Laplace Transform Questions and Answers – Convolution

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This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Convolution”.

1. Find the \(L^{-1} (\frac{1}{s(s^2+4)})\).
a) \(\frac{1-sin⁡(t)}{4}\)
b) \(\frac{1-cos⁡(t)}{4}\)
c) \(\frac{1-sin⁡(2t)}{4}\)
d) \(\frac{1-cos⁡(2t)}{4}\)
View Answer

Answer: d
Explanation: In the given question
Let p1(s) = \(\frac{1}{s^2+4}\) and p2(s) = \(\frac{1}{s}\) \(f_1 (t) = L^{-1} (\frac{1}{s^2+4}) = \frac{sin⁡(2t)}{2}\)
\(f_2 (t) = L^{-1} (\frac{1}{s}) = 1\)
By Convolution Theorem,
\(L^{-1} (p_1(s)×p_2(s)) = \int_{0}^{t}f_1(u) f_2(t-u) dt\)
\(L^{-1} (\frac{1}{s(s^2+4)}) = \int_{0}^{t}\frac{1}{2} sin⁡(2u) du\)
=\(\frac{1-cos⁡(2t)}{4}\)
Thus, the answer is \(\frac{1-cos⁡(2t)}{4}\).
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2. Find the \(L^{-1} (\frac{1}{s(s+4)^\frac{1}{2}})\), give the answer in terms of error function.
a) \(\frac{1}{2} erf⁡(2t)\)
b) \(\frac{1}{2} erf⁡(\sqrt{t})\)
c) \(\frac{1}{2} erf⁡(2\sqrt{t})\)
d) \(\frac{1}{2} erf⁡(4\sqrt{t})\)
View Answer

Answer: c
Explanation: In the given question,
Let \(p_1(s) = \frac{1}{(s+4)^{\frac{1}{2}}}\) and \(p_2(s) = \frac{1}{s}\)
\(f_1 (t) = L^{-1} (\frac{1}{(s+4)^{\frac{1}{2}}}) = \frac{e^{-4t}×\sqrt{t}}{\sqrt\pi}\)
\(f_2 (t) = L^{-1} \frac{1}{s} = 1\)
By Convolution Theorem,
\(L^{-1} (p_1(s)×p_2(s)) = \int_{0}^{t}f_1(u) f_2(t-u) dt\)
\(L^{-1} (\frac{1}{s(s+4)^{\frac{1}{2}}}) = \int_{0}^{t}\frac{\sqrt{u}}{\sqrt\pi} e^{-4u} \,du\)
\(=\frac{1}{\sqrt\pi} \int_{0}^{t}\sqrt{u} e^{-4u} du\)
We know that error function is given by –
\(erf⁡(x)=\frac{2}{\sqrt\pi} \int_{0}^{x}e^{-z^2} dz\)
Applying, 4u=z2 And setting the limits of u in terms of z, we get
\(L^{-1} (\frac{1}{s(s+4)^{\frac{1}{2}}}) = \frac{1}{2} erf⁡(2\sqrt{t})\)
Thus, the answer is \(\frac{1}{2} erf⁡(2\sqrt{t})\).

3. Find the \(L^{-1} \frac{s}{(s^2+4)^2}\).
a) \(\frac{1}{4} tcos(2t)\)
b) \(\frac{1}{4} tsin(t)\)
c) \(\frac{1}{4} tsin(2t)\)
d) \(\frac{1}{2} tsin(2t)\)
View Answer

Answer: c
Explanation: In the given question,
Let \(p_1(s) = \frac{1}{s^2+4}\) and \(p_2(s) = \frac{1}{s}\)
\(f_1 (t) = L^{-1} (\frac{1}{s^2+4}) = \frac{sin⁡(2t)}{2}\)
\(f_2 (t) = L^{-1} (\frac{s}{s^2+4}) = cos⁡(2t)\)
By Convolution Theorem,
\(L^{-1} (p_1(s)×p_2(s)) = \int_{0}^{t} f_1(u) f_2(t-u) dt\)
\(L^{-1} \left (\frac{s}{(s^2+4^2)^2}\right )=\int_{0}^{t} sin⁡(2u)×\frac{1}{2}×cos⁡(2(t-u))du\)
=\(\frac{1}{4}[tsin(2t)-\frac{cos⁡(2t)}{4}+\frac{cos⁡(2t)}{4}]\)
=\(\frac{1}{4} tsin(2t)\)
Thus, the correct answer is \(\frac{1}{4} tsin(2t)\)
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Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.

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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