# Signals & Systems Questions and Answers – Inverse Fourier Transform

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

1. Find the inverse Fourier transform of X(ω) = e-2ω u(ω).
a) $$\frac{1}{2π(2+jt)}$$
b) $$\frac{1}{2π(2-jt)}$$
c) $$\frac{1}{2(2+jt)}$$
d) $$\frac{1}{π(2+jt)}$$

Explanation: We know that x(t) = $$\frac{1}{2π} \int_{-∞}^∞ X(ω) e^{jωt} \,dω$$
x(t) = $$\frac{1}{2π} \int_{-∞}^∞ e^{-2ω} \,u(ω) e^{jωt} \,dω = \frac{1}{2π} \int_{-∞}^∞ e^{-2ω} e^{jωt} \, dω = \frac{1}{2π(2-jt)}$$.

2. Find the inverse Fourier transform of X(ω) = $$\frac{1+3(jω)}{(3+jω)^2}$$.
a) 3e-3t u(t) + 8e-3t u(t)
b) 3te-3t u(t) – 8e-8t u(t)
c) 3e-3t u(t) + 8te8t u(t)
d) 3e-3t u(t) – 8te-3t u(t)

Explanation: Given X(ω) = $$\frac{1+3(jω)}{(3+jω)^2} = \frac{A}{3+jω} + \frac{B}{(3+jω)^2} = \frac{3}{3+jω} – \frac{8}{(3+jω)^2}$$
Applying inverse Fourier transform, we get
x(t) = 3e-3t u(t) – 8te-3t u(t).

3. Find the inverse Fourier transform of δ(ω).
a) $$\frac{1}{2π}$$
b) 2π
c) $$\frac{1}{π}$$
d) π

Explanation: We know that x(t) = $$\frac{1}{2π} \int_{-∞}^∞ X(ω) e^{jωt} \,dω$$
= $$\frac{1}{2π} \int_{-∞}^∞ δ(ω) e^{jωt} \,dω = \frac{1}{2π}$$.

4. Find the inverse Fourier transform of u(ω).
a) $$\frac{1}{2} δ(t) + \frac{j}{2πt}$$
b) $$\frac{1}{2} δ(t) – \frac{j}{2πt}$$
c) δ(t) + $$\frac{j}{2πt}$$
d) δ(t) – $$\frac{j}{2πt}$$

Explanation: We know that u(ω) = $$\frac{1}{2}$$[1+sgn(ω)].
Applying linearity property,
u(ω) = -1 $$[\frac{1}{2}] + F^{-1} [\frac{1}{2} sgn(ω)]$$
u(ω) = $$\frac{1}{2} δ(t) + \frac{j}{2πt}$$.

5. Find the inverse Fourier transform of ej2t.
a) 2πδ(ω-2)
b) πδ(ω-2)
c) πδ(ω+2)
d) 2πδ(ω+2)

Explanation: We know that e0 t ↔ 2πδ(ω-ω0)
∴ ej2t ↔ 2πδ(ω-2).
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6. Find the inverse Fourier transform of jω.
a) δ(t)
b) $$\frac{d}{dt}$$ δ(t)
c) $$\frac{1}{δ(t)}$$
d) ∫δ(t)

Explanation: Time differentiation property, $$\frac{d}{dt}$$ x(t) ↔ jωX(ω) and we know that δ(t) ↔ 1
∴ $$\frac{d}{dt}$$ δ(t) ↔ jω.

7. Find the inverse Fourier transform of $$X(ω) = \frac{6+4(jω)}{(jω)^2 + 6(jω) + 8}$$.
a) e-2t u(t) – 5e-4t u(t)
b) e-2t u(t) + 5e-4t u(t)
c) -e-2t u(t) – 5e-4t u(t)
d) -e-2t u(t) + 5e-4t u(t)

Explanation: $$X(ω) = \frac{6+4(jω)}{(jω)^2+6(jω)+8} = \frac{A}{jω+2} + \frac{B}{jω+4} = -\frac{1}{jω+2} + \frac{5}{jω+4}$$
Applying inverse Fourier transform, we get
x(t) = -e-2t u(t) + 5e-4t u(t).

8. Find the convolution of the signals x1 (t) = e-2t u(t) and x2 (t) = e-3t u(t).
a) e-2t u(t) – e-3t u(t)
b) e-2t u(t) + e-3t u(t)
c) e2t u(t) – e3t u(t)
d) e2t u(t) – e-3t u(t)

Explanation: Convolution property, x1 (t)*x2 (t) ↔ X1 (ω) X2 (ω)
∴ x1 (t)*x2 (t) = F-1 [X1 (ω) X2 (ω)]
Given x1 (t) = e-2t u(t)
∴ X1 (ω) = $$\frac{1}{jω+2}$$
Given x2 (t) = e-3t u(t)
∴ X1 (ω) = $$\frac{1}{jω+3}$$
x1 (t)*x2 (t) = F-1 [X1 (ω) X2 (ω)] = F-1 $$[\frac{1}{jω+2} \frac{1}{jω+3}] = F^{-1} [\frac{1}{jω+2} – \frac{1}{jω+3}]$$
∴ x1 (t)*x2 (t) = e-2t u(t)-e-3t u(t).

9. Find the inverse Fourier transform of f(t)=1.
a) u(t)
b) δ(t)
c) e-t
d) $$\frac{1}{jω}$$

Explanation: We know that the Fourier transform of f(t) = 1 is F(ω) = 2πδ(ω).
Replacing ω with t
F(t) = 2πδ(t)
As per duality property F(t) ↔ 2πf(-ω), we have
2πδ(t) ↔ 2π(1)
δ(t) ↔ 1
Hence, the inverse Fourier transform of 1 is δ(t).

10. Find the inverse Fourier transform of sgn(ω).
a) $$\frac{1}{πt}$$
b) $$\frac{j}{πt}$$
c) $$\frac{j}{t}$$
d) $$\frac{1}{t}$$

Explanation: Given the function F(ω)=sgn(ω). The Fourier transform of a Signum function is sgn(ω) = $$\frac{2}{jω}$$.
Applying the duality property F(t) ↔ 2πf(-ω), we get
F($$\frac{2}{jt}$$) = 2πsgn(-ω).
As sgn(ω) is an odd function, sgn(-ω)=-sgn(ω).
Hence, $$\frac{2}{jt}$$ ↔ -sgn(ω)
Or $$\frac{2}{πt}$$ ↔ sgn(ω)
Therefore, the inverse Fourier transform of sgn(ω) is $$\frac{j}{πt}$$.

Sanfoundry Global Education & Learning Series – Signals & Systems.

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