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Given $H(z)$ is the z-transform of a signal, I know that $H(-z)$ results in shifting of frequencies in DTFT by $\pi$ or $-\pi$. Does it produce aliasing ? How ?

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closed as unclear what you're asking by Marcus Müller, A_A, Peter K. Feb 8 '17 at 13:54

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Aliasing is an artefact introduced by sampling. A transfer function cannot produce aliasing. Please clarify your question. $\endgroup$ – Matt L. Feb 8 '17 at 8:56
  • $\begingroup$ Also, please don't just randomly add tags. Removing all tags to which you don't even mention in your question. $\endgroup$ – Marcus Müller Feb 8 '17 at 9:23
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    $\begingroup$ Also, you have a bit of a history (2) of us having to ask you for hours until you described your problem sufficiently. Please don't assume we'll repeat that procedure every time; I've voted to close this question as unclear. $\endgroup$ – Marcus Müller Feb 8 '17 at 9:25
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This Wikipedia Article defines aliasing as follows:

In signal processing and related disciplines, aliasing is an effect that causes different signals to become indistinguishable (or aliases of one another) when sampled.

Really, to discuss aliasing, we must consider sampling or resampling a signal. There must be some possibility that two frequency components that were distinct before the sampling or resampling process will be mapped to the same frequency component afterwards. In this sense, the modulation operation that you are asking about (i.e. replacing $z$ with $-z$) does not result in aliasing.

Is $H(z)$ the z-transform of a signal or the transfer function of a system?

I'll assume $H(z)$ is a signal for simplicity (it really affects terminology more than anything else). The spectrum of $H(-z)$ does wrap-around from the negative to the positive part of the spectrum (compared to $H(z)$). A complex exponential at a positive frequency of $\omega_0$ will reside at $-\pi + \omega_0$ after the modulation operation. In other words, it re-centers the frequency response from being centered at zero (DC) to being centered at $\pi$ (the highest frequency in our system). Recall that both the input and output spectrum are periodic with period $2\pi$ radians. This shifting of frequencies is similar to what happens when you don't properly filter a signal prior to down conversion. Some frequencies are mapped to places where you wouldn't normally expect to find them. In this example, the mapping does not result in aliasing though. All of the frequencies that were representable before the transformation are still representable afterward, and they have a unique mapping.

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  • $\begingroup$ Yes, $H(z)$ is the z-transform of a signal. $\endgroup$ – Abdul Ateek Feb 8 '17 at 10:11
  • $\begingroup$ @AbdulAteek: I find your comment superfluous. $\endgroup$ – jojek Feb 8 '17 at 11:00

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