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

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.

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

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.

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.