Answer to Question #132267 in Differential Equations for Mithun

Question

Find the z – transform of z(cos tita) and z(sin tita)

Answer 1

since

$cos\theta=real \, part\, of \,e^{i\theta}$

sin$\theta$ =Imaginary part of $e^{i\theta}$

let calculate z-transform of $e^{i\theta}$

Z{$e^{i\theta}$ }= $\frac{z}{z-e^{i\theta}}$

=$\frac{z}{z-cos\theta-isin\theta}$

=$\frac{z}{z-cos\theta-isin\theta}\times \frac{z-cos\theta+isin\theta}{z-cos\theta+isin\theta}$

=$\frac{z(z-cos\theta)+izsin\theta}{(z-cos\theta)^2-i^2sin^2\theta}$

=$\frac{z(z-cos\theta)+izsin\theta}{(z^2+cos^2\theta-2zcos\theta+1sin^2\theta}$

=$\frac{z(z-cos\theta)+izsin\theta}{(z^2+1-2zcos\theta)}$

on separating real and imaginary parts we get,

we get

$Z( {cos\theta}) =\frac{z(z-cos\theta)}{(z^2+1-2zcos\theta)}$

$Z(sin\theta)=\frac{zsin\theta}{(z^2+1-2zcos\theta)}$