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

Step 1
Given:
The functions $f(t)=\cos 2t,g(t)=e^t$
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:
$(f\ast g)(t)={\int }_0^{t}f(\tau )g(t-\tau )d\tau$
Therefore, by definition
$(f\ast g)(t)={\int }_0^t\cos (2\tau )e^{t-\tau }d\tau$ Integrate using integration by parts:
$u=\cos (2\tau ),dv=e^{(}t-\tau )d\tau$
$du=-2\sin (2\tau )d\tau ,v=\int e^{t-\tau }d\tau =-e^{t-\tau }\int udv=uv-\int vdu$
$\Rightarrow I=\int \cos (2\tau )e^{t-\tau }d\tau =-cos(2\tau )e^{t-\tau }-\int (-e^{t-\tau })(-2\sin (2\tau )d\tau )$
$\Rightarrow I=-\cos (2\tau )e^{t-\tau }-2\int (e^{t-\tau })(\sin (2\tau ))d\tau \dots (1)$
Step 3
To simplify further, use parts of integration for the second term Compute the integral
$\int e^{t-\tau }\sin (2\tau )d\tau$
$u=\sin (2\tau ),dv=e^{t-\tau }d\tau$
$du=2\cos (2\tau ),v=-e^{t-\tau }$
$\Rightarrow \int e^{t-\tau }\sin (2\tau )d\tau =-e^{t-\tau }\sin (2\tau )-\int (-e^{t-\tau })(2\cos (2\tau ))d\tau$
$=-e^{t-\tau }\sin (2\tau )+2\int (e^{t-\tau })(\cos (2\tau ))d\tau$
$\Rightarrow \int e^{t-\tau }\sin (2\tau )d\tau =-e^{t-\tau }\sin (2\tau )+2I$
$\{\text{Because }I=\int (e^{t-\tau })(\cos (2\tau ))d\tau \}$
Equation (1) becomes:
$\Rightarrow I=-\cos (2\tau )e^{t-\tau }-2[-e^{t-\tau }\sin (2\tau )+2I]$
$\Rightarrow I=-\cos (2\tau )e^{t-\tau }+2e^{t-\tau }\sin (2\tau )-4I$
$\Rightarrow 5I=-\cos (2\tau )e^{t-\tau }+2e^{t-\tau }\sin (2\tau )$
$\Rightarrow I=-\frac{1}{5}\cos (2\tau )e^{t-\tau }+\frac{2}{5}{e}^{t-\tau }\sin (2\tau )$
$\int \cos (2\tau )e^{t-\tau }d\tau =-\frac{1}{5}\cos (2\tau )e^{t-\tau }+\frac{2}{5}{e}^{t-\tau }\sin (2\tau )$
Step 4
Substitute the upper and lower limit and simplify

Not exactly what you’re looking for?

Ask My Question

This is helpful 28