Given
$$H(z) = \frac{z^{-1}-0.8}{1-0.8z^{-1}}\text,$$
does the filter stay an All-Pass filter after quantization of coefficients?
If so, how can I prove it mathematically?
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$\begingroup$ I think this one is pretty easy: What's the requirement for all-passyness? $\endgroup$– Marcus MüllerJan 29, 2019 at 16:46
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$\begingroup$ That there is an equal number of zeros and poles and that each zero will be equal to its inverse complex conjugate. The thing is, will the single pole and single zero remain the same after quantization? $\endgroup$– Nadav TalmonJan 29, 2019 at 18:18
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$\begingroup$ Try it: quantize the coefficients to a, say, 3 bits, calculate pole and zero and see of they still match. $\endgroup$– HilmarJan 29, 2019 at 18:40
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1 Answer
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Hint:
- Write the transfer with a general coefficient $a$ (instead of the value $0.8$) in the numerator as well as in the denominator.
- Compute the magnitude of the numerator for $z=e^{j\omega}$.
- Compute the magnitude of the denominator for $z=e^{j\omega}$.
- Show that both magnitudes are equal, regardless of the value of the coefficient $a$, hence $H(e^{j\omega})=1$, i.e., the system is an allpass, even after coefficient quantization.
