In z-transform, if z means "delay", why do we talk about the value of z?

Question

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.

Answer 1

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.

Answer 2

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.