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ISRO Scientist/Engineer Electronics 2023 question paper

All 95 questions of ISRO's written test for Scientist/Engineer 'SC' (Electronics), held on 7 January 2024, with ISRO's answer key. ISRO excluded Q.23, Q.32 and Q.44 from evaluation.

Answers from ISRO's revised and final answer key of 16 February 2024.

The question paper (PDF, ISRO)ISRO’s answer key (PDF)

questions · 100 marks: 95questions · 100 marks
Part A: Electronics: 80Part A: Electronics
Part B: aptitude: 15Part B: aptitude

How the paper is marked

Part A (Q.1 to Q.80): 1 mark for a correct answer, minus 1/3 for a wrong answer.
Part B (Q.81 to Q.95): 1 or 2 marks as printed with each question, 20 marks in all, no negative marking.

Source: ISRO's recruitment advertisement ISRO:ICRB:02(EMC):2023 of 25 May 2023, item 16(b).

ISRO EC 2023 answer key

ISRO's answer key for ISRO Scientist/Engineer Electronics 2023
Question	Part	Marks	Answer
Q.1	A	1	C
Q.2	A	1	C
Q.3	A	1	C
Q.4	A	1	C
Q.5	A	1	D
Q.6	A	1	B
Q.7	A	1	B
Q.8	A	1	A
Q.9	A	1	A
Q.10	A	1	C
Q.11	A	1	B
Q.12	A	1	B
Q.13	A	1	A
Q.14	A	1	C
Q.15	A	1	B
Q.16	A	1	C
Q.17	A	1	A
Q.18	A	1	C
Q.19	A	1	D
Q.20	A	1	B
Q.21	A	1	B or C
Q.22	A	1	B
Q.23	A	1	Excluded
Q.24	A	1	A
Q.25	A	1	D
Q.26	A	1	C
Q.27	A	1	A
Q.28	A	1	D
Q.29	A	1	C
Q.30	A	1	A
Q.31	A	1	B
Q.32	A	1	Excluded
Q.33	A	1	C
Q.34	A	1	B
Q.35	A	1	A
Q.36	A	1	B
Q.37	A	1	A
Q.38	A	1	A
Q.39	A	1	A
Q.40	A	1	B
Q.41	A	1	B
Q.42	A	1	A
Q.43	A	1	C
Q.44	A	1	Excluded
Q.45	A	1	A
Q.46	A	1	D
Q.47	A	1	B
Q.48	A	1	D
Q.49	A	1	A
Q.50	A	1	D
Q.51	A	1	B
Q.52	A	1	C
Q.53	A	1	D
Q.54	A	1	A
Q.55	A	1	B
Q.56	A	1	C
Q.57	A	1	A
Q.58	A	1	C
Q.59	A	1	A
Q.60	A	1	C
Q.61	A	1	C
Q.62	A	1	A
Q.63	A	1	C
Q.64	A	1	A
Q.65	A	1	A
Q.66	A	1	A
Q.67	A	1	C
Q.68	A	1	A
Q.69	A	1	A
Q.70	A	1	A
Q.71	A	1	B
Q.72	A	1	B
Q.73	A	1	B
Q.74	A	1	A
Q.75	A	1	A
Q.76	A	1	C
Q.77	A	1	B
Q.78	A	1	C
Q.79	A	1	A
Q.80	A	1	D
Q.81	B	1	D
Q.82	B	1	D
Q.83	B	2	B
Q.84	B	1	A
Q.85	B	1	A
Q.86	B	2	A
Q.87	B	1	A
Q.88	B	1	C
Q.89	B	2	C
Q.90	B	1	C
Q.91	B	1	D
Q.92	B	2	D
Q.93	B	1	A
Q.94	B	1	C
Q.95	B	2	A

ISRO excluded Q.23, Q.32 and Q.44 from evaluation.

Part A: Electronics Q.1 to Q.80

Q.11 markOfficial answer: C
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Engineering Mathematics › Linear algebra
A square matrix is singular when its determinant is zero. Expand the determinant along a row, then solve the resulting linear equation for the unknown.
Full solution
Q.21 markOfficial answer: C
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Engineering Mathematics › Linear algebra
For any positive integer $n$ , if $λ$ is an eigenvalue of $A$ then $λ^{n}$ is an eigenvalue of $A^{n}$ , with the same eigenvector. Raise each eigenvalue to the power.
Full solution
Q.31 markOfficial answer: C
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Engineering Mathematics › Calculus
The given series is the Maclaurin series of $sin x$ , and differentiating $sin x$ (or the series term by term) gives $cos x$ .
Full solution
Q.41 markOfficial answer: C
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Signals and Systems › Signals basics
The sifting property of the unit impulse: $\int δ (t) f (t) d t = f (0)$ . Only the value of the function at the impulse location matters.
Full solution
Q.51 markOfficial answer: D
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Signals and Systems › Fourier series
Symmetry kills integrals: an even $x (t)$ makes every $b_{n}$ vanish (sine terms drop out), leaving $a_{0}$ and cosines; an odd $x (t)$ makes $a_{0}$ and every $a_{n}$ vanish, leaving only sines.
Full solution
Q.61 markOfficial answer: B
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Engineering Mathematics › Probability & statistics
A PDF must have total area 1. Normalising the two-sided exponential $2 a e^{- b ∣ x ∣}$ gives $4 a / b = 1$ .
Full solution
Q.71 markOfficial answer: B
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Analog Electronics › Noise & shielding (IN)
A noise spectrum has three regions: $1/ f$ (flicker) noise at low frequency, a flat white-noise floor in the middle, and a rise from transit-time effects at very high frequency.
Full solution
Q.81 markOfficial answer: A
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Q.91 markOfficial answer: A
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Q.101 markOfficial answer: C
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Q.111 markOfficial answer: B
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Q.121 markOfficial answer: B
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Electromagnetics › Transmission lines
A reflection from a load behind an attenuator crosses the attenuator twice, so the return loss seen at the input is the load's return loss plus twice the attenuation (in dB), provided the attenuator's own reflection is small.
Full solution
Q.131 markOfficial answer: A
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Electronic Devices › P-N junction
Thermally generated minority carriers make reverse saturation current grow rapidly with temperature, about doubling per $10^{\circ} C$ . Forward voltage falls with temperature at about $2 mV /^{\circ} C$ .
Full solution
Q.141 markOfficial answer: C
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Analog Electronics › Biasing
With a fixed base current, $I_{C} = β I_{B}$ grows with $β$ while $V_{C E} = V_{C C} - I_{C} R_{C}$ falls. The Q point slides along the dc load line toward saturation, which is the upper end.
Full solution
Q.151 markOfficial answer: B
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Electronic Devices › MOS devices
In strong-inversion saturation $I_{D} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{GS} - V_{t h})^{2}$ , so $I_{D}$ varies as the square of $V_{GS} - V_{t h}$ .
Full solution
Q.161 markOfficial answer: C
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Signals and Systems › LTI & convolution
Convolving any signal with the unit impulse returns the signal unchanged, since $x (n) * δ (n) = x (n)$ by the sifting property; the impulse plays the role that $1$ plays for ordinary multiplication.
Full solution
Q.171 markOfficial answer: A
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Engineering Mathematics › Probability & statistics
For a uniform distribution on $[a, b]$ , the mean is the midpoint $\frac{a + b}{2}$ and the variance is $\frac{( b - a ) ^{2}}{12}$ .
Full solution
Q.181 markOfficial answer: C
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Signals and Systems › Laplace
Poles are the roots of the denominator, and a squared factor $(s + a)^{2}$ gives a repeated pole at $s = - a$ , counted twice.
Full solution
Q.191 markOfficial answer: D
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Signals and Systems › z-transform
A minimum phase system and its inverse are both causal and stable, which puts every pole and every zero of $H (z)$ inside the unit circle.
Full solution
Q.201 markOfficial answer: B
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Signals and Systems › DTFT & DFT
For a real sequence, the $N$ -point DFT is conjugate-symmetric: $X (N - k) = X^{*} (k)$ . With $N = 8$ , bin 6 is the mirror of bin 2.
Full solution
Q.211 markOfficial answer: B or C
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Earlier, ISRO's key gave B.
Signals and Systems › DTFT & DFT
A delay by $n_{d}$ in time multiplies the DTFT by $e^{- j ω n_{d}}$ (a linear phase), while multiplying by $e^{j ω_{0} n}$ in time shifts the spectrum to $X (e^{j (ω - ω_{0})})$ .
Full solution
Q.221 markOfficial answer: B
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Signals and Systems › DTFT & DFT
Stage $m$ of a radix-2 DIT FFT uses $W_{2^{m}}^{k} = W_{N}^{k N / 2^{m}}$ for $k < 2^{m - 1}$ , so its $W_{N}$ exponents are multiples of $N / 2^{m}$ below $N /2$ . A given exponent appears only in the stages where it is such a multiple.
Full solution
Q.231 markExcluded from evaluation by ISRO
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Earlier, ISRO's key gave D.
Q.241 markOfficial answer: A
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Digital Electronics › ADC & DAC
Quantization noise power for a uniform quantizer is $Δ^{2} /12$ , where the step $Δ$ equals the full-scale range divided by $2^{N}$ .
Full solution
Q.251 markOfficial answer: D
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Digital Electronics › Boolean logic
For a POS answer, group the zeros of the function (maxterms) on the K-map, write $f^{'}$ as a sum of products, then complement it with De Morgan's theorem.
Full solution
Q.261 markOfficial answer: C
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Signals and Systems › z-transform
Cascaded systems multiply their $z$ -transforms. Write the target $H (z)$ with a leading 1 in the denominator, then match the coefficients of $z^{- 1}$ and $z^{- 2}$ and the numerator constant.
Full solution
Q.271 markOfficial answer: A
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Earlier, ISRO's key gave B.
Digital Electronics › Combinational
A 2:1 MUX gives both $S$ and $\overline{S}$ for free, and the constants 0 and 1 can feed its data inputs. So factor the function into nested forms $\overline{S} I_{0} + S I_{1}$ with $S$ a plain input, and use one MUX per nesting level.
Full solution
Q.281 markOfficial answer: D
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Q.291 markOfficial answer: C
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Digital Electronics › ADC & DAC
Convert the threshold to a DAC code ( $7.5 V$ is code 12, $110 0_{2}$ ). Then find which stuck bit changes code 8 ( $100 0_{2}$ ) to a code of at least 12 without causing an earlier crossing.
Full solution
Q.301 markOfficial answer: A
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Networks (Electric Circuits) › Laws & node/mesh
With a fixed source, the current through R1 is V divided by the total resistance, so the current ratio equals the ratio of the largest to the smallest total resistance over all ways of assigning the four values.
Full solution
Q.311 markOfficial answer: B
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Signals and Systems › z-transform
Stability of a causal discrete-time system depends only on the poles of $H (z)$ : all of them must lie strictly inside the unit circle. Find them from the roots of the denominator polynomial.
Full solution
Q.321 markExcluded from evaluation by ISRO
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Earlier, ISRO's key gave D.
Q.331 markOfficial answer: C
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Digital Electronics › Combinational
A 3:8 decoder produces every minterm, so the sum and carry of a full adder are each the OR of four minterms. A 3-input OR gate needs two gates to combine four signals, and the two outputs share no gate.
Full solution
Q.341 markOfficial answer: B
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Digital Electronics › Combinational
Split the function by the select variable of the final multiplexer. Each value of that select picks one first-stage multiplexer, and the function for that value of $B$ fixes that multiplexer's data inputs.
Full solution
Q.351 markOfficial answer: A
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Electromagnetics › Waveguides & fibres
In a waveguide, $λ_{g} = λ_{0} / 1 - (f_{c} / f)^{2}$ , so the guide wavelength always exceeds the free-space wavelength, and it grows without limit as $f$ approaches $f_{c}$ .
Full solution
Q.361 markOfficial answer: B
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Earlier, ISRO's key gave C.
Electromagnetics › Transmission lines
A matched port reflects nothing, so its reflection coefficient $S_{11} = b_{1} / a_{1}$ (with the other port matched-terminated) is zero. Matching says nothing about the transmission coefficient $S_{21}$ .
Full solution
Q.371 markOfficial answer: A
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Earlier, ISRO's key gave D.
Electromagnetics › Transmission lines
A stripline has a homogeneous dielectric and more than one conductor, so it carries a pure TEM wave with $v_{p} = 1/ μ ϵ$ and no cutoff frequency.
Full solution
Q.381 markOfficial answer: A
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Q.391 markOfficial answer: A
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Q.401 markOfficial answer: B
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Electromagnetics › Waveguides & fibres
A TEM mode has $k_{c} = 0$ , so it has no low-frequency cut-off and propagates from DC upward; only TE and TM modes in a hollow guide have a non-zero cut-off.
Full solution
Q.411 markOfficial answer: B
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Analog Electronics › Feedback & oscillators
An oscillator needs a loop gain of $1∠ 0^{\circ}$ or more at one frequency. Zero total phase shift is positive feedback, and it drives the closed-loop gain $\frac{A}{1 - A β}$ to infinity.
Full solution
Q.421 markOfficial answer: A
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Analog Electronics › Noise & shielding (IN)
Convert dB values to plain ratios, then apply Friis' formula: each later stage's excess noise $(F_{k} - 1)$ is divided by the total gain ahead of it.
Full solution
Q.431 markOfficial answer: C
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Electronic Devices › BJT
A BJT's high-frequency limit has two parts: the RC delay of the junction capacitances $C_{π}$ and $C_{μ}$ , and the carrier transit time across the base, which depends on mobility through $D_{n} = μ_{n} V_{T}$ .
Full solution
Q.441 markExcluded from evaluation by ISRO
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Earlier, ISRO's key gave D.
Q.451 markOfficial answer: A
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Q.461 markOfficial answer: D
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Electronic Devices › Semiconductors
GaAs has a lighter electron effective mass (higher mobility) and a wider direct bandgap (lower $n_{i}$ ), giving better high-frequency performance and lower leakage than Si.
Full solution
Q.471 markOfficial answer: B
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Q.481 markOfficial answer: D
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Q.491 markOfficial answer: A
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Q.501 markOfficial answer: D
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Q.511 markOfficial answer: B
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Q.521 markOfficial answer: C
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Q.531 markOfficial answer: D
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Analog Electronics › Filters & waveforms
In a Schmidt trigger the thresholds are the feedback fraction times each saturation level: $V_{T} = β V_{s a t}$ , with $β$ set by the resistor divider at the + input. The LTP uses the negative saturation level.
Full solution
Q.541 markOfficial answer: A
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Analog Electronics › Filters & waveforms
Find the angles where the sine wave crosses the reference level. The output is high only while the input exceeds the reference on the non-inverting side, and the duty cycle is that conduction angle divided by $36 0^{\circ}$ .
Full solution
Q.551 markOfficial answer: B
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Analog Electronics › Op-amps
A capacitor at the input with a feedback resistor gives an inverting differentiator, $V_{o} = - R C \frac{d V _{i}}{d t}$ . A triangular input has a constant slope on each segment, so the output is a square wave of opposite sign to the slope.
Full solution
Q.561 markOfficial answer: C
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Digital Electronics › ADC & DAC
Build an 8-bit DAC from two 4-bit DACs with an inverting summer. The upper nibble must be weighted 16 times the lower nibble, so the upper input resistor must be 1/16 of the lower one: $16 k Ω/16$ .
Full solution
Q.571 markOfficial answer: A
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Control Systems › Block diagrams & SFG
Write the error as $E = R - H C$ and eliminate $C$ using $C = GE$ ; this gives $E / R = 1/ (1 + G H)$ . The same denominator $1 + G H$ appears in every loop transfer function, and the numerator tells you which signal you are measuring.
Full solution
Q.581 markOfficial answer: C
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Control Systems › Routh–Hurwitz
Simple poles on the imaginary axis with none in the right half-plane make a system marginally stable. A row of zeros in the Routh array signals roots symmetric about the origin, found from the auxiliary polynomial.
Full solution
Q.591 markOfficial answer: A
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Control Systems › Bode & margins
A pure delay $e^{- s T}$ has constant gain ( $0 dB$ ) and a phase $- ω T$ that decreases linearly with frequency, with no corner, peak or resonance.
The question and its official answer
Q.601 markOfficial answer: C
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Earlier, ISRO's key gave B.
Control Systems › Nyquist
Each zero inside the contour adds $+ 1$ to the residue sum of $G^{'} / G$ and each pole adds $- 1$ , so the contour integral is $2 π j$ times (zeros minus poles).
Full solution
Q.611 markOfficial answer: C
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Control Systems › State space
Take $x_{1} = x$ and $x_{2} = \overset{x}{˙}$ , then move every term except the highest derivative to the right-hand side. The coefficients of $x_{1}$ and $x_{2}$ in that equation form the last row of $A$ .
Full solution
Q.621 markOfficial answer: A
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Control Systems › State space
Reachable states from the origin are spanned by $B, A B, A^{2} B, \dots, A^{n - 1} B$ ; the system is controllable iff this matrix has rank $n$ .
Full solution
Q.631 markOfficial answer: C
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Control Systems › Compensators & PID
In the Laplace domain, differentiation is multiplication by $s$ and integration is division by $s$ , so a PID controller is $K_{p} + K_{d} s + K_{I} / s$ .
Full solution
Q.641 markOfficial answer: A
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Control Systems › Routh–Hurwitz
For a second-order polynomial the Routh first column is $a_{2}, a_{1}, a_{0}$ . Stability needs no sign change, so all three coefficients must have the same sign, taken positive in the standard form.
Full solution
Q.651 markOfficial answer: A
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Electromagnetics › Magnetostatics
With currents in opposite directions, the fields of the two wires point the same way at any point between them, so their magnitudes $\frac{I}{2 π r}$ add.
Full solution
Q.661 markOfficial answer: A
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Electromagnetics › Transmission lines
For any uniform line, $Z_{oc} Z_{sc} = Z_{0}^{2}$ , so the characteristic impedance is the geometric mean of the open- and short-circuit input impedances and either one follows from the other.
Full solution
Q.671 markOfficial answer: C
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Control Systems › Time response
Read the damping ratio off the characteristic polynomial by comparing it with $s^{2} + 2 ζ ω_{n} s + ω_{n}^{2}$ . A value $0 < ζ < 1$ means complex-conjugate roots in the left half-plane and an under damped response.
Full solution
Q.681 markOfficial answer: A
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Networks (Electric Circuits) › Laws & node/mesh
With constant current and a voltage that changes linearly with time, the energy is current times the average voltage times the time: $E = I \frac{V _{1} + V _{2}}{2} T$ .
Full solution
Q.691 markOfficial answer: A
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Networks (Electric Circuits) › Laws & node/mesh
A circuit made of parts joined only through one common wire has no current between them. The right-hand loop therefore obeys KCL on its own, and the control voltage $V_{1}$ is fixed by $V_{s}$ alone.
Full solution
Q.701 markOfficial answer: A
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Networks (Electric Circuits) › Laws & node/mesh
Turn the current source with its parallel resistor into a voltage source with a series resistor, then write one KCL equation at the single remaining node.
Full solution
Q.711 markOfficial answer: B
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Electronic Devices › Optoelectronics
Power is $P = V I$ , which is zero at both ends of the V-I curve. It is largest at the knee, where $d P / d V = 0$ .
Full solution
Q.721 markOfficial answer: B
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Q.731 markOfficial answer: B
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Electromagnetics › Maxwell's equations
Induced EMF in an $N$ -turn coil is $e = - N d ϕ / d t$ : multiply the turns by the flux rate, and keep the minus sign from Lenz's law.
Full solution
Q.741 markOfficial answer: A
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Q.751 markOfficial answer: A
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Q.761 markOfficial answer: C
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Q.771 markOfficial answer: B
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Q.781 markOfficial answer: C
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Q.791 markOfficial answer: A
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Networks (Electric Circuits) › AC, power & resonance
Q measures how little energy a coil loses: $2 π$ times the peak stored energy divided by the energy lost per cycle. For a series $L$ - $R$ coil this reduces to $ω L / R = X_{L} / R$ .
Full solution
Q.801 markOfficial answer: D
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Part B: Aptitude Q.81 to Q.95

Q.811 markOfficial answer: D
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General Aptitude › Reasoning
In picture analogies, find the relation in the complete row (material to the tool that applies it) and carry the same relation to the incomplete row.
Full solution
Q.821 markOfficial answer: D
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Earlier, ISRO's key gave B.
General Aptitude › Spatial
In a figure analogy, list every change in the example pair separately (outline shape, shading direction, small symbol's fill, orientation and corner) and apply each one exactly as drawn, not as you expect it to behave.
The question and its official answer
Q.832 marksOfficial answer: B
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General Aptitude › Spatial
Work out which item each operator acts on. Both operators here change only the 1st item, so the 2nd and 3rd items pass through untouched.
Full solution
Q.841 markOfficial answer: A
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Q.851 markOfficial answer: A
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General Aptitude › Reasoning
To show a survey underestimates a group, find the option that moves members of that group out of the reported count (here, dishonest people who reported being honest). Anything that adds people to the count, or only changes the degree of those already in it, cannot cause an underestimate.
Full solution
Q.862 marksOfficial answer: A
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General Aptitude › Reasoning
In a circular seating puzzle, fix one person, number the seats, and decide which way a person's right hand points from the direction they face (towards the anticlockwise side when facing the centre). Then place people with the direct conditions and test each possible position for the others against the negative conditions.
Full solution
Q.871 markOfficial answer: A
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General Aptitude › Reasoning
The three groups of dominoes alternate between counting down and counting up. The last group must count down again, so with $[1∣0]$ at its end the missing dominoes are $[3∣2]$ and $[2∣1]$ .
Full solution
Q.881 markOfficial answer: C
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General Aptitude › Reasoning
In number-figure puzzles, test the simplest relation first: the sum of the outer numbers compared with the centre. Confirm that one rule fits every complete figure before applying it to the incomplete one.
Full solution
Q.892 marksOfficial answer: C
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General Aptitude › Spatial
Assign one operation to each symbol using the chains where it appears together with symbols already decoded. Then apply the operations to the query tile in order. Track three things separately: colour, the direction of the diagonal, and the position of the straight line.
The question and its official answer
Q.901 markOfficial answer: C
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General Aptitude › Reasoning
Combine an "all" statement with a "some" statement. The members picked out by "some" also fall under the "all" group, which gives a new "some" statement. A conclusion that brings in an idea the facts never mention (such as "better" or "simple") is not forced.
Full solution
Q.911 markOfficial answer: D
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General Aptitude › Reasoning
A course of action must be a sensible, proportionate response to the problem in the statement. An unverified allegation calls for an inquiry, not punishment, and the regulator's directive is not the thing at fault.
Full solution
Q.922 marksOfficial answer: D
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General Aptitude › Numerical
When each term is a multiple of the previous one plus a growing amount, find the rule from the early terms, then test whether the later terms continue from it. The wrong term is the one that fails while the terms after it still agree with the rule applied to the correct value.
Full solution
Q.931 markOfficial answer: A
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General Aptitude › Data interpretation
Read each point off the line graph, take the absolute difference for each pair of consecutive days given, and compare. A steep segment of the line marks a large change.
Full solution
Q.941 markOfficial answer: C
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General Aptitude › Data interpretation
Percentage change is the change divided by the starting value, times 100. Here the starting year is 1983.
Full solution
Q.952 marksOfficial answer: A
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General Aptitude › Geometry
For a fixed perimeter, the rectangle of largest area is a square: with $l + b$ fixed, $l b \leq (\frac{l + b}{2})^{2}$ .
Full solution

WHERE THE MARKS CAME FROM

Marks by subject in ISRO Scientist/Engineer Electronics 2023
Subject	Qs	Marks
General Aptitude	14	19
Signals and Systems	10	10
Control Systems	9	9
Electromagnetics	8	8
Analog Electronics	7	7
Digital Electronics	7	7
Electronic Devices	5	5
Engineering Mathematics	5	5
Networks (Electric Circuits)	5	5
Other questions	25	25

We sorted the questions into subjects ourselves; the exam paper does not label them.

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