Signals & Systems Questions and Answers – Properties of the Laplace Transform

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This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Properties of the Laplace Transform”.

1. Find the Laplace transform of x(t) = u(t+2) + u(t-2).
a) \(\frac{cos⁡2s}{s}\)
b) \(\frac{cosh⁡2s}{s}\)
c) \(\frac{sinh⁡2s}{s}\)
d) \(\frac{sin⁡2s}{s}\)
View Answer

Answer: b
Explanation: Given x(t) = u(t+2) + u(t-2)
We know that the Laplace transform u(t) ↔ \(\frac{1}{s}\)
Time shifting property states that L{x(t±t0)} = X(s)e±st0
L{u(t-2)}=\(e^{\frac{-2s}{s}}\); L{u(t+2)}=\(e^{\frac{2s}{s}}\)
∴X(s) = L{u(t+2)+u(t-2)} = \(\frac{e^{-2s}+e^{-2s}}{s} = \frac{cosh⁡2s}{s}\).

2. Find the Laplace transform of the signal x(t) = e-2t cos⁡(200πt)u(t).
a) \(\frac{s}{s^2+(200π)^2}\)
b) \(\frac{s}{s^2-(200π)^2}\)
c) \(\frac{s-2}{(s-2)^2+(200π)^2}\)
d) \(\frac{s+2}{(s+2)^2+(200π)^2}\)
View Answer

Answer: d
Explanation: Given x(t) = e-2t cos⁡(200πt)u(t)
We know that L{cos⁡ωt u(t)} = \(\frac{s}{s^2+ω^2}\)
∴L{cos⁡(200πt)u(t)} = \(\frac{s}{s^2+(200π)^2}\)
Frequency shifting property states that L{e-at x(t)} = X(s+a)
L{e-2t cos⁡(200πt)u(t)} = L{cos⁡(200πt)u(t)}|s=s+2 = \(\Big[\frac{s}{s^2+(200π)^2}\Big]_{s=s+2} = \frac{s+2}{(s+2)^2+(200π)^2}\).

3. Find the Laplace transform of the signal x(t) = sin⁡(\(\frac{t}{2}\))u(\(\frac{t}{2}\)).
a) \(\frac{1}{s^2+1}\)
b) \(\frac{s}{s^2+1}\)
c) \(\frac{2s}{(2s)^2+1}\)
d) \(\frac{2}{(2s)^2+1}\)
View Answer

Answer: d
Explanation: We know that sin⁡t u(t) ↔ \(\frac{1}{s^2+1}\)
Scaling property states that f(at) ↔ \(\frac{1}{a} F(\frac{s}{a})\)
\(sin⁡(\frac{t}{2})u(\frac{t}{2}) \leftrightarrow \frac{1}{(\frac{1}{2})} \frac{1}{\Big[(\frac{s}{1/2})^2+1\Big]} \leftrightarrow \frac{2}{(2s)^2+1}\).
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4. Find the Laplace transform of the signal x(t) = \(\frac{dδ(t)}{dt}\).
a) 1
b) s
c) \(\frac{1}{s}\)
d) s2
View Answer

Answer: b
Explanation: Given x(t) = \(\frac{dδ(t)}{dt}\)
We know that L{δ(t)} = 1
Time differentiation property, L{\(\frac{dδ(t)}{dt}\)} = sF(s)
L{\(\frac{dδ(t)}{dt}\)} = sL{δ(t)} = s × 1 = s.

5. Find the Laplace transform of the signal x(t) = te-αt.
a) \(\frac{1}{s^2}\)
b) \(\frac{1}{(s+α)^2}\)
c) \(\frac{1}{α}\)
d) \(\frac{1}{s+α}\)
View Answer

Answer: b
Explanation: We know that L{e-αt} = \(\frac{1}{s+α}\)
Differentiation in s-domain property states that (-t)n f(t) ↔ \(\frac{d^n F(s)}{ds^n}\)
L{te-αt} = –\(\frac{d}{ds} [\frac{1}{s+α}] = \frac{1}{(s+α)^2}\).

6. Find the Laplace transform for f(t) = \(\frac{1}{t}\) [e-2t – e-3t]u(t).
a) ln\(\left(\frac{s-2}{s-3}\right)\)
b) ln\(\left(\frac{s+2}{s+3}\right)\)
c) ln\(\left(\frac{s-2}{s+3}\right)\)
d) ln\(\left(\frac{s+2}{s-3}\right)\)
View Answer

Answer: b
Explanation: Given f(t) = \(\frac{1}{t}\) [e-2t – e-3t]u(t)
We know that L{e-2t) u(t)} = \(\frac{1}{s+2}\); L{-3t u(t)} = \(\frac{1}{s+3}\)
Integration in s-domain property states that \(\int_s^∞ F(s)ds \leftrightarrow \frac{f(t)}{t}\)
L{\(\frac{1}{t}\) [e-2t – e-3)]u(t)} = \(\int_s^∞ \left(\frac{1}{s+2} – \frac{1}{s+3}\right)ds = [ln⁡(s+2) – ln⁡(s+3)]|_s^∞ = ln\left(\frac{s+2}{s+3}\right)\).

7. Find the initial value of f(t) if F(s) = \(\frac{s}{(s+a)^2+ω^2}\).
a) 0
b) -1
c) ∞
d) 1
View Answer

Answer: d
Explanation: Given F(s) = \(\frac{s}{(s+a)^2+ω^2}\)
The initial value is f(0+) = lims→∞ sF(s)
= lims→∞⁡ s \(\frac{s}{(s+a)^2+ω^2} = lim_{s→∞} \frac{1}{(1+a/s)^2+(ω/s)^2} = 1\).
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8. Find the final value of the function F(s) given by \(\frac{(s-1)}{s(s^2-1)}\).
a) 1
b) 0
c) -1
d) ∞
View Answer

Answer: a
Explanation: Given F(s) = \(\frac{(s-1)}{s(s^2-1)}\)
The final value is f(∞)=lims→0⁡ sF(s)
= lims→0⁡ s \(\frac{(s-1)}{s(s^2-1)} = lim_{s→0} \frac{1}{s+1} = 1\).

9. Determine the initial value x(0+) for the Laplace transform X(s) = \(\frac{4}{s^2+3s-5}\).
a) -1
b) 0
c) 1
d) ∞
View Answer

Answer: b
Explanation: Given X(s) = \(\frac{4}{s^2+3s-5}\)
Initial value, x(0+) = lims→∞⁡ sX(s) = lims→∞⁡ s(\(\frac{4}{s^2+3s-5}\)) = limx→0⁡ \(\frac{4x}{1+3x-5x^2} = 0\) [let s = 1/x].
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10. Find x(∞) if X(s) is given by \(\frac{s-2}{s(s+4)}\).
a) 1
b) -1
c) \(\frac{1}{2}\)
d) –\(\frac{1}{2}\)
View Answer

Answer: d
Explanation: Given X(s) = \(\frac{s-2}{s(s+4)}\)
Final value, x(∞) = lims→0 sX(s) = lims→0 \(\frac{(s-2)}{s(s+4)} = -\frac{2}{4} = -\frac{1}{2}\).

Sanfoundry Global Education & Learning Series – Signals & Systems.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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