Suppose we have the z-transform of $x[n]$ is $X(z)$ and that of $y[n]$ is $Y(z)$. Then we know that the inverse z-transform of $G(z)=X(z)Y(z)$ is $g[n]=x[n]*y[n]$, where $*$ is convolution.

What will be the inverse z-transform of $G(z)=X(z)X^*(z)=|X(z)|^2$ if (i.e, $Y(z)=X^*(z)$, complex conjugate)?


What you're probably looking for is the inverse (discrete Fourier) transform of $|X(e^{j\omega})|^2$, i.e., of $|X(z)|^2$ for $|z|=1$. Otherwise, as pointed out in Fat32's answer, $|X(z)|^2$ is generally no valid $\mathcal{Z}$-transform.

Note that a valid $\mathcal{Z}$-transform must have the form of a power series:


Take as a simple example the $\mathcal{Z}$-transform of $x[n]=\delta[n]+\delta[n-1]$:


Then $|X(z)|^2$ is given by


which is not of the form $(1)$, hence it is no valid $\mathcal{Z}$-transform.

Since $|X(e^{j\omega})|^2=X(e^{j\omega})X^*(e^{j\omega})$, you need to find the IDFT of $X^*(e^{j\omega})$:


Consequently, the IDFT of $|X(e^{j\omega})|^2$ is given by $x[n]\star x^*[-n]$, where $\star$ denotes convolution.

It's also good to know what this means in terms of the $\mathcal{Z}$-transform. The $\mathcal{Z}$-transform of $x^*[-n]$ can be derived as follows:


So the $\mathcal{Z}$-transform equivalent of $|X(e^{j\omega})|^2$ is $X(z)X^*(1/z^*)$, as they correspond to the same time domain sequence. As shown above, the function $X(z)X^*(1/z^*)$ is a valid $\mathcal{Z}$-transform, unlike $|X(z)|^2$.

  • $\begingroup$ You've figured what I wanted to do, and you've pointed out exactly what the problem is with it and how to proceed correctly...a thousand thanks $\endgroup$ – Likely Dec 5 '18 at 11:00
  • $\begingroup$ @Likely: You're welcome :) $\endgroup$ – Matt L. Dec 5 '18 at 11:04

By definition, the Z-transform $X(z)$ of $x[n]$ is an analytic function; that's differentiable (and continuous) everywhere on its domain of convergence. Since the function $G(z)=|X(z)|^2$ is not differentiable everywhere, therefore (probably) there is no sequence $g[n]$ whose Z-transform would yield $G(z)=|X(z)|^2$.

Furthermore looking at Z-transform properties one can also see that

$$ x[n] \longleftrightarrow X(z) $$

$$ x[-n] \longleftrightarrow X(1/z) $$

$$ x[n]^* \longleftrightarrow X(z^*)^* $$

$$ x[-n]^* \longleftrightarrow X(1/z^*)^* $$

none of these basic combinations of $x[n]$ would yield $X(z) X(z)^* = |X(z)|^2$...

  • $\begingroup$ Why is it that the Z-transform is, by definition, analytic? I know that we can usually still extend the notion of differentiability of real functions of complex variables if we use Wirtinger calculus. Would you say that this would allow us finding the inverse Z-tranform for our function? $\endgroup$ – Likely Dec 5 '18 at 10:05
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    $\begingroup$ @Likely: The Wirtinger derivative has absolutely nothing to do with the Z-transform and with analytic functions. It's just a "trick" to find an extremum of a real-valued function of a complex variable. $\endgroup$ – Matt L. Dec 5 '18 at 10:15
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    $\begingroup$ @Likely The analytic property of X(z) stems from power series sum of its definition. $\endgroup$ – Fat32 Dec 5 '18 at 10:38
  • $\begingroup$ @Fat32, okay, I see. $\endgroup$ – Likely Dec 5 '18 at 10:46
  • $\begingroup$ I hate to decide on accepted answers, yours pointed out what the problem was, but I felt Matt L. answer was more complete, hope no hard feeling, thank you $\endgroup$ – Likely Dec 5 '18 at 11:03

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