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) \(\frac{1}{π(2+jt)}\)
View Answer

Answer: b
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)
View Answer

Answer: d
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
View Answer

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

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) – \(\frac{j}{2πt}\)
View Answer

Answer: a
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)
View Answer

Answer: a
Explanation: We know that e0 t ↔ 2πδ(ω-ω0)
∴ ej2t ↔ 2πδ(ω-2).
Free 30-Day Java Certification Bootcamp is Live. Join Now!

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)
View Answer

Answer: b
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)
View Answer

Answer: d
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)
View Answer

Answer: a
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) \(\frac{1}{jω}\)
View Answer

Answer: b
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).
advertisement

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

Answer: b
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.

To practice all areas of Signals & Systems, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
I’m Manish - Founder and CTO at Sanfoundry. I’ve been working in tech for over 25 years, with deep focus on Linux kernel, SAN technologies, Advanced C, Full Stack and Scalable website designs.

You can connect with me on LinkedIn, watch my Youtube Masterclasses, or join my Telegram tech discussions.

If you’re in your 40s–60s and exploring new directions in your career, I also offer mentoring. Learn more here.