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X(z) converges absolutely | X(z) converges absolutely | ||
<math>\leftrightarrow \sum_n |x[n]z^{-n}|</math> converges | <math>\leftrightarrow \sum_n |x[n]z^{-n}|</math> converges | ||
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<math>\leftrightarrow \sum_n |x[n]||z^{-n}|</math> converges | <math>\leftrightarrow \sum_n |x[n]||z^{-n}|</math> converges | ||
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Proof for z with <math>|z|> |z_0|</math>: | Proof for z with <math>|z|> |z_0|</math>: | ||
<math>\sum_{n=- \infty}^{\infty}|x[n]z^{-n}| = \sum_{n=0}^{\infty}|x[n]z^{-n}| = \sum_{n=0}^{\infty}|x[n]|{|z|}^{-n}</math> | <math>\sum_{n=- \infty}^{\infty}|x[n]z^{-n}| = \sum_{n=0}^{\infty}|x[n]z^{-n}| = \sum_{n=0}^{\infty}|x[n]|{|z|}^{-n}</math> | ||
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<math>\le \sum_{n=0}^{\infty}|x[n]|{|z_0|}^{-n} = \sum_{n=0}^{\infty}|x[n]z_0^{-n}|</math> which converges by assumption | <math>\le \sum_{n=0}^{\infty}|x[n]|{|z_0|}^{-n} = \sum_{n=0}^{\infty}|x[n]z_0^{-n}|</math> which converges by assumption | ||
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<math>\rightarrow</math> X(z) converges absolutely | <math>\rightarrow</math> X(z) converges absolutely | ||
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| + | '''Fact 3:''' If x[n] is anti-causal (i.e. x[n] = 0 for n > 0) and <math>z_0</math> is included in ROC, then any z with <math>|z|< z_0</math> is also in ROC. Has a similar proof. | ||
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| + | '''Fact 4:''' If x[n] is "mixed causal" (two-sided signal) and <math>z_0</math> is included in ROC then there exist <math>r_1, r_2 included in all real numbers \ge 0, r_1 <|z|< r_2</math> such that X(z) converges for all z with <math>r_1 <|z|< r_2</math>. | ||
Revision as of 17:45, 21 September 2009
Convergence of Z Transform
Definition: A series $ \sum_{\infty}^{n=0} a_n $ is said to converge to a value V if for every $ \epsilon > 0 $, there exists a positive integer M such that $ |\sum_{n=0}^{N-1} a_n - V | < \epsilon, for all N > M $
For the Z transform, it is customary to talk about the "region of absolute convergence."
Definition: A series $ \sum^{\infty}_{n=0} a_n $ is called "absolutely convergent" when $ \sum_{n=0}^{\infty} |a_n | $ converges.
Fact: If $ \sum|a_n| $ converges, then $ \sum a_n $ converges also, i.e. the region of absolute convergence is included in the region of convergence.
In the literature and here: ROC means "region of absolute convergence"
Reference for Z transform: Chapter 10 of the ECE 301 book
Facts about ROC
X(z) converges absolutely
$ \leftrightarrow \sum_n |x[n]z^{-n}| $ converges
$ \leftrightarrow \sum_n |x[n]||z^{-n}| $ converges
Fact 1: ROC is made of rings around the origin. If $ z_0 $ is in the ROC, then any other z with $ |z| = |z_0| $ is also in the ROC
Fact 2: If x[n] is "causal" (i.e. x[n] = 0 for all n < 0) and $ z_0 $ is in the ROC then any z with $ |z|>|z_0| $ is also in the ROC
Proof for z with $ |z|> |z_0| $: $ \sum_{n=- \infty}^{\infty}|x[n]z^{-n}| = \sum_{n=0}^{\infty}|x[n]z^{-n}| = \sum_{n=0}^{\infty}|x[n]|{|z|}^{-n} $
$ \le \sum_{n=0}^{\infty}|x[n]|{|z_0|}^{-n} = \sum_{n=0}^{\infty}|x[n]z_0^{-n}| $ which converges by assumption
$ \rightarrow $ X(z) converges absolutely
Fact 3: If x[n] is anti-causal (i.e. x[n] = 0 for n > 0) and $ z_0 $ is included in ROC, then any z with $ |z|< z_0 $ is also in ROC. Has a similar proof.
Fact 4: If x[n] is "mixed causal" (two-sided signal) and $ z_0 $ is included in ROC then there exist $ r_1, r_2 included in all real numbers \ge 0, r_1 <|z|< r_2 $ such that X(z) converges for all z with $ r_1 <|z|< r_2 $.
