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I know that, given the magnitude response $|H(e^{j \omega})|$ of a filter $H(z)$, it's minimum-phase response is given by $$ \phi(\omega) = -\mathscr{H}\Big\{ \log(|H(e^{j \omega})|) \Big\} \ . $$

I also know that this is related to the fact that

$$ H(e^{j \omega}) = \big| H(e^{j \omega}) \big| \, e^{j\phi(\omega)} $$

$$ \log(H(e^{j \omega})) = \log(|H(e^{j \omega})|) + j \phi(\omega), $$ i.e. the frequency domain representation $\log(H(\omega))$ of the complex cepstrum has $\log(|H(e^{j \omega})|)$ as it's real and $\phi(\omega)$ as it's imaginary part.

Now, if the time-domain representation of the cepstrum $\text{ifft}(\log(H(e^{j \omega})))$ is causal, its real and imaginary part need to be related by the Hilbert transform, as is the case for all causal signals, thus leading to the first equation.

However I'm not sure, why we know that the cepstrum is causal only for minimum phase filters. Furthermore, the fact that the phase of a filter is just given by the 90 degree phase shifted log spectrum seems utterly bizarre to me.

I know intuitive explanations are hard to come by for something like this, but maybe someone has an idea?

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Let's start with a causal impulse response $h[n]$. We know that the $\mathcal{Z}$-transform of a causal and stable sequence converges outside (and on) the unit circle, i.e., all singularities (poles in case of rational transfer functions) are inside the unit circle. The argument goes both ways, so if we are given a transfer function $H(z)$ with no poles outside the unit circle, we know that it corresponds to a causal (and stable) sequence:

$$h[n]\textrm{ causal and stable}\Longleftrightarrow H(z)\textrm{ has no poles on and outside the unit circle}$$

If we take the logarithm of $H(z)$

$$C(z)=\log H(z)$$

the function $C(z)$ generally has singularities outside the unit circle, because not only the poles but also the zeros of $H(z)$ cause singularities of $C(z)$. Only if $H(z)$ has neither poles nor zeros on and outside the unit circle will $C(z)$ have no singularities on and outside the unit circle. In that case, we have the same correspondence as above: if $C(z)$ has no singularities on and outside the unit circle, then its inverse transform $c[n]$ is causal (and stable). Of course, the sequence $c[n]$ is the cepstrum.

In sum, just as is the case with ordinary transfer functions, if $C(z)$ has no singularities outside the unit circle, its inverse transform (the cepstrum) $c[n]$ must be causal and stable. This is the case if the corresponding transfer function $H(z)$ has neither poles nor zeros outside the unit circle, i.e., if the system is minimum-phase.

Causality of a sequence implies Hilbert transform relations between the real and imaginary parts of its transform on the unit circle. Hence, for minimum-phase systems, we have two Hilbert transform relationships: first, since minimum-phase systems are causal, the real and imaginary parts of $H(e^{j\omega})$ are related by the Hilbert transform. Second, since also the cepstrum is causal, the real and imaginary parts of $\log H(e^{j\omega})$ are related by the Hilbert transform. The latter two are the log-magnitude and the phase. Consequently, one can be computed from the other, if we know that the system is minimum-phase.

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  • $\begingroup$ That's a great answer, thank you very much! $\endgroup$
    – herrzinter
    Oct 26, 2023 at 16:19

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