About this set. These are original practice questions written
in GATE style for the 2026 Signals and Systems syllabus, with fully worked
solutions. They are not reproductions of the official GATE 2026 question paper.
A continuous-time linear time-invariant system has the impulse response
\(h(t) = e^{-2t}u(t+1)\), where \(u(\cdot)\) is the unit step. The system is
- causal and BIBO stable
- causal and BIBO unstable
- non-causal and BIBO stable
- non-causal and BIBO unstable
Solution
The step \(u(t+1)\) switches on at \(t = -1\), so the impulse response is non-zero over the
interval \(-1 \le t \lt 0\). An LTI system is causal only if \(h(t) = 0\) for all
\(t \lt 0\), so this system is non-causal
(see Chapter 2).
BIBO stability requires the impulse response to be absolutely integrable:
Equation
\[\int_{-\infty}^{\infty}|h(t)|\,dt = \int_{-1}^{\infty}e^{-2t}\,dt = \left[\frac{e^{-2t}}{-2}\right]_{-1}^{\infty}\]
Equation
\[= 0 + \frac{e^{2}}{2} = 3.694 \lt \infty\]
The integral converges because the decaying exponential is only shifted, not made to grow.
The system is therefore stable but not causal.
C
Final Answer
Correct answer: (C) non-causal and BIBO stable.
The input \(x(t) = u(t) - u(t-2)\) is applied to a continuous-time LTI system whose impulse
response is \(h(t) = e^{-t}u(t)\). If \(y(t) = x(t) * h(t)\), the value of \(y(3)\) is
_____ (round off to four decimal places).
Solution
Write the convolution integral with the rectangular pulse as the sliding function. Since
\(x(\tau) = 1\) only for \(0 \le \tau \le 2\) and is zero elsewhere,
Equation
\[y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)\,d\tau = \int_{0}^{2} e^{-(t-\tau)}u(t-\tau)\,d\tau\]
At \(t = 3\) the argument \(t - \tau = 3 - \tau\) is positive over the whole range
\(0 \le \tau \le 2\), so the step is unity throughout and the pulse has fully passed:
Equation
\[y(3) = \int_{0}^{2} e^{-(3-\tau)}\,d\tau = e^{-3}\int_{0}^{2}e^{\tau}\,d\tau = e^{-3}\left[e^{\tau}\right]_{0}^{2}\]
Equation
\[y(3) = e^{-3}\left(e^{2} - 1\right) = e^{-1} - e^{-3}\]
Equation
\[y(3) = 0.367879 - 0.049787 = 0.318092\]
✓
Final Answer
Correct answer: 0.3181.
Two finite-length discrete-time sequences are \(x[n] = \{1, 2, 3\}\) and
\(h[n] = \{1, 1, 1\}\), both starting at \(n = 0\). If \(y[n] = x[n] * h[n]\), the value of
\(y[3]\) is
- 3
- 5
- 6
- 9
Solution
The convolution sum is
Equation
\[y[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k]\]
Only \(k = 0, 1, 2\) contribute, and \(h[n-k]\) is unity for \(0 \le n-k \le 2\). For
\(n = 3\) this requires \(1 \le k \le 3\), which combined with the support of \(x[k]\) leaves
\(k = 1\) and \(k = 2\):
Equation
\[y[3] = x[1]h[2] + x[2]h[1] = 2(1) + 3(1) = 5\]
The full output sequence, of length \(3 + 3 - 1 = 5\), is
Equation
\[y[n] = \{1,\; 3,\; 6,\; 5,\; 3\}, \quad n = 0,1,2,3,4\]
As a check, the sum of the output samples, 18, equals the product of the individual sums
\(6 \times 3\).
B
Final Answer
Correct answer: (B) 5.
A periodic signal of period \(T = 2\) s is defined over one period by \(x(t) = t\) for
\(0 \lt t \lt 2\). Its exponential Fourier series is
\(x(t) = \sum_{k} c_k e^{jk\pi t}\). The magnitude \(|c_1|\) is _____ (round off to four
decimal places).
Solution
With \(T = 2\) the fundamental radian frequency is \(\omega_0 = 2\pi/T = \pi\) rad/s, and
Equation
\[c_k = \frac{1}{T}\int_{0}^{T} x(t)e^{-jk\omega_0 t}\,dt = \frac{1}{2}\int_{0}^{2} t\,e^{-j\pi t}\,dt\]
Let \(a = -j\pi\) and integrate by parts:
Equation
\[\int_{0}^{2} t\,e^{at}\,dt = \left[\frac{t e^{at}}{a} - \frac{e^{at}}{a^{2}}\right]_{0}^{2} = \frac{2e^{2a}}{a} - \frac{e^{2a}}{a^{2}} + \frac{1}{a^{2}}\]
Here \(e^{2a} = e^{-j2\pi} = 1\), so the two \(1/a^{2}\) terms cancel exactly:
Equation
\[\int_{0}^{2} t\,e^{-j\pi t}\,dt = \frac{2}{a} = \frac{2}{-j\pi} = \frac{j2}{\pi}\]
Equation
\[c_1 = \frac{1}{2}\cdot\frac{j2}{\pi} = \frac{j}{\pi}, \qquad |c_1| = \frac{1}{\pi} = 0.3183\]
The coefficient is purely imaginary, which is expected: after the average value
\(c_0 = \tfrac{1}{2}\int_0^2 t\,dt = 1\) is subtracted, the remaining sawtooth is an odd
function about the midpoint of the period.
✓
Final Answer
Correct answer: 0.3183.
A signal is given by \(x(t) = e^{-2|t|}\) and its continuous-time Fourier transform is
\(X(j\omega)\). The value of the integral
Equation
\[\frac{1}{2\pi}\int_{-\infty}^{\infty}\left|X(j\omega)\right|^{2}d\omega\]
is
- 0.25
- 0.50
- 1.00
- 2.00
Solution
Parseval's relation for the Fourier transform states that the quantity asked for is simply
the energy of the time-domain signal:
Equation
\[E = \int_{-\infty}^{\infty}|x(t)|^{2}dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left|X(j\omega)\right|^{2}d\omega\]
So evaluate the time-domain integral, which is far easier. The integrand
\(|x(t)|^2 = e^{-4|t|}\) is even, so integrate over the positive axis and double:
Equation
\[E = 2\int_{0}^{\infty}e^{-4t}\,dt = 2\left[\frac{e^{-4t}}{-4}\right]_{0}^{\infty} = 2 \times \frac{1}{4} = 0.5\]
For confirmation, the transform of the two-sided exponential is
Equation
\[X(j\omega) = \frac{2 \times 2}{2^{2} + \omega^{2}} = \frac{4}{4 + \omega^{2}}\]
and direct integration of \(16/(4+\omega^2)^2\) over all \(\omega\) gives \(\pi\), which
divided by \(2\pi\) is again 0.5.
B
Final Answer
Correct answer: (B) 0.50.
A causal LTI system has the transfer function
Equation
\[H(s) = \frac{s+3}{s^{2} + 3s + 2}\]
The system is initially at rest and a unit step is applied at \(t = 0\). The value of the
output \(y(t)\) at \(t = 1\) s is _____ (round off to four decimal places).
Solution
With \(X(s) = 1/s\) and the denominator factored as \((s+1)(s+2)\)
(see Chapter 3):
Equation
\[Y(s) = \frac{s+3}{s(s+1)(s+2)} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+2}\]
Evaluate the residues by covering up each factor in turn:
Equation
\[A = \left.\frac{s+3}{(s+1)(s+2)}\right|_{s=0} = \frac{3}{2} = 1.5\]
Equation
\[B = \left.\frac{s+3}{s(s+2)}\right|_{s=-1} = \frac{2}{(-1)(1)} = -2\]
Equation
\[C = \left.\frac{s+3}{s(s+1)}\right|_{s=-2} = \frac{1}{(-2)(-1)} = 0.5\]
Inverting term by term gives the step response for \(t \ge 0\):
Equation
\[y(t) = 1.5 - 2e^{-t} + 0.5e^{-2t}\]
The expression passes two checks: \(y(0) = 1.5 - 2 + 0.5 = 0\), as required for a strictly
proper transfer function driven from rest, and \(y(\infty) = 1.5 = H(0)\). At \(t = 1\):
Equation
\[y(1) = 1.5 - 2(0.367879) + 0.5(0.135335) = 1.5 - 0.735759 + 0.067668 = 0.831909\]
✓
Final Answer
Correct answer: 0.8319.
A signal is given by \(x(t) = 10\cos(600\pi t)\cos(400\pi t)\). The Nyquist rate for
sampling \(x(t)\) without aliasing is
- 200 Hz
- 500 Hz
- 1000 Hz
- 1200 Hz
Solution
A product of two cosines is not a single-frequency signal. Expand it with the
product-to-sum identity \(\cos A\cos B = \tfrac{1}{2}[\cos(A-B) + \cos(A+B)]\):
Equation
\[x(t) = 5\cos(200\pi t) + 5\cos(1000\pi t)\]
The two components have frequencies
Equation
\[f_1 = \frac{200\pi}{2\pi} = 100~\text{Hz}, \qquad f_2 = \frac{1000\pi}{2\pi} = 500~\text{Hz}\]
The signal is therefore band-limited to \(f_{max} = 500\) Hz, and the Nyquist rate is twice
that maximum frequency:
Equation
\[f_{Nyq} = 2f_{max} = 2 \times 500 = 1000~\text{Hz}\]
C
Final Answer
Correct answer: (C) 1000 Hz.
A causal discrete-time LTI system has the transfer function
Equation
\[H(z) = \frac{z}{(z - 0.2)(z - 0.5)}, \qquad |z| \gt 0.5\]
The value of the impulse response sample \(h[2]\) is _____ (round off to two decimal
places).
Solution
Expand \(H(z)/z\) rather than \(H(z)\) itself, so that each term inverts to a plain
exponential (see Chapter 27):
Equation
\[\frac{H(z)}{z} = \frac{1}{(z-0.2)(z-0.5)} = \frac{A}{z-0.2} + \frac{B}{z-0.5}\]
Equation
\[A = \frac{1}{0.2 - 0.5} = -\frac{10}{3}, \qquad B = \frac{1}{0.5 - 0.2} = \frac{10}{3}\]
Equation
\[H(z) = \frac{10}{3}\left[\frac{z}{z-0.5} - \frac{z}{z-0.2}\right]\]
The region of convergence \(|z| \gt 0.5\) lies outside the outermost pole, so both terms
invert to right-sided sequences:
Equation
\[h[n] = \frac{10}{3}\left[(0.5)^{n} - (0.2)^{n}\right]u[n]\]
Equation
\[h[2] = \frac{10}{3}\left[0.25 - 0.04\right] = \frac{10}{3}(0.21) = 0.70\]
A check on the first two samples: \(h[0] = 0\), which agrees with
\(\lim_{z\to\infty}H(z) = 0\), and \(h[1] = \tfrac{10}{3}(0.5-0.2) = 1\).
✓
Final Answer
Correct answer: 0.70.
The four-point DFT of the sequence \(x[n] = \{1, 2, 3, 4\}\), \(n = 0,1,2,3\), is
\(X[k]\). The magnitude \(|X[1]|\) is
- 2.000
- 2.828
- 4.000
- 10.000
Solution
The DFT with \(N = 4\) uses the twiddle factor
Equation
\[W_4 = e^{-j2\pi/4} = -j, \qquad X[k] = \sum_{n=0}^{3}x[n]\,W_4^{nk}\]
For \(k = 1\) the required powers are \(W_4^{0}=1\), \(W_4^{1}=-j\), \(W_4^{2}=-1\) and
\(W_4^{3}=j\):
Equation
\[X[1] = 1(1) + 2(-j) + 3(-1) + 4(j) = (1 - 3) + j(-2 + 4) = -2 + j2\]
Equation
\[|X[1]| = \sqrt{(-2)^{2} + 2^{2}} = \sqrt{8} = 2\sqrt{2} = 2.828\]
The remaining bins follow the same way: \(X[0] = 10\), \(X[2] = -2\) and
\(X[3] = -2 - j2\). Since \(x[n]\) is real, \(X[3] = X^{*}[1]\), as conjugate symmetry
requires. Choice D is the DC bin \(X[0]\), not \(|X[1]|\).
B
Final Answer
Correct answer: (B) 2.828.
A first-order low-pass filter is built from a series resistance \(R = 1~\text{k}\Omega\)
feeding a shunt capacitance \(C = 1~\mu\text{F}\), with the output taken across the
capacitor. The frequency at which the output lags the input by \(45^\circ\) is
- 100.0 Hz
- 159.2 Hz
- 318.3 Hz
- 1000.0 Hz
Solution
By the voltage divider between \(R\) and the capacitive impedance \(1/(j\omega C)\):
Equation
\[H(j\omega) = \frac{1/(j\omega C)}{R + 1/(j\omega C)} = \frac{1}{1 + j\omega RC}\]
The phase of this transfer function is
Equation
\[\angle H(j\omega) = -\tan^{-1}(\omega RC)\]
A lag of \(45^\circ\) requires \(\tan^{-1}(\omega RC) = 45^\circ\), that is
\(\omega RC = 1\), which is exactly the half-power cut-off condition
(see Chapter 14):
Equation
\[RC = (1 \times 10^{3})(1 \times 10^{-6}) = 10^{-3}~\text{s}\]
Equation
\[f = \frac{1}{2\pi RC} = \frac{1}{2\pi \times 10^{-3}} = 159.2~\text{Hz}\]
At this frequency the gain magnitude is \(1/\sqrt{2} = 0.707\), confirming that the
\(45^\circ\) lag point and the \(-3\) dB point coincide for a single real pole.
B
Final Answer
Correct answer: (B) 159.2 Hz.