This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Trigonometric Fourier Series”.

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

View Answer

Explanation: The Fourier Transform of e

^{-t}u (t) = \(\frac{1}{1+jω}\)

∴ Fourier transform of e

^{-3|t|}= \(\frac{6}{9+ω^2}\)

So, x (t-1) <–> X {j (ω-2π)}

∴ X (jω) = \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\).

2. 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)

View Answer

Explanation: x (t) = \(∑_{k=-∞}^∞ X[k]e^{j2πkt}\)

= je

^{j2πt}– je

^{-j2πt}+ e

^{j6πt}+ e

^{-j6πt}

= 2(cos 6πt – sin 2πt).

3. 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

View Answer

Explanation: Impulse response = -4e

^{-t}+ 6e

^{-2t}

And step response = \(\int_{-∞}^t c(t)\,dt\)

= \(\int_0^t -4e^{-t} \,dt + \int_0^t 6e^{-2t} \,dt\)

= 4e

^{-t}– 3e

^{-2t}-1.

4. 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 – \(\frac{T}{2}\))

View Answer

Explanation: For an even symmetry, x (t) = x (t-T)

Thus no sine component will exist because b

_{n}=0 and by half wave symmetry condition odd harmonics will exist.

Now, x (t) = x (t-\(\frac{T}{2}\))

Combining the two conditions, we get, x (t) = x (t-T) = x (t-\(\frac{T}{2}\)).

5. 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 × 10^{3} t) – 1.06 cos (6π × 10^{3} t)

b) 2.5 – 3.18 cos (2π × 10^{3} t) – 1.06 cos (6π × 10^{3} t)

c) 2.5 + 3.18 cos (2π × 10^{3} t) + 1.06 cos (6π × 10^{3} t)

d) 2.5 – 3.18 cos (2π × 10^{3} t) + 1.06 cos (6π × 10^{3} t)

View Answer

Explanation: Period, T = \(\frac{1}{f} = \frac{1}{1×10^3} = 10^{-3}s\)

Ratio, \(\frac{τ}{T} = \frac{500 × 10^{-6}}{10^{-3}} \) = 0.5

Now, \(f (t) = \frac{Aτ}{T} + 2 \frac{Aτ}{T} \frac{sin \frac{πτ}{T}}{\frac{πτ}{T}} cos[\frac{2πt}{τ}] + \frac{2Aτ}{T} \frac{sin \frac{2πτ}{T}}{\frac{2πτ}{T}} cos \frac{4πt}{T}\)

= 5(0.5) + 2(5) (0.5)\(\frac{sin0.5π}{0.5π}\) cos(2π × 10

^{3}t) + 2(5) (0.5) \(\frac{sinπ}{π}\) cos(4π × 10

^{3}t)

= 2.5 + 3.18 cos (2π × 10

^{3}t) – 1.06 cos (6π × 10

^{3}t).

6. The lengths of two discrete time sequence x_{1}[n] and x_{2}[n] are 5 and 7 respectively. The maximum length of a sequence x_{1}[n] * x_{2}[n] is ____________

a) 5

b) 6

c) 7

d) 11

View Answer

Explanation: Maximum length of x

_{1}[n] * x

_{2}[n] is L (say).

Now, we know that, L = L

_{1}+ L

_{2}-1 where L

_{1}is length of x

_{1}[n] and L

_{2}is length of x

_{2}[n] So, the maximum length of a sequence x

_{1}[n] * x

_{2}[n] = 5 + 7 – 1

= 11.

7. 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

View Answer

Explanation: Given that v (t) = 10 sin2π 100 t

It holds the half-wave symmetry i.e. x (t) = -x (t ± \(\frac{T}{2}\))

And for half wave symmetry, a

_{0}= 0 and a

_{k}= 0 (k is even)

So, second harmonic component a

_{2}= 0.

8. If the FT of x(t) is \(\frac{2}{ω}\) sin(πω), then the FT of e^{j5t} 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)}

View Answer

Explanation: Given that the FT of x (t) is \(\frac{2}{ω}\) sin(πω)

Applying time shifting and scaling property, we get the FT of e

^{j5t}x (t) as \(\frac{2}{ω-5}\) sin{π(ω-5)}.

9. 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)

View Answer

Explanation: Applying the time shifting and scaling property, we get,

X [3(f+2/3)] = \(\frac{1}{3} x(\frac{t}{3})e^{-j4πt/3}\).

10. 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

View Answer

Explanation: For step response, the impulse response can be integrated.

So, y (t) = u (t-1) + u (t-3)

Hence, y (2) = u (1) + u (-1)

= 1 + 0 = 1.

11. 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

View Answer

Explanation: Signal is odd and ω = \(\frac{2π}{t}\) = π

Here, c

_{0}= a

_{0}= 0

So, c

_{n}= b

_{n}= \(2\int_0^t f(t).sinωt \,dt\)

∴ c

_{1}= b

_{1}= \(2\int_0^t sinωt \,dt\)

= \(\frac{2}{ω}[cosωt]_0^t\)

= \(-\frac{2}{π}\)[-1-1] = 4/π

So, the first two components are 0, 4/π.

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

a) (e^{t} – e^{2t}) u (t)

b) (e^{-2t} – e^{-t}) u (t)

c) (e^{-t} – e^{-2t}) u (t)

d) (e^{-t} + e^{-2t}) u (t)

View Answer

Explanation: Given x (t) = e

^{-t}u (t) and h (t) = e

^{-2t}u (t).

Now, y (t) = \(∫_{-∞}^∞ x(τ)h(t-τ)dτ\)

= e

^{-2t}(e

^{t}-1); t≥0

Or, y (t) = e

^{-t}– e

^{-2t}; t≥0

Since, y (t) = 0 for t<0

Therefore, y (t) = (e

^{-t}– e

^{-2t}) u (t).

13. A linear phase channel with phase delay T_{p} and group delay T_{g} must have ____________

a) T_{p} = T_{g} = constant

b) T_{p} ∝ f and T_{g} ∝ f

c) T_{p} = constant and T_{g} ∝ f

d) T_{p} ∝ f and T_{g} = constant

View Answer

Explanation: θ(ω) = -ωt

_{0}

T

_{p}= \(-\frac{θ(ω)}{ω}\) = t

_{0}

T

_{g}=\(-\frac{d θ(ω)}{dω}\) = t

_{0}

∴ T

_{p}= T

_{g}= t

_{0}= constant.

14. 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

View Answer

Explanation: Performing z-transform on a

^{n}u[n], we get \(\frac{z}{z-a}\)

∴ x[n] = \(\frac{1}{2} e^{j(\frac{π}{3})n} \,u[n] + \frac{1}{2} e^{-j(\frac{π}{3})n} \,u [n]\)

Hence, X (z) = 0.5 \(\left(\frac{1}{1-e^{j(\frac{π}{3})} z^{-1}} + \frac{1}{1-e^{-j(\frac{π}{3})} z^{-1}}\right)\)

Hence, X (z) = \(\frac{z}{2} \frac{(2z-1)}{(z^2-z+1)}\), |z|>1.

15. The Laplace transform of the function cosh^{2}(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) \(\frac{s^2+2}{s(s^2-4)}\)

View Answer

Explanation: L ((\(\frac{1}{2}\)(e

^{t}– e

^{-t}))

^{2})

= L \(\left(\frac{e^2t}{4} + \frac{1}{2} + \frac{e^{-2t}}{4}\right)\)

= \(\frac{1}{4} \frac{1}{s-2} + \frac{1}{2} \frac{1}{s} + \frac{1}{4} \frac{1}{s+2}\)

= \(\frac{s^2-2}{s(s^2-4)}\).

**Sanfoundry Global Education & Learning Series – Signals & Systems.**

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