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Consider a filter with real-valued impulse response $h[n]$. The filter is cascaded with another filter whose impulse response is $h'[n] = h[-n]$, i.e. whose impulse response is the time-reversed version of $h[n]$.

$h[n]$ cascaded with $h[-n]$

i think we have to perform convolution

if so what is phase of new cascaded impulse responseenter image description here

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HINT: See what happens in the frequency domain. Convolution corresponds to multiplication. You just have to figure out the Fourier transform of $h[-n]$, expressed in terms of the Fourier transform of $h[n]$. Multiplying the two functions results in the Fourier transform of the convolution of $h[n]$ with $h[-n]$. The result has a special property which makes it very easy to determine its phase.

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  • $\begingroup$ But we don't know waveform of h(n) . We can't get H(ejw) $\endgroup$ – Rohith Pandith May 21 at 2:46
  • $\begingroup$ I have uploaded question please have look ,help me answer it $\endgroup$ – Rohith Pandith May 21 at 2:57
  • $\begingroup$ @RohithPandith: I didn't tell you to compute $H(e^{j\omega})$. Instead you should express the FT of $h[-n]$ in terms of $H(e^{j\omega})$. $\endgroup$ – Matt L. May 21 at 6:55
  • $\begingroup$ i know fft of h(-n) is having same magnitude that of fft of h(n) so fft is complex conjugate of fft of h(n) so H(e-jw) will be fft of h(-n) $\endgroup$ – Rohith Pandith May 22 at 9:33
  • $\begingroup$ so is it autocorrelation???? $\endgroup$ – Rohith Pandith May 22 at 10:11
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filter is cascaded with another filter we have to perform convolution

so we have to perform convolution between signal with time reversed version of itself so it is nothing but auto-correlation.

so phase is zero

so we do not need any calculation

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