I am trying to solve my first problems at cepstrum calculation. I want to calculate the complex cepstrum $\hat{h}[n]$ of a signal $h[n]$ with Z-Transform: $$H(z)=\frac{(1-0.5z^{-1})(1+4z^{-2})}{(1-0.64z^{-2})}$$ I know that $\hat{H}[z]=\log(H(z))$ and $\hat{h}[n]=\frac{1}{2π}\int_{-π}^{π}\log[H(e^{jω})]e^{jωn}dω$

Which is the easiest way to approach this?

I started by applying $\log()$ to H(z) which gave me: $$\log(H(z))=\log(1-0.5z^{-1})+\log(1+4z^{-2})-\log(1-0.64z^{-2})\Rightarrow $$ $$\log(H(z))=\log(1-0.5z^{-1})+\log(4(z^{-2}+\frac{1}{4}))-\log((1-0.8z^{-1})(1+0.8z^{-1}))\Rightarrow$$ $$\log(H(z))=\log(1-0.5z^{-1})+\log(4(z^{-1}+\frac{j}{2})(z^{-1}-\frac{j}{2}))-\log((1-0.8z^{-1})(1+0.8z^{-1}))\Rightarrow$$ $$\log(H(z))=\log(1-0.5z^{-1})+\log(4)+\log(z^{-1}+\frac{j}{2})+\log(z^{-1}-\frac{j}{2}))-\log(1-0.8z^{-1})-\log(1+0.8z^{-1})$$

Is this apporach right? If so, how do I go on? Any help is appreciated. Thanks in advance!

  • $\begingroup$ Is this an exercise from a book? $\endgroup$
    – Matt L.
    Apr 28 '19 at 11:03
  • $\begingroup$ @MattL. No it's not from a book. Is my approach right? I also consider making use of the power series: $$\log(1-az^{-1})=-\sum_{n=1}^{\infty} \frac{a^{n}}{n}z^{-n}$$ but not every term is in such a form. $\endgroup$
    – MJ13
    Apr 28 '19 at 11:23
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    $\begingroup$ I asked because I think that you have to remove a linear phase term in order to compute the complex cepstrum. If it had been from a book I would have been surprised at the necessity to discover that fact. I'll write up an answer as soon as I have the time to do so. $\endgroup$
    – Matt L.
    Apr 28 '19 at 13:42
  • $\begingroup$ I am not used to computing complex cepstrums. I am now trying to solve my first problems. So, I am not sure about my approach. I just tried to use the definition and make use of some properties of the log() function. However, I am finding difficulty calculating the inverse Z -transform of the expression I end up to so that I get the complex cepstrum $\hat{h}[n]$. Thanks a lot for the response. I am looking forward for your answer. $\endgroup$
    – MJ13
    Apr 28 '19 at 13:57

We're looking for a series representation

$$\log H(z)=C(z)=\sum_{n=-\infty}^{\infty}c[n]z^{-n}\tag{1}$$

which converges in an annular region $r_1<|z|<r_2$ with $0<r_1<1<r_2$.

Note that the poles as well as the zeros of $H(z)$ lead to singularities of $C(z)$. All poles and zeros of $H(z)$ inside the unit circle contribute to the right-sided part of $c[n]$, whereas the poles and zeros of $H(z)$ outside the unit circle contribute to the left-sided part of $c[n]$. Consequently, a minimum-phase transfer function $H(z)$ has a right-sided cepstrum.

The given function $H(z)$ has all its poles inside the unit circle ($z^{\infty}_{1,2}=\pm 0.8$, $z^{\infty}_{3}=0$), one of its zeros is inside the unit circle ($z^{0}_{1}=0.5$), and two zeros are outside the unit circle ($z^{0}_{2,3}=\pm 2j$). Consequently, its complex cepstrum $c[n]$ is a two-sided sequence.

In order to find the sequence $c[n]$ we make use of the well-known Mercator series:

$$\log(1-z)=-\sum_{n=1}^{\infty}\frac{z^n}{n},\qquad |z|<1\tag{2}$$

From $(2)$ it is clear that terms of the form

$$\log(1-az^{-1})=-\sum_{n=1}^{\infty}\frac{a^n}{n}z^{-n},\qquad |z|>|a|\tag{3}$$

contribute to the right-sided part of $c[n]$, whereas terms of the form

$$\log(1-bz)=-\sum_{n=1}^{\infty}\frac{b^n}{n}z^{n}=\sum_{n=-\infty}^{-1}\frac{b^{-n}}{n}z^{-n},\qquad |z|<\frac{1}{|b|}\tag{4}$$

contribute to the left-sided part of $c[n]$. Note the argument can be easily generalized to higher powers of $z$ on the left-hand sides of $(3)$ and $(4)$.

The term $(1+4z^{-2})$ in the numerator of the given function $H(z)$ corresponds to two zeros outside the unit circle, and, consequently, it contributes to the left-sided part of $c[n]$. Hence, according to $(4)$, it must be rewritten in terms of positive powers of $z$:


We have


I think that due to the linear-phase term $z^{-2}$ in $(6)$ we actually cannot compute the complex cepstrum of $H(z)$, but we can compute the complex cepstrum of $\tilde{H}(z)$, which is just a shifted version of the original function $H(z)$.

From $(6)$ we obtain

$$\log \tilde{H}(z)=\log(4)+\log(1-0.5z^{-1})+\log(1+0.25z^{2})-\log(1-0.64z^{-2})\tag{7}$$

Using $(3)$ and $(4)$ we get

$$\log \tilde{H}(z)=\log(4)-\sum_{n=1}^{\infty}\frac{(0.5)^n}{n}z^{-n}+\sum_{n=-\infty}^{-1}\frac{(-0.25)^{-n}}{n}z^{-2n}+\sum_{n=1}^{\infty}\frac{(0.64)^n}{n}z^{-2n}\tag{8}$$

which is equivalent to

$$\log \tilde{H}(z)=\log(4)-\sum_{n=1}^{\infty}\frac{(0.5)^n}{n}z^{-n}+\sum_{n=-\infty\\n\textrm{ even}}^{-1}\frac{2(-0.5)^{-n}}{n}z^{-n}+\sum_{n=1\\n\textrm{ even}}^{\infty}\frac{2(0.8)^n}{n}z^{-n}\tag{9}$$

The region of convergence of the series representation $(9)$ is $0.8<|z|<2$, which includes the unit circle, as required by the definition of the complex cepstrum.

From $(9)$ we can directly write down the complex cepstrum of $\tilde{H}(z)$:

$$\tilde{c}[n]=\begin{cases}\log(4),&n=0\\ \frac{1}{n}\left[2(0.8)^n-(0.5)^n\right],&n>0 \land n\textrm{ even}\\-\frac{1}{n}(0.5)^n,&n>0 \land n\textrm{ odd}\\\frac{2}{n}(-0.5)^{-n},&n<0 \land n\textrm{ even}\\0,&n<0\land n \textrm{ odd}\end{cases}\tag{10}$$

  • $\begingroup$ Great explanation!If I wanted to calculate the real cepstrum $c[n]$ as well, should I use the following property? : $$c[n]=\frac{\tilde{c}[n]+\tilde{c}[-n]}{2} $$ $\endgroup$
    – MJ13
    Apr 28 '19 at 15:50
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    $\begingroup$ @MJ13: Yes, that should work. $\endgroup$
    – Matt L.
    Apr 28 '19 at 15:58
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    $\begingroup$ @MJ13: A right-sided sequence starts at some finite index (often $n=0$), and extends to $n\to\infty$. Same (but opposite) for a left-sided sequence. $\endgroup$
    – Matt L.
    Apr 28 '19 at 16:25
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    $\begingroup$ @MJ13: Yes, that's an option. $H_1(z)$ has all its poles and zeros inside the unit circle, so it's indeed minimum-phase. And the other filter is FIR but not (generalized) linear phase. $\endgroup$
    – Matt L.
    May 3 '19 at 15:25
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    $\begingroup$ @MJ13: Yes, thanks. Fixed. $\endgroup$
    – Matt L.
    May 9 '19 at 9:22

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