Question

The inverse z transform of ((e^-a)-3)/((z-e^-a)(z-3)),a>0 ROC:e^-a<z<3 is,​

Answer 1

Step-by-step explanation:

Recall that for a given discrete signal x[n] the Z-transform of that signal is:

X(z)=∑n=−∞∞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

1(z−a)=1z(11−az−1)=1z∑n=0∞anz−n=∑n=0∞anz−n−1

Now if we differentiate both sides twice we find:

2(z−a)3=∑n=0∞an(n+1)(n+2)z−(n+3)=∑n=3∞an−3(n−2)(n−1)z−n

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

x[n]=an−3(n−2)(n−1)2u[n−3]

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answered

Jun 2 '15 at 17:11

Joel

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

So the two examples mean

1(z−a)2=∑n=1∞a−2+n(−1+n)z−n1(z−a)3=12∑n=1∞a−3+n(−2+n)(−1+n)z−n

for z near ∞.

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

1z−a=∑n=1∞a−1+nz−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

∑j+k=na−1+j⋅a−2+k(−1+k)=a−3+n∑k=1n−1(−1+k)=a−3+n(−1+n)(−2+n)2,

as claimed.