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my question is about the Z- transform. My first question is what the title says. What does 'Z' in Z-transform represent ? Say in Fourier transform, 'w' (omega) represents frequency ? From Fourier transform, I can know what is the strength of the signal at a certain frequency component, right? Then what idea do I get from Z-transform ?

And, in Fourier transform, we can draw a graph with x-axis as the frequency and y-axis as the amplitude of a certain frequency component. Can we do the same for Z-transform, with x-axis as the different 'Z' values and y axis as the Z-transform for a particular Z ? My textbook says only about equations, but I don't get the significance of 'Z' in Z-transform. And about the graph thing, I haven't encountered any such graph in any textbook, only the Z-plane.

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    $\begingroup$ One more question that the author didn't choose a correct answer. $\endgroup$
    – JohnMarvin
    Dec 31, 2015 at 19:18

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The Fourier transform doesn't exist for every signal. For example, there is not a Fourier Transform of the signal $ (2)^{n} u(n) $. So, in order to get an useful toll for this signals, you can multiply the signal for a sequence $ r^{-n} $ in order to make it decay fast enough to converge:

$$ X(\omega) = \sum_{n=-\infty}^{\infty}{[x(n)r^{-n}]e^{-j \omega n}} $$

$$ X(\omega) = \sum_{n=-\infty}^{\infty}{x(n)(re^{j \omega})^{-n}} $$

Then, we can make $ z = re^{j \omega} $. So, in this case, z is a complex value that can be understood as a complex frequency.

It is important to verify each values of $ r $ the sum above converges. These values are called the Region of Convergence (ROC) of the Z transform. The z transform is equal to the Fourier transform when $ r = 1 $, but the Fourier transform will only exist if $ r = 1 $ is inside the ROC.

As $ z $ is a complex variable, you can't plot it in a Cartesian graph as you said. You will need to plot an 3D graph where x-axis and y-axis contain the real and imaginary values of $ z $, and the z-axis the values of the z-transform. I don't know if there is a real utility for a graph like that in any application, and that's why it is not common to see this graph. Normally, the z-plane is enough.

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  • $\begingroup$ The unit step sequence $u[n]$ does have a Fourier transform, if you allow a Dirac delta: $$U(\omega)=\frac{1}{1-e^{-j\omega}}+\pi\delta(\omega),\quad 0\le\omega < 2\pi$$ $\endgroup$
    – Matt L.
    Nov 21, 2015 at 10:47
  • $\begingroup$ Yes, you are right. I choose a poor example without thinking before. Thanks for noticing. I will correct my answer. $\endgroup$
    – JohnMarvin
    Nov 21, 2015 at 13:56
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    $\begingroup$ Actually, there were more mistakes in my answer. Corrected them as well. $\endgroup$
    – JohnMarvin
    Nov 21, 2015 at 14:13
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the Z transform is the Laplace transform applied to an ideally and uniformly sampled signal.

$$ \mathcal{Z} \left\{ x[n] \right\} \Bigg|_{z=e^{sT}} = \mathcal{L} \left\{ x_\text{s}(t) \right\} $$

where

$$ \begin{align} x_\text{s}(t) & \triangleq x(t) \sum_n \ \delta(t - nT) \\ & = \sum_n x(t) \ \delta(t - nT) \\ & = \sum_n x(nT) \ \delta(t - nT) \\ & = \sum_n x[n] \ \delta(t - nT) \\ \end{align} $$

$T$, of course, is the sampling period in the same units as continuous-time $t$. and $T = \frac{1}{f_\text{s}}$ where $f_\text{s}$ is the sample rate. and, by definition, $x[n] \triangleq x(nT)$.

if you evaluate $z$ at $e^{j \omega}$, you will get information relevant to frequency just as in the Laplace transform if you evaluate $s$ at $j\Omega$. the significance of "$z$" to the Z transform is the same as the significance of "$s$" is to the Laplace transform.

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  • $\begingroup$ Before evaluating the Z-transform at $z=e^{j\omega}$ it's important to check if the unit circle is actually part of the region of convergence. The analogy with the Laplace transform is of course correct and helpful, but the problem is that there seems to be a trend that students nowadays don't learn anything about the Laplace transform. $\endgroup$
    – Matt L.
    Nov 21, 2015 at 11:02

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