10/08 - Mechanics of Convolution & Fourier Transform

Mechanics of the Convolution

Remember from the game:

Input	LTI System	Output	Reason
$\delta (t)\,\!$	$\Longrightarrow$	$h(t)\,\!$	Given
$\delta (t-\lambda )\,\!$	$\Longrightarrow$	$h(t-\lambda )\,\!$	Time Invarience
$x(\lambda )\delta (t-\lambda )\,\!$	$\Longrightarrow$	$x(\lambda )h(t-\lambda )\,\!$	Proportionality
$x(t)=\int _{-\infty }^{\infty }x(\lambda )\delta (t-\lambda )\,dx$	$\Longrightarrow$	$\underbrace {\int _{-\infty }^{\infty }x(\lambda )h(t-\lambda )\,dx} _{ConvolutionIntegral}$	Superposition

We will also denote the convolution as $x(t)*h(t)\equiv \int _{-\infty }^{\infty }x(\lambda )h(t-\lambda )\,dx$

Communative Property

$x(t)*h(t)\,\!$	$=\int _{-\infty }^{\infty }x(\lambda )h(t-\lambda )\,d\lambda$	Let $t-\lambda =u\,\!$ thus $du=-d\lambda \,\!$
	$=-\int _{\infty }^{-\infty }x(t-u)h(u)\,du$	The order of integration switched due to changing from $-\lambda =u\,\!$
	$=\int _{-\infty }^{\infty }h(u)x(t-u)\,du$
	$=h(t)*x(t)\,\!$

Example 1

$\delta (t)*x(t)\,\!$	$=\int _{-\infty }^{\infty }\delta (\lambda )x(t-\lambda )\,d\lambda$
	$=x(t)\int _{-\infty }^{\infty }\,d\lambda$
	$=x(t)\,\!$

Example 2

$\overbrace {[u(t)-u(t-1)]} ^{x(t)}*\overbrace {[u(t-1)-u(t-3)]} ^{h(t)}$	$=\int _{-\infty }^{\infty }[u(\lambda )-u(\lambda -1)]\cdot [u(t-\lambda -1)-u(t-\lambda -3)]\,d\lambda$
	$=$

• How can we do this without using the flip shift multiply integrate? If $u(t)\,\!$ isn't involved, then you can plug n chug with the integral. The u(t) will change the limits, which can be impractical to evaulate if you have more than 2.