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Poles and Zeros of the Cepstrum

The complex cepstrum of a sequence $ h(n)$ is typically defined as the inverse Fourier transform of its log spectrum [60]

$\displaystyle {\tilde h}(n)\isdef \frac{1}{2\pi}\int_{-\pi}^\pi \ln[H(\ejo )] e^{j\omega n}d\omega,

where $ \ln(x)$ denotes the natural logarithm (base $ e$ ) of $ x$ , and $ H(\ejo )$ denotes the Fourier transform (DTFT) of $ h(n)$ ,

$\displaystyle H(\ejo )\isdef \sum_{n=-\infty}^\infty h(n)e^{-j\omega n}.

(The real cepstrum, in contrast, is the inverse Fourier transform of the log-magnitude spectrum.) An equivalent definition (when the DTFT exists) is to define the complex cepstrum of $ h$ as the inverse z transform of $ \ln H(z)$ . The cepstrum has numerous applications in digital signal processing including speech modeling [60] and pitch detection [34]. In §11.7, we use the cepstrum to compute minimum-phase spectra corresponding to a given spectral magnitude--an important tool in digital filter design.

From Eq.(8.2), the log z transform can be written in terms of the factored form as

$\displaystyle \ln H(z)$ $\displaystyle =$ $\displaystyle \ln(g)$  
    $\displaystyle + \ln(1-q_1z^{-1}) + \ln(1-q_2z^{-1}) + \cdots + \ln (1-q_Mz^{-1})$  
    $\displaystyle - \ln(1-p_1z^{-1}) - \ln(1-p_2z^{-1}) - \cdots - \ln(1-p_Nz^{-1})$  
  $\displaystyle =$ $\displaystyle \ln(g) + \sum_{m=1}^M\ln(1-q_mz^{-1}) - \sum_{n=1}^N\ln(1-p_nz^{-1})$  
  $\displaystyle =$ $\displaystyle \ln(g) - \sum_{m=1}^M\ln\left(\frac{1}{1-q_mz^{-1}}\right) + \sum_{n=1}^N\ln\left(\frac{1}{1-p_nz^{-1}}\right),\qquad{} % force eqn no. to next line
\protect$ (9.9)

where $ q_m$ denotes the $ m$ th zero and $ p_n$ denotes the $ n$ th pole of the z transform $ H(z)$ . Applying the Maclaurin series expansion

$\displaystyle \ln\left(\frac{1}{1-x}\right) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^k}{k} + \cdots,\qquad \left\vert x\right\vert<1,

we have that each pole and zero term in Eq.(8.9) may be expanded as

\ln\left(\frac{1}{1-q_iz^{-1}}\right) &=& \sum_{n=1}^\infty \frac{q_i^n}{n} z^{-n}, \qquad \left\vert z\right\vert>\left\vert q_i\right\vert\\ [5pt]
\ln\left(\frac{1}{1-p_iz^{-1}}\right) &=& \sum_{n=1}^\infty \frac{p_i^n}{n} z^{-n}, \qquad \left\vert z\right\vert>\left\vert p_i\right\vert.

Since the region of convergence of the z transform must include the unit circle (where the spectrum (DTFT) is defined), we see that the Maclaurin expansion gives us the inverse z transform of all terms of Eq.(8.9) corresponding to poles and zeros inside the unit circle of the $ z$ plane. Since the poles must be inside the unit circle anyway for stability, this restriction is normally not binding for the poles. However, zeros outside the unit circle--so-called ``non-minimum-phase zeros''--are used quite often in practice.

For a zero (or pole) outside the unit circle, we may rewrite the corresponding term of Eq.(8.9) as

\ln\left(\frac{1}{1-q_iz^{-1}}\right) &=& \ln\left(\frac{q_i^{-1}z}{q_i^{-1}z - 1}\right)
= \ln\left(\frac{-z}{q_i}\right) + \ln\left(\frac{1}{1-q_i^{-1}z}\right)\\
&=& \ln\left(\frac{-z}{q_i}\right) + \sum_{n=1}^\infty \frac{q_i^{-n}}{n}z^n,\qquad
\left\vert z\right\vert<\left\vert q_i\right\vert,

where we used the Maclaurin series expansion for $ \ln(1/(1-x))$ once again with the region of convergence including the unit circle. The infinite sum in this expansion is now the bilateral z transform of an anticausal sequence, as discussed in §8.7. That is, the time-domain sequence is zero for nonnegative times ($ n\geq 0$ ) and the sequence decays in the direction of time minus-infinity. The factored-out terms $ -z/q_i$ and $ -z/p_i$ , for all poles and zeros outside the unit circle, can be collected together and associated with the overall gain factor $ g$ in Eq.(8.9), resulting in a modified scaling and time-shift for the original sequence $ h(n)$ which can be dealt with separately [60].

When all poles and zeros are inside the unit circle, the complex cepstrum is causal and can be expressed simply in terms of the filter poles and zeros as

$\displaystyle {\tilde h}(n) = \left\{\begin{array}{ll}
\ln(g), & n=0 \\ [5pt]
\displaystyle\sum_{i=1}^N \frac{p_i^n}{n} -\displaystyle\sum_{k=1}^M \frac{q_k^n}{n}, & n=1,2,3,\ldots\,, \\
\end{array} \right.

where $ N$ is the number of poles and $ M$ is the number of zeros. Note that when $ N > M$ , there are really $ N - M$ additional zeros at $ z=0$ , but these contribute zero to the complex cepstrum (since $ q_i=0$ ). Similarly, when $ M>N$ , there are $ M - N$ additional poles at $ z=0$ which also contribute zero (since $ p_i=0$ ).

In summary, each stable pole contributes a positive decaying exponential (weighted by $ 1/n$ ) to the complex cepstrum, while each zero inside the unit circle contributes a negative weighted-exponential of the same type. The decaying exponentials start at time 1 and have unit amplitude (ignoring the $ 1/n$ weighting) in the sense that extrapolating them to time 0 (without the $ 1/n$ weighting) would use the values $ p_i^0 = 1$ and $ -q_i^0 = -1$ . The decay rates are faster when the poles and zeros are well inside the unit circle, but cannot decay slower than $ 1/n$ .

On the other hand, poles and zeros outside the unit circle contribute anticausal exponentials to the complex cepstrum, negative for the poles and positive for the zeros.

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``Introduction to Digital Filters with Audio Applications'', by Julius O. Smith III, (September 2007 Edition)
Copyright © 2024-05-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University