When my book introduces the relationship between z-transform and Fourier transform, it starts from the following sequence ($x[n]$ generic sequence):

$$\widetilde{x}[n]=x[n] \space \left(\frac{1}{r}\right)^n$$

and it says that the above sequence convergs when $r>0$ and $n\rightarrow +\infty$. Why? If $0<r<1$, the sequence divergs.

Thank you very much.

  • $\begingroup$ Which book is that? $\endgroup$ – bertozzijr Aug 30 '17 at 19:26
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    $\begingroup$ Luise-Vitetta Teoria dei segnali $\endgroup$ – Gennaro Arguzzi Aug 30 '17 at 19:41

Teoria dei segnali, 1999 - Page 317

(Sorry for my rusty Italian, in case I miss something on the translation, ok?)

6.3 - Overview of the Z-Transform of a sequence

6.3.1 - Definition of the Z Transform and Region of Convergence

There is the sequence you provided:

$$\widetilde{x}[n]=x[n] \space \left(\frac{1}{r}\right)^n$$

Then he says that it is defined for $r>0$ and that it should quickly go to zero as $n \to \infty$. Then he uses as an example with Heaviside Step and then he states:

[...] However, the sequence $\widetilde{x}[n]$ defined as above is exponentially damped that has a transform if $r>1$

So, I also think there should be a better explanation about the values between $0$ and $1$, but he uses this "modified version" of the sequence just to introduce and define the Z-Transform from Fourier...



Let us do the inverse analysis and try to crack it down a little more...

We know that the definition of the Z-transform is:


But, we know thta $z\in \mathbb{C}$, so that $z=re^{j\omega}$ and $r=|z|$. That way, we can rewrite:

$$ \begin{align} X(z)&=\sum_{n=-\infty}^{+\infty}x[n](re^{j\omega})^{-n}\\ &=\sum_{n=-\infty}^{+\infty}x[n]r^{-n}e^{-j\omega n}\\ &=\sum_{n=-\infty}^{+\infty}x[n]\left(\frac{1}{r}\right)^{n}e^{-j\omega n}\\ &=\sum_{n=-\infty}^{+\infty}\widetilde{x}[n]e^{-j\omega n}=DTFT\{\widetilde{x}[n]\} \end{align} $$

So, we get to the relation:


With special case when $r=1$, having:


This is the Discrete-Time equivalent analogy to Continuous-Time Fourier Transform incapability of dealing witrh unlimited energy signals, which causes the Laplace Transform to take place:


We know that $s \in \mathbb{C}$ and $s=\sigma + j\omega$,giving us:

$$ \begin{align} X(s)&=\int_{-\infty}^{+\infty}x(t)e^{-(\sigma + j\omega)t}dt\\ &=\int_{-\infty}^{+\infty}x(t)e^{-\sigma t}e^{-j\omega t}dt\\ &=\int_{-\infty}^{+\infty}\widetilde{x}(t)e^{-j\omega t}dt=\mathcal{F}\{\widetilde{x}(t)\} \end{align} $$

So, for a given continuous-time, exponentially damped signal

$$\widetilde{x}(t)=x(t)e^{-\sigma t}$$

We have the relation:


With special case when $\sigma=0$, giving:


Which is widely used to trace Bode Diagrams of systems described in Frequency Domain.

One more thing: I believe the book just defined $r>0$ because $r\stackrel{\Delta}=|z|$. Of course it does not imply that is convergent for all values of $r$ in that interval, after all, a divergent series $x[n]$ has a Z-transform, with pole(s) outside the unit circle. The convergence of $\widetilde{x}[n]$ is useful on the analysis of the $ROC$, which I can add to a further Edit if you need/want so.

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    $\begingroup$ Hi @bertozzijr. Thank you for your efforts! My goal is not to understand the z transform, but is to understand the particular above formula. $\endgroup$ – Gennaro Arguzzi Sep 1 '17 at 6:51
  • $\begingroup$ @GennaroArguzzi I made a major edit to the answer, and I hope it helps. What I try to explain there is that that expression was induced just to show the relationship between the two transforms and that the Z-Transform helps in cases where the DTFT can't handle the signal just as the Laplace Transform comes in rescue for the Fourier Transform. Further analysis of $r$ is useful for finding the ROC, which I can add to the answer if you need. $\endgroup$ – bertozzijr Sep 1 '17 at 15:43

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