We received the following task:
Determine without the maple the numeric row for which the Z image is rendered by the function f(z) = z^2/(z^2-4), and where is the Z-image valaible.
I started working on this question and I figured out that you can rewrite the row as a sum (via the geometric row):
sum((4^n) / (z^(2*n), n = 0..infinity)
And z must be |z|>|2|. My question is now: how can you see the numeric row by using the summation I found? I know that the first term must be 1 but I can't get any further. The row you get must be:
1 0 4 0 16 0 64 0 ...
Can anyone help me?
The inverse Z-transform is defined as $$a_n=\frac{1}{2 \pi i}\oint \frac{z^2 }{z^2-4}\,z^{n-1} dz$$ From residue method to compute the contour integral we get, calling $f(z)=\frac{z^2 }{z^2-4}\,z^{n-1}$ $$a_n=\frac{1}{2 \pi i}\oint f(z) dz=\left(\operatorname {Res} _{z=-2}f(z)+\operatorname {Res} _{z=2}f(z)\right)=2^{n-1}+(-1)^n 2^{n-1}$$ which gives the sequence you asked for
$\{a_n\}=1,\,0,\,4,\,0,\,16,\,0,\,64,\,0,\,256,\,0,\,1024,\,0,\,4096,\,0,\,16384,\,\ldots$
hope this helps
Edit
the same result can be achieved considering
$\sum _{n=0}^{\infty } 4^n z^{2 n}=\dfrac{1}{1-4 z^2}$
as generating function of $\{a_n\}$. If you develop in MacLaurin series the previous function then the coefficients give the sequence
$\dfrac{1}{1-4 z^2}=1+4 z^2+16 z^4+64 z^6+256 z^8+1024 z^{10}+4096 z^{12}+16384 z^{14}+\ldots$
The series is a formal series so we don't care about its convergence