10/08 - Mechanics of Convolution & Fourier Transform: Difference between revisions

Revision as of 00:26, 14 November 2008

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 $x(t)*h(t)\equiv \int _{-\infty }^{\infty }x(\lambda )h(t-\lambda )\,dx$

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

@@ Line 41: / Line 41: @@
 |<math>=-\int_{\infty}^{-\infty} x(t-u)h(u)\, du</math>
 |The order of integration switched due to changing from <math>-\lambda = u\,\!</math>
+|-
+|
+|<math>=\int_{-\infty}^{\infty} h(u)x(t-u)\, du</math>
+|
+|-
+|
+|<math>=h(t)*x(t) \,\!</math>
+|
 |}