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Daniele Tampieri
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The derivative of a filter with respect to a output singalsignal

Post Closed as "Not suitable for this site" by Alexandre Eremenko, kodlu, Dmitri Pavlov, Eric Peterson, Adam P. Goucher

TwoI have two signals:, $d(t)$ and $p(t)$, respectively the input and the output of the matching filter $w(t)$, they have $d(t)*w(t)=p(t)$ (i.e. $$ d(t)*w(t)=p(t) $$ where $*$ denotes convolution),.The impulse response $w(t)$ may be calculated inby going into the frequency domain: $w(t)=F^{-1}\left[\frac{F[p(t)]\overline{F[d(t)]}}{F[d(t)]\overline{F[d(t)]}+\epsilon}\right]$ $$ w(t)=F^{-1}\left[\frac{F[p(t)]\overline{F[d(t)]}}{F[d(t)]\overline{F[d(t)]}+\epsilon}\right] $$ How can I get the derivative of the the filter $w(t)$ with respect to $p(t)$ :, i.e. $$\frac{\partial{w}}{\partial{p}}=?$$

Two signals: $d(t)$ and $p(t)$, the matching filter $w(t)$, they have $d(t)*w(t)=p(t)$ ( $*$ denotes convolution), $w(t)$ may be calculated in frequency domain: $w(t)=F^{-1}\left[\frac{F[p(t)]\overline{F[d(t)]}}{F[d(t)]\overline{F[d(t)]}+\epsilon}\right]$ How can I get the derivative of the the filter $w(t)$ with respect to $p(t)$ : $$\frac{\partial{w}}{\partial{p}}=?$$

I have two signals, $d(t)$ and $p(t)$, respectively the input and the output of the matching filter $w(t)$, i.e. $$ d(t)*w(t)=p(t) $$ where $*$ denotes convolution.The impulse response $w(t)$ may be calculated by going into the frequency domain: $$ w(t)=F^{-1}\left[\frac{F[p(t)]\overline{F[d(t)]}}{F[d(t)]\overline{F[d(t)]}+\epsilon}\right] $$ How can I get the derivative of the the filter $w(t)$ with respect to $p(t)$, i.e. $$\frac{\partial{w}}{\partial{p}}=?$$

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The derivative of the a filter with respect to a output singal

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