assuming h[n] is real...

If frequency response $H(e^{j\omega})$ is Conjugate Symmetric:

$$ H(e^{-j\omega}) = H^*(e^{j\omega}) $$

$$ H(e^{j\omega}) = H^*(e^{-j\omega}) $$

Then, what symmetry does the System Function $H(z)$ have? Is it the same symmetry as the frequency response $H(e^{j\omega})$, or is some type of symmetry like conjugate reciprocal symmetry?

$$ H(z) = H^*\left(\frac{1}{z^*}\right) $$

This is confusing the heck out of me, because they start with something that is conjugate symmetric as a frequency response then convert it to a z transform by substituting $z=e^{j\omega}$, then, suddenly by changing variable to z its "conjugate reciprocal symmetric", instead of just plain "conjugate symmetric"... except it doesn't actually tell you this anywhere... that the system function has this symmetry...

Would this be some weird property that "Conjugate reciprocal symmetry" is the same as "conjugate symmetry" on the unit circle. and the frequency response is really a degenerate case of "conjugate reciprocal symmetry"

  • 1
    $\begingroup$ This answer to your other question should also answer this question. $\endgroup$
    – Matt L.
    Jan 20 '19 at 18:17

it's because $e^{j\omega}$ is such a signal that has the property:

$$ (e^{j\omega})^* = e^{-j\omega} = \frac{1}{e^{j\omega}}$$

then going back from the DTFT to Z-transform is not obvious. So it's better to start from definition of Z-transform to define the symmetry:

$$ X(z) = \sum_{n} x[n] z^{-n} $$

and apply this to a signal which is real; $x[n] = x^*[n]$

$$ x[n] \xrightarrow{ Z-transform} X(z) \implies x^*[n] \xrightarrow{ Z-transform} X^*(z^*) $$

and conclude for real $x[n]$ : $$ X(z) = X^*(z^*) $$


$$\begin{align} \mathcal{Z} \{x^*[n] \} &= \sum_n x^*[n] z^{-n} \\ \\ & = (\sum_n x[n] (z^*)^{-n})^* \\ \\ & = X^*(z^*) \\ \end{align} $$

As you can see, on the unit circle where $z = e^{j \omega}$, this becomes:

$$ X(e^{j\omega}) = X(e^{-j\omega})^* $$ the conjugate-symmetry property...

  • $\begingroup$ So basically the system function H(z) is also "conjugate symmetric"... $\endgroup$ Jan 20 '19 at 18:23
  • $\begingroup$ Those symmetry idenfiers are generally preferred for Fourier transforms, For Z-transform, I don't rmember such usage, please check your DSP books what they call such property. Nevertheless, we re-state: for real x[n] we have $X(z) = X^*(z^*)$ . You might call it conjuage symmetric due to an analogy between the associated DTFT but I don't recall seeing it. $\endgroup$
    – Fat32
    Jan 20 '19 at 18:51

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