1

I'm trying to compute this Inverse Z Transform:

$\displaystyle X(z) = \sin\left(\frac{1}{z}\right)$

Suppose that the sequences are right handed and one sided

Any help would be appreciated.

Iuli
  • 6,790
Hernan
  • 11
  • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Davide Giraudo Nov 23 '13 at 12:58
  • I forgot to mention that the sequences are right handed and one sided. Thanks for your attention. – Hernan Nov 23 '13 at 13:42

1 Answers1

2

I'm going to compute the Inverse Z Transform with use of the power series for sine

$$ \ X(z)=sin(1/z)\ $$ The power series for sine is $$ \ sin(x)=x+x^3/3!+x^5/5!+... =\sum_{0}^{\infty} \frac{x^{(2i+1)}}{(2i+1)!}\ $$ We changing variables $$ \ x=1/z\ $$ $$ \ sin(1/z)=\sum_{0}^{\infty} \frac{1}{(2i+1)!z^{(2i+1)}}\ $$

$$ \ x[n]=Z^{-1}{(X(z))}=\frac{1}{2\pi j}\oint sin(1/z)z^{n-1}dz = \sum_{0}^{\infty} \frac{1}{2\pi j}\oint \frac{z^{n-1}}{(2i+1)!z^{(2i+1)}}dz $$

the last integral is not zero iff it equals $n = 2i+1$. Then

$$ \ x[n]= \sum_{0}^{\infty}\frac{1}{(2i+1)!}\ $$

D Wiggles
  • 2,818
user114056
  • 31