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z-Transform Examples

Module by: Mark A. Davenport. E-mail the author

zz-transform examples

Example 1

Consider the zz-transform given by H(z)=zH(z)=z, as illustrated below.

Figure 1
A three-dimensional graph displaying a cone-shaped z-transform.

The corresponding DTFT has magnitude and phase given below.

Figure 2
A graph of horizontal axis ω and vertical axis, the absolute value of H(e^jω). There is a horizontal blue line on the graph above the horizontal axis.
Figure 3
A graph of horizontal axis ω and vertical axis, H(e^jω). There is a diagonal blue line on the graph that crosses the origin.

What could the system HH be doing? It is a perfect all-pass, linear-phase system. But what does this mean?

Suppose h[n]=δ[n-n0]h[n]=δ[n-n0]. Then

H(z)=∑n=-∞∞h[n]z-n=∑n=-∞∞δ[n-n0]z-n=z-n0.H(z)=∑n=-∞∞h[n]z-n=∑n=-∞∞δ[n-n0]z-n=z-n0.
(1)

Thus, H(z)=z-n0H(z)=z-n0 is the zz-transform of a system that simply delays the input by n0n0. H(z)=z-1H(z)=z-1 is the zz-transform of a unit-delay.

Example 2

Now consider x[n]=αnu[n]x[n]=αnu[n]

Figure 4
A graph of horizontal axis n and vertical axis x[n]. There are evenly-spaced, decreasing vertical lines extending from the horizontal axis to a point in the first quadrant. At the top of each line segment is a small blue circle.
X(z)=∑n=-∞∞x[n]z-n=∑n=0∞αnz-n=∑n=0∞αzn=11-αz(if|α/z|<1)(GeometricSeries)=zz-αX(z)=∑n=-∞∞x[n]z-n=∑n=0∞αnz-n=∑n=0∞αzn=11-αz(if|α/z|<1)(GeometricSeries)=zz-α
(2)

What if az≥1?az≥1? Then ∑n=0∞(αn)n∑n=0∞(αn)n does not converge! Therefore, whenever we compute a zz-transform, we must also specify the set of zz's for which the zz-transform exists. This is called the region of convergence (ROC). In the above example, the ROC={z:z>α}{z:z>α}.

Figure 5
A graph of horizontal axis Re[z] and vertical axis Im[z], labeled ROC. The graph contains a blue circle centered at the origin, with a line from the origin to a point on the circle in the first quadrant, labeled α.

Example 3

What about the “evil twin” x[n]=-αnu[-1-n]x[n]=-αnu[-1-n]?

X(z)=∑n=-∞∞-αnu[-1-n]z-n=∑n=-∞-1-αnz-n=-∑n=-∞-1zα-n=-∑n=1∞zαn=1-∑n=0∞zαn(convergesif|z/α|<1)=1-11-zα=α-z-αα-z=zz-αX(z)=∑n=-∞∞-αnu[-1-n]z-n=∑n=-∞-1-αnz-n=-∑n=-∞-1zα-n=-∑n=1∞zαn=1-∑n=0∞zαn(convergesif|z/α|<1)=1-11-zα=α-z-αα-z=zz-α
(3)

We get the exact same result but with ROC={z:z<α}{z:z<α}.

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