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This might be a weird question but here's the setup: I have a few biquads that filter a signal $x[n]$ and output the filtered signal $y[n]$. I can calculate the frequency response of those biquads with the z-transform. Now I mix $x[n]$ and the filtered signal $y[n]$ such that the mixed signal $y'[n]= a \cdot x[n] + (1 - a) \cdot y[n]$ where $0\leq a\leq 1$. How would I go about calculating the frequency response of $y'[n]$?

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The Z transform is linear. So if $$ Y(z) = H(z) \cdot X(z) $$ then $$ Y^{'}(z) = a \cdot X(z) + (1-a) \cdot Y(z) = a \cdot X(z) + (1-a) \cdot H(z) \cdot X(z) = (a + (1-a) \cdot H(z)) \cdot X(z) $$

Hence $$ H'(z) = a + (1-a) \cdot H(z) $$

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  • $\begingroup$ Thanks! So I just need to combine the real parts and do the magnitude response, right? $\endgroup$
    – Agiltohr
    May 23 '20 at 1:32
  • $\begingroup$ For arguments sake, assume that H(z) is a 10 sample delay. Then mixing (1-a) H(z) with a «bypass chain» of a is not going to produce a «watered out» frequency response, but a comb filter. If H(z) is a linear phase FIR filter and you put a pure delay in the bypass chain, you can get a more predictable blend between full on H(z) magnitude and no filtering, controlled by a. $\endgroup$
    – Knut Inge
    May 23 '20 at 6:35

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