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I'm a novice in DSP and I have few doubts regarding the $\mathcal Z$-transform and its region of convergence (ROC).

I know what a $\mathcal Z$-transform is. But I'm having trouble with understanding the ROC. First of all I have some confusion with $X(z)$ and $x(z)$. I easily get caught by exchanging these terms. I know the ROC defines the region of where the $\mathcal Z$-transform exists. From the web and my books states that:

If $x[n]$ is a finite-duration sequence, then the ROC is the entire $z$-plane, except possibly $z = 0$ or $\lvert z\rvert = \infty$. A finite-duration sequence is a sequence that is nonzero in a finite interval $n_1 \le n \le n_2$

And later it says:

When $n_2 > 0 $ there will be a $z^{-1}$ term and thus the ROC will not include $z=0$. When $n_1 < 0$ then the sum will be infinite and thus the ROC will not include $\lvert z\rvert=\infty$.

This is where I get stuck!. What they try to say in the above line "When $n_2 > 0$ there will be a $z^{-1}$ term and thus the ROC will not include $z=0$" What do they mean by $z=0$? Are they substituting $z$ as $0$, if so in which equation?

How do we calculate the region of convergence for an infinite sequence?

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    $\begingroup$ It will be nice to get a couple different perspectives on this... $\endgroup$ – Matt M. Oct 31 '11 at 17:17
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To be completely honest, I thought the theory behind the Z-transform was kinda opaque in college too. In hindsight, taking a course in complex analysis would have made it clearer. And I too dislike the notational conventions that seem to be used for this stuff. Strictly speaking, the usual convention here is that

  • $x[n]$ denotes a discrete-time sequence
    • $n\in \mathbb{Z}$
    • the brackets denote a discrete argument
  • $X(z)$ denotes a continuous-valued transformed function
    • $z\in\mathbb{C}$ (is a complex number)
    • the parentheses denote a function accepting a continuous-valued parameter
    • the capital $X$ denotes a transformed version of some other function/sequence $x$ (a similar notation is used for Fourier transforms: $F(j\omega)\leftrightarrow f(t)$

What do they mean by z=0? Are they substituting z as 0, if so in which equation?

They mean, just plug $z=0$ into your usual definition of the Z-transform.

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

Generally (more precisely, when $x[n] \ne 0$ for some $n \ne 0$), this sum will diverge (to infinity) for some complex $z$. For instance, let $x[0] = 1, x[1] = 1$, and $x[n]=0$ for $n<0$ and $n>1$. Then $X(z) = 1 + z^{-1}$. The ROC does not includes $z=0$, for $\lim_{z\rightarrow 0} X(z)=\infty$

When your text says "When $n_2>0$ there will be a $z^{−1}$ term and thus the ROC will not include $z=0$", what they mean by that is, when $x[n]$ is nonzero for some $n>0$, it is unavoidable for the z-transform to include a $z^{-n}$ term, which diverges to infinity at $z=0$. That's all.

How do we calculate the region of convergence for an infinite sequence?

Lots of math. Ha!

srsly, the way this is done is to obtain an algebraic formulation for the sequence in question, plug it into the Z-transform definition, and use the tools available from the analysis of geometric series (and complex power series) to determine where this Z-transform converges/diverges. In practice, determining whether $|z|=1$ converges is the most important question to answer, because that determines stability, and whether or not you can obtain a frequency response from the system, etc. But causality might matter too, depending on what you're doing.

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  • $\begingroup$ what do you mean by The ROC does not includes z=0, for limz→0X(z)=∞ Since z^-0 haven't came in X(z), this what the statement say's? $\endgroup$ – Ant's Nov 1 '11 at 2:40
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    $\begingroup$ @Ant's (I think what the OP is asking is what is 'z' exactly?) So basically Ant, AFAIK, $ \Large{z = e^{(j\frac {2\pi f}{f_s} )}} $. Basically, the z-transform is analogous to the discrete fourier transform. (DFT). For a lot of control analysis where they want to look at stability they usually just replace that complex exponential with 'z' to make it easier to work with. $\endgroup$ – Spacey Nov 5 '11 at 21:49

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