12.6: Region of Convergence for the Z-Transform (2024)

Properties of the Region of Convergencec

The Region of Convergence has a number of properties that are dependent on the characteristics of the signal, \(x[n]\).

• The ROC cannot contain any poles. By definition a pole is a where \(X(z)\) is infinite. Since \(X(z)\) must be finite for all \(z\) for convergence, there cannot be a pole in the ROC.
• If \(\bf{x[n]}\) is a finite-duration sequence, then the ROC is the entire z-plane, except possibly \(\bf{z=0}\) or \(\bf{|z|=\infty}\). A finite-duration sequence is a sequence that is nonzero in a finite interval \(n_1≤n≤n_2\). As long as each value of \(x[n]\) is finite then the sequence will be absolutely summable. 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 \(|z|=\infty\). On the other hand, when \(n_2≤0\) then the ROC will include \(z=0\), and when \(n_1≥0\) the ROC will include \(|z|=\infty\). With these constraints, the only signal, then, whose ROC is the entire z-plane is \(x[n]=c \delta[n]\).

The next properties apply to infinite duration sequences. As noted above, the z-transform converges when \(|X(z)|<\infty\). So we can write

\[|X(z)|=\left|\sum_{n=-\infty}^{\infty} x[n] z^{-n}\right| \leq \sum_{n=-\infty}^{\infty}\left|x[n] z^{-n}\right|=\sum_{n=-\infty}^{\infty}|x[n]|(|z|)^{-n} \nonumber \]

We can then split the infinite sum into positive-time and negative-time portions. So

\[|X(z)| \leq N(z)+P(z) \nonumber \]

where

\[N(z)=\sum_{n=-\infty}^{-1}|x[n]|(|z|)^{-n} \nonumber \]

and

\[P(z)=\sum_{n=0}^{\infty}|x[n]|(|z|)^{-n} \nonumber \]

In order for \(|X(z)|\) to be finite, \(|x[n]|\) must be bounded. Let us then set

\[|x(n)| \leq C_{1} r_{1}^{n} \nonumber \]

for

\[n<0 \nonumber \]

and

\[|x(n)| \leq C_{2} r_{2}^{n} \nonumber \]

for

\[n≥0 \nonumber \]

From this some further properties can be derived:

• If \(\bf{x[n]}\) is a right-sided sequence, then the ROC extends outward from the outermost pole in \(\bf{X(z)}\). A right-sided sequence is a sequence where \(x[n]=0\) for \(n<n_1<\infty\). Looking at the positive-time portion from the above derivation, it follows that

  \[P(z) \leq C_{2} \sum_{n=0}^{\infty} r_{2}^{n}(|z|)^{-n}=C_{2} \sum_{n=0}^{\infty}\left(\frac{r_{2}}{|z|}\right)^{n} \nonumber \]

  Thus in order for this sum to converge, \(|z|>r_2\), and therefore the ROC of a right-sided sequence is of the form \(|z|>r_2\).

• If \(\bf{x[n]}\) is a left-sided sequence, then the ROC extends inward from the innermost pole in \(\bf{X(z)}\). A left-sided sequence is a sequence where \(x[n]=0\) for \(n>n_2>−\infty\). Looking at the negative-time portion from the above derivation, it follows that

  \[N(z) \leq C_{1} \sum_{n=-\infty}^{-1} r_{1}^{n}(|z|)^{-n}=C_{1} \sum_{n=-\infty}^{-1}\left(\frac{r_{1}}{|z|}\right)^{n}=C_{1} \sum_{k=1}^{\infty}\left(\frac{|z|}{r_{1}}\right)^{k} \nonumber \]

  Thus in order for this sum to converge, \(|z|<r_1\), and therefore the ROC of a left-sided sequence is of the form \(|z|<r_1\).

• If \(\bf{x[n]}\) is a two-sided sequence, the ROC will be a ring in the z-plane that is bounded on the interior and exterior by a pole. A two-sided sequence is an sequence with infinite duration in the positive and negative directions. From the derivation of the above two properties, it follows that if \(-r_2<|z|<r_2\) converges, then both the positive-time and negative-time portions converge and thus \(X(z)\) converges as well. Therefore the ROC of a two-sided sequence is of the form \(-r_2<|z|<r_2\).

Examples

Example \(\PageIndex{1}\)

Let's take

\[x_{1}[n]=\left(\frac{1}{2}\right)^{n} u[n]+\left(\frac{1}{4}\right)^{n} u[n] \nonumber \]

The z-transform of \(\left(\frac{1}{2}\right)^{n} u[n]\) is \(\frac{z}{z-\frac{1}{2}}\) with an ROC at \(|z|>\frac{1}{2}\).

The z-transform of \(\left(\frac{-1}{4}\right)^{n} u[n]\) is \(\frac{z}{z+\frac{1}{4}}\) with an ROC at \(|z|>\frac{-1}{4}\).

Due to linearity,

\[\begin{align}
X_{1}[z] &=\frac{z}{z-\frac{1}{2}}+\frac{z}{z+\frac{1}{4}} \nonumber \\
&=\frac{2 z\left(z-\frac{1}{8}\right)}{\left(z-\frac{1}{2}\right)\left(z+\frac{1}{4}\right)}
\end{align} \nonumber \]

By observation it is clear that there are two zeros, at \(0\) and \(\frac{1}{8}\), and two poles, at \(\frac{1}{2}\), and \(\frac{−1}{4}\). Following the obove properties, the ROC is \(|z|>\frac{1}{2}\).

Example \(\PageIndex{2}\)

Now take

\[x_{2}[n]=\left(\frac{-1}{4}\right)^{n} u[n]-\left(\frac{1}{2}\right)^{n} u[(-n)-1] \nonumber \]

The z-transform and ROC of \(\left(\frac{-1}{4}\right)^{n} u[n]\) was shown in the example above. The z-transorm of \(\left(-\left(\frac{1}{2}\right)^{n}\right) u[(-n)-1]\) is \(\frac{z}{z-\frac{1}{2}}\) with an ROC at \(|z|>\frac{1}{2}\).

Once again, by linearity,

\[\begin{align}
X_{2}[z] &=\frac{z}{z+\frac{1}{4}}+\frac{z}{z-\frac{1}{2}} \nonumber \\
&=\frac{z\left(2 z-\frac{1}{8}\right)}{\left(z+\frac{1}{4}\right)\left(z-\frac{1}{2}\right)}
\end{align} \nonumber \]

By observation it is again clear that there are two zeros, at \(0\) and \(\frac{1}{16}\), and two poles, at \(\frac{1}{2}\), and \(\frac{−1}{4}\). in ths case though, the ROC is \(|z|<\frac{1}{2}\).