1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Which book is that? $\endgroup$ – bertozzijr Aug 30 '17 at 19:26
  • 1
    $\begingroup$ Luise-Vitetta Teoria dei segnali $\endgroup$ – Gennaro Arguzzi Aug 30 '17 at 19:41
1
$\begingroup$

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...

-------------------------------------------------------------------------------------------

EDIT:

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:

$$X(z)=\mathcal{Z}\{x[n]\}=\sum_{n=-\infty}^{+\infty}x[n]z^{-n}$$

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:

$$\mathcal{Z}\{x[n]\}=DTFT\{\widetilde{x}[n]\}$$

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

$$\mathcal{Z}\{x[n]\}=DTFT\{x[n]\}$$

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:

$$X(s)=\mathcal{L}\{x(t)\}=\int_{0}^{+\infty}x(t)e^{-st}dt$$

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:

$$\mathcal{L}\{x(t)\}=\mathcal{F}\{\widetilde{x}(t)\}$$

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

$$\mathcal{L}\{x(t)\}=\mathcal{F}\{x(t)\}$$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

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.