# 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) $$\frac{1-cos⁡(2t)}{4}$$

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 $$\frac{1-cos⁡(2t)}{4}$$.

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) $$\frac{1}{2} erf⁡(4\sqrt{t})$$

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) $$\frac{1}{2} tsin(2t)$$

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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