# Signals & Systems Questions and Answers – Exponential Fourier Series and Fourier Transforms

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) = $$\frac{1}{jω}$$

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) = $$\frac{(b-a) e^{jω}}{e^{-j2ω}-(a+b) e^{jω} + ab}$$, |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]

Explanation: X (e) = $$\frac{(b-a) e^{jω}}{e^{-j2ω}-(a+b) e^{jω} + ab}$$
= $$\frac{(b-a) e^{jω}}{1-(a+b)e^{-jω}+ab e^{-j2ω}}$$
= $$\frac{1}{1-be^{-jω}} + \frac{(-1)}{1-ae^{-jω}}$$
∴ 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) $$\frac{dy(t)}{dt} – 2y(t) = \frac{dx(t)}{dt} + 3x(t)$$

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) $$\frac{e^{3(2+jω)}}{2+jω}$$

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) $$\frac{-4}{16+ω^2}$$

Explanation: X (jω) = $$\int_{-∞}^∞ e^{-4|t|} e^{-jωt} \,dt$$
Now, e-4|t| = e-4t, when t>0 and
e4t, when t<0
∴X (jω) = $$\int_{-∞}^0 e^{4t} e^{-jωt} \,dt + \int_0^∞ e^{-4t} e^{-jωt} \,dt$$
= $$\frac{8}{16+ω^2}$$.

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) $$\frac{1}{(4+t^2)}$$

Explanation: e-2|ω| = e-2ω, ω > 0 and e, 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) 2ek (t-b) u (t-b)
d) 2ek (t-b) u (t+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) $$\frac{(s+2)-15}{(s+2)^2+9}$$

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

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 6e5tcos(2t) – e7t 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) $$\frac{6(s+5)}{(s+5)^2+4} + \frac{1}{s-7}$$

Explanation: We know that, Laplace transform of eat = $$\frac{1}{s-a}$$
Here, a=7, so L {e7t} = $$\frac{1}{s-7}$$
And the Laplace transform of eat cos (b t) = $$\frac{(s-a)}{(s-a)^2+b^2}$$
Here, a=5 and b=2, so L {6e5t cos (2t)} = $$\frac{6(s-5)}{(s-5)^2+4}$$
∴ L {6e5tcos(2t) – e7t} = $$\frac{6(s-5)}{(s-5)^2+4} – \frac{1}{s-7}$$.

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)

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) $$\frac{e^{2jω}}{1-2ω}$$

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) $$\frac{6e^{-jω}}{9+ω^2} + \frac{12jωe^{-jω}}{9+ω^2}$$

Explanation: e-3|t| = e-3t, when t>0 and
e3t, 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) $$\frac{1}{(1-jω)^2}$$

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) $$\frac{1}{1-ae^{-jω}}$$

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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