In Z-Transform,

For a real signal, $x(n)$ =$x^*(n)$ . Taking Z-transform on both sides,

$X(z)$=$X^*(z^*)$ , which gives certain pole/zero condition

similarly for a purely imaginary signal ,

$x(n)$ =$-x^*(n)$ , if we take Z-Transform on both sides , $X(z)$=$-X^*(z^*)$

which gives similar pole/zero condition as that of real signal(poles and zeros are not effected by negative sign).

But My professor says if the poles/zeros are symmetric with respect to real axis, then it is a real signal. But i am getting same condition for both real and imaginary signal.

Can anyone help me? (examples might be helpful)

  • $\begingroup$ For imaginary time domain signals the positive frequencies and negative frequencies would have a very different symmetry. As such I would have to say the answer is no, as the pole locations don't change the symmetry. But hopefully someone else can provide a more robust/credible answer. Also you should specify if you are talking specifically about minimum phase systems as limiting to minimum phase cases has great effect on the final answers regarding these types of questions. $\endgroup$ – Dole Feb 13 '16 at 21:17
  • $\begingroup$ I am not restricting it to minimum phase and i don't understand how it will effect the answer(if you can provide any example, it would be helpful).And what i am guessing is " if signal is real , above pole/zero condition is true, but converse is not true" .Which means pole/zero locations willnot give any specific information about the signal. $\endgroup$ – spectre Feb 15 '16 at 4:23
  • $\begingroup$ For example, the ROC(region of convergence) is different for minimum phase and linear phase. Convergence is usually of great interest with zero/pole analysis. Anyway, you may want to add a new question regarding the specific features, since the original question was already answered. $\endgroup$ – Dole Feb 15 '16 at 23:24

I think there is a confusion between signals and systems here. A signal doesn't have any poles and zeros. Only systems have poles and zeros.

A system is a thing that turns one signal into another one. For example a loudspeaker turns a voltage signal into a sound signal. The voltage going in is a signal, the sound coming out is a signal, but the loudspeaker itself is a system.

Linear Time Invariant Systems can be characterized by their transfer function or their impulse response. The transfer function in the Z domain can be written as a rationale function, which in turn can be characterized through poles and zeros.

If the system has only a real valued input and only a real valued output, the impulse response must be real as well and that in turn will require the poles and zeros to be either real or be conjugate complex pairs.

  • $\begingroup$ thank you for the answer.Though i mentioned as 'signal' in the question what i real meant was system only.Sorry for the confusion.What i was asking was , can we tell the nature of signals by looking at the pole/zero locations of it?.Can we say anything(like odd,even,real,imaginary) about the time domain signal looking at the pole locations? $\endgroup$ – spectre Dec 16 '15 at 10:40

You're right, for a real-valued signal as well as for a purely imaginary signal the zeros (and poles) are symmetrical with respect to the real axis, i.e. if $z_0$ is a zero, then also $z_0^*$ is a zero. This is obvious, because you can construct a purely imaginary signal simply by multiplying a real-valued signal by the imaginary unit $j$, which won't change the pole and zero locations.

What your professor probably meant is that for general complex signals where neither the real part nor the imaginary part equals zero, the poles and zeros are not symmetrical anymore.

  • $\begingroup$ Can we say anything(like odd,even,real,imaginary) about the time domain signal looking at the pole locations? $\endgroup$ – spectre Sep 21 '15 at 15:40
  • $\begingroup$ So, both real,even and imaginary,even and real , odd and imaginary, odd all must have same pole/zero conditions right? $\endgroup$ – spectre Feb 15 '16 at 4:28

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