

MCQOPTIONS
Saved Bookmarks
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. \(\frac{1}{2j} \left(\frac{1}{1+j(ω+2π)} – \frac{1}{1+j(ω-2π)}\right)\) | |