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How to find the $z$-transform of $\left[a^{n}\sin\left(bn\right)\right]/n!$ where "!" denotes factorial of a number and b is constant??

Felix Marin
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rock321987
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    $\large{\tt Hint}:, a^{n}\sin\left(bn\right) = \Im\left[\left(a,{\rm e}^{{\rm i}b}\right)^{n}\right]$ assuming $\large a \in {\mathbb R}$. – Felix Marin Nov 15 '13 at 05:18
  • @Felix Marin can u please elaborate your hint..i am not able to understand it..thanks – rock321987 Nov 15 '13 at 05:32

2 Answers2

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Hints:

i)

$$\sin( bn )=\frac{1}{2i}(e^{ibn}-e^{-ibn}).$$

ii) Z-Transform of $a^ne^{ibn}$ is given by

$$F(z) = \sum_{n=0}^{\infty} a^n (e^{ib})^nz^{-n}.$$

iii) The following is known as the geometric series which you need to find a closed form for $F(z)$

$$ \sum_{n=0}^{\infty} t^n = \frac{1}{1-t}. $$

I think you can finish it now.

Mhenni Benghorbal
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I'm going to find Z Transform with the use of Euler's formula $$ sin(x)= \frac{e^{ix} - e^{-ix}}{2i} $$ $$ x(n)= \frac{a^n sin(bn)}{n!}=\frac{a^n (e^{ibn} - e^{-ibn})}{(2i)n!} $$ $$ X(z) = \sum_{0}^{\infty}\frac{(ae^{ib})^n - (ae^{-ib})^n}{(2i)n!}z^{-n}=\frac{1}{2i}(\sum_{0}^{\infty} \frac{1}{n!}(\frac{ae^{ib}}{z})^n -\sum_{0}^{\infty} \frac{1}{n!}(\frac{ae^{-ib}}{z})^n ) $$ $$ X(z) = \frac{1}{2i}(exp{(\frac{ae^{ib}}{z})} -exp{(\frac{ae^{-ib}}{z})} ) $$ $$ X(z) = \frac{e^{\frac{a}{z}}}{2i}(exp{(cos(b) + i sin(b))} -exp{(cos(b) - i sin(b))} ) $$ $$ X(z) = \frac{e^{\frac{a}{z}}exp{(cos(b))}}{2i}(exp{(i sin(b))} -exp{(- i sin(b))} ) $$ $$ X(z) = {e^{\frac{a}{z}}exp{(cos(b))}}sin{( sin(b))} $$

user114056
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