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Connexions module: m10622

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Region of Convergence for the Z-transform

Benjamin Fite Dan CalderonThis work is produced by The Connexions Project and licensed under the Creative Commons Attribution License

1 IntroductionWith the z-transform , the s-plane represents a set of signals (complex exponentials ). For any given LTI system, some of these signals may cause the output of the system to converge, while others cause the output to diverge ("blow up"). The set of signals that cause the system's output to converge lie in the region of convergence (ROC). This module will discuss how to nd this region of convergence for any discrete-time, LTI system.1 2 3

2 The Region of ConvergenceThe region of convergence, known as the ROC, is important to understand because it denes the region where the z-transform exists. The z-transform of a sequence is dened as4

X (z) =n=

x [n] z n

(1)

The ROC for a given x [n] , is dened as the range of z for which the z-transform converges. Since the z-transform is a power series, it converges when x [n] z n is absolutely summable. Stated dierently,

|x [n] z n | < n=

(2)

must be satised for convergence.

2.1 Properties of the Region of ConvergencecThe Region of Convergence has a number of properties that are dependent on the characteristics of the signal, x [n]. Version

1 "The Z Transform: Denition" 2 "Continuous Time Complex Exponential" 3 "System Classications and Properties" 4 "The Z Transform: Denition"

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2.8: Jul 20, 2011 2:10 pm GMT-5

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The ROC cannot contain any poles. By denition a pole is a where X (z) is innite. Since X (z)x [n] is a nite-duration sequence, then the ROC is the entire z-plane, except possibly z = 0 or |z| = . A nite-duration sequence is a sequence that is nonzero in a nite interval n1 n n2 . As long as each value of x [n] is nite then the sequence will be absolutely summable. When n2 > 0 there will be a z 1 term and thus the ROC will not include z = 0. When n1 < 0 then the sum will be innite and thus the ROC will not include |z| = . On the other hand, when n2 0 then the ROC will include z = 0, and when n1 0 the ROC will include |z| = . With these constraints, the only signal, then, whose ROC is the entire z-plane is x [n] = c [n].

If

must be nite for all z for convergence, there cannot be a pole in the ROC.

Figure 1: An example of a nite duration sequence.

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

|X (z) | = |n=

x [n] z

n

|n=

|x [n] z

n

|=n=

|x [n] |(|z|)

n

(3)

We can then split the innite sum into positive-time and negative-time portions. So|X (z) | N (z) + P (z)

(4) (5)

whereN (z) =

1

|x [n] |(|z|)n=

n

andP (z) =

|x [n] |(|z|)n=0

n

(6)

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In order for |X (z) | to be nite, |x [n] | must be bounded. Let us then set|x (n) | C1 r1 n

(7)

forn r2 , and therefore the ROC of a right-sided sequence is of the form |z| > r2 .

Figure 2: A right-sided sequence.

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Figure 3: The ROC of a right-sided sequence.

negative-time portion from the above derivation, it follows that1 1

If x [n] is a left-sided sequence, then the ROC extends inward from the innermost pole in X (z). A left-sided sequence is a sequence where x [n] = 0 for n > n2 > . Looking at theN (z) C1n=

r1 n (|z|)

n

= C1n=

r1 |z|

n

= C1k=1

|z| r1

k

(10)

Thus in order for this sum to converge, |z| < r1 , and therefore the ROC of a left-sided sequence is of the form |z| < r1 .

Figure 4: A left-sided sequence.

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Figure 5: The ROC of a left-sided sequence.

in the positive and negative directions. From the derivation of the above two properties, it follows that if r2 < |z| < r2 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 r2 < |z| < r2 .

If 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 innite duration

Figure 6: A two-sided sequence.

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Figure 7: The ROC of a two-sided sequence.

2.2 Examples Example 1Lets takex1 [n] = 1 2n

u [n] +

1 4

n

u [n]

(11)

The z-transform of

1 n u [n] 2

is

z 1 z 2

with an ROC at |z| > 1 . 2

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Figure 8: The ROC of

` 1 n2

u [n]

The z-transform of

1 n u [n] 4

is

z z+ 1 4

with an ROC at |z| >

1 4

.

Figure 9: The ROC of

` 1 n4

u [n]

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Due to linearity,X1 [z] = =z z + z+ 1 z 1 2 4 1 2z (z 8 )

(12)1 4

(

z 1 2

)(

z+ 1 4

)

1 By observation it is clear that there are two zeros, at 0 and 8 , and two poles, at 1 , and 2 1 Following the obove properties, the ROC is |z| > 2 .

.

Figure 10: The ROC of x1 [n] =

` 1 n2

u [n] +

` 1 n4

u [n]

Example 2Now takex2 [n] =n

1 4

n

u [n]

1 2

n

u [(n) 1]

(13)

The z-transform and ROC of 1 u [n] was shown in the example above (Example 1). The 4 n z z-transorm of 1 u [(n) 1] is z 1 with an ROC at |z| > 1 . 2 22

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Figure 11: The ROC of

` ` 1 n 2

u [(n) 1]

Once again, by linearity,X2 [z] = =z z + z 1 z+ 1 4 2 1 z (2z 8 )

(14)1 16

(z+ 1 )(z 1 ) 4 2

By observation it is again clear that there are two zeros, at 0 and 1 in ths case though, the ROC is |z| < 2 .

, and two poles, at 1 , and 2

1 4

.

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Figure 12: The ROC of x2 [n] =

` 1 n4

u [n]

` 1 n2

u [(n) 1].

3 Graphical Understanding of ROCUsing the demonstration, learn about the region of convergence for the Laplace Transform.

4 ConclusionClearly, in order to craft a system that is actually useful by virtue of being causal and BIBO stable, we must ensure that it is within the Region of Convergence, which can be ascertained by looking at the pole zero plot. The Region of Convergence is the area in the pole/zero plot of the transfer function in which the function exists. For purposes of useful lter design, we prefer to work with rational functions, which can be described by two polynomials, one each for determining the poles and the zeros, respectively.

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