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This is a problem from the book Discrete-Time Speech Signal Processing by Thomas Quatieri, Chapter 6 Problem 13:

Suppose $h[n]$ is all-pole (minimum-phase) of order $p$. Show that the all-pole (predictor) coefficients can be obtained recursively from the first $p$ complex cepstral coefficients of $h[n]$.

I was only able to solve for the opposite. i.e. complex cepstral coefficients from the pole coefficients.

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If $h[n]$ is the impulse response of an all-pole minimum-phase filter of order $p$, then we have

$$\mathcal{Z}\{h[n]\}=H(z)=\frac{G}{1-\displaystyle\sum_{k=1}^pa_kz^{-k}}\tag{1}$$

where $G$ is some gain value, and $a_k$ are the predictor coefficients. Let $\hat{h}[n]$ be the complex cepstral coefficients of $h[n]$. If you know how to compute $\hat{h}[n]$ from $a_k$, then you're probably familiar with the following recursion:

$$\hat{h}[n]=a_n+\frac{1}{n}\sum_{k=1}^{n-1}ka_{n-k}\hat{h}[k],\quad 0<n\le p\tag{2}$$

From $(2)$ you can formulate a recursion for $a_k$:

$$a_n=\hat{h}[n]-\frac{1}{n}\sum_{k=1}^{n-1}ka_{n-k}\hat{h}[k],\quad 1<n\le p\tag{3}$$

with the initial value $a_1=\hat{h}[1]$.

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  • $\begingroup$ Isn't $a_k$ an LPC, rather than a pole coefficient? I thought that the pole coefficients should be of the form: $\prod_{k=1}^p (1-x_k z^{-1})$ My approach was to convert the denominator, or the inverse filter into this pole form, then applied $\log$, which gives $\hat{H}$, then applied Laurent series, then took the inverse z-transform, which gave me the complex cepstral coefficients from pole coefficients. $\endgroup$ – stock username Nov 30 '15 at 18:58
  • $\begingroup$ @5422time: Those $x_k$s are usually called pole positions, not coefficients. And the text refers to "all-pole" coefficients, not pole coefficients. That just reinforces that the transfer function is AR / all-pole. $\endgroup$ – Peter K. Nov 30 '15 at 19:01
  • $\begingroup$ @PeterK.: Thanks for the clarification! I appreciate it. $\endgroup$ – stock username Nov 30 '15 at 19:03

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