# Solve both a)using the integral definition , find the convolution f*g text{ of } f(t)=cos 2t , g(t)=e^t b) Using above answer , find the Laplace Transform of f*g

Solve both
a)using the integral definition , find the convolution

b) Using above answer , find the Laplace Transform of f*g
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d2saint0
Step 1
Given:
The functions $f\left(t\right)=\mathrm{cos}2t,g\left(t\right)={e}^{t}$
Step 2
a) Definition of convolution:
The convolution of piecewise continuous functions f, g : $\mathbb{R}\to \mathbb{R}$ is the function f * g : $\mathbb{R}\to \mathbb{R}$ given by:
$\left(f\ast g\right)\left(t\right)={\int }_{0}^{t}f\left(\tau \right)g\left(t-\tau \right)d\tau$
Therefore, by definition
$\left(f\ast g\right)\left(t\right)={\int }_{0}^{t}\mathrm{cos}\left(2\tau \right){e}^{t-\tau }d\tau$ Integrate using integration by parts:
$u=\mathrm{cos}\left(2\tau \right),dv={e}^{\left(}t-\tau \right)d\tau$
$du=-2\mathrm{sin}\left(2\tau \right)d\tau ,v=\int {e}^{t-\tau }d\tau =-{e}^{t-\tau }\int udv=uv-\int vdu$
$⇒I=\int \mathrm{cos}\left(2\tau \right){e}^{t-\tau }d\tau =-cos\left(2\tau \right){e}^{t-\tau }-\int \left(-{e}^{t-\tau }\right)\left(-2\mathrm{sin}\left(2\tau \right)d\tau \right)$
$⇒I=-\mathrm{cos}\left(2\tau \right){e}^{t-\tau }-2\int \left({e}^{t-\tau }\right)\left(\mathrm{sin}\left(2\tau \right)\right)d\tau \dots \left(1\right)$
Step 3
To simplify further, use parts of integration for the second term Compute the integral
$\int {e}^{t-\tau }\mathrm{sin}\left(2\tau \right)d\tau$
$u=\mathrm{sin}\left(2\tau \right),dv={e}^{t-\tau }d\tau$
$du=2\mathrm{cos}\left(2\tau \right),v=-{e}^{t-\tau }$
$⇒\int {e}^{t-\tau }\mathrm{sin}\left(2\tau \right)d\tau =-{e}^{t-\tau }\mathrm{sin}\left(2\tau \right)-\int \left(-{e}^{t-\tau }\right)\left(2\mathrm{cos}\left(2\tau \right)\right)d\tau$
$=-{e}^{t-\tau }\mathrm{sin}\left(2\tau \right)+2\int \left({e}^{t-\tau }\right)\left(\mathrm{cos}\left(2\tau \right)\right)d\tau$
$⇒\int {e}^{t-\tau }\mathrm{sin}\left(2\tau \right)d\tau =-{e}^{t-\tau }\mathrm{sin}\left(2\tau \right)+2I$

Equation (1) becomes:
$⇒I=-\mathrm{cos}\left(2\tau \right){e}^{t-\tau }-2\left[-{e}^{t-\tau }\mathrm{sin}\left(2\tau \right)+2I\right]$
$⇒I=-\mathrm{cos}\left(2\tau \right){e}^{t-\tau }+2{e}^{t-\tau }\mathrm{sin}\left(2\tau \right)-4I$
$⇒5I=-\mathrm{cos}\left(2\tau \right){e}^{t-\tau }+2{e}^{t-\tau }\mathrm{sin}\left(2\tau \right)$
$⇒I=-\frac{1}{5}\mathrm{cos}\left(2\tau \right){e}^{t-\tau }+\frac{2}{5}{e}^{t-\tau }\mathrm{sin}\left(2\tau \right)$
$\int \mathrm{cos}\left(2\tau \right){e}^{t-\tau }d\tau =-\frac{1}{5}\mathrm{cos}\left(2\tau \right){e}^{t-\tau }+\frac{2}{5}{e}^{t-\tau }\mathrm{sin}\left(2\tau \right)$
Step 4
Substitute the upper and lower limit and simplify