Exponential Fourier Series and Fourier Transforms Multiple Choice Questions

This set of Signals & Systems Question Paper focuses on “Exponential Fourier Series and Fourier Transforms”.

1. The Fourier transform of u (t) is B (jω) and the Laplace transform of u (t) is A(s). Which of the following is correct?
a) B(jω) = A(s)
b) A(s) = \(\frac{1}{s}\) but B(jω) ≠ \(\frac{1}{jω}\)
c) A(s) ≠ \(\frac{1}{s}\) but B(jω) ≠ \(\frac{1}{jω}\)
d) A(s) ≠ \(\frac{1}{s}\) but B(jω) = \(\frac{1}{jω}\)
View Answer

Answer: b
Explanation: Laplace transform of u (t) is given by u (t) –> A(s) = \(\frac{1}{s}\)
Fourier transform of u (t) is given by, u (t) = B (jω) = (\(\frac{1}{jω}\)) + π δ (ω)
Therefore A(s) = \(\frac{1}{s}\) but B(jω) ≠ \(\frac{1}{jω}\) is satisfied.

2. Given, X (e^jω) = \(\frac{(b-a) e^{jω}}{e^{-j2ω}-(a+b) e^{jω} + ab}\), |b|<1<|a|
The value of x[n] is __________
a) bⁿ u [n] + aⁿ u [n-1]
b) bⁿ u [n] – aⁿ u [-n-1]
c) bⁿ u [n] + aⁿ u [-n-1]
d) bⁿ u [n] – aⁿ u [n+1]
View Answer

Answer: c
Explanation: X (e^jω) = \( \frac{(b-a) e^{jω}}{e^{-j2ω}-(a+b) e^{jω} + ab}\)
= \( \frac{(b-a) e^{jω}}{1-(a+b)e^{-jω}+ab e^{-j2ω}}\)
= \(\frac{1}{1-be^{-jω}} + \frac{(-1)}{1-ae^{-jω}}\)
∴ x [n] = bⁿ u [n] + aⁿ u [-n-1].

3. The input and output of an LTI system are x (t) = e^-3t u (t) and y (t) = e^-t u (t). The differential equation which characterizes the system is ___________
a) \( \frac{dy(t)}{dt} + y(t) = \frac{dx(t)}{dt} + 3x(t)\)
b) \( \frac{dy(t)}{dt} + 2y(t) = \frac{dx(t)}{dt} + 3x(t)\)
c) \( \frac{dy(t)}{dt} – y(t) = \frac{dx(t)}{dt} + 3x(t)\)
d) \( \frac{dy(t)}{dt} – 2y(t) = \frac{dx(t)}{dt} + 3x(t)\)
View Answer

Answer: a
Explanation: X (s) = \( \frac{1}{s+3}\), Y (s) = \( \frac{1}{s+1}\)
∴ H(s) = \(\frac{Y(s)}{X(s)} = \frac{1/(s+1)}{1/(s+3)} = \frac{s+3}{s+1}\)
Now, s Y(s) + Y(s) = s X(s) + 3 X(s)
So, the differential equation together with the condition of initial rest that characterizes the system is \( \frac{dy(t)}{dt} + y(t) = \frac{dx(t)}{dt} + 3x(t)\).

4. The Fourier transform of signal e^-2t u(t-3) is ___________
a) \(\frac{e^{-3(2-jω)}}{2-jω}\)
b) \(\frac{e^{-3(2+jω)}}{2+jω}\)
c) \(\frac{e^{3(2-jω)}}{2-jω}\)
d) \(\frac{e^{3(2+jω)}}{2+jω}\)
View Answer

Answer: b
Explanation: X (jω) = \(\int_{-∞}^∞ x(t) e^{-jωt} \,dt\)
Given u (t-3), hence value of this will be equal to 1 when t>=3
∴ X (jω) = \(\int_3^∞ e^{-2t} e^{-jωt} \,dt \)
= \(\frac{e^{-3(2+jω)}}{2+jω}\).

5. The Fourier transform of the signal e^-4|t| is ____________
a) \(\frac{8}{16+ω^2}\)
b) \(\frac{-8}{16+ω^2}\)
c) \(\frac{4}{16+ω^2}\)
d) \(\frac{-4}{16+ω^2}\)
View Answer

Answer: a
Explanation: X (jω) = \(\int_{-∞}^∞ e^{-4|t|} e^{-jωt} \,dt\)
Now, e^-4|t| = e^-4t, when t>0 and
e^4t, when t<0
∴X (jω) = \(\int_{-∞}^0 e^{4t} e^{-jωt} \,dt + \int_0^∞ e^{-4t} e^{-jωt} \,dt\)
= \(\frac{8}{16+ω^2}\).

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6. The Inverse Fourier transform of the signal e^-2|ω| is ____________
a) \(\frac{2}{π(4+t^2)}\)
b) \(\frac{1}{2π(4+t^2)}\)
c) \(\frac{1}{π(4+t^2)}\)
d) \(\frac{1}{(4+t^2)}\)
View Answer

Answer: a
Explanation: e^-2|ω| = e^-2ω, ω > 0 and e^2ω, ω<0
Hence, x (t) = \(\frac{1}{2π} \int_{-∞}^∞ e^{-2|ω|} e^{-jωt} \,dω\)
= \(\frac{1}{2π} \int_{-∞}^0 e^{2ω} e^{-jωt} \,dω + \frac{1}{2π} \int_0^∞ e^{-2ω} e^{-jωt} \,dω\)
= \(\frac{2}{π(4+t^2)}\).

7. The inverse Laplace transform of F(s) = \(\frac{2}{s+c}e^{-bs}\) is _____________
a) 2e^{-k (t+b)} u (t+b)
b) 2e^{-k (t-b)} u (t-b)
c) 2e^{k (t-b)} u (t-b)
d) 2e^{k (t-b)} u (t+b)
View Answer

Answer: b
Explanation: Let G(s) = \(\frac{2}{s+c}\)
Or, G (t) = L^-1{G(s)} = 2e^-ct
∴ F (t) = L^-1{G(s) e^-bs}
= 2e^{-k (t-b)} u (t-b).

8. The Laplace transform of the function e^-2tcos(3t) + 5e^-2tsin(3t) is ____________
a) \(\frac{(s+2)-15}{(s+2)^2-9}\)
b) \(\frac{(s+2)+15}{(s+2)^2-9}\)
c) \(\frac{(s+2)+15}{(s+2)^2+9}\)
d) \(\frac{(s+2)-15}{(s+2)^2+9}\)
View Answer

Answer: c
Explanation: L {e^-2tcos(3t) + 5e^-2tsin(3t)}
= \(\frac{(s+2)}{(s+2)^2+9} + \frac{53}{(s+2)^2+9}\)
= \(\frac{(s+2)+15}{(s+2)^2+9}\).

9. A band pass signal extends from 1 KHz to 2 KHz. The minimum sampling frequency that is needed to retain all information of the sampled signal is ___________
a) 1 KHz
b) 2 KHz
c) 3 KHz
d) 4 KHz
View Answer

Answer: b
Explanation: We know that the minimum sampling frequency is twice the maximum bandwidth.
Here, maximum bandwidth = 2-1 = 1 KHz
So, Minimum sampling frequency = 2(Bandwidth) = 2(2-1) = 2 KHz.

10. The Laplace transform of the function 6e^5tcos(2t) – e^7t is ______________
a) \(\frac{6(s-5)}{(s-5)^2+4} – \frac{1}{s-7}\)
b) \(\frac{6(s-5)}{(s-5)^2+4} + \frac{1}{s-7}\)
c) \(\frac{6(s+5)}{(s+5)^2+4} – \frac{1}{s-7}\)
d) \(\frac{6(s+5)}{(s+5)^2+4} + \frac{1}{s-7}\)
View Answer

Answer: a
Explanation: We know that, Laplace transform of e^at = \(\frac{1}{s-a}\)
Here, a=7, so L {e^7t} = \(\frac{1}{s-7}\)
And the Laplace transform of e^at cos (b t) = \(\frac{(s-a)}{(s-a)^2+b^2}\)
Here, a=5 and b=2, so L {6e^5t cos (2t)} = \(\frac{6(s-5)}{(s-5)^2+4}\)
∴ L {6e^5tcos(2t) – e^7t} = \(\frac{6(s-5)}{(s-5)^2+4} – \frac{1}{s-7}\).

11. The inverse Laplace transform of F(s) = \(\frac{e^{-3s}}{s(s^2+3s+2)}\) is ______________
a) {0.5 + 0.5e^-2(t+3)-e^-(t+3)} u (t+3)
b) {0.5 + 0.5e^-2(t-3)-e^-(t-3)} u (t-3)
c) {0.5 – 0.5e^-2(t-3)-e^-(t-3)} u (t-3)
d) 0.5 + 0.5e^-2t-e^-t)
View Answer

Answer: b
Explanation: Let G(s) = \(\frac{1}{s(s^2+3s+2)}\)
Or, F(s) = G(s) e^-3s
G (t) = L^-1{G(s)}
= L^-1{\(\frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+1}\)}
Solving we get, A=0.5, B=0.5, C=-1
So, G (t) = 0.5 + 0.5e^-2t-e^-t
The inverse Laplace transform is F (t) = {0.5 + 0.5e^-2(t-3)-e^-(t-3)} u (t-3).

12. The Fourier transform of the signal e^-t+2 u (t-2) is ___________
a) \(\frac{e^{-2jω}}{1-2ω}\)
b) \(\frac{e^{2jω}}{1+2ω}\)
c) \(\frac{e^{-2jω}}{1+2ω}\)
d) \(\frac{e^{2jω}}{1-2ω}\)
View Answer

Answer: c
Explanation: Fourier transform of e^-t u(t) = \(\frac{1}{1+jω}\)
∴ The Fourier transform of x (t-2) = e^-2jω X (jω) [Using Shifting property]
Hence, X (jω) = \(\frac{e^{-2jω}}{1+2ω}\).

13. The Fourier transform of the signal te^-3|t-1| is _____________
a) \(\frac{6e^{-jω}}{9+ω^2} – \frac{12jωe^{-jω}}{9+ω^2}\)
b) \(\frac{6e^{-jω}}{9+ω^2}\)
c) –\(\frac{12jωe^{-jω}}{9+ω^2}\)
d) \(\frac{6e^{-jω}}{9+ω^2} + \frac{12jωe^{-jω}}{9+ω^2}\)
View Answer

Answer: a
Explanation: e^-3|t| = e^-3t, when t>0 and
e^3t, when t<0
Now, Fourier transform of e^-3|t| = \(\frac{6}{9+ω^2}\)
Again, x (t-1) ↔ e^-jω X(jω)
And t x (t) ↔ \(j\frac{d}{dω}\) X(jω)
∴ X (jω)= \(j\frac{d}{dω} [e^{-jω} \frac{6}{9+ω^2}]\)
= \(\frac{6e^{-jω}}{9+ω^2} – \frac{12jωe^{-jω}}{9+ω^2}\).

14. The Fourier transform of the signal te^-t u(t) is _____________
a) \(\frac{1}{1+ω^2}\)
b) \(\frac{-1}{1+ω^2}\)
c) \(\frac{1}{(1+jω)^2}\)
d) \(\frac{1}{(1-jω)^2}\)
View Answer

Answer: c
Explanation: Fourier transform of te^-at u (t) is \(\frac{1}{(a+jω)^2}\)
∴ X (jω) = \(\int_0^∞ te^{-t} e^{-jωt} \,dt\)
= \(\int_0^∞ te^{-t(1+jω)} \,dt\)
= \(\frac{1}{(1+jω)^2}\).

15. The Fourier transform of the signal \(∑_{m=0}^∞ a^m δ(t-m)\) is _____________
a) \(\frac{1}{1+ae^{-jω}}\)
b) \(\frac{1}{1+ae^jω}\)
c) \(\frac{1}{1-ae^jω}\)
d) \(\frac{1}{1-ae^{-jω}}\)
View Answer

Answer: d
Explanation: X (jω) = \(\int_0^∞ ∑_{m=0}^∞ a^m δ(t-m) e^{-jωt} \,dt\)
= \(∑_{m=0}^∞ (ae^{-jω})^m \)
= \(\frac{1}{1-ae^{-jω}}\).

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