Why a filter of the form

$$ 1+a_0 z^{-1} +a_1 z^{-2}+ a_2 z^{-3} $$ has the same amplitude response when in the reversed form $$ a_2+a_1 z^{-1} +a_0 z^{-2}+z^{-3} $$ but the phase response is different. I don't get it, and can't understand what the differences in terms of the relative group delays mean.


1 Answer 1


If you have a causal length $N$ FIR filter with impulse response $h[n]$, and with frequency response $H(e^{j\omega})$, and you invert it on the time axis, you get a new filter


This filter is non-causal, but since it is an FIR filter, it can be made causal by shifting it to the left by $N-1$ samples:


This is what you're doing. Now since


you have




From (3) you can see that for real-valued $h[n]$ you have $H(e^{-j\omega})=H^*(e^{j\omega})$ (where $*$ denotes complex conjugation). So you finally get


and since $|e^{-j(N-1)\omega}|=1$, and $|H^*(e^{j\omega})|=|H(e^{j\omega})|$ you have


So the magnitude responses of the original filter and the time-reversed filter are identical. The phase responses are different: from (4) you have


from which


follows. Since the group delay is the negative derivative of the phase, the group delays are related by



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