2
$\begingroup$

I've been watching MIT's signals course and trying to understand $z$-transform. The course introduces $z^{-1}$ as an operator that delays the signal by $1$ time unit (which works very well with the given examples), but directly after that goes on to talk about values of $z$, regions of convergence, etc. If $z^{-1}$ is an operator, why/how does it have a value?? And what does that value signify??

Note: I tried other textbooks, but most just define the transform, without any interpretation. The only interpetations I can see are that

  • the transform correlates a signal by another complex exponential signal (described by $z^{-k}$)
  • the transform of the impulse response is the eigenvalue of a complex exponential input signal

But I'm an absolute beginner on the subject and my understanding might be wrong. Also, I can't see a direct connection to the "delay operator" approach.

$\endgroup$
  • $\begingroup$ $z^{-1}$ denotes a delay element but it is not an operator like + or - which do not have any values. $\endgroup$ – DSP Novice Mar 12 at 2:53
2
$\begingroup$

It is a bit to wrap your head around.

Mathematically, if you denote the z transform as $X(z) = \mathcal{Z}\left \lbrace x_k\right \rbrace$, then when you take the transform of $x_k$ after it's been delayed by one sample you get $\mathcal{Z}\left \lbrace x_{k-1}\right \rbrace = z^{-1}X(z)$. Except in strange corner cases* this always works, and you don't need to know $X(z)$ to know it works.

So $z^{-1}$ (and thus $z$) are frequency-domain variables, just like the $k$ that gets lost in the transform from $x_k$ to $X(z)$ is a time-domain variable. But $z^{-1}$ acts exactly like a delay operator, so in most circumstances you can treat it as such.

* Which I can't even call out -- you simply don't run into them in practice.

| improve this answer | |
$\endgroup$
1
$\begingroup$

The Z-transform, defined as $$X(z) = \sum_{n=-\infty}^{+\infty}x[n]z^{-n}$$ expresses a discrete time signal as a sum of complex exponential signals $z^{-n}$, with non unitary amplitudes, that is $$z = re^{j\omega}, \: \: r \in \Re_+$$ Just to compare with, the Discrete Time Fourier Transform (DTFT) expresses a discrete time signal as a sum of complex exponential signals $re^{-j\omega n}$, with unitary amplitudes, that is $r=1$.

So, $|z| = r$ is the amplitude of the complex exponentials. In general, $z$ represents any complex number on the complex plane with $r$ being its modulus.

Since $X(z)$ is a function of $z$, in some cases, the transform converges for very specific values of $z$, which all together constitute a region of the complex plane named Region of Convergence - ROC. Thus, indeed, $z$ can take any complex value on the complex plane but only a specific subset of these values make sense for the Z Transform, that is, the transform converges ($|X(z)|<+\infty$) for these values of $z$.

Now, a simple one-sample delay operator is defined as $\delta[n-1]$, since its convolution with any signal $x[n]$ gives $$x[n]*\delta[n-1]= x[n-1]$$

It can be easily shown that the Z Transform of such a delay is simply $$Z\{\delta[n-1]\} = z^{-1}$$

So finally, $z^{-1}$ is simply a "shortcut" for naming a one-sample delay operator. What's really happening behind this symbol is what I have already explained.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.