Signals & Systems Questions and Answers – Common Fourier Transforms

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

1. The Fourier series of an odd periodic function, contains __________
a) Only odd harmonics
b) Only even harmonics
c) Only cosine terms
d) Only sine terms
View Answer

Answer: d
Explanation: We know that, for a periodic function, if the dc term i.e. a0 = 0, then it is an odd function. Also, we know that an odd function consists of sine terms only since sine is odd. Hence the Fourier series of an odd periodic function contains only sine terms.
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2. The trigonometric Fourier series of a periodic time function can have only ___________
a) Only cosine terms
b) Only sine terms
c) Both cosine and sine terms
d) Dc and cosine terms
View Answer

Answer: d
Explanation: The Fourier series of a periodic function () is given by,
X (t) = \(∑_{n=0}^∞ a_n \,cos⁡nωt + ∑_{n=1}^∞ b_n \,sin⁡nωt\)
Thus the series has cosine terms of all harmonics i.e., n = 0,1,2,…..
The 0th harmonic which is the DC term = a0.
So, the trigonometric Fourier series of a periodic time function can have only Dc and cosine terms.

3. The Fourier transform of the signal δ(t+1) + δ(t-1) is ____________
a) \(\frac{2}{1+jω}\)
b) \(\frac{2}{1-jω}\)
c) 2 cos ω
d) 2 sin ω
View Answer

Answer: c
Explanation: Fourier transform of x (t+t0) = ejωtX (jω)
Hence, X (jω) = \( \int_{-∞}^∞ [δ(t + 1) + δ(t – 1)]^1 e^{-jωt} \,dt\)
= e + e-jω
= 2 cos ω.
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4. The trigonometric Fourier series of an even function of time does not have ___________
a) The dc term
b) The cosine terms
c) The sine terms
d) The odd harmonic terms
View Answer

Answer: c
Explanation: For periodic even function, the trigonometric Fourier series does not contain the sine terms since sine terms are in odd functions. The function only has dc term and cosine terms of all harmonics. So, the sine terms are absent in the trigonometric Fourier series of an even function.

5. 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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6. For a periodic signal () = 30 sin 1000 + 10 cos 3000 + 6 sin(5000 + 4), the fundamental frequency in rad/s is __________
a) 1000
b) 3000
c) 5000
d) 15000
View Answer

Answer: a
Explanation: First term has w1 = 1000 rad/s
Second term has w2 = 3000 rad/s
Third term has w3 = 500 rad/s
Now, w1 is the fundamental frequency, w2 is the third harmonic and w3 is the 5th harmonic. So, fundamental frequency = 1000 rad/s.

7. The Fourier transform of the signal 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

Answer: b
Explanation: The Fourier Transform of e-tu (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) = \(\frac{1}{2j} \left(\frac{1}{1+j(ω-2π)} – \frac{1}{1+j(ω+2π)}\right)\).
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8. Which of the following cannot be the Fourier series expansion of a periodic signal?
a) x1(t) = 2 cos + 3 cos 3
b) x2(t) = 2 cos + 7 cos
c) x3(t) = cos + 0.5
d) x4(t) = 2 cos 1.5 + sin 3.5t
View Answer

Answer: b
Explanation: x1(t) = 2 cost + 3 cost is periodic signal with fundamental frequency w0 = 1.
x2(t) = 2 cos πt + 7 cos t The frequency of first term w1 = π frequency of 2nd term is w2 = 1. Since, \(\frac{ω_1}{ω_2}\) = π, which is not rational. So, x(t) is not periodic.
x3(t) = cos t + 0.5 is a periodic function with w0 = 1
x4(t) = 2 cos(1.5)t + sin(3.5)t first term has frequency w1 = 1.5π and 2nd term has frequency w2 = 3.5π. Since, \(\frac{ω_1}{ω_2} = \frac{3}{7}\), which is rational.
Since x2(t) is not periodic, so it cannot be expanded in Fourier series.

9. The Fourier transform of the signal sgn (t) is ____________
a) \(\frac{-2}{jω}\)
b) \(\frac{4}{jω}\)
c) \(\frac{2}{jω}\)
d) \(\frac{1}{jω} + 1\)
View Answer

Answer: c
Explanation: sgn (t) = 1, 0≤t<∞ and
-1, -∞<t<0
X (ω) = \(\int_{-∞}^0 (-1)e^{-jωt} \,dt + \int_0^∞ (1)e^{-jωt} \,dt\)
= \(\frac{e^{-jωt}}{jω} + \frac{e^{-jωt}}{-jω}\)
= \(\frac{2}{jω}\).
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10. The Fourier transform of the signal u (t) is ____________
a) π δ(ω)
b) \(\frac{1}{jω}\)
c) π δ(ω) + \(\frac{1}{jω}\)
d) –\(\frac{1}{jω}\)
View Answer

Answer: c
Explanation: Fourier transform of 1 = 2π δ(ω)
And Fourier transform of \(\frac{1}{2}\) = π δ(ω)
Hence, Fourier transform of sgn (t) = \(\frac{2}{jω}\)
Or, \(\frac{1}{2}\) sgn (t) = \(\frac{1}{jω}\)
∴ \(\frac{1}{2} + \frac{1}{2}\) sgn (t) = w (t)
Fourier transform of w (t) = π δ(ω) + \(\frac{1}{jω}\).

11. The Fourier transform of the signal \(\frac{1}{a^2+w^2}\) is _____________
a) \(\frac{π}{a} e^{a|ω|}\)
b) \(\frac{π}{a} e^{-a|ω|}\)
c) \(\frac{2}{a} e^{-a|ω|}\)
d) \(\frac{π}{a} e^{a|ω|}\)
View Answer

Answer: b
Explanation: Fourier transform of e-a|t| = \(\frac{2a}{a^2+ω^2}\)
Now, e-a|t| = e-at, t>0 and eat, t<0
Using Duality property, \(\frac{2a}{a^2+w^2}\) ↔ 2πe-a|ω|
Or, \(\frac{1}{a^2+w^2} \leftrightarrow \frac{π}{a} e^{-a|ω|}\).

12. The Fourier transform of the signal e-|t| in [-2,2] is _____________
a) \(\frac{2 + 2e^{-2} cos⁡2ω + 2ωe^{-2} \,sin⁡2ω}{1-ω^2}\)
b) \(\frac{2 – 2e^{-2} cos⁡2ω – 2ωe^{-2} \,sin⁡2ω}{1+ω^2}\)
c) \(\frac{2 – 2e^{-2} cos⁡2ω + 2ωe^{-2} \,sin⁡2ω}{1+ω^2}\)
d) \(\frac{2 + 2e^{-2} cos⁡2ω + 2ωe^{-2} \,sin⁡2ω}{1-ω^2}\)
View Answer

Answer: c
Explanation: X (jω) = \(\int_{-2}^0 e^t e^{-jωt} \,dt + \int_0^2 e^{-t} e^{-jωt} \,dt\)
= \(\frac{1-e^{-(1-jω)^2}}{1-jω} + \frac{1-e^{-(1+jω)^2}}{1+jω}\)
= \(\frac{2 – 2e^{-2} cos⁡2ω + 2ωe^{-2} \,sin⁡2ω}{1+ω^2}\).

13. The Fourier transform of the signal u(t+1) – 2u(t) + u(t-1) is ____________
a) \(\frac{2 sin⁡ω-2}{jω}\)
b) \(\frac{2 cos⁡ω-2}{jω}\)
c) \(\frac{2 cos⁡ω+2}{jω}\)
d) \(\frac{2 sin⁡ω+2}{jω}\)
View Answer

Answer: b
Explanation: X (jω) = \(\int_{-1}^0 e^{-jωt} \,dt – \int_0^1 e^{-jωt} \,dt\)
= \(\frac{2 cos⁡ω-2}{jω}\).

14. The Inverse Fourier transform of the signal 2πδ(ω) + πδ(ω-4π) + πδ(ω+4π) is ______________
a) 2π(1 – cos 4πt)
b) π (1 – cos 4πt)
c) 1 + cos 4πt
d) 2 π (1 + cos 4πt)
View Answer

Answer: c
Explanation: x (t) = \(\frac{1}{2π} \int_{-∞}^∞\) [2πδ(ω) + πδ(ω-4π) + πδ(ω+4π)] ejωt
= 1 + \(\frac{1}{2} e^{j4πt} + \frac{1}{2} e^{-j4πt}\)
= 1 + cos 4πt

15. Given x (t) = 1, |t|<T1; 0, T1<|t|<\(\frac{T_0}{2}\). The DC component of x(t) is ___________
a) \(\frac{T_1}{T_0}\)
b) \(\frac{2T_1}{T_0}\)
c) \(\frac{T_1}{2T_0}\)
d) \(\frac{T_0}{T_1}\)
View Answer

Answer: b
Explanation: Fourier series the function () can be written as, x(t) = a0 + \(∑_{n=1}^∞ a_n \,cos⁡nωt + b_n \,sin⁡nωt\)
The DC component is given by, a0 = \(\frac{1}{T_0} \int_{T_0}^ x(t) \,dt\)
Or, a0 = \(\frac{1}{T} \int_{-T_0/2}^{T_0/2} \,x(t) \,dt\)
= \(\frac{1}{T_0}\) [0 + 2T1 + 0]
= \(\frac{2T_1}{T_0}\).

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter