GATE EC Networks, Signals and Systems Previous Year Questions with Solutions

Full question on page 15 of the official paper (PDF)

GATE 2026, Q.15

Networks, Signals and Systems1 markMCQ

$x_1(t) = \cos(20\pi t)$ and $x_2(t) = \cos(100\pi t)$ , for $t \ge 0$ , are both sampled at $F_s = 40$ Hz. Compare their first ten samples.

Option A: Each of the first three samples of $x_1$ is greater than that of $x_2$ .
Option B: Each of the last three samples of $x_1$ is greater than that of $x_2$ .
Option C: Every sample of $x_2$ is greater than that of $x_1$ .
Option D: The fourth to seventh samples of $x_1$ equal those of $x_2$ .

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Key idea: 50 Hz sampled at 40 Hz aliases onto 10 Hz.

Official answer: D

At $F_s = 40$ Hz, the sample times are $t = n/40$ :

x_1[n] = \cos\!\left(\frac{\pi n}{2}\right), \qquad x_2[n] = \cos\!\left(\frac{5\pi n}{2}\right) = \cos\!\left(\frac{\pi n}{2} + 2\pi n\right) = x_1[n]

\begin{aligned} x_1[n] &= \cos\!\left(\frac{\pi n}{2}\right), \\ x_2[n] &= \cos\!\left(\frac{5\pi n}{2}\right) \\ &= \cos\!\left(\frac{\pi n}{2} + 2\pi n\right) \\ &= x_1[n] \end{aligned}

The 50 Hz signal aliases onto 10 Hz, so every sample is identical. Option (D) is the only statement that holds; (A), (B) and (C) all need one set of samples to be greater.

Full question on page 16 of the official paper (PDF)

GATE 2026, Q.16

Networks, Signals and Systems1 markMCQ

A system obeys $y[n] = \tfrac{5}{6}y[n-1] - \tfrac{1}{6}y[n-2] + x[n]$ , with input $x[n] = \delta[n] - \tfrac{1}{3}\delta[n-1]$ . Describe the response $y[n]$ .

Option A: Stable and causal
Option B: Stable and non-causal
Option C: Unstable and causal
Option D: Unstable and non-causal

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Key idea: The input's zero cancels one of the system's poles.

Official answer: A

H(z) = \frac{1}{(1 - \tfrac{1}{2}z^{-1})(1 - \tfrac{1}{3}z^{-1})}, \qquad X(z) = 1 - \tfrac{1}{3}z^{-1}

\begin{aligned} H(z) &= \frac{1}{(1 - \tfrac{1}{2}z^{-1})(1 - \tfrac{1}{3}z^{-1})}, \\ X(z) &= 1 - \tfrac{1}{3}z^{-1} \end{aligned}

Y(z) = \frac{1}{1 - \tfrac{1}{2}z^{-1}} \;\Rightarrow\; y[n] = \left(\tfrac{1}{2}\right)^n u[n]

\begin{aligned} Y(z) &= \frac{1}{1 - \tfrac{1}{2}z^{-1}} \\ \Rightarrow\; y[n] &= \left(\tfrac{1}{2}\right)^n u[n] \end{aligned}

The input cancels the pole at 1/3. The response starts at $n = 0$ (causal) and decays (stable). Option (A).

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Full question on page 23 of the official paper (PDF)

GATE 2026, Q.23

Networks, Signals and Systems1 markMCQ

A circuit obeys $C\dfrac{dV}{dt} = I(t) - m(t)$ and is driven by $I(t) = q\,\delta(t)$ , with $V(0^-) = V_0$ . Which source-free problem is equivalent?

Option A: $C\dfrac{dV}{dt} = -m(t)$ with $V(0^-) = V_0 + q/C$
Option B: $C\dfrac{dV}{dt} = -m(t)$ with $V(0^-) = V_0 + q/C + m(0^-)$
Option C: $C\dfrac{dV}{dt} = -m(t)$ with $V(0^-) = V_0 - q/C + m(0^-)$
Option D: $C\dfrac{dV}{dt} = -m(t)$ with $V(0^-) = q/C$

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Key idea: An impulse of current delivers its charge instantly, so it only shifts the initial voltage.

Official answer: A

Integrate the equation across $t = 0$ . The impulse delivers charge $q$ in zero time, while $m(t)$ (finite) contributes nothing over an instant:

C\,[V(0^+) - V(0^-)] = q \;\Rightarrow\; V(0^+) = V_0 + \frac{q}{C}

\begin{aligned} &C\,[V(0^+) - V(0^-)] \\ &\quad = q \\ \Rightarrow\; V(0^+) &= V_0 + \frac{q}{C} \end{aligned}

After that, $C\,dV/dt = -m(t)$ . So the impulse is equivalent to starting the source-free equation from $V_0 + q/C$ . Option (A), which writes this initial value at $t = 0^-$ for the equivalent problem.

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Full question on page 30 of the official paper (PDF)

GATE 2026, Q.30

Networks, Signals and Systems1 markNAT

A 1 mA source feeds two parallel branches, each of two 1 kΩ resistors in series, and a 6 V battery joins the two branch midpoints. Find the output voltage across the source.

A 1 mA current source feeds two parallel branches, each of two 1 kΩ resistors in series. A 6 V battery connects the two branch midpoints. The output Vo is taken across the source, positive at the top. — Figure from the official GATE 2026 EC paper, IIT Guwahati

Answer format: Volts, rounded off to two decimal places

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Key idea: The battery's current leaves one midpoint and enters the other, so it cancels from the sum of the node equations.

Official answer: 0.95 to 1.05

Let the bottom rail be 0 V, the top rail $T$ , and the midpoints $L$ and $R$ (volts, kΩ, mA). Whatever current $i$ the battery carries leaves one midpoint and enters the other, so it cancels when the two midpoint equations are added:

(T - 2L) + (T - 2R) = 0 \;\Rightarrow\; T = L + R

\begin{aligned} &(T - 2L) + (T - 2R) \\ &\quad = 0 \\ \Rightarrow\; T &= L + R \end{aligned}

At the top rail, the source current leaves through the two upper resistors:

1 = (T - L) + (T - R) = 2T - (L + R) = T

\begin{aligned} 1 &= (T - L) + (T - R) \\ &= 2T - (L + R) \\ &= T \end{aligned}

So $V_o = 1.00$ V, whatever the battery voltage.

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Full question on page 31 of the official paper (PDF)

GATE 2026, Q.31

Networks, Signals and Systems1 markNAT

A source $12\sin(\omega t)$ V drives 1 µF, 2.2 mH and 1 kΩ, all in parallel across it. Find the average power in the 1 kΩ resistor at resonance.

A source vi(t) = 12 sin(ωt) with a 1 µF capacitor, a 2.2 mH inductor and a 1 kΩ resistor all connected in parallel across it. — Figure from the official GATE 2026 EC paper, IIT Guwahati

Answer format: Milliwatts, rounded off to two decimal places

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Key idea: The resistor sits directly across the ideal source, so resonance doesn't change its voltage.

Official answer: 71.50 to 72.50

The resistor sees the full 12 V peak, so

P = \frac{V_m^2}{2R} = \frac{12^2}{2 \times 1000} = 72\text{ mW}

\begin{aligned} P &= \frac{V_m^2}{2R} \\ &= \frac{12^2}{2 \times 1000} \\ &= 72\text{ mW} \end{aligned}

Resonance only makes the L and C currents cancel; it doesn't change the resistor's voltage.

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Full question on page 34 of the official paper (PDF)

GATE 2026, Q.37

Networks, Signals and Systems2 marksMCQ

A real periodic signal with period $T$ and Fourier coefficients $a_n$ satisfies $x(t - T/2) = -x(t)$ . For any integer $m$ , which is correct?

Option A: $a_{2m} = 0$
Option B: $a_{2m} = 1$
Option C: $a_{2m} = a_{2m+1}$
Option D: $a_{2m} = -1$

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Key idea: Half-wave symmetry removes the even harmonics.

Official answer: A

A shift by $T/2$ multiplies the $n$ th coefficient by $e^{-j\pi n} = (-1)^n$ . The condition $x(t - T/2) = -x(t)$ then says, coefficient by coefficient,

(-1)^n a_n = -a_n \;\Rightarrow\; \left(1 + (-1)^n\right) a_n = 0

\begin{aligned} (-1)^n a_n &= -a_n \\ &\Rightarrow\; \left(1 + (-1)^n\right) a_n \\ &\quad = 0 \end{aligned}

For even $n$ the bracket is 2, so $a_{2m} = 0$ . This is half-wave symmetry: only odd harmonics survive. Option (A).

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Full question on page 36 of the official paper (PDF)

GATE 2026, Q.39

Networks, Signals and Systems2 marksMCQ

An $R_C$ - $C$ branch and an $R_L$ - $L$ branch ( $L = 1$ µH, $C = 1$ µF) are in parallel across $V$ . At 159.15 kHz their currents are $90^\circ$ apart ( $\theta_1 + \theta_2 = 90^\circ$ ). Find the nearest integer to $R_C R_L$ .

Left: across a voltage V, a branch of RC in series with C carries IC, in parallel with a branch of RL in series with L carrying IL. Right: a phasor diagram with V horizontal, IC leading V by θ1 and IL lagging V by θ2. — Figure from the official GATE 2026 EC paper, IIT Guwahati

Option A: 0
Option B: 1
Option C: 2
Option D: 10

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Key idea: Two angles add to $90^\circ$ when the product of their tangents is 1.

Official answer: B

At 159.15 kHz, $\omega = 2\pi f = 10^6$ rad/s, so both reactances are 1 Ω:

X_C = \frac{1}{\omega C} = 1\ \Omega, \qquad X_L = \omega L = 1\ \Omega

\begin{aligned} X_C &= \frac{1}{\omega C} \\ &= 1\ \Omega, \\ X_L &= \omega L \\ &= 1\ \Omega \end{aligned}

$I_C$ leads $V$ by $\theta_1$ and $I_L$ lags it by $\theta_2$ , with

\tan\theta_1 = \frac{X_C}{R_C}, \qquad \tan\theta_2 = \frac{X_L}{R_L}

\begin{aligned} \tan\theta_1 &= \frac{X_C}{R_C}, \\ \tan\theta_2 &= \frac{X_L}{R_L} \end{aligned}

Two angles add to 90° exactly when the product of their tangents is 1:

\frac{X_C X_L}{R_C R_L} = 1 \;\Rightarrow\; R_C R_L = X_C X_L = \frac{L}{C} = 1\ \Omega^2

\begin{aligned} \frac{X_C X_L}{R_C R_L} &= 1 \\ \Rightarrow\; R_C R_L &= X_C X_L \\ &= \frac{L}{C} \\ &= 1\ \Omega^2 \end{aligned}

Option (B).

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Full question on page 40 of the official paper (PDF)

GATE 2026, Q.43

Networks, Signals and Systems2 marksMCQ

$x(t) = 2$ for $0 \le t < 1$ and 0 elsewhere. Find the ratio of the energy of $x(t)$ to the energy of $3x(-3t + 5)$ .

Option A: 3
Option B: 1
Option C: $\tfrac{1}{3}$
Option D: $\tfrac{1}{9}$

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Key idea: Energy goes with the square of the amplitude and with the duration; shifting or reversing a signal leaves it unchanged.

Official answer: C

E[x] = \int_0^1 2^2\,dt = 4

Scaling the amplitude by 3 multiplies the energy by 9. Compressing time by a factor of 3 divides it by 3. Shifting and reversing don't change it:

E[3x(-3t + 5)] = \frac{9}{3} \times 4 = 12

\begin{aligned} &E[3x(-3t + 5)] \\ &\quad = \frac{9}{3} \times 4 \\ &\quad = 12 \end{aligned}

The ratio is $4/12 = 1/3$ . Option (C).

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Full question on page 43 of the official paper (PDF)

GATE 2026, Q.46

Networks, Signals and Systems2 marksMCQ

For $x(t) = e^{-2t}$ , $t \ge 0$ , 99% of the energy lies within $B$ Hz. Which is true of $B$ ?

Option A: $B > 1$ kHz
Option B: $\tfrac{63}{\pi} < B < \tfrac{64}{\pi}$ Hz
Option C: $\tfrac{126}{\pi} < B < \tfrac{128}{\pi}$ Hz
Option D: $B < 1$ Hz

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Key idea: Integrate $|X(f)|^2$ from $-B$ to $B$ and set it to 99% of the total energy.

Official answer: B

X(f) = \frac{1}{2 + j2\pi f}, \qquad |X(f)|^2 = \frac{1}{4\left(1 + \pi^2 f^2\right)}

\begin{aligned} X(f) &= \frac{1}{2 + j2\pi f}, \\ |X(f)|^2 &= \frac{1}{4\left(1 + \pi^2 f^2\right)} \end{aligned}

The total energy is $\int_0^\infty e^{-4t}\,dt = \tfrac{1}{4}$ . The energy in $-B < f < B$ is

\int_{-B}^{B} \frac{df}{4\left(1 + \pi^2 f^2\right)} = \frac{1}{2\pi}\tan^{-1}(\pi B)

\begin{aligned} &\int_{-B}^{B} \frac{df}{4\left(1 + \pi^2 f^2\right)} \\ &\quad = \frac{1}{2\pi}\tan^{-1}(\pi B) \end{aligned}

As a fraction of the total, that is $\tfrac{2}{\pi}\tan^{-1}(\pi B)$ . Setting it to 0.99:

\tan^{-1}(\pi B) = 0.495\pi \;\Rightarrow\; \pi B = \tan(89.1^\circ) \approx 63.66

\begin{aligned} \tan^{-1}(\pi B) &= 0.495\pi \\ \Rightarrow\; \pi B &= \tan(89.1^\circ) \\ &\approx 63.66 \end{aligned}

So $B \approx 63.66/\pi \approx 20.3$ Hz, between $63/\pi$ and $64/\pi$ . Option (B).

Full question on page 44 of the official paper (PDF)

GATE 2026, Q.47

Networks, Signals and Systems2 marksMCQ

A system S is built from two LTI subsystems $h_1[n]$ and $h_2[n]$ , two adders and two unit delays, as in the figure. Which statement is necessarily true?

x[n] feeds h1[n], whose output goes to an output adder producing y[n]. The h1 output, delayed by one sample, is subtracted from x[n] at a second adder; that result passes through h2[n], is delayed by one sample and is subtracted at the output adder. — Figure from the official GATE 2026 EC paper, IIT Guwahati

Option A: S is causal.
Option B: S is linear and time invariant.
Option C: S is linear and time varying.
Option D: S is non-linear.

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Key idea: Any interconnection of LTI blocks is LTI, but causality depends on the blocks.

Official answer: B

Reading the figure, the output is

y[n] = (h_1 * x)[n] - (h_2 * v)[n-1], \qquad v[n] = x[n] - (h_1 * x)[n-1]

\begin{aligned} y[n] &= (h_1 * x)[n] \\ &\quad - (h_2 * v)[n-1], \\ v[n] &= x[n] \\ &\quad - (h_1 * x)[n-1] \end{aligned}

Every block (the two subsystems, the delays and the adders) is linear and time invariant, and any interconnection of LTI blocks is LTI. In the z-domain the whole system is a single transfer function:

H(z) = H_1(z) - z^{-1}H_2(z) + z^{-2}H_1(z)H_2(z)

\begin{aligned} H(z) &= H_1(z) \\ &\quad - z^{-1}H_2(z) \\ &\quad + z^{-2}H_1(z)H_2(z) \end{aligned}

So (B) holds, and (C) and (D) are false. Causality (A) isn't guaranteed: nothing says $h_1[n]$ or $h_2[n]$ is zero for $n < 0$ , and if $h_1[n]$ isn't, neither is the overall impulse response. Option (B).

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Full question on page 48 of the official paper (PDF)

GATE 2026, Q.51

Networks, Signals and Systems2 marksMCQ

Find the ABCD parameters of a T network with 1 kΩ in each series arm and 1 kΩ in the shunt arm.

T network between ports P1 and P2: a 1 kΩ resistor in each series arm and 1 kΩ in the shunt arm to the common bottom line. — Figure from the official GATE 2026 EC paper, IIT Guwahati

Option A: $A = \tfrac{1}{2}$ , $B = 3$ kΩ, $C = 10^{-3}$ S, $D = \tfrac{1}{2}$
Option B: $A = 2$ , $B = 3$ Ω, $C = 1$ S, $D = 2$
Option C: $A = 2$ , $B = 6$ kΩ, $C = 2 \times 10^{-3}$ S, $D = 2$
Option D: $A = 2$ , $B = 3$ kΩ, $C = 10^{-3}$ S, $D = 2$

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Key idea: Find A and C with the output open, and B and D with it shorted.

Official answer: D

For a T network with series arms $Z_1$ , $Z_2$ and shunt arm $Z_3$ :

A = 1 + \frac{Z_1}{Z_3}, \quad B = Z_1 + Z_2 + \frac{Z_1 Z_2}{Z_3}, \quad C = \frac{1}{Z_3}, \quad D = 1 + \frac{Z_2}{Z_3}

\begin{aligned} A &= 1 + \frac{Z_1}{Z_3}, \\ B &= Z_1 + Z_2 + \frac{Z_1 Z_2}{Z_3}, \\ C &= \frac{1}{Z_3}, \\ D &= 1 + \frac{Z_2}{Z_3} \end{aligned}

With all three at 1 kΩ: $A = D = 2$ , $B = 3 \times 10^3$ Ω and $C = 10^{-3}$ Ω⁻¹. As a check, $AD - BC = 4 - 3 = 1$ , as it must be for a reciprocal network. Option (D).

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Full question on page 55 of the official paper (PDF)

GATE 2026, Q.58

Networks, Signals and Systems2 marksMSQ

Let $x_1(t) = \cos(2\pi n t)$ and $x_2(t) = 2\sin(4\pi n t)$ for a positive integer $n$ . Which statements are valid?

One or more options are correct.

Option A: $x_1$ and $x_2$ are orthogonal over $0 \le t < 1/n$ .
Option B: $x_1$ and $x_2$ are orthonormal over $0 \le t < 1/n$ .
Option C: $x_2$ is a harmonic of $x_1$ .
Option D: $x_1$ and $x_2$ are not orthogonal over $0 \le t < 1/(2n)$ .

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Key idea: $x_2$ is at twice $x_1$ 's frequency; orthogonality depends on the interval you integrate over.

Official answer: A, C and D

$x_1$ has frequency $n$ and $x_2$ has frequency $2n$ . Multiply them with the product-to-sum identity

2\sin a\cos b = \sin(a+b) + \sin(a-b)

and the product is

x_1(t)\,x_2(t) = 2\cos(2\pi nt)\sin(4\pi nt) = \sin(6\pi nt) + \sin(2\pi nt)

\begin{aligned} &x_1(t)\,x_2(t) \\ &\quad = 2\cos(2\pi nt)\sin(4\pi nt) \\ &\quad = \sin(6\pi nt) + \sin(2\pi nt) \end{aligned}

Over $0 \le t < 1/n$ , both sines run through whole cycles, so the inner product is 0: orthogonal. True, (A).
Orthonormal also needs unit energy, but $\int_0^{1/n}\cos^2(2\pi nt)\,dt = 1/(2n)$ , which is never 1 for a positive integer $n$ (and $x_2$ 's energy is $2/n$ ). False, (B).
$x_2$ 's frequency $2n$ is twice $x_1$ 's, so $x_2$ is the second harmonic of $x_1$ . True, (C).
Over $0 \le t < 1/(2n)$ , $\sin(6\pi nt)$ runs through 1.5 cycles and $\sin(2\pi nt)$ through half a cycle, so neither integral cancels:

\int_0^{1/(2n)} x_1 x_2\,dt = \frac{1}{3\pi n} + \frac{1}{\pi n} = \frac{4}{3\pi n} \neq 0

\begin{aligned} &\int_0^{1/(2n)} x_1 x_2\,dt \\ &\quad = \frac{1}{3\pi n} + \frac{1}{\pi n} \\ &\quad = \frac{4}{3\pi n} \\ &\quad \neq 0 \end{aligned}

So they are not orthogonal over that interval. True, (D). Options (A), (C) and (D).

Note: Option (C) uses harmonic in its usual sense: a sinusoid at a whole-number multiple of a fundamental frequency. Here $x_2$ , at $2n$ , is the second harmonic of the fundamental $n$ of $x_1$ , so the key includes (C).

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Full question on page 16 of the official paper (PDF)

GATE 2025, Q.17

Networks, Signals and Systems1 markMCQ

In the figure, $x[n]$ passes through an LTI subsystem with impulse response $h_1[n]$ , then $b\,\delta[n]$ is added, then it passes through an LTI subsystem $h_2[n]$ , and $b\,\delta[n]$ is subtracted to give $y[n]$ , with $b > 0$ . If $h_2[n] \neq \delta[n]$ , the overall system is

Block diagram inside a dashed box: x[n] enters Subsystem 1, h1[n]; an adder adds b delta[n]; the sum enters Subsystem 2, h2[n]; a second adder adds minus b delta[n] to give y[n]. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: linear and time invariant
Option B: linear and time variant
Option C: nonlinear and time invariant
Option D: nonlinear and time variant

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Key idea: An output term that doesn't depend on the input breaks linearity, and one fixed at $n = 0$ breaks time invariance.

Official answer: D

Follow a signal through the box:

y[n] = h_2[n] * \big(h_1[n] * x[n] + b\,\delta[n]\big) - b\,\delta[n] = (h_1 * h_2 * x)[n] + b\big(h_2[n] - \delta[n]\big)

\begin{aligned} y[n] &= h_2[n] * \big(h_1[n] * x[n] + b\,\delta[n]\big) \\ &\quad - b\,\delta[n] \\ &= (h_1 * h_2 * x)[n] \\ &\quad + b\big(h_2[n] - \delta[n]\big) \end{aligned}

The second term is the same for every input, and it is not zero because $h_2[n] \neq \delta[n]$ . A zero input therefore gives a non-zero output, so the system is not linear.

That term also stays put at $n = 0$ when the input is delayed, so a delayed input doesn't give a delayed output: the system is time variant. Option (D). (If $h_2[n] = \delta[n]$ , the two $b\,\delta[n]$ terms would cancel and the system would be LTI.)

Full question on page 17 of the official paper (PDF)

GATE 2025, Q.18

Networks, Signals and Systems1 markMCQ

$f(t)$ is a real continuous-time signal whose Fourier transform $F(\omega) = \displaystyle\int_{-\infty}^{\infty} f(t)\, e^{-j\omega t}\, dt$ exists. Which statement is always TRUE?

Option A: $|F(\omega)| \le \displaystyle\int_{-\infty}^{\infty} |f(t)|\, dt$
Option B: $|F(\omega)| > \displaystyle\int_{-\infty}^{\infty} |f(t)|\, dt$
Option C: $|F(\omega)| \le \displaystyle\int_{-\infty}^{\infty} f(t)\, dt$
Option D: $|F(\omega)| \ge \displaystyle\int_{-\infty}^{\infty} f(t)\, dt$

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Key idea: The magnitude of an integral is at most the integral of the magnitude.

Official answer: A

Since $|e^{-j\omega t}| = 1$ ,

|F(\omega)| = \left|\int_{-\infty}^{\infty} f(t)\,e^{-j\omega t}\,dt\right| \le \int_{-\infty}^{\infty} |f(t)|\,dt

\begin{aligned} |F(\omega)| &= \left|\int_{-\infty}^{\infty} f(t)\,e^{-j\omega t}\,dt\right| \\ &\le \int_{-\infty}^{\infty} |f(t)|\,dt \end{aligned}

So (A) holds for every $f$ , and (B) never does. (C) and (D) compare with the signed integral, which can fail either way:

(C) fails for $f(t) = -1$ on $|t| < \tfrac{1}{2}$ : $|F(0)| = 1$ but $\int f\,dt = -1$ .
(D) fails for $f(t) = 1$ on $|t| < \tfrac{1}{2}$ : at $\omega = \pi$ , $|F| = 2/\pi \approx 0.64 < 1$ .

Option (A).

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Full question on page 18 of the official paper (PDF)

GATE 2025, Q.19

Networks, Signals and Systems1 markMCQ

In the network in the figure, 1 A flows through the 3 Ω resistor from the 9 V node into the junction. Node X is joined to an 8 V node through 2 Ω and to that junction through 1 Ω. What is the voltage at X, in volts?

A network fragment. An 8 V node connects through 2 ohms to node X, and X through 1 ohm to a junction. A 9 V node feeds the junction through a 3 ohm resistor carrying 1 A downward. From the junction, a 2 ohm resistor leads right to a triangle of three 2 ohm resistors that continues off the figure. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: $\tfrac{20}{3}$
Option B: $\tfrac{32}{3}$
Option C: $\tfrac{22}{3}$
Option D: $\tfrac{2}{3}$

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Key idea: The known current through the 3 Ω fixes the junction voltage, and then the rest of the network doesn't matter for node X.

Official answer: A

The 3 Ω resistor drops $1 \times 3 = 3$ V, so the junction is at $9 - 3 = 6$ V.

Node X has only two connections, so the current in from the 8 V node equals the current out to the junction:

\frac{8 - V_X}{2} = \frac{V_X - 6}{1} \;\Rightarrow\; 8 - V_X = 2V_X - 12 \;\Rightarrow\; V_X = \frac{20}{3}\ \text{V}

\begin{aligned} \frac{8 - V_X}{2} &= \frac{V_X - 6}{1} \\ \Rightarrow\; 8 - V_X &= 2V_X - 12 \\ \Rightarrow\; V_X &= \frac{20}{3}\ \text{V} \end{aligned}

Option (A). The resistors to the right of the junction only carry whatever current is left over there.

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Full question on page 19 of the official paper (PDF)

GATE 2025, Q.20

Networks, Signals and Systems1 markMCQ

A source of $1\angle 90°$ V is connected across a capacitor of admittance $j0.25$ S, an inductor of $-j0.1$ S and a resistor of $0.2$ S, all in parallel, as in the figure. Which set of branch currents is correct?

A 1 at 90 degrees volt AC source in parallel with three branches: a capacitor of admittance j0.25 S carrying iC, an inductor of admittance -j0.1 S carrying iL, and a resistor of 0.2 S carrying iR, each current marked downward. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: $i_C = 0.25\angle 180°$ A, $i_L = 0.1\angle 0°$ A, $i_R = 0.2\angle 90°$ A
Option B: $i_C = 4\angle 180°$ A, $i_L = 10\angle 0°$ A, $i_R = 5\angle 90°$ A
Option C: $i_C = 0.25\angle 270°$ A, $i_L = 0.1\angle 90°$ A, $i_R = 0.2\angle 90°$ A
Option D: $i_C = 4\angle 90°$ A, $i_L = 10\angle 270°$ A, $i_R = 5\angle 0°$ A

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Key idea: With admittances, each branch current is $I = YV$ : multiply, don't divide.

Official answer: A

Every branch has the full source voltage $V = 1\angle 90° = j$ across it, so

i_C = j0.25 \times j = -0.25 = 0.25\angle 180°\ \text{A}

\begin{aligned} i_C &= j0.25 \times j \\ &= -0.25 \\ &= 0.25\angle 180°\ \text{A} \end{aligned}

i_L = -j0.1 \times j = 0.1 = 0.1\angle 0°\ \text{A}

\begin{aligned} i_L &= -j0.1 \times j \\ &= 0.1 \\ &= 0.1\angle 0°\ \text{A} \end{aligned}

i_R = 0.2 \times j = 0.2\angle 90°\ \text{A}

\begin{aligned} i_R &= 0.2 \times j \\ &= 0.2\angle 90°\ \text{A} \end{aligned}

Option (A). Option (B) divides by the admittances, as if they were impedances.

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Full question on page 26 of the official paper (PDF)

GATE 2025, Q.27

Networks, Signals and Systems1 markMSQ

$x[n]$ has z-transform $X(z)$ . Which statements are TRUE?

One or more options are correct.

Option A: The DTFT of $x[n]$ always exists
Option B: The region of convergence (RoC) of $X(z)$ contains neither poles nor zeros
Option C: The DTFT exists if the RoC contains the unit circle
Option D: If $x[n] = \alpha\,\delta[n]$ for a scalar $\alpha$ , the RoC is the entire $z$ -plane

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Key idea: The DTFT is the z-transform on the unit circle, and the RoC can never contain a pole but may contain zeros.

Official answer: C and D

(A) is false: $x[n] = 2^n u[n]$ has $X(z) = \frac{1}{1 - 2z^{-1}}$ for $|z| > 2$ , but no DTFT.
(B) is false: the RoC has no poles, but it can contain zeros. $x[n] = \delta[n] - \delta[n-1]$ has $X(z) = 1 - z^{-1}$ , with a zero at $z = 1$ inside its RoC.
(C) is true: on the unit circle, $X(e^{j\omega})$ is the DTFT, and it converges there when the circle is in the RoC.
(D) is true: $X(z) = \alpha$ for every $z$ , so the RoC is the whole plane.

Options (C) and (D).

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Full question on page 33 of the official paper (PDF)

GATE 2025, Q.38

Networks, Signals and Systems2 marksMCQ

$f(t)$ has finite energy and a Fourier transform that is zero outside $[-\omega_c, \omega_c]$ . It is sampled as $y(t) = f(t)\,p(t)$ with $p(t) = \sum_{n=-\infty}^{\infty} \delta(t - \tau - nT_s)$ , where $T_s > 0$ and $\tau > 0$ , and $y(t)$ passes through an ideal low-pass filter with cutoff $\omega_c$ and passband gain $T_s$ . What is the filter's output?

Option A: $f(t)$ if $T_s < \pi/\omega_c$
Option B: $f(t - \tau)$ if $T_s < \pi/\omega_c$
Option C: $f(t - \tau)$ if $T_s < 2\pi/\omega_c$
Option D: $T_s f(t)$ if $T_s < 2\pi/\omega_c$

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Key idea: Shifting the sampling instants changes where the samples are taken, not which signal they are samples of.

Official answer: A

The impulses sit at $t = \tau + nT_s$ and carry $f$ 's own values there, $f(\tau + nT_s)$ . In frequency, the shifted impulse train is

P(\omega) = \frac{2\pi}{T_s} \sum_{k} e^{-jk\omega_s \tau}\, \delta(\omega - k\omega_s), \qquad \omega_s = \frac{2\pi}{T_s}

\begin{aligned} P(\omega) &= \frac{2\pi}{T_s} \sum_{k} e^{-jk\omega_s \tau}\, \delta(\omega - k\omega_s), \\ \omega_s &= \frac{2\pi}{T_s} \end{aligned}

Y(\omega) = \frac{1}{T_s} \sum_{k} e^{-jk\omega_s \tau}\, F(\omega - k\omega_s)

The $k = 0$ copy is $F(\omega)/T_s$ with no phase factor: the offset only multiplies the other copies by a phase. If the copies don't overlap, $\omega_s > 2\omega_c$ , that is $T_s < \pi/\omega_c$ , the filter keeps just the $k = 0$ copy and its gain $T_s$ restores $F(\omega)$ . The output is $f(t)$ . Option (A).

Option (B) is the usual slip: the samples are taken later, but they are still samples of $f(t)$ , not of $f(t - \tau)$ .

Full question on page 34 of the official paper (PDF)

GATE 2025, Q.39

Networks, Signals and Systems2 marksMCQ

An unknown linear time-invariant circuit, driven by $v_s(t) = 100\cos(200t)$ V, feeds a branch of 5 Ω, a variable capacitor $C$ and 1 H in series, as in the figure. An ideal AC voltmeter $M_1$ across the branch reads 25 V RMS, and an ideal AC ammeter $M_2$ in it reads 5 A RMS. What should $C$ be?

A source vs(t) drives a box labelled Linear Time Invariant Circuit. Across its output is voltmeter M1. From the top output terminal, ammeter M2 leads to a series branch of 5 ohms, a variable capacitor C and a 1 H inductor, back to the bottom terminal. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: 25 μF
Option B: 4 μF
Option C: 0.25 μF
Option D: Insufficient information to find $C$

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Key idea: The two meter readings give the branch's impedance magnitude without needing anything about the rest of the circuit.

Official answer: A

The branch has 25 V across it and 5 A through it, so

|Z| = \frac{25}{5} = 5\ \Omega = \sqrt{5^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}

\begin{aligned} |Z| &= \frac{25}{5} \\ &= 5\ \Omega \\ &= \sqrt{5^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \end{aligned}

So the reactance is zero: the branch is at resonance at $\omega = 200$ rad/s.

C = \frac{1}{\omega^2 L} = \frac{1}{200^2 \times 1} = 25\ \mu\text{F}

\begin{aligned} C &= \frac{1}{\omega^2 L} \\ &= \frac{1}{200^2 \times 1} \\ &= 25\ \mu\text{F} \end{aligned}

Option (A). The unknown circuit doesn't matter, because the meters measure the branch directly.

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Full question on page 35 of the official paper (PDF)

GATE 2025, Q.40

Networks, Signals and Systems2 marksMCQ

Find the Z-parameter matrix, in ohms, of the two-port in the figure: from port 1, a 2 Ω series resistor, a 2 Ω shunt, a 2 Ω series resistor, a 2 Ω shunt, and a 2 Ω series resistor to port 2.

A resistor ladder between port 1 (V1, I1 in) and port 2 (V2, I2 in): a 2 ohm series resistor, a 2 ohm shunt to the bottom rail, another 2 ohm series resistor, another 2 ohm shunt, and a final 2 ohm series resistor to port 2. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: $\begin{bmatrix} 10/3 & 2/3 \\ 2/3 & 10/3 \end{bmatrix}$
Option B: $\begin{bmatrix} 2/3 & 10/3 \\ 10/3 & 2/3 \end{bmatrix}$
Option C: $\begin{bmatrix} 10 & 2 \\ 2 & 10 \end{bmatrix}$
Option D: $\begin{bmatrix} 10/3 & 1/3 \\ 1/3 & 10/3 \end{bmatrix}$

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Key idea: Each Z-parameter is a ratio measured with the other port open, so the open port's series resistor carries no current.

Official answer: A

Call the two shunt nodes A (near port 1) and B. With port 2 open, the last 2 Ω carries nothing, so $V_2 = V_B$ . From node A the network looks like 2 Ω in parallel with $2 + 2 = 4$ Ω:

Z_{11} = 2 + \frac{2 \times 4}{2 + 4} = \frac{10}{3}\ \Omega

\begin{aligned} Z_{11} &= 2 + \frac{2 \times 4}{2 + 4} \\ &= \frac{10}{3}\ \Omega \end{aligned}

The current into node A splits so that $V_A = \tfrac{4}{3} I_1$ , and the divider of 2 Ω and 2 Ω gives $V_B = \tfrac{1}{2} V_A$ :

Z_{21} = \frac{V_2}{I_1} = \frac{2}{3}\ \Omega

The network is symmetric and reciprocal, so $Z_{22} = \tfrac{10}{3}$ and $Z_{12} = \tfrac{2}{3}$ . Option (A).

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Full question on page 47 of the official paper (PDF)

GATE 2025, Q.55

Networks, Signals and Systems2 marksMSQ

$f(t)$ is periodic with fundamental period $T_0 > 0$ , and $y(t) = f(\alpha t)$ with $\alpha > 1$ . Their Fourier series are $f(t) = \sum_{k} c_k\, e^{j 2\pi k t / T_0}$ and $y(t) = \sum_{k} d_k\, e^{j 2\pi \alpha k t / T_0}$ . Which statements are TRUE?

One or more options are correct.

Option A: $c_k = d_k$ for all $k$
Option B: $y(t)$ is periodic with fundamental period $\alpha T_0$
Option C: $c_k = d_k / \alpha$ for all $k$
Option D: $y(t)$ is periodic with fundamental period $T_0 / \alpha$

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Key idea: Compressing a signal in time shrinks its period and raises every harmonic's frequency, but leaves the weight of each harmonic unchanged.

Official answer: A and D

Substitute $\alpha t$ for $t$ in the series for $f$ :

y(t) = f(\alpha t) = \sum_{k} c_k\, e^{j 2\pi k \alpha t / T_0}

\begin{aligned} y(t) &= f(\alpha t) \\ &= \sum_{k} c_k\, e^{j 2\pi k \alpha t / T_0} \end{aligned}

That is exactly the given form for $y$ , so $d_k = c_k$ : (A) is TRUE and (C) is FALSE. $y$ repeats when $\alpha t$ advances by $T_0$ , so its fundamental period is $T_0 / \alpha$ : (D) is TRUE and (B) is FALSE. Options (A) and (D).

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Full question on page 18 of the official paper (PDF)

GATE 2024, Q.21

Networks, Signals and Systems1 markMCQ

A causal, stable LTI system with impulse response $h(t)$ gives output $y(t)$ for input $x(t)$ . The input $x(0.5t)$ is applied to another causal, stable LTI system with impulse response $h(0.5t)$ . What is the output?

Option A: $2y(0.5t)$
Option B: $4y(0.5t)$
Option C: $0.25y(2t)$
Option D: $0.25y(0.25t)$

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Key idea: Stretching both signals in a convolution stretches the result by the same factor and scales its area.

Official answer: A

\int_{-\infty}^{\infty} x(0.5\tau)\,h\big(0.5(t - \tau)\big)\,d\tau

Substitute $u = 0.5\tau$ , so $d\tau = 2\,du$ :

2\int_{-\infty}^{\infty} x(u)\,h(0.5t - u)\,du = 2\,y(0.5t)

\begin{aligned} &2\int_{-\infty}^{\infty} x(u)\,h(0.5t - u)\,du \\ &\quad = 2\,y(0.5t) \end{aligned}

Option (A).

Full question on page 21 of the official paper (PDF)

GATE 2024, Q.24

Networks, Signals and Systems1 markMSQ

A causal discrete-time LTI system has $H(z) = \dfrac{2z^2 + 3}{(z + \frac{1}{3})(z - \frac{1}{3})}$ . Which statements are true?

One or more options are correct.

Option A: The system is stable.
Option B: The system is a minimum phase system.
Option C: The initial value of the impulse response is 2.
Option D: The final value of the impulse response is 0.

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Key idea: For a causal system, the poles decide stability, the zeros decide minimum phase, and $h[0] = \lim_{z\to\infty} H(z)$ .

Official answer: A, C and D

(A) The poles $\pm\frac{1}{3}$ are inside the unit circle, so the causal system is stable: true.
(B) The zeros satisfy $z^2 = -\frac{3}{2}$ , so $|z| = \sqrt{1.5} \approx 1.22 > 1$ . Zeros outside the unit circle mean it isn't minimum phase: false.
(C) $h[0] = \lim_{z\to\infty} H(z) = 2$ : true.
(D) Stable means $h[n] \to 0$ : true.

Options (A), (C) and (D).

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Full question on page 25 of the official paper (PDF)

GATE 2024, Q.31

Networks, Signals and Systems1 markNAT

In the circuit in the figure, a 5 mA source and a 1 kΩ resistor carrying $I_0$ sit on the left node. A current-controlled voltage source of $1000\,I_0$ (positive on the right) joins it to the right node, which has a 1 kΩ resistor in series with a 2 mA source, and a 1 kΩ resistor carrying $I_x$ , all to the common bottom rail. What is $I_x$ , in mA?

A circuit with a common bottom rail. On the left node: a 5 mA source pointing up and a 1 kilohm resistor carrying I0 down. A dependent voltage source 1000 I0, minus on the left and plus on the right, joins the top of the left node to the right node. On the right node: a 1 kilohm resistor in series with a 2 mA source pointing down, and a 1 kilohm resistor carrying Ix down. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

Answer format: mA, an integer

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Key idea: The dependent source fixes the right node at the left node's voltage plus $1000\,I_0$ , so both node voltages are multiples of $I_0$ .

Official answer: 2

With currents in mA and resistances in kΩ, the left node is at $V_1 = I_0$ volts, and the dependent source adds $1000\,I_0$ A $= I_0$ (mA) volts, so the right node is at $V_2 = 2I_0$ and $I_x = V_2 / 1 = 2I_0$ .

Current into the right node is what the left node doesn't send through its 1 kΩ, and it leaves through the 2 mA branch and $I_x$ :

5 - I_0 = 2 + 2I_0 \;\Rightarrow\; I_0 = 1\ \text{mA}, \qquad I_x = 2\ \text{mA}

\begin{aligned} 5 - I_0 &= 2 + 2I_0 \\ \Rightarrow\; I_0 &= 1\ \text{mA}, \\ I_x &= 2\ \text{mA} \end{aligned}

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Full question on page 25 of the official paper (PDF)

GATE 2024, Q.32

Networks, Signals and Systems1 markNAT

In the circuit in the figure, a 5 A source with 2 Ω across it feeds a 3 H inductor into a node with 4 Ω to ground. The switch S, open for a long time, closes at $t = 0$ and puts a second 4 Ω across that node. What is the time constant for $t > 0$ , in seconds?

A 5 A current source in parallel with a 2 ohm resistor feeds a 3 H inductor in the top rail. After the inductor, a 4 ohm resistor goes to the bottom rail, and a switch S, closing at t = 0, connects a second 4 ohm resistor in parallel with it. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

Answer format: Seconds

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Key idea: The time constant is $L$ over the resistance the inductor sees with the independent source turned off.

Official answer: 0.75

With the 5 A source open-circuited, the inductor sees the 2 Ω on one side and the two 4 Ω resistors in parallel on the other, all in series:

R = 2 + (4 \parallel 4) = 2 + 2 = 4\ \Omega, \qquad \tau = \frac{L}{R} = \frac{3}{4} = 0.75\ \text{s}

\begin{aligned} R &= 2 + (4 \parallel 4) \\ &= 2 + 2 \\ &= 4\ \Omega, \\ \tau &= \frac{L}{R} \\ &= \frac{3}{4} \\ &= 0.75\ \text{s} \end{aligned}

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Full question on page 34 of the official paper (PDF)

GATE 2024, Q.44

Networks, Signals and Systems2 marksMCQ

The figure shows $x(t)$ , a downward triangle between $t = -1$ and $t = 0$ with minimum $-1$ , and $y(t)$ , an upward triangle between $t = -2$ and $t = 2$ with peak 1. If $X(f)$ is the Fourier transform of $x(t)$ , which is the Fourier transform of $y(t)$ ?

Left: x(t), a downward triangle from t = -1 to t = 0 with its minimum of -1 marked below a tick labelled 0.5. Right: y(t), an upward triangle from t = -2 to t = 2 with its peak of 1 at t = 0. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

Option A: $-4X(4f)\,e^{-j\pi f}$
Option B: $-4X(4f)\,e^{-j4\pi f}$
Option C: $-\tfrac{1}{4}X(f/4)\,e^{-j\pi f}$
Option D: $-\tfrac{1}{4}X(f/4)\,e^{-j4\pi f}$

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Key idea: A question given marks to all scores full marks for every candidate, whatever they answered.

Official answer: Marks to all

The final key awards marks to all for this question and doesn't say why, so we don't pick an option.

The method it tests: write $y$ as a flipped, stretched and shifted copy of $x$ , then apply the scaling and shift properties, $x(at - b) \leftrightarrow \frac{1}{|a|}X\!\left(\frac{f}{a}\right)e^{-j2\pi f b/a}$ . One likely reason for the ruling is the figure: the tick under the minimum of $x(t)$ is labelled 0.5, though the minimum sits between $-1$ and 0.

Ruling in the key: The final key gives marks to all (MTA) for this question: every candidate received its 2 marks.

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Full question on page 36 of the official paper (PDF)

GATE 2024, Q.48

Networks, Signals and Systems2 marksMCQ

$x(t) = 2\cos(8\pi t + \pi/3)$ is sampled at 15 Hz. The sampled signal $x_s(t)$ passes through an LTI system with $h(t) = \left(\dfrac{\sin 2\pi t}{\pi t}\right)\cos(38\pi t - \pi/2)$ , giving $x_o(t)$ . What is $x_o(t)$ ?

Option A: $15\sin(38\pi t + \pi/3)$
Option B: $15\sin(38\pi t - \pi/3)$
Option C: $15\cos(38\pi t - \pi/6)$
Option D: $15\cos(38\pi t + \pi/6)$

Show answerHide answer

Key idea: A question given marks to all scores full marks for every candidate, whatever they answered.

Official answer: Marks to all

The final key awards marks to all for this question and doesn't say why, so we don't pick an option.

The method it tests: sampling at 15 Hz copies the 4 Hz tone to $15k \pm 4$ Hz, the filter passes a narrow band round 19 Hz with its own phase shift, and the output is the 19 Hz copy. One likely reason for the ruling is that the answer's amplitude and phase depend on how the impulse sampling and the filter's gain are normalised, which the question leaves open.

Ruling in the key: The final key gives marks to all (MTA) for this question: every candidate received its 2 marks.

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Full question on page 43 of the official paper (PDF)

GATE 2024, Q.56

Networks, Signals and Systems2 marksMSQ

For which radian frequencies $\Omega$ does $x[n] = A\cos(\Omega n + \pi/3)$ have a period of 40?

One or more options are correct.

Option A: $0.15\pi$
Option B: $0.225\pi$
Option C: $0.3\pi$
Option D: $0.45\pi$

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Key idea: $\cos(\Omega n)$ repeats after $N$ samples when $\Omega N$ is a whole number of turns, $2\pi m$ .

Official answer: A and D, or A, C and D

Write $\Omega = 2\pi m/N$ in lowest terms; the fundamental period is then $N$ :

$0.15\pi = 2\pi \cdot \frac{3}{40}$ : fundamental period 40.
$0.225\pi = 2\pi \cdot \frac{9}{80}$ : period 80, and 40 is not a period.
$0.3\pi = 2\pi \cdot \frac{3}{20}$ : fundamental period 20, so 40 is a period but not the fundamental one.
$0.45\pi = 2\pi \cdot \frac{9}{40}$ : fundamental period 40.

If "a period of 40" means the fundamental period, the answer is (A) and (D); if it means any period, (C) qualifies too. The key accepts both: A and D, or A, C and D.

Full question on page 44 of the official paper (PDF)

GATE 2024, Q.59

Networks, Signals and Systems2 marksNAT

For an $N$ -point sequence, $X[k] = \mathcal{F}\{x[n]\}$ is its $N$ -point DFT. Let $y[n] = \mathcal{F}\{\mathcal{F}\{\mathcal{F}\{\mathcal{F}\{x[n]\}\}\}\}$ . For $x[n] = \{1, 2, 1, 3\}$ , what is $Y[0]$ ?

Answer format: An integer

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Key idea: Applying the DFT twice gives $N$ times the time-reversed sequence, so four times gives $N^2$ times the original.

Official answer: 112

\mathcal{F}^2\{x\}[n] = N\,x[(-n) \bmod N] \;\Rightarrow\; \mathcal{F}^4\{x\} = N^2\,x

\begin{aligned} \mathcal{F}^2\{x\}[n] &= N\,x[(-n) \bmod N] \\ \Rightarrow\; \mathcal{F}^4\{x\} &= N^2\,x \end{aligned}

So $y[n] = 16\,x[n]$ , and $Y[0] = \sum_n y[n] = 16(1 + 2 + 1 + 3) = 112$ .

Full question on page 45 of the official paper (PDF)

GATE 2024, Q.60

Networks, Signals and Systems2 marksNAT

In the two-port in the figure, port 1 connects through 2 Ω to a node that is wired straight to port 2. From that node a shunt branch runs through 4 Ω and a dependent voltage source $8V_1$ (negative at the top, positive at the bottom) to the common rail. What is $Y_{21}$ , in siemens?

Port 1 (V1, I1 in) connects through a 2 ohm resistor to a node that continues straight to port 2 (V2, I2 in). From the node, a 4 ohm resistor in series with a dependent voltage source 8V1, minus at the top and plus at the bottom, runs to the common bottom rail. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

Answer format: Siemens

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Key idea: $Y_{21} = I_2/V_1$ with port 2 short-circuited.

Official answer: 1.5

With port 2 shorted the node is at 0 V, so $I_1 = V_1/2$ . The source holds the bottom of the 4 Ω at $-8V_1$ , so the shunt branch draws $\frac{0 - (-8V_1)}{4} = 2V_1$ from the node. KCL at the node:

I_1 + I_2 = 2V_1 \;\Rightarrow\; I_2 = 2V_1 - 0.5V_1 = 1.5V_1, \qquad Y_{21} = 1.5\ \text{S}

\begin{aligned} I_1 + I_2 &= 2V_1 \\ \Rightarrow\; I_2 &= 2V_1 - 0.5V_1 \\ &= 1.5V_1, \\ Y_{21} &= 1.5\ \text{S} \end{aligned}

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Full question on page 45 of the official paper (PDF)

GATE 2024, Q.62

Networks, Signals and Systems2 marksNAT

In the network in the figure, a 50 V source feeds the load $R_L$ through two parallel paths: 2 Ω then 3 Ω, with $v_x$ across the 3 Ω (positive at the load end); and a dependent voltage source $v_x$ (positive towards the load) then 2 Ω. What value of $R_L$ , in ohms, receives maximum power?

A 50 V source, positive at the top, feeds two parallel paths to the top of the load RL: a 2 ohm resistor then a 3 ohm resistor carrying vx (minus on the left, plus on the right), and a dependent voltage source vx (minus on the left, plus on the right) then a 2 ohm resistor. RL returns to the source's negative terminal. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

Answer format: Ohms

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Key idea: With a dependent source, find the Thevenin resistance by turning off the independent source and driving the terminals with a test voltage.

Official answer: 2.5

Short the 50 V source and apply a test voltage $V$ across the load terminals. The top path is 5 Ω, carrying $V/5$ , and its 3 Ω takes $v_x = \tfrac{3}{5}V$ .

In the lower path the dependent source lifts its right end to $v_x = 0.6V$ , so its 2 Ω carries $(V - 0.6V)/2 = 0.2V$ . The total test current is $0.2V + 0.2V = 0.4V$ :

R_{th} = \frac{V}{0.4V} = 2.5\ \Omega = R_L

\begin{aligned} R_{th} &= \frac{V}{0.4V} \\ &= 2.5\ \Omega \\ &= R_L \end{aligned}

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Full question on page 12 of the official paper (PDF)

GATE 2023, Q.18

Networks, Signals and Systems1 markMCQ

A series $RLC$ circuit has a quality factor $Q$ of 1000 at a centre frequency of $10^6$ rad/s. What are possible values of $R$ , $L$ and $C$ ?

Option A: $R = 1\ \Omega$ , $L = 1\ \mu$ H and $C = 1\ \mu$ F
Option B: $R = 0.1\ \Omega$ , $L = 1\ \mu$ H and $C = 1\ \mu$ F
Option C: $R = 0.01\ \Omega$ , $L = 1\ \mu$ H and $C = 1\ \mu$ F
Option D: $R = 0.001\ \Omega$ , $L = 1\ \mu$ H and $C = 1\ \mu$ F

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Key idea: For a series RLC, $\omega_0 = 1/\sqrt{LC}$ and $Q = \omega_0 L / R$ .

Official answer: D

Every option has $L = 1\ \mu$ H and $C = 1\ \mu$ F, so $\omega_0 = 1/\sqrt{10^{-12}} = 10^6$ rad/s, as required. Then

R = \frac{\omega_0 L}{Q} = \frac{10^6 \times 10^{-6}}{1000} = 0.001\ \Omega

\begin{aligned} R &= \frac{\omega_0 L}{Q} \\ &= \frac{10^6 \times 10^{-6}}{1000} \\ &= 0.001\ \Omega \end{aligned}

Option (D).

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Full question on page 18 of the official paper (PDF)

GATE 2023, Q.26

Networks, Signals and Systems1 markMCQ

A system has input $x(t)$ and output $y(t) = x(e^t)$ . The system is

Option A: Causal and time invariant.
Option B: Non-causal and time varying.
Option C: Causal and time varying.
Option D: Non-causal and time invariant.

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Key idea: Causal means $y(t)$ uses no input later than $t$ ; time invariant means a delayed input gives the same output, delayed.

Official answer: B

Causality: $e^t > t$ for every $t$ , so $y(t)$ always uses the input at a later time. At $t = 0$ , $y(0) = x(1)$ . The system is non-causal.

Time invariance: delaying the input by $t_0$ gives $x(e^t - t_0)$ , but delaying the output gives $y(t - t_0) = x(e^{t - t_0})$ . These differ, so the system is time varying. Option (B).

Full question on page 19 of the official paper (PDF)

GATE 2023, Q.28

Networks, Signals and Systems1 markMCQ

What is the Fourier transform $X(\omega)$ of $x(t) = e^{-t^2}$ ? (Note: $\int_{-\infty}^{\infty} e^{-y^2}\, dy = \sqrt{\pi}$ .)

Option A: $\sqrt{\pi}\, e^{\omega^2/2}$
Option B: $\dfrac{e^{-\omega^2/4}}{2\sqrt{\pi}}$
Option C: $\sqrt{\pi}\, e^{-\omega^2/4}$
Option D: $\sqrt{\pi}\, e^{-\omega^2/2}$

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Key idea: Complete the square in the exponent, then use the given integral.

Official answer: C

X(\omega) = \int_{-\infty}^{\infty} e^{-t^2 - j\omega t}\, dt = e^{-\omega^2/4} \int_{-\infty}^{\infty} e^{-(t + j\omega/2)^2}\, dt = \sqrt{\pi}\, e^{-\omega^2/4}

\begin{aligned} X(\omega) &= \int_{-\infty}^{\infty} e^{-t^2 - j\omega t}\, dt \\ &= e^{-\omega^2/4} \int_{-\infty}^{\infty} e^{-(t + j\omega/2)^2}\, dt \\ &= \sqrt{\pi}\, e^{-\omega^2/4} \end{aligned}

A quick check: $X(0)$ must equal $\int x(t)\, dt = \sqrt{\pi}$ , which rules out (B), and (A) grows with $\omega$ . Option (C).

Full question on page 20 of the official paper (PDF)

GATE 2023, Q.29

Networks, Signals and Systems1 markMCQ

Match each signal type with its spectral characteristics. Signal types: (i) continuous, aperiodic; (ii) continuous, periodic; (iii) discrete, aperiodic; (iv) discrete, periodic. Spectral characteristics: (a) continuous, aperiodic; (b) continuous, periodic; (c) discrete, aperiodic; (d) discrete, periodic.

Option A: (i) → (a), (ii) → (b), (iii) → (c), (iv) → (d)
Option B: (i) → (a), (ii) → (c), (iii) → (b), (iv) → (d)
Option C: (i) → (d), (ii) → (b), (iii) → (c), (iv) → (a)
Option D: (i) → (a), (ii) → (c), (iii) → (d), (iv) → (b)

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Key idea: Periodic in one domain means discrete in the other, and discrete in one domain means periodic in the other.

Official answer: B

(i) Continuous, aperiodic: the Fourier transform, continuous and aperiodic, (a).
(ii) Continuous, periodic: the Fourier series, discrete lines, aperiodic, (c).
(iii) Discrete, aperiodic: the DTFT, continuous and periodic in $2\pi$ , (b).
(iv) Discrete, periodic: the DFS, discrete and periodic, (d).

Option (B).

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Full question on page 22 of the official paper (PDF)

GATE 2023, Q.32

Networks, Signals and Systems1 markNAT

In the circuit in the figure, what is the current $i$ flowing through the 200 Ω resistor, in mA?

Node A at the top left connects to ground through a 2 V source (positive terminal up) in series with 2 kilohms, through a separate 2 kilohm resistor, and through a 1 mA current source pointing up into node A. A 1 kilohm resistor joins node A to node B, with a 1 mA current source in parallel pointing from A to B. Node B connects to ground through 200 ohms, carrying the current i downwards. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: mA, rounded off to two decimal places

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Key idea: Two unknown node voltages: write KCL at each, working in volts, mA and kΩ.

Official answer: 1.30 to 1.40

Call the top-left node A and the node above the 200 Ω resistor B. In mA and kΩ, KCL at A (the 1 mA into A from below and the 1 mA out of A to B cancel):

\frac{V_A - 2}{2} + \frac{V_A}{2} + \frac{V_A - V_B}{1} = 1 - 1 = 0 \quad\Rightarrow\quad 2V_A - V_B = 1

\begin{aligned} &\frac{V_A - 2}{2} \\ &\qquad + \frac{V_A}{2} \\ &\qquad + \frac{V_A - V_B}{1} \\ &\quad = 1 - 1 \\ &\quad = 0 \\ \Rightarrow\; \\ 2V_A - V_B &= 1 \end{aligned}

KCL at B, with the 1 mA source feeding it:

\frac{V_B - V_A}{1} + \frac{V_B}{0.2} = 1 \quad\Rightarrow\quad -V_A + 6V_B = 1

\begin{aligned} &\frac{V_B - V_A}{1} + \frac{V_B}{0.2} \\ &\quad = 1 \\ \Rightarrow\; \\ -V_A + 6V_B &= 1 \end{aligned}

So $V_A = (1 + V_B)/2$ , and $5.5\,V_B = 1.5$ gives $V_B = 0.2727$ V. Then $i = V_B / 0.2\ \text{k}\Omega = 1.36$ mA.

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Full question on page 22 of the official paper (PDF)

GATE 2023, Q.33

Networks, Signals and Systems1 markNAT

For the two-port network in the figure, $[Y] = \dfrac{1}{100}\begin{bmatrix} 2 & -1 \\ -1 & 4/3 \end{bmatrix}$ S. It is driven by a 120 V source through 10 Ω at port 1. What load impedance $Z_L$ , in Ω, takes the maximum power?

A 120 V source in series with 10 ohms drives port 1 of a two-port network labelled [Y]; port 2 is loaded by Z_L. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: Ohms, rounded off to the nearest integer

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Key idea: Maximum power needs $Z_L$ equal to the Thevenin impedance seen at port 2, with the source shorted.

Official answer: 80

With the source shorted, port 1 sees $I_1 = -V_1/10$ , so $-0.1V_1 = 0.02V_1 - 0.01V_2$ , which gives $V_1 = V_2/12$ . Then

I_2 = -0.01 V_1 + \tfrac{4}{300} V_2 = \left(-\tfrac{0.01}{12} + 0.01333\right) V_2 = 0.0125\, V_2

\begin{aligned} I_2 &= -0.01 V_1 + \tfrac{4}{300} V_2 \\ &= \left(-\tfrac{0.01}{12} + 0.01333\right) V_2 \\ &= 0.0125\, V_2 \end{aligned}

So $Z_{th} = 1/0.0125 = 80$ Ω, and $Z_L = 80$ Ω. Equivalently, $Y_{out} = Y_{22} - Y_{12}Y_{21}/(Y_{11} + 0.1)$ . Answer: 80.

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Full question on page 23 of the official paper (PDF)

GATE 2023, Q.35

Networks, Signals and Systems1 markNAT

In the circuit in the figure, a 2 V source drives, through switch S and 1 Ω, a 2 Ω resistor in parallel with a 1 H inductor. S was closed for a long time and is opened at $t = 0$ . What is the maximum magnitude of the voltage $V_R$ across the 2 Ω resistor, in volts?

A source Vs = 2 V in series with switch S and a 1 ohm resistor, feeding a 2 ohm resistor (voltage V_R across it) in parallel with a 1 H inductor, back to the source. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: volts, rounded off to the nearest integer

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Key idea: The inductor current can't change instantly, and once S opens it has nowhere to go but the 2 Ω resistor.

Official answer: 4

Before $t = 0$ the inductor is a short, so $V_R = 0$ and the whole current, $2/1 = 2$ A, flows through the inductor.

After S opens, the inductor and the 2 Ω resistor form a loop, and the 2 A must flow through the resistor: $|V_R(0^+)| = 2 \times 2 = 4$ V. It then decays as $4e^{-2t}$ , so the maximum magnitude is 4 V.

Full question on page 31 of the official paper (PDF)

GATE 2023, Q.44

Networks, Signals and Systems2 marksMCQ

In the circuit in the figure, switch $S_1$ was closed and $S_2$ was open for a long time. At $t = 0$ , $S_1$ is opened and $S_2$ is closed, simultaneously. $S_2$ shorts the 100 Ω resistor that is in series with the 1 H inductor. What is $i_C(0^+)$ , in amperes?

A 1 A current source connects through switch S1 to a top node. From the top node to the bottom rail: a 0.01 F capacitor carrying i_C downwards; a 100 ohm resistor in series with a 1 H inductor, with switch S2 across the 100 ohm resistor; and a 25 ohm resistor. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: 1
Option B: −1
Option C: 0.2
Option D: 0.8

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Key idea: Find the capacitor voltage and inductor current before switching; both carry over to $t = 0^+$ .

Official answer: B

Before $t = 0$ the capacitor is open and the inductor a short, so the 1 A splits between the 100 Ω branch and the 25 Ω resistor:

i_L(0^-) = 1 \times \frac{25}{125} = 0.2\ \text{A}, \qquad v_C(0^-) = 25 \times 0.8 = 20\ \text{V}

\begin{aligned} i_L(0^-) &= 1 \times \frac{25}{125} \\ &= 0.2\ \text{A}, \\ v_C(0^-) &= 25 \times 0.8 \\ &= 20\ \text{V} \end{aligned}

At $t = 0^+$ the source is gone, $v_C = 20$ V and $i_L = 0.2$ A. With the 100 Ω shorted, the inductor still carries 0.2 A, and the 25 Ω carries $20/25 = 0.8$ A. The capacitor supplies both, so the current down into it is $i_C(0^+) = -(0.2 + 0.8) = -1$ A. Option (B).

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Full question on page 33 of the official paper (PDF)

GATE 2023, Q.46

Networks, Signals and Systems2 marksMCQ

The h-parameter model of a two-port network is shown in the figure, driven by $V_s$ and loaded by $R_L$ . What is the condition for the maximum small-signal voltage gain $V_{out}/V_s$ ?

h-parameter two-port: source Vs drives the input through h11 into a voltage source h12 v2; at the output, a dependent current source labelled h21 v1 in parallel with h22 and the load R_L, across which Vout is taken. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $h_{11} = 0$ , $h_{12} = 0$ , $h_{21}$ = very high and $h_{22} = 0$
Option B: $h_{11}$ = very high, $h_{12} = 0$ , $h_{21}$ = very high and $h_{22} = 0$
Option C: $h_{11} = 0$ , $h_{12}$ = very high, $h_{21}$ = very high and $h_{22} = 0$
Option D: $h_{11} = 0$ , $h_{12} = 0$ , $h_{21}$ = very high and $h_{22}$ = very high

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Key idea: A question given marks to all scores full marks for every candidate, whatever they answered.

Official answer: Marks to all

The key awards marks to all for this question and doesn't say why, so we don't pick an option.

The idea behind it: with $V_s$ driving port 1 directly, the gain of an h-parameter two-port is $-h_{21}R_L / \left(h_{11} + \Delta h\, R_L\right)$ with $\Delta h = h_{11}h_{22} - h_{12}h_{21}$ , so a large $h_{21}$ and small $h_{11}$ , $h_{12}$ and $h_{22}$ raise it. One likely reason for the ruling is the figure: it labels the output source $h_{21}v_1$ where the h-parameter model has $h_{21}i_1$ , and with $h_{11} = 0$ and $h_{12} = 0$ the input current isn't set by $V_s$ at all.

Ruling in the key: The key gives marks to all (MTA) for this question: every candidate received its 2 marks.

Full question on page 34 of the official paper (PDF)

GATE 2023, Q.47

Networks, Signals and Systems2 marksMCQ

A discrete-time periodic signal has period $N = 5$ and DTFS $x[n] = \sum_{k=0}^{4} a_k e^{jk2\pi n/5}$ , with $a_0 = 1$ , $a_1 = 3j$ , $a_2 = 2j$ , $a_3 = -2j$ and $a_4 = -3j$ . What is $\sum_{n=0}^{4} x[n]\sin\dfrac{4\pi n}{5}$ ?

Option A: −10
Option B: 10
Option C: −2
Option D: 2

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Key idea: $\sum_n x[n] e^{-jk2\pi n/N} = N a_k$ , so write the sine as two exponentials.

Official answer: A

\sin\frac{4\pi n}{5} = \frac{e^{j4\pi n/5} - e^{-j4\pi n/5}}{2j}

$\sum_n x[n] e^{-j2(2\pi n/5)} = 5a_2 = 10j$ , and $\sum_n x[n] e^{+j2(2\pi n/5)} = 5a_{-2} = 5a_3 = -10j$ . So

\sum_{n=0}^{4} x[n]\sin\frac{4\pi n}{5} = \frac{-10j - 10j}{2j} = -10

\begin{aligned} &\sum_{n=0}^{4} x[n]\sin\frac{4\pi n}{5} \\ &\quad = \frac{-10j - 10j}{2j} \\ &\quad = -10 \end{aligned}

Option (A).

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Full question on page 34 of the official paper (PDF)

GATE 2023, Q.48

Networks, Signals and Systems2 marksMCQ

An input $x[n]$ with DTFT $X(e^{j\Omega}) = 1 - e^{-j\Omega} + 2e^{-j3\Omega}$ is passed through an LTI system with frequency response $H(e^{j\Omega}) = 1 - \tfrac{1}{2}e^{-j2\Omega}$ . What is the output $y[n]$ ?

Option A: $\delta[n] + \delta[n-1] - \tfrac{1}{2}\delta[n-2] - \tfrac{5}{2}\delta[n-3] + \delta[n-5]$
Option B: $\delta[n] - \delta[n-1] - \tfrac{1}{2}\delta[n-2] - \tfrac{5}{2}\delta[n-3] + \delta[n-5]$
Option C: $\delta[n] - \delta[n-1] - \tfrac{1}{2}\delta[n-2] + \tfrac{5}{2}\delta[n-3] - \delta[n-5]$
Option D: $\delta[n] + \delta[n-1] + \tfrac{1}{2}\delta[n-2] + \tfrac{5}{2}\delta[n-3] + \delta[n-5]$

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Key idea: Multiply the two polynomials in $e^{-j\Omega}$ ; each power $e^{-jk\Omega}$ is a $\delta[n-k]$ .

Official answer: C

Y = \left(1 - e^{-j\Omega} + 2e^{-j3\Omega}\right)\left(1 - \tfrac{1}{2}e^{-j2\Omega}\right) = 1 - e^{-j\Omega} - \tfrac{1}{2}e^{-j2\Omega} + \tfrac{5}{2}e^{-j3\Omega} - e^{-j5\Omega}

\begin{aligned} Y &= \left(1 - e^{-j\Omega} + 2e^{-j3\Omega}\right)\left(1 - \tfrac{1}{2}e^{-j2\Omega}\right) \\ &= 1 \\ &\quad - e^{-j\Omega} \\ &\quad - \tfrac{1}{2}e^{-j2\Omega} \\ &\quad + \tfrac{5}{2}e^{-j3\Omega} \\ &\quad - e^{-j5\Omega} \end{aligned}

The $e^{-j3\Omega}$ term collects $2$ from $X$ and $+\tfrac{1}{2}$ from $(-e^{-j\Omega})(-\tfrac{1}{2}e^{-j2\Omega})$ . So $y[n] = \delta[n] - \delta[n-1] - \tfrac{1}{2}\delta[n-2] + \tfrac{5}{2}\delta[n-3] - \delta[n-5]$ . Option (C).

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Full question on page 35 of the official paper (PDF)

GATE 2023, Q.49

Networks, Signals and Systems2 marksMCQ

$x(t) = 10\cos(10.5Wt)$ is passed through an LTI system with impulse response $h(t) = \pi\left(\dfrac{\sin Wt}{\pi t}\right)^2 \cos 10Wt$ . What is the output?

Option A: $\left(\dfrac{15W}{4}\right)\cos(10.5Wt)$
Option B: $\left(\dfrac{15W}{2}\right)\cos(10.5Wt)$
Option C: $\left(\dfrac{15W}{8}\right)\cos(10.5Wt)$
Option D: $(15W)\cos(10.5Wt)$

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Key idea: $\sin Wt/(\pi t)$ is an ideal low-pass of height 1 and cut-off $W$ ; squaring it makes a triangle in frequency.

Official answer: A

$\dfrac{\sin Wt}{\pi t}$ transforms to a rectangle of height 1 on $|\omega| < W$ . Its square transforms to $\tfrac{1}{2\pi}$ times the rectangle convolved with itself: a triangle of peak $\tfrac{2W}{2\pi} = \tfrac{W}{\pi}$ , reaching zero at $|\omega| = 2W$ . Times $\pi$ , the peak is $W$ :

T(\omega) = W\left(1 - \frac{|\omega|}{2W}\right), \quad |\omega| < 2W

\begin{aligned} T(\omega) &= W\left(1 - \frac{|\omega|}{2W}\right), \\ |\omega| &< 2W \end{aligned}

Multiplying by $\cos 10Wt$ shifts half of it to each of $\pm 10W$ : $H(\omega) = \tfrac{1}{2}[T(\omega - 10W) + T(\omega + 10W)]$ . At $\omega = 10.5W$ , 0.5W from the centre:

H(10.5W) = \frac{1}{2} \cdot W\left(1 - \frac{0.5W}{2W}\right) = \frac{3W}{8}

\begin{aligned} H(10.5W) &= \frac{1}{2} \cdot W\left(1 - \frac{0.5W}{2W}\right) \\ &= \frac{3W}{8} \end{aligned}

$H$ is real and positive, so there is no phase shift: $y(t) = 10 \cdot \tfrac{3W}{8}\cos(10.5Wt) = \tfrac{15W}{4}\cos(10.5Wt)$ . Option (A).

Full question on page 36 of the official paper (PDF)

GATE 2023, Q.50

Networks, Signals and Systems2 marksMCQ

$x_1(t)$ and $x_2(t)$ are band-limited signals, each with bandwidth $B = 4\pi \times 10^3$ rad/s. In the figure, $y(t) = x_1(t)\cos(4\pi \times 10^3 t) + x_2(t)\cos(12\pi \times 10^3 t)$ . What Nyquist sampling frequency, in rad/s, is required to sample $y(t)$ ?

x1(t) is multiplied by cos(4 pi 10^3 t) and x2(t) by cos(12 pi 10^3 t); the two products are added to give y(t). — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $20\pi \times 10^3$
Option B: $40\pi \times 10^3$
Option C: $8\pi \times 10^3$
Option D: $32\pi \times 10^3$

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Key idea: Find the highest frequency in $y(t)$ and double it.

Official answer: D

Multiplying by a cosine shifts the spectrum to the carrier. $x_1$ reaches $4\pi + 4\pi = 8\pi \times 10^3$ rad/s, and $x_2$ reaches $12\pi + 4\pi = 16\pi \times 10^3$ rad/s. The highest frequency is $16\pi \times 10^3$ , so

\omega_s = 2 \times 16\pi \times 10^3 = 32\pi \times 10^3\ \text{rad/s}

\begin{aligned} \omega_s &= 2 \times 16\pi \times 10^3 \\ &= 32\pi \times 10^3\ \text{rad/s} \end{aligned}

Option (D).

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Full question on page 42 of the official paper (PDF)

GATE 2023, Q.58

Networks, Signals and Systems2 marksNAT

Let $x_1(t) = u(t + 1.5) - u(t - 1.5)$ , and let $x_2(t)$ be as in the figure. For $y(t) = x_1(t) * x_2(t)$ , what is $\int_{-\infty}^{\infty} y(t)\, dt$ ?

x2(t): an impulse of strength 1 at t = -3, a rectangular pulse of height 1 from t = -1 to t = 1, and an impulse of strength 2 at t = 2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: rounded off to the nearest integer

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Key idea: The area of a convolution is the product of the two areas.

Official answer: 15

\int y\, dt = \left(\int x_1\, dt\right)\left(\int x_2\, dt\right)

$x_1$ is a pulse of height 1 and width 3, area 3. $x_2$ has area $1 + (1 \times 2) + 2 = 5$ : the two impulses and the pulse. So $\int y\, dt = 3 \times 5 = 15$ .