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I would like to know how one would calculate the inverse Z-transform of $1/(z-a)^3$. Mathematica says this:

In[0]:= InverseZTransform[1/(z - a)^2, z, n]

Out[0]:= (a^(-2 + n)) (-1 + n) UnitStep[-1 + n]

In[0]:= InverseZTransform[1/(z - a)^3, z, n]

Out[0]:= (1/2) (a^(-3 + n)) (-2 + n) (-1 + n) UnitStep[-1 + n]

I've calculated the first one by convolution. How to would I do it for the second?

Thanks.

Joel
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Lokesh Jaddu
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2 Answers2

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Recall that for a given discrete signal $x[n]$ the Z-transform of that signal is: $$X(z) = \sum_{n=-\infty}^\infty x[n] z^{-n}.$$

Therefore the game is to determine a series expansion for $1/(z-a)^3$ in terms of $z^{-1}$.

Note that $$\frac{1}{(z-a)} = \frac{1}{z} \left(\frac{1}{1-az^{-1}}\right) = \frac{1}{z} \sum_{n=0}^\infty a^n z^{-n} = \sum_{n=0}^\infty a^n z^{-n-1}$$

Now if we differentiate both sides twice we find:

$$\frac{2}{(z-a)^3} = \sum_{n=0}^\infty a^n (n+1)(n+2) z^{-(n+3)} = \sum_{n=3}^\infty a^{n-3} (n-2)(n-1) z^{-n}$$

Hence the inverse Z-transform of $1/(z-a)^3$ gives:

$$x[n] = \frac{a^{n-3}(n-2)(n-1)}{2} u[n-3]$$

Joel
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Note: "inverse z transform" is called "Laurent series" by mathematicians.

So the two examples mean $$ \frac{1}{(z-a)^2} = \sum_{n=1}^\infty a^{-2+n}(-1+n)z^{-n} \\ \frac{1}{(z-a)^3} = \frac{1}{2}\sum_{n=1}^\infty a^{-3+n}(-2+n)(-1+n) z^{-n} $$ for $z$ near $\infty$.

And the question asks us to do the second one "by convolution". Like Joel, I would do it by differentiation. But that is not what he asks...

I assume also known, this one $$ \frac{1}{z-a} = \sum_{n=1}^\infty a^{-1+n}z^{-n} $$ So the "convolution method" will multiply the $1/(z-a)$ series by the $1/(z-a)^2$ series to get the $1/(z-a)^3$ series. In the product, the coefficient of $z^{-n}$ is the convolution $$ \sum_{j+k=n} a^{-1+j}\cdot a^{-2+k}(-1+k) = a^{-3+n}\sum_{k=1}^{n-1} (-1+k) =a^{-3+n}\frac{(-1+n)(-2+n)}{2} , $$ as claimed.

Community
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GEdgar
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