmit data error-free through the channel.

Question 18

Topic: Electromagnetics

Consider a standard rectangular and a circular waveguide having same dominant mode cut-off frequency, determine the ratio of area of circular waveguide to rectangular waveguide

  • A. $\frac{1}{\pi}$
  • B. $\frac{6.78}{\pi}$
  • C. $\frac{0.69}{\pi}$
  • D. $\pi$

Answer: B. $\frac{6.78}{\pi}$

Explanation: For circular waveguide $\lambda_c = \frac{2\pi r}{1.841}$.
For rectangular waveguide $\lambda_c = 2a$.
Equating them: $\frac{2\pi r}{1.841} = 2a \Rightarrow a = \frac{\pi}{1.841}r$.
For standard rectangular waveguide, $b = a/2$.
Ratio of Area = $\frac{\pi r²}{ab} = \frac{\pi r²}{a(a/2)} = \frac{2\pi r²}{a²} = \frac{2\pi r²}{(\frac{\pi r}{1.841})^2} = \frac{2\pi(1.841)^2}{\pi²} = \frac{2(1.841)^2}{\pi} \approx \frac{6.78}{\pi}$.

Question 19

Topic: Digital Electronics

What is the lowest level of abstraction in any hardware description language?

  • A. Behavioral
  • B. Gate level
  • C. Structural
  • D. Register-Transfer level

Answer: B. Gate level

Explanation: Among the commonly described abstraction levels in standard HDL contexts (Behavioral, Dataflow/RTL, Structural/Gate), Gate level abstraction is considered the lowest level abstraction in a hardware description language.

Question 20

Topic: Analog Circuits

With the introduction of negative feedback, the gain-bandwidth product of an amplifier?

  • A. Becomes infinity
  • B. Increases
  • C. Decreases
  • D. Remains constant

Answer: D. Remains constant

Explanation: With negative feedback, the gain of the system decreases and the bandwidth increases by the same factor $(1+A\beta)$, but the overall gain-bandwidth product remains constant.

Question 21

Topic: Communication Systems

For a BER of $10^{-5}$ which modulation scheme will require minimal power?

  • A. QPSK
  • B. 8-PSK
  • C. 16-PSK
  • D. 16-QAM

Answer: A. QPSK

Explanation: The distance between the symbols in the constellation diagram determines the probability of error. As the distance is maximum in QPSK among the given higher-order schemes, for a given BER of $10^{-5}$, QPSK requires minimal power.

Question 22

Topic: Digital Electronics

Which one of following is not synthesizable VHDL statement?

  • A. 'case'
  • B. 'wait until'
  • C. 'wait for'
  • D. 'generate'

Answer: C. 'wait for'

Explanation: The 'wait for' statement specifies a time delay and is used exclusively for simulation purposes. It cannot be synthesized into actual hardware.

Question 23

Topic: Analog Circuits

Subsystem A requires two voltages $V_{1}$ and $V_{2}$ with equal power ratings. Power conversion scheme adapted for the power supply design is shown in the following figure. What is the overall power conversion efficiency?
(The primary converter with efficiency $\eta_1$ supplies $V_1$ directly to Subsystem A and also feeds a secondary converter with efficiency $\eta_2$ which supplies $V_2$ to Subsystem A).

  • A. $\frac{\eta_1\cdot\eta_2}{\eta_1+\eta_2}$
  • B. $\eta_1*\eta_2$
  • C. $\frac{2\eta_1\cdot\eta_2}{1+\eta_2}$
  • D. $\frac{2\eta_1\cdot\eta_2}{1+\eta_1}$

Answer: C. $\frac{2\eta_1\cdot\eta_2}{1+\eta_2}$

Explanation: Let Subsystem A require power $P$ for $V_1$ and power $P$ for $V_2$. Total output power = $2P$.
The secondary converter delivers $P$ with efficiency $\eta_2$, so it requires an input power of $P/\eta_2$.
The primary converter delivers $P$ directly and $P/\eta_2$ to the secondary converter, so its total output power is $P + P/\eta_2$.
The primary converter has efficiency $\eta_1$, so its input power is $(P + P/\eta_2) / \eta_1 = \frac{P(1+1/\eta_2)}{\eta_1}$.
Overall efficiency = $\frac{\text{Total Output Power}}{\text{Total Input Power}} = \frac{2P}{\frac{P(1+\eta_2)}{\eta_1\eta_2}} = \frac{2\eta_1\eta_2}{1+\eta_2}$.

Question 24

Topic: Communication Systems

Which of the following modulation scheme is most bandwidth efficient?

  • A. AM
  • B. FM
  • C. PM
  • D. SSB-SC

Answer: D. SSB-SC

Explanation: Bandwidth requirements: $BW_{AM} = 2f_m$, $BW_{PM} = 2(\beta_p+1)f_m$, $BW_{FM} = 2(\beta_f+1)f_m$, $BW_{SSB} = f_m$. Thus, SSB-SC requires the minimum bandwidth and is the most bandwidth efficient.

Question 25

Topic: Analog Circuits

A Colpitts Oscillator is having tank capacitances of 1 nF and 10 nF, and inductance of 0.1 $\mu$H. The gain required by the circuit to start oscillating is

  • A. 10
  • B. 100
  • C. 1
  • D. 1000

Answer: A. 10

Explanation: For a Colpitts oscillator, the minimum gain required to sustain oscillations is $A \ge \frac{C_2}{C_1}$.
Given $C_1 = 1\text{ nF}$ and $C_2 = 10\text{ nF}$.
$A \ge \frac{10}{1} = 10$.

Question 26

Topic: Digital Electronics

A digital CMOS IC operating at 10 MHz clock frequency consumes 100 mW power; the same IC operating at 15 MHz clock frequency consumes 140 mW power. What is the static power consumption of the IC?

  • A. 10 mW
  • B. 15 mW
  • C. 20 mW
  • D. 40 mW

Answer: C. 20 mW

Explanation: Total power $P = P_{static} + P_{dynamic}$. Dynamic power is proportional to frequency ($P_{dynamic} = Kf$).
At $10\text{ MHz}$: $100 = P_{static} + 10K$ (Eq. 1)
At $15\text{ MHz}$: $140 = P_{static} + 15K$ (Eq. 2)
Subtracting Eq. 1 from Eq. 2 gives $40 = 5K \Rightarrow K = 8$.
Substitute $K$ back into Eq. 1: $100 = P_{static} + 10(8) \Rightarrow P_{static} = 100 - 80 = 20\text{ mW}$.

Question 27

Topic: Communication Systems

An amplifier has a gain of 50 dB, and noise figure of 10 dB. Assuming an ambient temperature of 300 K, what will be the total equivalent noise temperature at the input of the amplifier?

  • A. 760 K
  • B. 360 K
  • C. 2700 K
  • D. 3000 K

Answer: D. 3000 K

Explanation: Noise figure $F = 10\text{ dB}$, which corresponds to a linear ratio of $10^{10/10} = 10$.
Equivalent noise temperature of the amplifier $T_e = (F-1)T_0 = (10-1)300 = 9 \times 300 = 2700\text{ K}$.
Total equivalent noise temperature at the input $= T_{ambient} + T_e = 300\text{ K} + 2700\text{ K} = 3000\text{ K}$.

Question 28

Topic: Analog Circuits

Diodes in the following circuit are ideal. Which is the correct waveform across $R_{L}$?
(The circuit shows a transformer with $N_p=3$ and three secondary windings $N_{s1}=3$, $N_{s2}=6$, $N_{s3}=3$ with a multi-diode full-wave rectifying arrangement).

  • A. Full-wave rectified sine wave with peak 10
  • B. Full-wave rectified sine wave with equal peaks of 5
  • C. Full-wave rectified sine wave with alternating peaks of 10 and 5
  • D. Unrectified sine wave with peak 10

Answer: B. Full-wave rectified sine wave with equal peaks of 5

Explanation: During the positive half cycle, $D_1$ is ON, and current flows from $D_1$ through $R_L$ to point A. During the negative half cycle, $D_4$ is ON, and current flows to $R_L$ through $D_4$ and through point A. In both cases, the active secondary turns utilized relative to the primary ($N_p=3$) are effectively symmetric (turns ratio is 1:1 for the active loop), yielding a full-wave rectified output with equal peaks of 5V.

Question 29

Topic: Communication Systems

Identify the modulation schemes for the shown signal constellation diagrams:
(A) 8 points connected to center and square perimeter,
(B) 4 points forming a square,
(C) 4 points on a circle.

  • A. (A)-QPSK (B)-BPSK (C)-QAM
  • B. (A)-QPSK (B)-OQPSK (C)-MSK
  • C. (A)-OQPSK (B)-QAM (C)-QPSK
  • D. (A)-OQPSK (B)-QPSK (C)-MSK

Answer: B. (A)-QPSK (B)-OQPSK (C)-MSK

Explanation: Constellation (A) shows phase transitions through the origin, which is characteristic of QPSK. Constellation (B) shows phase transitions that avoid the origin (no diagonal crossing lines), which is characteristic of Offset-QPSK (OQPSK). Constellation (C) depicts a constant envelope with 4 states, characteristic of MSK.

Question 30

Topic: Signals and Systems

The Fourier Transform of a discrete and aperiodic signal is

  • A. Continuous and aperiodic
  • B. Continuous and periodic
  • C. Discrete and aperiodic
  • D. Discrete and periodic

Answer: B. Continuous and periodic

Explanation: The discrete nature of a signal in one domain leads to a periodic nature in the other domain. The aperiodic nature in one domain leads to a continuous nature in the other domain. Therefore, the Fourier transform (DTFT) of a discrete and aperiodic signal is continuous and periodic.

Question 31

Topic: Digital Electronics

In the following circuit, Y can be expressed as
(The diagram shows a 4-to-1 MUX with inputs $I_0=0, I_1=1, I_2=0, I_3=A$ and select lines $S_1=A, S_0=B$, whose output is connected to the '0' input of a 2-to-1 MUX. The '1' input of the 2-to-1 MUX is connected to B, and its select line is $S=C$. The output is Y.)

  • A. $Y=BC+A$
  • B. $Y=C$
  • C. $Y=AC'+BC$
  • D. $Y=B$

Answer: D. $Y=B$

Explanation: For the 4-to-1 MUX, the output is $\overline{A}\overline{B}(0) + \overline{A}B(1) + A\overline{B}(0) + AB(A) = \overline{A}B + AB = B(\overline{A} + A) = B$.
For the 2-to-1 MUX, the inputs are $0 \rightarrow B$ and $1 \rightarrow B$, with select line $C$. The output $Y = \overline{C}B + CB = B(\overline{C} + C) = B$.

Question 32

Topic: Control Systems

$s^{2}-0.75s+0.25$ is the characteristic equation of a second order system. Its response to unit step function will be,

  • A. Over damped
  • B. Under damped
  • C. Critically damped
  • D. Exponential growth

Answer: D. Exponential growth

Explanation: The characteristic equation $= s² - 0.75s + 0.25 = 0$. Since the characteristic equation contains a negative coefficient, the roots (poles) will lie on the right half of the s-plane. Thus, the system is unstable and its response will be an exponential growth.

Question 33

Topic: Communication Systems

Which modulation scheme would you suggest for high speed data transmission over a fading channel?

  • A. MSK
  • B. BPSK
  • C. FSK
  • D. OFDM

Answer: D. OFDM

Explanation: In OFDM (Orthogonal Frequency Division Multiplexing), the high-speed data stream is divided into several parallel low-rate data streams. Because the subcarriers are orthogonal and have a longer symbol duration, it is highly robust to multipath fading.

Question 34

Topic: Electromagnetics

For a two port network to be reciprocal, its Scattering parameters must satisfy the following condition, for $i=1,2; j=1,2$

  • A. $S_{ij}=S_{ji}(i\ne j)$
  • B. $S_{ij}=S_{ji}(i=j)$
  • C. $S_{ij}=1/S_{ji}(i\ne j)$
  • D. $S_{ij}=S_{ji}*(i=j)$

Answer: A. $S_{ij}=S_{ji}(i\ne j)$

Explanation: For a two-port network to be reciprocal, the transmission coefficients in both directions must be equal. Therefore, the scattering matrix must be symmetric, which means $S_{ij} = S_{ji}$ for $i \ne j$ (i.e., $S_{12} = S_{21}$).

Question 35

Topic: Digital Electronics

In the following circuit, what sequence is followed by A and B on rising edge of CLK after reset is de-asserted?
(The circuit shows two D flip-flops. $D_1$ is connected to the output of an XOR gate. The XOR inputs are A (which is $\overline{Q}_2$) and B (which is $Q_1$). $D_2$ is connected to $Q_1$. They share a common CLK.)

  • A. $AB=10, 11, 00, 10, 11, ...$
  • B. $AB=10, 01, 00, 11, 10, ...$
  • C. $AB=10, 00, 01, 10, 00, ...$
  • D. $AB=11, 01, 00, 10, 11, ...$

Answer: A. $AB=10, 11, 00, 10, 11, ...$

Explanation: When the flip-flops are reset, $Q_1=0$ and $Q_2=0$, meaning $\overline{Q}_1=1$ and $\overline{Q}_2=1$.
$A = \overline{Q}_2$ and $B = Q_1$.
Initial state after reset: $A=1$, $B=0$ ($AB=10$).
$D_1 = A \oplus B = \overline{Q}_2 \oplus Q_1$, and $D_2 = Q_1$.
State 1: $D_1 = 1 \oplus 0 = 1$, $D_2 = 0$. After CLK, $Q_1=1$, $Q_2=0$. $A=\overline{Q}_2=1$, $B=Q_1=1$ ($AB=11$).
State 2: $D_1 = 1 \oplus 1 = 0$, $D_2 = 1$. After CLK, $Q_1=0$, $Q_2=1$. $A=\overline{Q}_2=0$, $B=Q_1=0$ ($AB=00$).
State 3: $D_1 = 0 \oplus 0 = 0$, $D_2 = 0$. After CLK, $Q_1=0$, $Q_2=0$. $A=\overline{Q}_2=1$, $B=Q_1=0$ ($AB=10$).
The sequence of AB is 10, 11, 00, 10, 11...

Question 36

Topic: Electromagnetics

In an axial mode operation of a helical antenna, maximum field radiated by the antenna is

  • A. along the axis of the helical antenna
  • B. in a plane normal to the helix axis
  • C. in both axial and normal planes
  • D. none of the above

Answer: A. along the axis of the helical antenna

Explanation: In the axial mode of operation of a helical antenna, the radiation pattern is highly directional and the maximum field is radiated along the axis of the helix. (Note: the official explanation contained a typo mentioning normal mode, but the answer key (a) corresponds to the axial direction as defined in standard antenna theory).

Question 37

Topic: Network Theory

Which of the following theorem is applicable for both linear and non-linear circuits?

  • A. Superposition theorem
  • B. Thevenin's theorem
  • C. Norton's theorem
  • D. None of these

Answer: D. None of these

Explanation: Superposition, Thevenin's, and Norton's theorems are strictly applicable only to linear networks. Tellegen's theorem is valid for any linear or non-linear circuit because it is based on the principles of energy conservation (KCL and KVL), but it is not listed among the options. Hence, "None of these" is the correct answer.

Question 38

Topic: Communication Systems

What is the maximum data rate that can be transmitted using a QPSK modulation with a roll-off factor of 0.2 for a 36 MHz transponder?

  • A. 7.2 Mbps
  • B. 360 Mbps
  • C. 43.2 Mbps
  • D. 60 Mbps

Answer: D. 60 Mbps

Explanation: The bandwidth for M-ary PSK with a raised cosine filter is given by $BW = \frac{R_b}{\log_2 M}(1+\alpha)$.
For QPSK, $M = 4$, so $\log_2 4 = 2$. Thus, $BW = \frac{R_b}{2}(1+\alpha)$.
Given $BW = 36\text{ MHz}$ and $\alpha = 0.2$.
$36\text{ MHz} = \frac{R_b}{2}(1 + 0.2) = \frac{R_b}{2}(1.2) = 0.6 R_b$.
$R_b = \frac{36}{0.6} = 60\text{ Mbps}$.

Question 39

Topic: Analog Circuits

Consider a single stage tuned amplifier having 3dB bandwidth of 100 kHz. Determine the bandwidth if two such single stage tuned amplifiers are cascaded.

  • A. 1000 kHz
  • B. 10 kHz
  • C. 41.4 kHz
  • D. 64.3 kHz

Answer: D. 64.3 kHz

Explanation: When $n$ identical single-stage tuned amplifiers are cascaded, the overall bandwidth is given by $BW_{overall} = BW_{single} \times \sqrt{2^{1/n}-1}$.
For $n=2$, $BW_{overall} = 100\text{ kHz} \times \sqrt{2^{1/2}-1} = 100 \times \sqrt{1.414-1} = 100 \times \sqrt{0.414} \approx 100 \times 0.643 = 64.3\text{ kHz}$.

Question 40

Topic: Communication Systems

500 Bits of data needs to be transmitted from an UART every 100 ms. What should be the minimum transmission baud rate of UART, assuming 1 start bit, 8 data bits and 2 stop bits?

  • A. 50000
  • B. 55000
  • C. 40000
  • D. 45000

Answer: B. 55000

Explanation: Assuming the question intended 500 Bytes based on the standard UART 8-bit frame (typographical error in the source material):
Each byte requires $N_{start} + N_{data} + N_{stop} = 1 + 8 + 2 = 11\text{ bits}$.
Total number of bits to transmit $= 500\text{ bytes} \times 11\text{ bits/byte} = 5500\text{ bits}$.
Time allowed $T = 100\text{ ms} = 0.1\text{ s}$.
Baud rate (bits per second) $= \frac{5500}{0.1} = 55000\text{ bps}$.

Question 41

Topic: Electromagnetics

What is the value of the major cross-sectional dimension(width) of a rectangular waveguide with dominant $TE_{10}$ mode propagation, if its cut off frequency is 10 GHz?

  • A. 15 mm
  • B. 30 mm
  • C. 7.5 mm
  • D. 45 mm

Answer: A. 15 mm

Explanation: For the dominant $TE_{10}$ mode in a rectangular waveguide, the cutoff frequency is $f_c = \frac{c}{2a}$.
Given $f_c = 10\text{ GHz} = 10 \times 10^9\text{ Hz}$ and $c = 3 \times 10^8\text{ m/s}$.
$a = \frac{c}{2f_c} = \frac{3 \times 10^8}{2 \times 10 \times 10^9} = \frac{3 \times 10^8}{20 \times 10^9} = 1.5 \times 10^{-2}\text{ m} = 15\text{ mm}$.

Question 42

Topic: Microwave Engineering

In a Traveling-wave Tube, the velocity of electron beam with respect to the axial velocity of RF field is kept

  • A. lower
  • B. higher
  • C. equal
  • D. any of these

Answer: B. higher

Explanation: The velocity of the electron beam should ideally equal the axial velocity of the RF field for synchronization. However, to achieve net energy transfer from the electron beam to the RF field, the electron beam velocity is kept slightly higher.

Question 43

Topic: Communication Systems

In satellite communication, scrambling is mainly used for

  • A. clock recovery
  • B. encryption
  • C. limiting power spectral density
  • D. bandwidth efficiency

Answer: C. limiting power spectral density

Explanation: In satellite communications, data scrambling is employed to randomize the bitstream. This prevents long strings of 1s or 0s, thereby limiting the concentrated power spectral density (preventing sharp spectral peaks) and ensuring energy is spread evenly, which mitigates interference with adjacent channels.

Question 44

Topic: Electromagnetics

Scattering Matrix for a microwave network, which is matched at all the ports, will satisfy the following condition

  • A. All non-diagonal elements will be zero
  • B. All diagonal elements will be zero
  • C. None of the matrix elements will be zero
  • D. All the matrix elements will be zero

Answer: B. All diagonal elements will be zero

Explanation: For a microwave network that is perfectly matched at all its ports, there are no reflections at any port when the other ports are terminated in matched loads. This implies that the reflection coefficients (the diagonal elements $S_{ii}$ of the scattering matrix) are all zero.

Question 45

Topic: Digital Electronics

Following waveform shows output of a 4 bit DAC with 5 V reference voltage. The 4 bit digital input of DAC is connected to 4 bit up counter, the one bit input of DAC is stuck at '0', which is this bit?
(The waveform shows a staircase signal starting from 0V, stepping up to 0.9375V, then dropping to 0V, stepping up to 0.9375V, then jumping to 2.5V, stepping up to 3.4375V, and dropping to 2.5V repeatedly.)

  • A. Bit-0 (LSB)
  • B. Bit-1
  • C. Bit-2
  • D. Bit-3 (MSB)

Answer: C. Bit-2

Explanation: For a 4-bit DAC with a 5V reference, the step size is $5/16 = 0.3125\text{ V}$.
The expected output sequence for a normal up-counter would increase by 0.3125V per step. The sequence shown goes 0V, 0.3125V, 0.625V, 0.9375V (inputs 0000 to 0011).
The next expected input is 0100 (which corresponds to $4 \times 0.3125 = 1.25\text{ V}$), but the output drops back to 0V, indicating the DAC receives 0000 instead.
Similarly, when the counter reaches 1000 to 1011, the output is 2.5V to 3.4375V, but drops to 2.5V when it should be 1100. This indicates that the 3rd bit from the right (Bit-2, where LSB is Bit-0) is stuck at 0.

Question 46

Topic: Electromagnetics

Which of the following has the highest skin depth?

  • A. Aluminium
  • B. Gold
  • C. Sliver
  • D. Copper

Answer: A. Aluminium

Explanation: Skin depth $\delta$ is inversely proportional to the square root of conductivity ($\delta \propto 1/\sqrt{\sigma}$). Among the given options, Aluminium has the lowest electrical conductivity ($\sim 3.5 \times 10^7$ S/m), which results in the highest skin depth.

Question 47

Topic: Network Theory

A network contains linear resistors and ideal voltage sources. If values of all the resistors are doubled, then voltage across each resistor is

  • A. Halved
  • B. Doubled
  • C. Increases by 4 times
  • D. Remains the same

Answer: D. Remains the same

Explanation: Since the network is linear and contains only ideal voltage sources, doubling the value of all resistors decreases the current in each individual branch to half ($I/2$). However, the voltage across each resistor is $V = I_{new} \times R_{new} = (I/2) \times (2R) = IR$. Thus, the voltage across each resistor remains the same.

Question 48

Topic: Communication Systems

A coherent QPSK demodulator is required on ground for receiving data from a LEO satellite. What should be optimum order of PLL in the carrier phase tracking loop?

  • A. $1^{st}$ order
  • B. $2^{nd}$ order
  • C. $3^{rd}$ order
  • D. None of the above

Answer: C. $3^{rd}$ order

Explanation: Low Earth Orbit (LEO) satellites experience high and rapidly changing Doppler shifts (Doppler rate or acceleration of phase). A 3rd order Phase-Locked Loop (PLL) is required to track phase acceleration with zero steady-state error.

Question 49

Topic: Electromagnetics

VSWR of a purely resistive load of normalized value $n+j0$ for $n<1$ is

  • A. n
  • B. $1/n$
  • C. 1
  • D. infinite

Answer: B. $1/n$

Explanation: For a purely resistive network, VSWR = $Z_L/Z_0$ if $Z_L > Z_0$ and VSWR = $Z_0/Z_L$ if $Z_L < Z_0$. Here, the normalized load is $n < 1$, which means $Z_L < Z_0$. Therefore, $\text{VSWR} = 1/n$.

Question 50

Topic: Signals and Systems

Two discrete time systems have impulse response $h_{1}[n]=\delta[n-3]$ and $h_{2}[n]=\delta[n-5]$. If these two systems are cascaded then the overall impulse response of the cascaded system is

  • A. $\delta[n-8]$
  • B. $\delta[n-3]+\delta[n-5]$
  • C. $\delta[n-1]*\delta[n-2]$
  • D. $\delta[n-2]$

Answer: A. $\delta[n-8]$

Explanation: The overall impulse response of cascaded systems is the convolution of their individual impulse responses.
$h[n] = h_1[n] * h_2[n] = \delta[n-3] * \delta[n-5] = \delta[n - (3+5)] = \delta[n-8]$.

Question 51

Topic: Microwave Engineering

Frequency range of Ka-band EM waves is

  • A. 18.0 GHz-26.5 GHz
  • B. 26.5 GHz-40.0 GHz
  • C. 12.0 GHz-18.0 GHz
  • D. 40.0 GHz-60.0 GHz

Answer: B. 26.5 GHz-40.0 GHz

Explanation: The standard frequency band designations are: Ku-band is 12-18 GHz, K-band is 18-27 GHz, and Ka-band is 26.5-40 GHz.

Question 52

Topic: Network Theory

Following four resistances are connected in parallel; the largest power is dissipated by
1. 2.4 $\Omega$, 1 W
2. 5.1 $\Omega$, 2 W
3. 10 $\Omega$, 5 W
4. 22 $\Omega$, 10 W

  • A. 2.4 $\Omega$, 1 W
  • B. 5.1 $\Omega$, 2 W
  • C. 10 $\Omega$, 5 W
  • D. 22 $\Omega$, 10 W

Answer: A. 2.4 $\Omega$, 1 W

Explanation: In a parallel combination, the voltage ($V$) across each branch remains the same. Power dissipated is $P = V²/R$. Therefore, the branch with the minimum resistance will have the largest power dissipation. The 2.4 $\Omega$ resistor has the smallest resistance.

Question 53

Topic: Communication Systems

Eye diagram gives us an idea of

  • A. Modulation scheme
  • B. Clock jitter
  • C. Signal-to-noise ratio
  • D. All of the above

Answer: D. All of the above

Explanation: An eye diagram is created by taking the time domain signal and overlapping the traces for a certain number of symbols. It provides visual information about clock jitter, signal-to-noise ratio, the amount of distortion (intersymbol interference), and the modulation scheme.

Question 54

Topic: Signals and Systems

Determine the correlation coefficient between the pulses $x(t)$ and $y(t)$ shown in the figure below:
(The figure shows $x(t)$ as a rectangular pulse from $t=0$ to $t=2$ with amplitude 1, and $y(t)$ as a sine wave $\sin(2\pi t)$ from $t=0$ to $t=2$).

  • A. 1
  • B. -1
  • C. 0
  • D. $2\pi$

Answer: C. 0

Explanation: The cross-correlation at $\tau=0$ is given by $\int_{-\infty}^{\infty} x(t)y(t) dt$.
Here, $\int_{0}^{2} (1)\sin(2\pi t) dt = \left[ \frac{-\cos(2\pi t)}{2\pi} \right]_0^2 = \frac{-\cos(4\pi) + \cos(0)}{2\pi} = \frac{-1 + 1}{2\pi} = 0$. Because the integral of their product is zero, the signals are orthogonal, and the correlation coefficient is 0.

Question 55

Topic: Digital Signal Processing

What is 4 point Discrete Fourier Transform of signal $x(n)=\{1,2,3,4\}$?

  • A. $[10, 2j, 2, -2 ]$
  • B. $[10, -2, -2, -2]$
  • C. $[10, -2+2j, -2, -2-2j]$
  • D. $[10, -2, -2, -10 ]$

Answer: C. $[10, -2+2j, -2, -2-2j]$

Explanation: $X(k) = \sum_{n=0}^{3} x(n) e^{-j\frac{2\pi}{4}kn} = \sum_{n=0}^{3} x(n) (-j)^{kn}$.
$X(0) = 1 + 2 + 3 + 4 = 10$.
$X(1) = 1(1) + 2(-j) + 3(-1) + 4(j) = -2 + 2j$.
$X(2) = 1(1) + 2(-1) + 3(1) + 4(-1) = 1 - 2 + 3 - 4 = -2$.
$X(3) = 1(1) + 2(j) + 3(-1) + 4(-j) = -2 - 2j$.
Therefore, the 4-point DFT is $[10, -2+2j, -2, -2-2j]$.

Question 56

Topic: Electromagnetics

The value of axial ratio of a $45^{\circ}$ linearly polarized EM wave in dB is

  • A. infinite
  • B. 0
  • C. 1
  • D. 2

Answer: A. infinite

Explanation: Axial ratio is defined as the ratio of maximum to minimum electric field ($E_{max}/E_{min}$). For any linearly polarized EM wave, the minimum electric field component in the polarization ellipse is 0. So, Axial ratio = $E_{max}/0 = \infty$. In dB, it remains infinite.

Question 57

Topic: Analog Circuits

If an input signal with non-zero dc component is applied to a low pass RC network with single resistor R and single capacitor C as shown below, the dc component in the output will be
(The diagram shows $V_{in}$ passing through series R, with C connected in parallel to the output $V_{o}$).

  • A. the same as the input
  • B. less than the input
  • C. more than the input
  • D. zero

Answer: A. the same as the input

Explanation: A capacitor acts as an open circuit to DC (frequency $f=0$). Because no direct current flows through the capacitor, there is no voltage drop across the series resistor R for the DC component. Hence, the DC component at the output is the same as that of the input.

Question 58

Topic: Communication Systems

The thermal noise spectral density generated in a resistor kept at $t^{\circ}C$ is given by

  • A. $k(t+273)$
  • B. $ktB$
  • C. $k(t+273)B$
  • D. $2k(t+273)$

Answer: A. $k(t+273)$

Explanation: Thermal noise power is $P_n = kTB$. The thermal noise power spectral density is the ratio of thermal noise power to bandwidth, which is $P_n/B = kT$ Watts/Hz. Converting temperature from Celsius to Kelvin gives $T = (t+273)$ K. Therefore, the spectral density is $k(t+273)$.

Question 59

Topic: Signals and Systems

What are poles and zeros of a system having following transfer function?
$H(Z)=\frac{(1-Z^{-2})}{(1+1.3Z^{-1}+0.36Z^{-2})}$

  • A. $Zeros=-1, 1$; $Poles=-0.4, -0.9$
  • B. $Zeros=-1, 1$; $Poles=0.4, 0.9$
  • C. $Zeros=1$; $Poles=0.4, 0.9$
  • D. $Zeros=1, 1$; $Poles=0.4, 0.9$

Answer: A. $Zeros=-1, 1$; $Poles=-0.4, -0.9$

Explanation: Multiply the numerator and denominator by $Z²$:
$H(Z) = \frac{Z²-1}{Z²+1.3Z+0.36} = \frac{(Z-1)(Z+1)}{(Z+0.9)(Z+0.4)}$.
Zeros are obtained by equating the numerator to zero: $Z²-1 = 0 \Rightarrow Z = 1, -1$.
Poles are obtained by equating the denominator to zero: $Z²+1.3Z+0.36 = 0 \Rightarrow Z = -0.9, -0.4$.

Question 60

Topic: Electromagnetics

The change in phase of an electromagnetic wave normally incident on a perfect conducting wall is?

  • A. $90^{\circ}$
  • B. $180^{\circ}$
  • C. $0^{\circ}$
  • D. $270^{\circ}$

Answer: B. $180^{\circ}$

Explanation: The reflection coefficient for the electric field of an electromagnetic wave normally incident on a perfect conducting wall is -1. This negative sign indicates a phase change or phase shift of $180^{\circ}$ upon reflection.

Question 61

Topic: Electromagnetics

What is the turns ratio of the transformer needed to match 1 k$\Omega$ source resistance to a 50 $\Omega$ load?

  • A. 4.47:1
  • B. 400:1
  • C. 20:1
  • D. 0.05:1

Answer: A. 4.47:1

Explanation: For a transformer matching a load, the impedance relationship is $Z_L' = n² Z_L$.
Given $Z_L = 50~\Omega$ and required $Z_L' = 1~\text{k}\Omega = 1000~\Omega$.
$n² = \frac{Z_L'}{Z_L} = \frac{1000}{50} = 20$.
$n = \sqrt{20} \approx 4.47$.
Thus, the required turns ratio is 4.47:1.

Question 62

Topic: Communication Systems

A satellite communication link has both uplink and downlink $C/N_{0}$ of 50 dB-Hz. What would be overall link $C/N_{0}$?

  • A. 25 dB-Hz
  • B. 47 dB-Hz
  • C. 53 dB-Hz
  • D. 100 dB-Hz

Answer: B. 47 dB-Hz

Explanation: $CNR_{uplink} = 50~\text{dB-Hz} = 10^5$
$CNR_{downlink} = 50~\text{dB-Hz} = 10^5$
$\frac{1}{CNR_{overall}} = \frac{1}{CNR_{uplink}} + \frac{1}{CNR_{downlink}}$
$CNR_{overall} = \frac{10^5}{2} = 5 \times 10^4$
$CNR_{overall}(\text{in dB}) = 10\log_{10}(5 \times 10^4) \approx 47~\text{dB-Hz}$.
Note: If the uplink and downlink CNR are same, the overall CNR decreases by a factor of 2, i.e., 3 dB.

Question 63

Topic: Electromagnetics

Determine polarization loss factor for a circularly polarized signal upon reception by linearly polarized antenna

  • A. $\frac{1}{\sqrt{2}}$
  • B. $\frac{1}{2}$
  • C. 1
  • D. 0

Answer: B. $\frac{1}{2}$

Explanation: When a circularly polarized signal is received by a linearly polarized antenna, exactly half of the signal power is received because the linear antenna can only pick up one of the two orthogonal linear components of the circularly polarized wave. Thus, the Polarization Loss Factor (PLF) is $1/2$ (or 3 dB).

Question 64

Topic: Microprocessors

Which instruction of 8051 microcontroller is used for reading data from code memory?

  • A. MOV
  • B. MOVC
  • C. MOVX
  • D. XCH

Answer: B. MOVC

Explanation: MOVC (Move Code) is used to move a byte from the program (code) memory to the accumulator.

Question 65

Topic: Electromagnetics

If frequency is decreased 4 times, then skin depth of an Electromagnetic wave inside a conductor will be

  • A. halved
  • B. quadrupled
  • C. doubled
  • D. no change

Answer: C. doubled

Explanation: Skin depth is given by $\delta = \frac{1}{\sqrt{\pi f \mu \sigma}}$.
It is inversely proportional to the square root of frequency ($\delta \propto 1/\sqrt{f}$). If the frequency is decreased by 4 times, the skin depth increases by a factor of $\sqrt{4} = 2$. Hence, it is doubled.

Question 66

Topic: Network Theory

In the circuit shown below, the current through 10 $\Omega$ resistor is
(The circuit shows two parallel branches with 150 V sources and 20 $\Omega$ resistors each, connected across a 10 $\Omega$ resistor. The polarities of the 150 V sources oppose each other relative to the common node).

  • A. 5 A
  • B. 10 A
  • C. -5 A
  • D. None of these

Answer: D. None of these

Explanation: Using nodal analysis at node A:
$\frac{V_A - 150}{20} + \frac{V_A + 150}{20} + \frac{V_A}{10} = 0$
$V_A = 0~\text{V}$.
The current through the 10 $\Omega$ resistor is $I_{10\Omega} = V_A / 10 = 0~\text{A}$. Therefore, the correct answer is 'None of these'.

Question 67

Topic: Analog Circuits

If two tones $f_{1}$ and $f_{2}$ are amplified by a non-linear amplifier, which frequency components would be present in output?

  • A. $f_{1}$, $f_{2}$
  • B. $f_{1}, f_{2}, f_{1}+f_{2}, f_{1}-f_{2}$
  • C. $f_{1}+f_{2}$, $f_{1}-f_{2}$
  • D. $nf_{1}\pm mf_{2}$ where n and m are integers

Answer: D. $nf_{1}\pm mf_{2}$ where n and m are integers

Explanation: When input frequencies $f_1$ and $f_2$ are passed through a non-linear amplifier, intermodulation distortion occurs. The output will contain multiple harmonics and intermodulation products of the form $nf_1 \pm mf_2$, where n and m are integers.

Question 68

Topic: Analog Circuits

When a sinusoidal voltage wave drives a Schmitt Trigger, the output is a

  • A. Triangular wave
  • B. Rectified sine wave
  • C. Rectangular wave
  • D. Trapezoidal wave

Answer: C. Rectangular wave

Explanation: A Schmitt trigger is a regenerative comparator. When driven by an analog signal such as a sinusoidal wave, it triggers at defined upper and lower threshold levels, converting the input into a rectangular (or square) wave at the output.

Question 69

Topic: Microprocessors

What will be content of Stack Pointer if following assembly code of 8051 microcontroller is executed immediately after reset?
PUSH 00h
POP 01h
POP 02h

  • A. 07h
  • B. 06h
  • C. 0Ah
  • D. 31h

Answer: B. 06h

Explanation: After a reset instruction, the stack pointer (SP) content in an 8051 microcontroller is 07H.
- PUSH 00h increments SP to 08H.
- POP 01h decrements SP to 07H.
- POP 02h decrements SP to 06H.
Thus, the stack pointer content is 06H.

Question 70

Topic: Communication Systems

Consider the sampling of the band-pass signal, whose spectrum is as shown below. What minimum sampling frequency would you use from the options given below, so as to avoid aliasing?
(The spectrum is a triangle from 94 Hz to 106 Hz with peak at 100 Hz).

  • A. 24 Hz
  • B. 26.5 Hz
  • C. 48 Hz
  • D. 212 Hz

Answer: B. 26.5 Hz

Explanation: $F_H = 106~\text{Hz}$, $F_L = 94~\text{Hz}$.
Bandwidth $BW = F_H - F_L = 106 - 94 = 12~\text{Hz}$.
$K = \lfloor \frac{F_H}{BW} \rfloor = \lfloor \frac{106}{12} \rfloor = 8$.
Minimum sampling frequency $F_s = \frac{2F_H}{K} = \frac{2(106)}{8} = \frac{212}{8} = 26.5~\text{Hz}$.

Question 71

Topic: Network Theory

A resistor and a capacitor are connected in series with a sine wave generator having frequency such that the resistance and the capacitive reactance are same i.e. voltage across capacitance, $V_{C}$ is equal to Voltage across Resistance, $V_{R}$. If the frequency of the sine wave generator is doubled, then

  • A. $V_{R}<V_{C}$
  • B. $V_{R}=V_{C}$
  • C. $V_{R}>V_{C}$
  • D. None of the above

Answer: C. $V_{R}>V_{C}$

Explanation: Initially, $V_R = V_C$, which means $R = X_C = \frac{1}{2\pi f C}$.
If the frequency $f$ is doubled, the capacitive reactance $X_C$ halves, while the resistance $R$ remains the same. Since the same current flows through both, $V_R = I \times R$ and $V_C = I \times X_{C_{new}}$. Because $R > X_{C_{new}}$, it follows that $V_R > V_C$.

Question 72

Topic: Digital Electronics

In CMOS designs, why size of PMOS is kept larger than size of NMOS?

  • A. To get higher drive strength
  • B. To reduce power dissipation
  • C. To get balanced rise/fall time
  • D. All of above

Answer: C. To get balanced rise/fall time

Explanation: Hole mobility is lower than electron mobility (typically by a factor of 2 to 3). To achieve symmetric transfer characteristics and balanced rise/fall times, the PMOS transistor is sized appropriately larger (wider) than the NMOS transistor.

Question 73

Topic: Electromagnetics

A plane wave with a frequency of 3 GHz is propagating in an unbounded medium with relative permittivity $\epsilon_{r}=6$ and permeability $\mu_{r}=1.5$, the phase velocity for this wave in this medium is

  • A. $1\times 10^{8}\text{ m/s}$
  • B. $2\times 10^{8}\text{ m/s}$
  • C. $3\times 10^{8}\text{ m/s}$
  • D. $1.5\times 10^{8}\text{ m/s}$

Answer: A. $1\times 10^{8}\text{ m/s}$

Explanation: Phase velocity $v_p = \frac{c}{\sqrt{\mu_r \epsilon_r}}$.
Given $\epsilon_r = 6$ and $\mu_r = 1.5$, $\mu_r \epsilon_r = 9$.
$v_p = \frac{3 \times 10^8}{\sqrt{9}} = \frac{3 \times 10^8}{3} = 1 \times 10^8~\text{m/s}$.

Question 74

Topic: Radar and Navigation

Which of the following statements is NOT true?

  • A. A high PRF pulse Doppler radar has no ambiguities in Doppler frequency.
  • B. A high PRF pulse Doppler radar has no range ambiguities.
  • C. A high PRF pulse Doppler radar provides an accurate measurement of target's radial velocity.
  • D. An MTI radar has many Doppler ambiguities.

Answer: B. A high PRF pulse Doppler radar has no range ambiguities.

Explanation: A high PRF (Pulse Repetition Frequency) pulse Doppler radar has no ambiguity in Doppler frequency (velocity), but it suffers from severe range ambiguities. Therefore, the statement "A high PRF pulse Doppler radar has no range ambiguities" is not true.

Question 75

Topic: Digital Electronics

The Boolean equation for Y in the following circuit is
(The diagram shows two transmission gates connected in series with inputs B and B'. The control lines are A and A').

  • A. Y = XNOR (A, B)
  • B. Y = OR (A, B)
  • C. Y = AND (A, B)
  • D. Y = XOR (A, B)

Answer: A. Y = XNOR (A, B)

Explanation: When A=0 and A'=1, the first transmission gate is ON and Y=B.
When A=1 and A'=0, the second transmission gate is ON and Y=B'.
This logic yields $Y = \overline{A}B + A\overline{B}$, which is the XOR logic equation $Y = A \oplus B$. Note: the official ISRO answer key listed (a) (which textually says XNOR) despite deriving XOR logic in standard solutions. The provided answer reflects the official answer key.

Question 76

Topic: Electromagnetics

A wave is incident from free-space onto a dielectric region of dielectric constante $\epsilon_{r}=4$. The wave impedance in the dielectric is

  • A. 55.5 $\Omega$
  • B. 100.5 $\Omega$
  • C. 108.6 $\Omega$
  • D. 188.5 $\Omega$

Answer: D. 188.5 $\Omega$

Explanation: The wave impedance in a dielectric medium is $Z = \sqrt{\frac{\mu}{\epsilon}} = \frac{\eta_0}{\sqrt{\epsilon_r}}$.
Given $\eta_0 \approx 377~\Omega$ and $\epsilon_r = 4$.
$Z = \frac{377}{\sqrt{4}} = \frac{377}{2} = 188.5~\Omega$.

Question 77

Topic: Radar and Navigation

Range resolution of a Linear Frequency Modulated pulse compression radar depends on which of the following factors

  • A. Radiated power
  • B. Antenna size
  • C. Center frequency of the radar
  • D. Bandwidth of the transmitted pulse

Answer: D. Bandwidth of the transmitted pulse

Explanation: Range resolution is inversely proportional to the bandwidth of the transmitted pulse. For a Linear Frequency Modulated (LFM) pulse compression radar, the resolution improves as the swept bandwidth of the chirp increases.

Question 78

Topic: Digital Electronics

Which logical function is implemented by following Transmission Gate based circuit?
(The circuit displays a standard CMOS complex logic gate structure with a pull-up PMOS network and a pull-down NMOS network taking inputs A, B, C, and D).

  • A. $Y=(A(B+C)+D)$
  • B. $Y=A(B+C)+D$
  • C. $Y=(A+BC)D$
  • D. $Y=\overline{((A+BC)D)}$

Answer: D. $Y=\overline{((A+BC)D)}$

Explanation: The image represents a static CMOS logic gate, not a transmission gate. Observing the pull-down NMOS network: transistor D is in series with a group composed of A in parallel with the series combination of B and C. This forms the logic $D(A+BC)$. Because of the inherent inversion in static CMOS gates, the output is $Y = \overline{D(A+BC)} = \overline{((A+BC)D)}$.

Question 79

Topic: Communication Systems

The orbital period of a satellite in circular orbit of radius R from the centre of the earth, is proportional to

  • A. $R^{1/2}$
  • B. $R^{3/2}$
  • C. $R^{2}$
  • D. $R$

Answer: B. $R^{3/2}$

Explanation: According to Kepler's Third Law, the square of the orbital period ($T²$) is directly proportional to the cube of the orbital radius ($R³$). Therefore, $T \propto R^{3/2}$.

Question 80

Topic: Electronic Devices and Circuits

Which of the following meters is based on the principle of Hall Effect?

  • A. Ammeter
  • B. Gaussmeter
  • C. Voltmeter
  • D. All of the above

Answer: B. Gaussmeter

Explanation: A Gaussmeter measures magnetic field strength and is fundamentally based on the principle of the Hall effect. Voltmeters and ammeters typically work on electromagnetic or electrostatic principles.