Attempt the ISRO Electronics & Communication (ECE) 2013 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Electronics & Communication (ECE) papers
80 questions.
Topic: Analog Circuits
Input voltage applied to a circuit is 1 V rms and the output is 1 mV rms. Net gain of the circuit is
Answer: C. -60 dB
Explanation: Voltage gain in dB is given by $Gain(dB) = 20 \log_{10}\left(\frac{V_{o}}{V_{i}}\right)$.
Here, $V_{i} = 1 \text{ V}$ and $V_{o} = 1 \text{ mV} = 10^{-3} \text{ V}$.
$Gain(dB) = 20 \log_{10}\left(\frac{10^{-3}}{1}\right) = 20 \log_{10}(10^{-3}) = 20 \times (-3) = -60 \text{ dB}$.
Topic: Communication Systems
Two ideal quantizers A and B have following specifications:
A: 5 bit quantizer with input dynamic range of -1 V to +1 V with $Q_{1}$ as quantization noise power.
B: 8 bit quantizer with input dynamic range of -0.5 V to +0.5 V with $Q_{2}$ as quantization noise power.
Then $Q_{1}/Q_{2}$ will be
Answer: B. 256
Explanation: Quantization noise power is given by $Q = \frac{s^{2}}{12}$, where step size $s = \frac{V_{p-p}}{2^{n}}$.
For Quantizer A: Dynamic range is from -1 V to +1 V, so $V_{p-p} = 2 \text{ V}$. Since it is a 5-bit quantizer, $n=5$. Step size $s_{1} = \frac{2}{2^{5}} = \frac{1}{2^{4}}$. The quantization noise power is $Q_{1} = \frac{(1/2^{4})^{2}}{12}$.
For Quantizer B: Dynamic range is from -0.5 V to +0.5 V, so $V_{p-p} = 1 \text{ V}$. Since it is an 8-bit quantizer, $n=8$. Step size $s_{2} = \frac{1}{2^{8}}$. The quantization noise power is $Q_{2} = \frac{(1/2^{8})^{2}}{12}$.
The ratio $\frac{Q_{1}}{Q_{2}} = \frac{(1/2^{4})^{2}}{(1/2^{8})^{2}} = \frac{2^{16}}{2^{8}} = 2^{8} = 256$.
Topic: Electromagnetics
The divergence of magnetic intensity is
Answer: C. zero
Explanation: According to Gauss's law in magnetostatics, $\nabla \cdot \vec{B} = 0$ or $\iint \vec{B} \cdot \vec{ds} = 0$. This implies that the divergence of a magnetic field is always zero, which mathematically represents that isolated magnetic poles (magnetic monopoles) do not exist.
Topic: Communication Systems
A UART is configured to transmit 8 bit data, 1 start bit and 1 stop bit. The serial data output is observed on oscilloscope, which looks like a square wave with frequency of 9600 Hz. What is the baud rate transmitted data?
Answer: B. Baud rate = 19200, Data=FFh
Explanation: Baud rate is defined as the number of symbols or pulses per second. If the serial data output looks like a square wave with a frequency of $9600 \text{ Hz}$, one full cycle of the square wave contains two symbols (one high, one low). Therefore, the Baud rate is $2 \times 9600 = 19200$. (Note: The official key selects (b), indicating Data=FFh, though standard derivations often point to alternating bits like 55H producing a continuous square wave. The provided answer reflects the official document).
Topic: Electromagnetics
A transmission line having characteristic impedance of 50 $\Omega$ has to deliver 10 kW power at 100 kHz to a load. Maximum permissible value of rms current anywhere along the line is 20 A. What is the maximum VSWR that can be tolerated on this line?
Answer: A. 2
Explanation: Power delivered $P = I^{2} Z_{L}$. Given $P = 10 \times 10^{3} \text{ W}$ and maximum $I_{rms} = 20 \text{ A}$, we can find $Z_{L} = \frac{10 \times 10^{3}}{20 \times 20} = 25 \text{ } \Omega$.
The reflection coefficient $\Gamma = \frac{Z_{L}-Z_{o}}{Z_{L}+Z_{o}} = \frac{25-50}{25+50} = -\frac{1}{3}$.
The Voltage Standing Wave Ratio (VSWR) is $\frac{1+|\Gamma|}{1-|\Gamma|} = \frac{1+1/3}{1-1/3} = \frac{4/3}{2/3} = 2$.
Topic: Control Systems
Routh Hurwitz criterion is used to determine
Answer: C. absolute stability of the system
Explanation: The Routh Hurwitz Criterion is an analytical method used to determine the absolute stability of a linear time-invariant control system by examining the roots of its characteristic equation.
Topic: Microprocessors
The decoding circuit shown in the figure has been used to generate active low chip select signal of memory interfaced to 8 bit microprocessor with 16 bit address bus. What is address range and size of memory?
(The circuit shows $A_{15}$ and $A_{14}$ passing directly to a NAND gate. $A_{13}$ passes through an inverter to the NAND gate, and $A_{12}$ passes directly to the NAND gate. The output is CS'.)
Answer: A. D000H to EFFFH, size = 8 k bytes
Explanation: Since the chip select (CS') is active low, the output of the NAND gate must be 0. This requires all inputs to the NAND gate to be 1. Thus, $A_{15} = 1$, $A_{14} = 1$, the inverted $A_{13} = 1 \Rightarrow A_{13} = 0$, and $A_{12} = 1$.
Wait, the official explanation provides a slightly different logic: "The output of NAND gate will always be 1 when both the inputs at it are 0. Thus $A_{15}$ and $A_{14}$ are always 1, but $A_{13}$ and $A_{12}$ should always be opposite". Following the truth table provided in the solution, the valid address combinations for the higher bits ($A_{15}-A_{12}$) are 1101 (D) and 1110 (E).
Therefore, the address range is D000H to EFFFH, which corresponds to a size of 8 k bytes.
Topic: Electromagnetics
If the waveguide cross-section of a square waveguide with $TE_{11}$ propagation mode is gradually deformed into a circle, then the corresponding circular waveguide mode will be
Answer: C. $TE_{21}$
Explanation: Based on the official answer key, the correct option is (c) $TE_{21}$. No detailed derivation is provided in the reference.
Topic: Analog Circuits
What could be the output current rating of following shunt regulator?
(The circuit shows a 10 V source connected to a 50 $\Omega$ resistor, which is in series with a parallel combination of a 5 V, 0.4 W Zener diode and a load $I_L$.)
Answer: B. $20 \text{ mA}<I_{L}<100 \text{ mA}$
Explanation: Given Zener power $P_{Z} = 0.4 \text{ W}$ and Zener voltage $V_{Z} = 5 \text{ V}$.
The maximum Zener current is $I_{Z(max)} = \frac{P_{Z}}{V_{Z}} = \frac{0.4}{5} = 0.08 \text{ A} = 80 \text{ mA}$.
The total current supplied by the source is $I = \frac{V_{in} - V_{Z}}{R_{s}} = \frac{10 - 5}{50} = \frac{5}{50} = 0.1 \text{ A} = 100 \text{ mA}$.
To maintain regulation without exceeding the Zener's power rating, the load current $I_{L}$ must satisfy $I = I_{Z} + I_{L}$ where $I_{Z} \le 80 \text{ mA}$.
Thus, $100 \text{ mA} \le 80 \text{ mA} + I_{L} \Rightarrow I_{L} \ge 20 \text{ mA}$.
When $I_{Z} = 0 \text{ mA}$, the maximum load current is $I_{L} = 100 \text{ mA}$.
Therefore, $I_{L}$ lies in the range of 20 mA to 100 mA.
Topic: Radar and Navigation
An FM-CW (Frequency Modulated-Continuous Wave) radar is essentially
Answer: C. can operate either as monostatic or as bistatic
Explanation: An FM-CW radar typically has both a transmitting antenna and a receiving antenna. While it can use separate antennas (making it bistatic), it can also be designed to operate with a single shared antenna (monostatic) using a circulator. Note: The official explanation diagram depicts separate $T_{x}$ and $R_{xr}$ antennas, but the official answer key marks (c), indicating it can operate as either.
Topic: Electromagnetics
The flux in a magnetic core is sinusoidally varying at 200 Hz. The maximum flux density is 2 Tesla and eddy current loss is 15 W. If the frequency is raised to 400 Hz maximum flux density reduced to 1 Tesla, the eddy current loss will
Answer: C. reduce to one-fourth
Explanation: Eddy current loss in a magnetic core is proportional to the square of both frequency and maximum flux density, given by $W_{e} = K_{e} f^{2} B_{m}^{2}$.
Initially, $W_{e1} = 15 \text{ W} = K_{e} (200)^{2} (2)^{2}$.
When frequency is 400 Hz and $B_m$ is 1 Tesla, the new eddy current loss is $W_{e2} = K_{e} (400)^{2} (1)^{2}$.
Comparing the two, the term $f² B_m^2$ changes from $(200\times2)^2 = 400^2$ to $(400\times1)^2 = 400^2$. Wait, the calculation in the reference says: "$K_e = \frac{15}{200 \times 200 \times 4}$... $W_{e2} = 15 \text{ W}$. Thus, Eddy current loss remains same to 15 W". However, the official answer key specifies option (c). This suggests a discrepancy in the provided material, but based strictly on the provided answer key, the answer is (c).
Topic: Electromagnetics
The electric field intensity E and magnetic field intensity H are coupled and propagating in free space in x and y direction respectively, the Poynting vector is given by
Answer: D. none of the above
Explanation: The Poynting Vector represents the directional energy flux density of an electromagnetic field, given by $\vec{P} = \vec{E} \times \vec{H} \text{ Watt/m}^{2}$.
If $\vec{E}$ is in the x-direction ($E_{x}\hat{a}_{x}$) and $\vec{H}$ is in the y-direction ($E_{y}\hat{a}_{y}$), then $\vec{P} = E_{x}\hat{a}_{x} \times E_{y}\hat{a}_{y} = E H \hat{a}_{z}$.
Since the result is in the z-direction, none of the provided options (a, b, c) accurately describe the vector $EH\hat{a}_z$, making (d) the correct choice.
Topic: Communication Systems
If x and y are two random signals with zero mean Gaussian distribution having identical standard deviation, the phase angle between them is
Answer: B. uniform between -$\pi$ and $\pi$
Explanation: When x and y are two independent zero-mean Gaussian random variables with identical standard deviations, they form a circularly symmetric complex Gaussian random variable. In polar coordinates, the phase angle $\theta = \tan^{-1}\left(\frac{y}{x}\right)$ of such a distribution is uniformly distributed between $-\pi$ to $\pi$.
Topic: Electromagnetics
The current flowing through a capacitor in an AC circuit is
Answer: C. displacement current
Explanation: In a capacitor, there is no direct physical connection between the two plates; instead, an insulator (dielectric) of definite permittivity is placed between them. When an AC field is applied, the changing electric field establishes a path for current known as displacement current. Displacement current is the result of permittivity and exists whenever the electric field changes with time, according to Ampere-Maxwell's law $\nabla \times \vec{H} = J_C + J_D$.
Topic: Digital Electronics
Which of the following is the Boolean function for Majority Voting, assuming A, B, C are inputs and Y is output?
Answer: C. $Y=ABC$
Explanation: Majority voting means the output is 1 only when a majority (two or more) of the inputs are 1. By constructing a truth table for inputs A, B, C, the output Y is 1 for input combinations 011, 101, 110, and 111. Using a Karnaugh map, this simplifies to $Y = BC + AC + AB$. (Note: The official answer key marks (c) as the answer, which is $Y=ABC$, despite the provided derivation pointing to $BC+AC+AB$. The answer key is followed here.)
Topic: Electromagnetics
For broadside antenna array, the largest possible spacing between the antenna elements without any grating lobes is
Answer: B. $\lambda$
Explanation: In general, as the element spacing is increased, the main lobe beam width is decreased. However, grating lobes (maxima in direction other than the main lobe direction) are introduced when the array element spacing is greater than or equal to one wavelength. Thus, $\lambda$ represents the spacing boundary.
Topic: Microprocessors
For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is
3000 MVI A, 45H
3002 MOV B, A
3003 STC
3004 CMC
3005 RAR
3006 XRA B
Answer: C. 67H
Explanation: MVI A, 45H loads A with 45H (0100 0101). MOV B, A copies it to B, so B is 45H. STC sets the carry flag (CY=1). CMC complements the carry flag (CY=0). RAR rotates the accumulator right through carry: A becomes 0010 0010 (22H) and CY becomes 1. XRA B performs XOR operation of B with accumulator: 22H XOR 45H = 0010 0010 XOR 0100 0101 = 0110 0111 = 67H.
Topic: Analog Circuits
Conduction angle of a Class AB amplifier is
Answer: B. between $180^{\circ}$ and $360^{\circ}$
Explanation: In class AB amplifier, the collector current flows for more than $180^{\circ}$ but less than $360^{\circ}$.
Topic: Electromagnetics
For non dispersive medium
Answer: C. Phase velocity = Group velocity
Explanation: In non-dispersive media, the speed of wave is independent of frequency. Thus, both the phase velocity and group velocity are same.
Topic: Digital Electronics
Schottky clamping is resorted in TTL gates
Answer: A. to reduce propagation delay
Explanation: The Schottky diode prevents the BJT from going into deep saturation by diverting excess base current through the Schottky diode and around the BJT. Thus, it reduces the propagation delay i.e. the time needed for the output transistor to switch from the ON state to OFF state is decreased.
Topic: Electromagnetics
At cut-off frequency, the phase velocity of a waveguide is
Answer: B. infinite
Explanation: Phase velocity $v_{p} = \frac{c}{\sqrt{1-(f_{c}/f_{o})^{2}}}$. As the wave frequency $f_{o}$ approaches the cut-off frequency $f_{c}$, then $f_{c}/f_{o}$ reaches 1. Therefore $1-(f_{c}/f_{o})^{2} \approx 0$, so $v_{p} = \frac{c}{0} \rightarrow \infty$.
Topic: Analog Circuits
A Zener diode, when used in voltage stabilization circuits, is biased in
Answer: B. reverse breakdown region
Explanation: The diode in the reverse biased region above breakdown voltage will regulate the load voltage against variations in load current and against the variations in supply voltage.
Topic: Control Systems
The closed loop frequency response of a dc-dc converter is shown in following figure. What are the gain and phase margins?
(The graph plots Gain in dB and Phase in degrees against frequency. The phase plot crosses 0 degrees at a point where the gain is roughly 20 dB, and the gain crosses 0 dB where the phase is roughly 120 degrees.)
Answer: C. 20 dB, $120^{\circ}$
Explanation: Since it is a closed loop system, i.e. $1+G(s)H(s)=0$, the phase margin is calculated above $0^{\circ}$ rather than $180^{\circ}$ (for $G(s)H(s)=-1$). Similarly, the gain margin is calculated for $1+G(s)H(s)=0$. From the plot, the Gain margin = $0 - (-20) = 20 \text{ dB}$ and phase margin = $0 + 120^{\circ} = 120^{\circ}$.
Topic: Electronic Devices and Circuits
If for a silicon npn transistor, the base-to-emitter voltage $(V_{BE})$ is 0.7 V and the collector-to-base voltage $(V_{CB})$ is 0.2 V, then the transistor is operating in the
Answer: A. normal active mode
Explanation: $V_{BE} = 0.7\text{ V}$, meaning the base-emitter junction is forward biased. $V_{CB} = 0.2\text{ V}$. Since $V_{C} - V_{B} = 0.2$, it implies $V_{C} > V_{B}$. This means the collector-base junction is reverse biased. Thus, the transistor is in normal active-mode.
Topic: Microwave Engineering
3-port circulator is
Answer: C. non reciprocal, matched
Explanation: A circulator is a passive non-reciprocal three or 4 port device, in which a microwave or radio frequency entering any port is transmitted to the next port in rotation only. Thus, it is perfectly matched and non reciprocal.
Topic: Digital Electronics
An 8 bit ripple counter and an 8 bit synchronous counter are made using flip flops having a propagation delay of 10 ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S respectively, then
Answer: D. $R=80\text{ ns}, S=10\text{ ns}$
Explanation: In an 8 bit ripple counter, the clock is applied to the first flip-flop only and remaining flip-flops receive the clock from their previous flip-flop. So it requires $8 \times 10\text{ ns} = 80\text{ ns}$. In a synchronous counter, the clock is applied simultaneously to all F/Fs, so the output is obtained after one clock pulse only i.e., $1 \times 10\text{ ns} = 10\text{ ns}$.
Topic: Analog Circuits
Gain of an RC low pass filter having a time constant '$\tau$' and frequency '$\omega$' is
Answer: B. $\frac{1}{\sqrt{1+(\omega\tau)^{2}}}$
Explanation: For an RC low pass filter, $H(s) = \frac{1}{RCs+1}$. The frequency response is $H(j\omega) = \frac{1}{1+j\omega RC}$. The magnitude or gain is $|H(j\omega)| = \left|\frac{1}{1+j\omega\tau}\right| = \frac{1}{\sqrt{1+\omega^{2}\tau^{2}}}$, where $\tau=RC$ is the time constant.
Topic: Microwave Engineering
For a directional coupler, the quantities I (isolation in dB), D (directivity in dB), C (coupling in dB) are related by
Answer: C. $I=D+C$
Explanation: For a directional coupler, Isolation $I = 10 \log_{10}\frac{P_i}{P_b}$, Directivity $D = 10 \log_{10}\frac{P_f}{P_b}$, and Coupling $C = 10 \log_{10}\frac{P_i}{P_f}$. Therefore, $I(\text{dB}) = C(\text{dB}) + D(\text{dB})$.
Topic: Digital Electronics
The two numbers represented in signed 2's complement form are $P=11101101$ and $Q=11100110$. If Q is subtracted from P, the value obtained in signed 2's complement is
Answer: B. 00000111
Explanation: To perform $P - Q$, we take the 2's complement of the subtrahend $Q$ and add it to the minuend $P$.
$P = 11101101$
$Q = 11100110$, so the 2's complement of $Q = 00011010$.
Adding them: $11101101 + 00011010 = 1 00000111$.
Since there is a carry out of the sign bit, the answer is positive, and the carry is neglected. The result is $00000111$.
Topic: Electronic Devices and Circuits
Depletion type MOSFET operates in
Answer: C. both depletion and enhancement mode
Explanation: Depletion type MOSFET operates in both depletion as well as enhancement mode by changing the gate voltage applied.
Topic: Microwave Engineering
Electric field and magnetic field are perpendicular to each other in
Answer: A. Klystron
Explanation: Based on the official answer key, the correct option is (a). In a standard Klystron, the electromagnetic fields in the cavity resonators are generally transverse/orthogonal, although linear beam tubes typically rely on axial electric fields for electron bunching. The given official key provides (a) as the answer.
Topic: Analog Circuits
A linear regulator is attempted using discrete components as shown below. What is the output voltage?
(The circuit shows a 10 V supply, an NPN transistor, a 2.5 V source at the base, and resistors. Output $V_o$ is measured across the load.)
Answer: A. 9.3 V
Explanation: Based on the official answer key, the correct option is (a). No detailed mathematical derivation is provided in the reference.
Topic: Radar and Navigation
If range of a radar is to be doubled, the peak transmit power of the radar has to be
Answer: D. increased by a factor of 16
Explanation: The maximum radar range is given by the Radar Equation: $R_{\text{max}} = \left[\frac{P_{t}G\sigma A_{e}}{(4\pi)^{2}S_{\text{min}}}\right]^{1/4}$.
From this, $R_{\text{max}} \propto [P_{t}]^{1/4}$.
So, if the range of a radar is to be doubled ($2R_{\text{max}}$), the peak transmit power of the radar has to be increased by a factor of $2^4 = 16$.
Topic: Electromagnetics
The electric field measured in the far field of an antenna at a distance of 50 m is $1~\text{V/m}$. The average power densities at a distance of 500 m from the antenna is
Answer: D. $13.3~\mu\text{W/m}^{2}$
Explanation: In the far field, the electric field intensity $E \propto \frac{1}{r}$, so $E = \frac{K}{r}$.
At $r = 50~\text{m}$, $E_{1} = \frac{K}{50} \Rightarrow 1 = \frac{K}{50} \Rightarrow K = 50$.
At $r = 500~\text{m}$, $E_{2} = \frac{50}{500} = 0.1~\text{V/m}$.
The average power density is $P_{\text{avg}} = \frac{1}{2}\frac{E^{2}}{\eta}$ where $\eta = 120\pi~\Omega$.
$P_{\text{avg}} = \frac{(0.1)^{2}}{2 \times 120\pi} = \frac{0.01}{240\pi} \approx 13.3~\mu\text{W/m}^{2}$.
Topic: Digital Electronics
If a counter having 10 flip-flops is initially at 0, what count will it hold after 2060 pulses?
Answer: A. 000 000 1100
Explanation: A 10 flip-flop counter counts from 0 up to $2^{10}-1 = 1023$, and resets on the $1024^{\text{th}}$ clock pulse.
At the $1024^{\text{th}}$ pulse, the count is 00 0000 0000.
At the $2048^{\text{th}}$ pulse, the count is 00 0000 0000.
The distance between the $2060^{\text{th}}$ clock pulse and the $2048^{\text{th}}$ clock pulse is $2060 - 2048 = 12$.
So, it will count 12 clock pulses from zero, which corresponds to binary 1100. Thus the counter state is 00 0000 1100.
Topic: Communication Systems
For a frequency modulated signal represented by $s(t)=10\sin(6\times 10^{8}t+2\sin 100\pi t)$. The maximum frequency deviation in the carrier from its unmodulated frequency is
Answer: B. 100 Hz
Explanation: The standard equation of the FM modulated wave is $x(t) = A_{c}\sin\left(\omega_{c}t + \frac{K_{f}A_{m}}{\omega_{m}}\sin \omega_{m}t\right)$.
Comparing this with $s(t) = 10\sin(6\times 10^{8}t+2\sin 100\pi t)$, we get:
$\frac{K_{f}A_{m}}{\omega_{m}} = 2$
$\omega_{m} = 100\pi \Rightarrow 2\pi f_{m} = 100\pi \Rightarrow f_{m} = 50~\text{Hz}$.
Maximum frequency deviation is $\Delta f = K_{f}A_{m} = 2 \times f_{m} = 2 \times 50 = 100~\text{Hz}$.
Topic: Analog Circuits
For which of the following conditions, the circuit shown below will function as precision full wave rectifier?
(The circuit is a standard precision full wave rectifier using two op-amps, two diodes $D_1$ and $D_2$, and resistors $R_1, R_2, R_3, R$.)
Answer: C. $R_{2}=2R_{1}$
Explanation: For $V_{in} > 0$, the diode $D_{1}$ is forward biased and diode $D_{2}$ is reverse biased. Applying KCL at the input of the second op-amp yields $\frac{V_{o}}{R_{3}} = \frac{V_{in}}{R_{1}} - \frac{V_{in}}{R_{2}}$.
For it to act as a full-wave rectifier, the gain magnitude must be proportional, e.g., using condition (c) $R_{2}=2R_{1}$:
$\frac{V_{o}}{R_{3}} = \frac{V_{in}}{R_{1}} - \frac{V_{in}}{2R_{1}} = \frac{V_{in}}{2R_{1}}$.
$\Rightarrow V_{o} = \frac{R_{3}}{2R_{1}}V_{in}$.
Because $V_{o} \propto V_{in}$ in a linear fashion for this half cycle matching the other, this condition is correct.
Topic: Radar and Navigation
In a monostatic radar, if the antenna aperture is doubled, then the radar range will
Answer: D. increase by a factor of $\sqrt{2}$
Explanation: $R_{\text{max}} = \left(\frac{P_{t}G\sigma A_{e}}{(4\pi)^{2}S_{\text{min}}}\right)^{1/4}$.
Assuming antenna aperture here relates to the cross-sectional area derived from diameter D, where $A_{e} = \frac{\pi D^{2}}{4}$.
So, $R_{\text{max}} = \left(\frac{P_{t}G\sigma \pi D^{2}}{4(4\pi^{2})S_{\text{min}}}\right)^{1/4}$.
Thus, $R_{\text{max}} \propto (D^{2})^{1/4} \propto D^{1/2}$.
If the antenna aperture (diameter D) is doubled, the radar range will be increased by a factor of $\sqrt{2}$.
Topic: Electromagnetics
The disadvantage of single stub matching is that
Answer: A. every load needs a new stub position
Explanation: Single stub matching is popular in matching fixed impedances at microwave frequencies. But it is not suitable for matching variable impedances. A change in load impedance results in a change in the length as well as the location of the stub, so every load needs a new stub position.
Topic: Electromagnetics
A certain antenna with an efficiency of 95% has maximum radiation intensity of 0.5 W/sr. The directivity of the antenna fed by input power of 0.4 W.
Answer: A. 16.53
Explanation: Given Efficiency $\eta = 0.95$, Maximum radiation intensity $U_{\text{max}} = 0.5~\text{W/sr}$, Input power $P_{\text{in}} = 0.4~\text{W}$.
$\text{Gain} = \frac{4\pi U_{\text{max}}}{P_{\text{in}}} = \frac{4\pi \times 0.5}{0.4} = \frac{2\pi}{0.4} = 5\pi$.
Since $\text{Gain} = \eta \times \text{Directivity (D)}$,
$D = \frac{\text{Gain}}{\eta} = \frac{5\pi}{0.95} = 16.53$.
Topic: Microprocessors
A memory system of size 16 k bytes is required to be designed using memory chips which have 12 address lines and 4 data lines each. Then number of such chips required to design the memory system is
Answer: C. 8
Explanation: Size of each memory chip = $2^{12} \times 4$ bits.
Required memory system of size = $16~\text{k bytes} = 16 \times 1024 \times 8 \text{ bits} = 2^{14} \times 8$ bits.
Number of chips required = $\frac{\text{Total required memory}}{\text{Capacity of one chip}} = \frac{2^{14} \times 8}{2^{12} \times 4} = 2^{2} \times 2 = 8$.
Topic: Control Systems
In the asymptotic bode plot of a transfer function of a closed loop system shown below, the number of poles and zeros are,
(The Bode plot starts from +40 dB at low frequency, slopes down to 10 dB at a frequency between 1K and 10K, then slopes up to 20 dB at 10K, and finally slopes down crossing 0 dB.)
Answer: A. 3 pole, 2 zero
Explanation: Since the plot is starting from 40 dB and decreasing at low frequencies, it indicates there are 2 poles at the origin (slope of -40 dB/dec).
At the frequency between 1k and 10k, the slope turns positive (+20 dB/decade), which means 2 zeros are added to change the slope by +40 dB/dec.
At the frequency 10k, a pole is added so that the slope is again -20 dB/decade.
Thus, there are $2+1=3$ poles and 2 zeros in total.
Topic: Electromagnetics
Plane wave travelling in free space has an average Poynting vector of $3~\text{W/m}^{2}$. Average energy density $\text{nJ/m}^{3}$ of the wave is
Answer: A. 10
Explanation: Average Poynting vector $P_{\text{avg}} = \frac{1}{2}\frac{E^{2}}{\eta}$.
$3 = \frac{E^{2}}{2 \times 120\pi} \Rightarrow E^{2} = 240 \times 3 \times \pi = 720\pi$.
Average energy density $= \frac{1}{2}\epsilon_{0} E^{2} = \frac{1}{2} \times \frac{1}{36\pi \times 10^{9}} \times 720\pi$.
$= \frac{720}{72 \times 10^{9}} = 10 \times 10^{-9}~\text{J/m}^{3} = 10~\text{nJ/m}^{3}$.
Topic: Electronic Devices and Circuits
For two identical n-channel JFET's connected in parallel as shown in figure below, the pinch-off voltage of equivalent JFET is
Answer: A. doubled
Explanation: Note: According to the official answer key, the answer is option (a). However, the detailed explanation in the source document states: "When the two identical JFETs are connected in parallel, then pinch-off voltage of equivalent JFET is some as that of individual JFET", which corresponds to option (c). The JSON extraction captures the options exactly as presented.
Topic: Electromagnetics
The velocity at which a sinusoidal signal of $10^{9}~\text{rad/s}$ travels down a loss-less transmission line for which $L=0.4~\mu\text{H/m}$ and $C=40~\text{pF/m}$ is
Answer: B. $2.5\times 10^{8}~\text{m/s}$
Explanation: The phase velocity for a lossless transmission line is $v_{p} = \frac{1}{\sqrt{LC}}$.
$v_{p} = \frac{1}{\sqrt{0.4 \times 10^{-6} \times 40 \times 10^{-12}}}$
$= \frac{1}{\sqrt{16 \times 10^{-18}}} = \frac{1}{4 \times 10^{-9}}$
$= 0.25 \times 10^{9} = 2.5 \times 10^{8}~\text{m/sec}$.
Topic: Electromagnetics
The Maxwell's equation $\nabla\times\overline{E}=-\frac{\partial\overline{B}}{\partial t}$ is obtained from
Answer: D. Both (b) and (c)
Explanation: Faraday discovered that the induced emf in a closed circuit is equal to the time rate of change of the magnetic flux linkage. $V_{emf} = -\frac{\partial\psi}{\partial t}$. The negative sign shows that the induced voltage acts in such a way to oppose the flux producing it, as given by Lenz's law. This gives the differential form $\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$.
Topic: Electromagnetics
A lossless line having characteristic impedance $Z_{o}$ is terminated with a load impedance of $jZ_{0}$. VSWR of the line will be
Answer: C. infinite
Explanation: Reflection coefficient $\Gamma = \frac{Z_{L}-Z_{o}}{Z_{L}+Z_{o}} = \frac{jZ_{o}-Z_{o}}{jZ_{o}+Z_{o}} = \frac{j-1}{j+1}$.
The magnitude of the reflection coefficient is $|\Gamma| = 1$.
$\text{VSWR} = \frac{1+|\Gamma|}{1-|\Gamma|} = \frac{1+1}{1-1} = \infty$.
Topic: Communication Systems
A signal $1+\cos(2\pi ft)+\cos(6\pi ft)$ where $f=1\text{ MHz}$ is sampled at 3 MHz and Fourier Transform is carried out. How many lines will be seen in Fourier Transform?
Answer: C. 3
Explanation: The signal has frequency components at DC ($0\text{ MHz}$), $f = 1\text{ MHz}$, and $3f = 3\text{ MHz}$. When sampled at $3\text{ MHz}$, the $3\text{ MHz}$ component aliases to $0\text{ MHz}$ (DC). The $1\text{ MHz}$ component remains at $1\text{ MHz}$. Thus, in the baseband spectrum, there will be spectral lines at $-1\text{ MHz}$, $0\text{ Hz}$, and $+1\text{ MHz}$, totaling 3 spectral lines.
Topic: Electromagnetics
The array factor of an array antenna depends on
Answer: C. phase of the applied signal
Explanation: Note: According to the official answer key, the answer is option (c). However, the detailed explanation in the source document states: "Thus, the array factor depends on the number of radiating elements, the elements spacing, amplitude and phase of the applied signal to each other", which fundamentally points to "all of the above". The JSON extraction captures the option strictly as presented in the key.
Topic: Digital Electronics
Which of the following parameter is improved by introducing pipelining in digital design?
Answer: B. Maximum clock frequency
Explanation: Pipelined approach divides an entire process into small and roughly equal time consumable sub-processes. The main advantage is that throughput increases considerably as the clock is applied simultaneously to all the blocks, so maximum clock frequency increases. Note that chip area and delay also increase as a trade-off.
Topic: Electromagnetics
A transmission line having characteristic impedance '$Z_{1}$' of varying length in series with a load impedance '$Z_{L}$' appears in a Smith Chart on
Answer: B. Constant VSWR Circle
Explanation: Varying the length of a lossless transmission line changes the phase of the reflection coefficient but keeps its magnitude constant. Since the VSWR depends only on the magnitude of the reflection coefficient, the locus of impedance points maps onto a constant VSWR circle on the Smith Chart.
Topic: Electromagnetics
Impedance characteristics on a Smith Chart repeat after a distance of
Answer: C. $\lambda/2$
Explanation: On a transmission line, the impedance pattern (including maxima and minima of the standing wave) repeats every half wavelength ($\lambda/2$). Therefore, one full rotation around the Smith chart corresponds to a physical distance of $\lambda/2$.
Topic: Analog Circuits
If $\tau$ is the time constant and w is the applied frequency, a low pass RC filter acts as a pure integrator when
Answer: B. $w\tau>>1$
Explanation: For a low pass RC filter, the transfer function is $\frac{V_{0}(s)}{V_{i}(s)} = \frac{1}{1+RCs} = \frac{1}{1+j\omega RC}$. For it to act as a pure integrator, the term $j\omega RC$ must dominate the denominator, which requires $\omega\tau >> 1$ (or $\omega >> \frac{1}{RC}$).
Topic: Network Theory
10 $\mu\text{F}$ capacitor is connected across secondary winding of a high frequency transformer having primary to secondary turns ratio 5:2. What is the value of capacitance seen across primary?
Answer: D. $1.6~\mu\text{F}$
Explanation: For a transformer, the reflected impedance is $Z_{1} = Z_{2}\left(\frac{N_{1}}{N_{2}}\right)^{2}$. Since $Z = \frac{1}{j\omega C}$, this implies $C_{1} = C_{2}\left(\frac{N_{2}}{N_{1}}\right)^{2}$.
Given $C_2 = 10~\mu\text{F}$ and $\frac{N_1}{N_2} = \frac{5}{2}$, we have $C_{1} = 10 \times 10^{-6}\left(\frac{2}{5}\right)^{2} = 10 \times 10^{-6} \times \frac{4}{25} = 1.6~\mu\text{F}$.
Topic: Digital Electronics
What will be the output of the following circuit, if point-P is stuck at 1?
(The circuit shows A and B feeding an OR gate. B is inverted and fed with C into an AND gate. The outputs of these two gates feed an OR gate to produce intermediate signal x. Then x and P feed an XOR gate to produce Y.)
Answer: B. $A^{\prime} B^{\prime} C^{\prime}$
Explanation: From the logic diagram, the intermediate signal $x = (A+B) + \bar{B}C = A + B + \bar{B}C = A+B+C$.
Since point P is stuck at 1, the input $C_{XOR}=1$. When any input of an XOR gate is 1, it behaves as an inverter.
Thus, $Y = x \oplus 1 = \bar{x} = \overline{A+B+C} = \bar{A}\bar{B}\bar{C}$ (or $A' B' C'$).
Topic: Analog Circuits
For the current mirror circuit shown below, if the emitter area of $Q_{2}$ is thrice of $Q_{1}$ the current I is
(The circuit is a basic BJT current mirror. $Q_1$ is diode-connected. $V_C = +10\text{ V}$, $R = 20\text{ k}\Omega$ in series with $Q_1$. The emitters of $Q_1$ and $Q_2$ are connected to $V_E = -10\text{ V}$. Current $I$ flows into the collector of $Q_2$.)
Answer: B. 2.955 mA
Explanation: Assuming the transistor is Ge with cut-off $V_{BE} = 0.3\text{ V}$ (as per official solution logic).
$V_{B1} = V_{E} + V_{BE1} = -10\text{ V} + 0.3\text{ V} = -9.7\text{ V}$.
The reference current $I_{1} = \frac{10 - (-9.7)}{20\text{ k}\Omega} = \frac{19.7}{20} = 0.985\text{ mA}$.
Assuming very high $\beta$, $I_{C1} \approx I_{1} = 0.985\text{ mA}$.
Since the emitter area of $Q_{2}$ is three times that of $Q_{1}$, $I_{C2} = 3 I_{C1} = 3 \times 0.985\text{ mA} = 2.955\text{ mA}$.
Topic: Analog Circuits
Output of an op-amp is 1 V peak, and slew rate is 5 V/$\mu$s. The maximum frequency of input sinusoidal signal that can be reproduced is
Answer: C. 796 kHz
Explanation: Slew rate $SR = \frac{dV_{o}}{dt} = V_m 2\pi f_{m}$.
$f_{m} = \frac{SR}{V_m 2\pi} = \frac{5 \times 10^{6}}{1 \times 2\pi} \approx 795774\text{ Hz} \approx 796\text{ kHz}$.
Topic: Digital Electronics
What is the division factor of the following clock divider circuit?
(The circuit contains two D flip-flops and an AND gate. The clock input feeds the first flip-flop directly and the second flip-flop with an inverted clock. The $\bar{Q}$ outputs loop back to their D inputs. The Q outputs feed into the AND gate producing clk out.)
Answer: C. 1.5
Explanation: By tracing the timing diagram of the dual flip-flop arrangement (one triggered on positive edge, one on negative edge, with outputs ANDed), the output waveform has a time period that is $1.5$ times the input clock period ($T_{out} = 1.5 T_{in}$). Thus, the frequency division factor is $1.5$.
Topic: Analog Circuits
For the circuit given below, the voltage $V_{0}$ across the op-amp output is
(The circuit shows an inverting amplifier with $R_{in}=10~\Omega$ and $R_{f}=50~\Omega$ connecting to node B. Node B is the center of a voltage divider formed by two $10~\Omega$ resistors from the op-amp output $V_{o}$ to ground.)
Answer: C. -11 $V_{i}$
Explanation: Due to virtual ground at the inverting terminal (node A), $V_A = 0$.
Applying KCL at node A: $\frac{0-V_A}{50} + \frac{0-V_i}{10} = 0 \Rightarrow V_B = -5V_i$ (Note: Official solution denotes the node as $V_R$ or $V_B$).
Applying KCL at node B: $\frac{V_B}{10} + \frac{V_B-0}{50} + \frac{V_B-V_o}{10} = 0$.
$\Rightarrow \frac{V_B}{5} + 2V_B = V_o \Rightarrow \frac{11V_B}{5} = V_o$.
Substitute $V_B = -5V_i$: $V_o = \frac{11(-5V_i)}{5} = -11V_i$.
Topic: Network Theory
The rms value of current $i(t)=I_{1}\sin\omega t+I_{2}\sin\omega t$ is
Answer: B. $(\frac{I_{1}^{2}}{2}+\frac{I_{2}^{2}}{2})^{1/2}$
Explanation: Note: According to the official answer key, the answer is option (b). The original question likely contained a typographical error and meant to represent two orthogonal components (e.g., different frequencies or a sine and cosine), for which the RMS value is $\sqrt{\frac{I_{1}^{2}}{2} + \frac{I_{2}^{2}}{2}}$. The text is transcribed exactly as provided.
Topic: Analog Circuits
Given the output for the following non-inverting summing amplifier, the relation between $R_{f}$ and R in the circuit is
(The diagram shows a non-inverting summer with 4 input voltages $V_{1}$ to $V_{4}$ connected via identical resistors $R$ to the non-inverting terminal. The output is $V_{o}=(V_{1}+V_{2}+V_{3}+V_{4})/2$.)
Answer: A. $R_{f}=R$
Explanation: Voltage at the non-inverting node A is $V_{A} = \frac{R/3}{R + R/3} \dots$ wait, by KCL at the non-inverting node: $\frac{V_{1}-V_{A}}{R} + \frac{V_{2}-V_{A}}{R} + \frac{V_{3}-V_{A}}{R} + \frac{V_{4}-V_{A}}{R} = 0 \Rightarrow 4V_{A} = V_{1}+V_{2}+V_{3}+V_{4} \Rightarrow V_{A} = \frac{V_{1}+V_{2}+V_{3}+V_{4}}{4}$.
For a non-inverting amplifier, $V_{o} = \left(1 + \frac{R_{f}}{R}\right) V_{A}$.
$V_{o} = \left(1 + \frac{R_{f}}{R}\right) \frac{V_{1}+V_{2}+V_{3}+V_{4}}{4}$.
Given $V_{o} = \frac{V_{1}+V_{2}+V_{3}+V_{4}}{2}$, we have $\frac{1 + R_{f}/R}{4} = \frac{1}{2} \Rightarrow 1 + \frac{R_{f}}{R} = 2 \Rightarrow R_{f} = R$.
Topic: Electronic Measurements and Instrumentation
A pulse signal having 100 kHz frequency and 70 nsec rise time is to be measured on an oscilloscope. The minimum required bandwidth of the oscilloscope is,
Answer: C. 5 MHz
Explanation: The relationship between rise time ($t_{r}$) and bandwidth ($B.W.$) is given by $t_{r} = \frac{0.35}{B.W.}$.
Therefore, $B.W. = \frac{0.35}{t_{r}} = \frac{0.35}{70 \times 10^{-9}} = 5 \times 10^{6}\text{ Hz} = 5\text{ MHz}$.
Topic: Microwave Engineering
Multiple collectors are used in Travelling Wave Tube (TWT) to
Answer: B. to increase the overall efficiency
Explanation: TWT is basically a wideband amplifier in which the interaction space is extended and the electron beam exchanges energy with the RF wave. The efficiency offered (dc to RF) is about 10%. TWT efficiency can be increased by the use of multistage collectors at the end of the electron beam's path that are biased at voltages lower than the accelerating anode voltage.
Topic: Electromagnetics
A lossless antenna has directional gain $G(\theta,\varphi)$ then $\int_{\varphi=-\pi/2}^{\varphi=\pi/2}\int_{\theta=-\pi/2}^{\theta=\pi/2}G(\theta,\varphi)d\theta~d\varphi$ is
Answer: C. $\le 4\pi$
Explanation: The integral of the directional gain over all solid angles (the entire sphere, i.e., $4\pi$ steradians) is equal to $4\pi$. Since the given integration limits ($-\pi/2$ to $\pi/2$) cover only a portion of the entire sphere, the integral evaluates to a value less than or equal to $4\pi$.
Topic: Engineering Mathematics
$\sqrt[3]{\cos x-j~\sin x}$ is equal to
Answer: C. $\cos\frac{x}{3}-j~\sin\frac{x}{3}$
Explanation: Using Euler's formula, $\cos x - j\sin x = e^{-jx}$.
Taking the cube root: $(e^{-jx})^{1/3} = e^{-jx/3}$.
Applying Euler's formula again: $e^{-jx/3} = \cos\left(\frac{-x}{3}\right) + j\sin\left(\frac{-x}{3}\right) = \cos\frac{x}{3} - j\sin\frac{x}{3}$.
Topic: Communication Systems
If $f(t)$ is 1 MHz sinusoid with 1 $V_{p-p}$ and sampling frequency $f_{s}$ is 25 kHz, the output of an A/D followed by D/A converter will be
Answer: C. DC value anywhere between -0.5 V and +0.5 V
Explanation: The signal frequency is $1\text{ MHz}$ ($1000\text{ kHz}$) and the sampling frequency is $25\text{ kHz}$. Since $1000\text{ kHz}$ is an exact integer multiple of $25\text{ kHz}$ ($1000/25 = 40$), the samples are taken at the exact same phase of the sine wave in every cycle. This causes the sampled signal to be a constant DC value. Since the input is a $1\text{ V}_{p-p}$ sinusoid (ranging from -0.5 V to +0.5 V), the resulting DC value can be anywhere in the range of -0.5 V to +0.5 V.
Topic: Signals and Systems
For a practical low pass filter, the ideal amplitude and phase response will be
Answer: A. Amplitude is a flat rectangular pulse, Group Delay is a flat horizontal line
Explanation: For an ideal low pass filter, the amplitude response should be flat in the passband (a rectangular pulse shape) and zero outside. The phase response should be perfectly linear, which corresponds to a constant (flat) group delay in the passband. The official key marks option (a) which visually corresponds to flat amplitude and flat group delay.
Topic: Communication Systems
A 10 dB attenuator is put at the input of a low noise amplifier having 3 dB noise figure. Now the noise figure of the cascaded amplifier will be
Answer: D. none of the above
Explanation: The noise figure of the attenuator is equal to its attenuation, so $NF_1 = 10\text{ dB}$. The noise figure of the cascaded system is approximated by $NF_{total} \approx NF_1 + NF_2 = 10\text{ dB} + 3\text{ dB} = 13\text{ dB}$. Since 13 dB is not in the options, the correct choice is 'none of the above'.
Topic: Signals and Systems
A 33.33% duty cycle rectangular wave is fed to the input of an Spectrum Analyzer. What will be observed?
Answer: B. $3^{rd}$, $6^{th}$, $9^{th}$... harmonics missing
Explanation: A rectangular wave with a duty cycle $d$ has Fourier series coefficients containing a $\text{sinc}(n\pi d)$ term. For $d = 1/3$ (33.33%), the harmonics are zero when $n\pi/3 = k\pi$, which means $n = 3k$. Therefore, the 3rd, 6th, 9th, etc., harmonics will be missing.
Topic: Signals and Systems
Phase function of a filter is $\phi(f)=kf^{2}$, $k>0$. The group delay of the filter has the shape
Answer: A. Straight line with negative slope
Explanation: Group delay is defined as $\tau_{g} = -\frac{d\phi(f)}{df}$ (or with respect to $\omega$). Taking the derivative of the phase function, $\tau_{g} = -\frac{d(kf^{2})}{df} = -2kf$. This represents a straight line with a negative slope passing through the origin.
Topic: Communication Systems
A $1~V_{p-p}$ sinusoid is digitized by a 4 bit A-to-D converter with input dynamic range of $2V_{p-p}$. The signal to noise ratio of the digitized signal is
Answer: B. 96
Explanation: The step size is $S = \frac{2V_{m}}{M} = \frac{2V_{p-p}/2}{2^n} = \frac{2 \times 1}{2^4} = \frac{1}{8}\text{ V}$.
Quantization noise power $P_{n} = \frac{S²}{12} = \frac{(1/8)^2}{12} = \frac{1}{64 \times 12}$.
Signal power $P_{s} = \frac{V_{m}^2}{2} = \frac{(0.5)^2}{2} = \frac{1}{8}\text{ W}$.
SNR = $\frac{P_{s}}{P_{n}} = \frac{1/8}{1/(64 \times 12)} = \frac{64 \times 12}{8} = 8 \times 12 = 96$.
Topic: Communication Systems
Determine the matched filter response $h(t)$ for signal $s(t)$ shown below:
(The signal $s(t)$ is a rectangular pulse of duration T starting from $t=0$).
Answer: D. Rectangular pulse from t=-T to 0
Explanation: For a real-valued signal, the impulse response of the matched filter is the time-reversed version of the input signal, i.e., $h(t) = s(-t)$ (for non-causal representation). Thus, a rectangular pulse from 0 to T becomes a rectangular pulse from -T to 0.
Topic: Electromagnetics
A signal source with 100 m wavelength is connected to the input terminals of a 150 m long transmission line terminated in its characteristic impedance. The phase difference between the voltages at two ends of the transmission line in steady-state condition is
Answer: B. $3\pi$
Explanation: The phase difference is $\phi = \beta d = \frac{2\pi}{\lambda} \times d$.
Given $\lambda = 100\text{ m}$ and $d = 150\text{ m}$.
$\phi = \frac{2\pi}{100} \times 150 = 3\pi$.
Topic: Electromagnetics
A charge $Q_{2}=8.854\times10^{-9}\text{C}$ is located in vacuum at $P_{2}(2,3,1)$. The force on $Q_{2}$ due to a charge $Q_{1}=4\pi\times10^{-3}\text{C}$ at $P_{1}(2,2,1)$ is
(Note: All the coordinates are measured in metres. $a_{x}$, $a_{y}$ and $a_{z}$ are unit vectors in X, Y and Z direction respectively.)
Answer: A. $a_{y}\text{N}$
Explanation: The distance vector is $\vec{R} = P_2 - P_1 = (2-2)\hat{a}_x + (3-2)\hat{a}_y + (1-1)\hat{a}_z = \hat{a}_y$.
Magnitude $|\vec{R}| = 1\text{ m}$.
The force is $\vec{F} = \frac{Q_1 Q_2}{4\pi\epsilon_0 R²} \hat{a}_y = \frac{4\pi\times 10^{-3} \times 8.854\times 10^{-9}}{4\pi (8.854\times 10^{-12}) \times 1^2} \hat{a}_y = 1 \hat{a}_y \text{ N}$.
Topic: Analog Circuits
A low pass filter as shown in following figure is built using an operational amplifier having unity gain bandwidth of 1 MHz. What is the bandwidth of this circuit?
(The circuit is a practical active low pass filter/integrator with $R_{in} = 100\text{ }\Omega$, $R_{f} = 10\text{ k}\Omega$ and $C = 160\text{ pF}$ in parallel.)
Answer: C. 100 kHz
Explanation: This is a practical integrator or active low pass filter. Its 3-dB bandwidth is determined by the feedback resistor and capacitor: $B.W. = \frac{1}{2\pi R_f C} = \frac{1}{2\pi \times 10\times 10^3 \times 160\times 10^{-12}} = \frac{1}{2\pi \times 1.6\times 10^{-6}} \approx 99.47\text{ kHz} \approx 100\text{ kHz}$.
Topic: Digital Electronics
What is the frequency and duty cycle of output Y, when clk frequency is 1 MHz @ 50% duty cycle?
(The circuit shows a Johnson counter formed by 2 D-flip-flops, where $\overline{Q}_2$ feeds $D_1$, and $Q_1$ feeds $D_2$. The output Y is the AND of $Q_1$ and $\overline{Q}_2$.)
Answer: B. 500 kHz @ 25% duty cycle
Explanation: Assuming initially both flip-flops are 0 ($Q_1=0, Q_2=0$), the sequence of states ($Q_1Q_2$) is 00 -> 10 -> 11 -> 01 -> 00.
The output Y is $Q_1 \cdot \overline{Q}_2$.
For 00, Y=0. For 10, Y=1. For 11, Y=0. For 01, Y=0.
According to the official ISRO key and explanation, the output time period is taken as twice the input clock (despite a full 4-state cycle generating one pulse, the official solution asserts "Output frequency = 500 kHz, Duty cycle = 1/4 = 25%"). The answer key records (b).
Topic: Electromagnetics
For an isotropic radiator, electric field intensity at a distance R is measured as $3~\text{V/m}$. What will be the electric field intensity at a distance 3 R?
Answer: A. $1~\text{V/m}$
Explanation: Electric field intensity $E$ is inversely proportional to the distance $r$ ($E \propto \frac{1}{r}$).
At distance R, $E = 3\text{ V/m} = \frac{K}{R} \Rightarrow K = 3R$.
At distance 3R, $E' = \frac{K}{3R} = \frac{3R}{3R} = 1\text{ V/m}$.
Topic: Digital Electronics
The logic function implemented by following 4:1 MUX is
(The diagram shows a MUX with inputs $I_0=X$, $I_1=Y$, $I_2=X$, $I_3=0$ and select lines $S_1=X$ (MSB), $S_0=Y$ (LSB). The output is Z.)
Answer: C. Z=X XOR Y
Explanation: The logical expression for Z is given by evaluating the MUX logic:
$Z = \overline{X}\overline{Y}I_0 + \overline{X}Y I_1 + X\overline{Y}I_2 + XY I_3$
$Z = \overline{X}\overline{Y}(X) + \overline{X}Y(Y) + X\overline{Y}(X) + XY(0)$
$Z = 0 + \overline{X}Y + X\overline{Y} + 0 = X \oplus Y = X \text{ XOR } Y$.
Topic: Electromagnetics
Characteristic impedance of a two-wire transmission line at 10 kHz is 200-j50 $\Omega$. Line is terminated in its characteristic impedance, and a 28.28 $V_{p-p}$ signal is measured at its input. Determine the real power supplied by the signal source to the line?
Answer: B. 0.485 W
Explanation: The actual real power mathematically should be derived using $|Z_{in}|^2$, but according to the official ISRO solution, the real power supplied is calculated as $P = \frac{V_{rms}^2}{2|Z_{L}|}$ or similar incorrect simplification $P = \frac{V_{p}^2}{2|Z_L|}$. Following their method: $|Z_L| = \sqrt{200^2 + 50^2} = 206.15\text{ }\Omega$. Using $V_p = 14.14\text{ V}$, the Real power $= \frac{14.14^2}{2 \times 206.15} = 0.485\text{ W}$. Option (b) is marked correct in the official key.
Topic: Analog Circuits
Which is the correct waveform across capacitor in the following circuit?
(The circuit is a half-wave rectifier with a capacitive filter, driven by a 5 Vrms 50 Hz AC source).
Answer: D. Half-wave rectified signal with exponential capacitor discharge (ripple)
Explanation: The circuit functions as a half wave rectifier with a capacitive filter. During the positive half cycle, the capacitor charges to the peak value. During the negative half cycle, the diode is reverse biased and the capacitor discharges slowly through the load resistor. The resulting waveform shows this characteristic charging and discharging ripple pattern.