Explore topic-wise MCQs in Signals Systems.

This section includes 15 Mcqs, each offering curated multiple-choice questions to sharpen your Signals Systems knowledge and support exam preparation. Choose a topic below to get started.

1.

The Laplace transform of the function cosh2(t) is ____________

A. \(\frac{s^2+2}{s(s^2+4)}\)
B. \(\frac{s^2-2}{s(s^2-4)}\)
C. \(\frac{s^2-2}{s(s^2+4)}\)
D. \(\frac{s^2+2}{s(s^2-4)}\)
Answer» C. \(\frac{s^2-2}{s(s^2+4)}\)
2.

The z-transform of cos(\(\frac{π}{3}\) n) u[n] is __________

A. \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), 0<|z|<1
B. \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), |z|>1
C. \(\frac{z}{2} \frac{(1-2z)}{(z^2-z+1)}\), 0<|z|<1
D. \(\frac{z}{2} \frac{(1-2z)}{(z^2-z+1)}\), |z|>1
Answer» C. \(\frac{z}{2} \frac{(1-2z)}{(z^2-z+1)}\), 0<|z|<1
3.

A linear phase channel with phase delay Tp and group delay Tg must have ____________

A. Tp = Tg = constant
B. Tp ∝ f and Tg ∝ f
C. Tp = constant and Tg ∝ f
D. Tp ∝ f and Tg = constant
Answer» B. Tp ∝ f and Tg ∝ f
4.

The convolution of two continuous time signals x(t) = e-t u(t) and h(t) = e-2t u(t) is ________________

A. (et – e2t) u (t)
B. (e-2t – e-t) u (t)
C. (e-t – e-2t) u (t)
D. (e-t + e-2t) u (t)
Answer» D. (e-t + e-2t) u (t)
5.

The first two components of trigonometric Fourier series of the given signal is ____________

A. 0, 4/π
B. 4/π, 0
C. 0, -4/π
D. -4/π, 1
Answer» B. 4/π, 0
6.

The impulse response of a continuous time system is given by h(t) = δ(t-1) + δ(t-3). The value of the step response at t=2 is _______________

A. 0
B. 1
C. 2
D. 3
Answer» C. 2
7.

A signal x (t) has its FT as X (f). The inverse FT of X(3f +2) is _____________

A. \(\frac{1}{2} x(\frac{t}{2})e^{j3πt}\)
B. \(\frac{1}{3} x(\frac{t}{3})e^{-j4πt/3}\)
C. 3x(3t)e-j4πt
D. x(3t + 2)
Answer» C. 3x(3t)e-j4πt
8.

If the FT of x(t) is \(\frac{2}{ω}\) sin⁡(πω), then the FT of ej5t x(t) is ____________

A. \(\frac{2}{ω-5}\) sin⁡(πω)
B. \(\frac{2}{ω+5}\) sin⁡{π(ω-5)}
C. \(\frac{2}{ω+5}\) sin⁡(πω)
D. \(\frac{2}{ω-5}\) sin⁡{π(ω-5)}
Answer» E.
9.

A waveform is given by v(t) = 10 sin2π 100 t. The magnitude of the second harmonic in its Fourier series representation is ____________

A. 0 V
B. 20 V
C. 100 V
D. 200 V
Answer» B. 20 V
10.

The lengths of two discrete time sequence x1[n] and x2[n] are 5 and 7 respectively. The maximum length of a sequence x1[n] * x2[n] is ____________

A. 5
B. 6
C. 7
D. 11
Answer» E.
11.

Given a periodic rectangular waveform of frequency 1 kHz, symmetrical about t=0 and having a pulse width of 500 µs and amplitude 5 V. The Fourier series is ___________

A. 2.5 + 3.18 cos (2π × 103 t) – 1.06 cos (6π × 103 t)
B. 2.5 – 3.18 cos (2π × 103 t) – 1.06 cos (6π × 103 t)
C. 2.5 + 3.18 cos (2π × 103 t) + 1.06 cos (6π × 103 t)
D. 2.5 – 3.18 cos (2π × 103 t) + 1.06 cos (6π × 103 t)
Answer» B. 2.5 – 3.18 cos (2π × 103 t) – 1.06 cos (6π × 103 t)
12.

Given a real valued function x (t) with period T. Its trigonometric Fourier series expansion contains no term of frequency ω = 2π \(\frac{(2k)}{T}\); where, k = 1, 2….. Also no terms are present. Then, x(t) satisfies the equation ____________

A. x (t) = x (t+T) = -x (t + \(\frac{T}{2}\))
B. x (t) = x (t+T) = x (t + \(\frac{T}{2}\))
C. x (t) = x (t-T) = -x (t – \(\frac{T}{2}\))
D. x (t) = x (t-T) = x (t – \(\frac{T}{2}\))
Answer» E.
13.

The unit impulse response of the system c (t) = -4e-t + 6e-2t for t>0. The step response of the same system for t ≥ 0 is _______________

A. -3e-2t – 4e-t + 1
B. -3e-2t + 4e-t – 1
C. -3e-2t – 4e-t – 1
D. 3e-2t + 4e-t – 1
Answer» C. -3e-2t – 4e-t – 1
14.

The Fourier series coefficient of time domain signal x (t) is X[k] = jδ[k-1] – jδ[k+1] + δ[k+3] + δ[k-3], the fundamental frequency of the signal is ω=2π. The signal is ___________

A. 2(cos 3πt – sin πt)
B. -2(cos 3πt – sin πt)
C. 2(cos 6πt – sin 2πt)
D. -2(cos 6πt – sin 2πt)
Answer» D. -2(cos 6πt – sin 2πt)
15.

The Fourier transform of sin(2πt) e-t u (t) is ____________

A. \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} + \frac{1}{1+j(ω+2π)}\right)\)
B. \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\)
C. \(\frac{1}{2j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)
D. \(\frac{1}{j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)
Answer» C. \(\frac{1}{2j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\)