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| − | <center> | + | [[Category:ECE301Spring2013JVK]] [[Category:ECE]] [[Category:ECE301]] [[Category:probability]] [[Category:problem solving]] [[Category:convolution]] |
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| + | ==<center>Convolution</center>== | ||
| + | <center>[[Green26|alec green]]</center> | ||
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: | 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: | ||
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<math>\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</math> | <math>\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</math> | ||
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| + | [[Bonus_point_1_ECE301_Spring2013|Back to first bonus point opportunity, ECE301 Spring 2013]] | ||
Revision as of 19:55, 10 February 2013
Convolution
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. $
$ \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 $
