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I need to take the inverse z transform of $X(z) = \dfrac{.5}{(z-1)(z-.5)}$. I've used partial fractions to split this into $X(z) = \dfrac{1}{z-.5} - \dfrac{1}{z-1}$

But Here I'm stuck. This isn't in a table, and I'm not sure how to solve it. I've found pages with similar problems, e.g. $\dfrac{1}{(z-1)^2}$, but they claim that "this is simple" and just stop once they've handled the square. How would I go about deriving the inverse Z transform of $\dfrac{1}{z-a}$?

Daniel B.
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3 Answers3

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Perhaps the "inverse z-transform" in signal processing is known as "Laurent series" in mathmatics? In other words: find coefficients $a_n$ so that $$ \frac{1/2}{(z-1)(z-1/2)} = \sum_{n=-\infty}^{+\infty} a_n z^{-n} $$ I guess this series is supposed to be valid near $\infty$? If so, Maple says $$ \frac{1}{2}\;z^{-2}+\frac{3}{4}\;z^{-3}+\frac{7}{8}\;z^{-4}+\frac{15}{16}\,z^{-5}+\dots $$ Yes, it can be done by using the partial fraction expansion found in the question. Each of those series expansions is a geometric series. Then add the two series.

GEdgar
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  • Yes, though for the unilateral transforms we're using, the sum is from n = 0 -> infinity and the expression is x[n] rather than a_n, where x[n] is a discrete expression. From wikipedia: "From a mathematical view the Z-transform can also be viewed as a Laurent series where one views the sequence of numbers under consideration as the (Laurent) expansion of an analytic function." – Daniel B. Apr 08 '15 at 02:03
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I completely understand your question; I was looking for the answer to this online, and found nothing.

The answer to this actually stems from the time-shift property of the z-transform, which states that $x(n-n_0) = (z^{-n_0})X(z)$. You can rewrite $\frac{1}{z-a}$ as $z^{-1} \frac{z}{z-a}$, which is in the form of the time-shifting formula. Therefore, the inverse z-transform of $\frac{1}{z-a}$ is as follows:

$a^{n-1}u(n-1)$, $|z| > |a|$

-or-

$-a^{n-1}u(-n+(a-1))$

Hope this helps.

irowe
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bob loblaw
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    This is a fairly terse Answer to offer on a three year old post. Brevity and haste are of little value to a Question which already has an upvoted Answer (as well as the OP's self-answer). Perhaps you could carry through on your Answer to get an actual result, and then highlight differences you see with existing Answers. See how to post mathematical notation. – hardmath Apr 29 '18 at 17:04
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    I actually like this answer because it uses tools that are already familiar to the asker- the properties of the z-transform. No need to bring Laurent series into it. Calculating the coefficients of the series is the long way of doing it. This is much simpler. – irowe Mar 08 '21 at 13:39
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My professor responded to an e-mail and reminded me of a trick from class ... If you do the partial fraction expansion with $\dfrac{F(z)}{z}$ then you get a third fraction: $\dfrac{C}{z}$. Then multiply through by z and you get z in the numerator to match the tables of common transforms.

Daniel B.
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