Revision as of 20:53, 10 February 2013 by Green26 (Talk | contribs)

==Convolution==
alec green

Convolution is often presented in a manner that emphasizes efficient calculation over comprehension of the convolution itself. To calculate in a pointwise fashion, we're told: "flip one of the input signals, and perform shift+multiply+add operations until the signals no longer overlap." This is numerically valid, but you could in fact calculate the convolution without flipping either signals (we'll do that here). Consider the convolution of the following constant input and causal impulse reponse:

$ x[n] \ast h[n] $

$ x[n] \,=\, 1 \;\;\;\;\; \forall \, n $

$ h[n] = \left\{ \begin{array}{lr} \mathrm{e}^{-n} & : n \geq 0\\ 0 & : n < 0 \end{array} \right. $


Ec1 2.PNGEc1 1.PNGEc1 3.PNG



$ \int_{-\infty}^{\infty} h(\tau)\,\mathrm{d}\tau \,=\, \int_{0}^{\infty} h(\tau)\,\mathrm{d}\tau \;\;\;\;\; \because h(t)=0 \;\;\; \forall \, t<0 $

$ \Rightarrow \int_{0}^{\infty} \mathrm{e}^{-\tau}\,\mathrm{d}\tau \,=\, \left.-\mathrm{e}^{-\tau}\right|_{0}^{\infty} \,=\, -(\mathrm{e}^{-\infty} - \mathrm{e}^{0}) \,=\, -(0 - 1) \,=\, 1 $


$ \int_{-\infty}^{\infty} h(t-\tau)\,\mathrm{d}\tau \,=\, \int_{-\infty}^{t} h(t-\tau)\,\mathrm{d}\tau \;\;\;\;\; \because h(t)=0 \;\;\; \forall \, t<0 $

$ \Rightarrow \int_{-\infty}^{t} \mathrm{e}^{-(t-\tau)}\,\mathrm{d}\tau \,=\, \int_{-\infty}^{t} \mathrm{e}^{\tau-t}\,\mathrm{d}\tau \,=\, \left.\mathrm{e}^{\tau-t}\right|_{-\infty}^{t} \,=\, \mathrm{e}^{0} - \mathrm{e}^{-\infty-t} \;\; (\forall \, t>0) \,=\, 1 - 0 \,=\, 1 $

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Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin