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Determine the $z$-transform of the signals (a) $x(n)=\left(\cos \omega_0 n\right) u(n)$ (b) $x(n)=\left(\sin \omega_{k, n}\right) u(n)$
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Solution:

(a) By using Euler's identity,

the signal x(n) can be expressed as,

$x(n)=\left(\cos \omega_0 n\right) \mu(n)=\frac{1}{2} e^{j \operatorname{man} n} u(n)+\frac{1}{2} e^{-j \omega p n} u(n)\\$

that,

$X(z)=\frac{1}{2} Z\left\{e^{j \operatorname{mov} n} u(n)\right\}+\frac{1}{2} Z\left\{e^{-j \operatorname{mov} n} u(n)\right\}\\$

If we set $\alpha=e^{\pm j \omega_0}\left(|\alpha|=\left|e^{\pm j a_0}\right|=1\right)$ in (3.2.2).

we obtain,

$e^{j \operatorname{man})^n} u(n) \stackrel{\therefore}{\longleftrightarrow} \frac{1}{1-e^{/ \omega_0} z^{-1}} \quad \text { ROC: }|z|\gt1\\$

and

Thus

$X(z)=\frac{1}{2} \frac{1}{1-e^{j \omega_0 z^{-1}}}+\frac{1}{2} \frac{1}{1-e^{-j \omega_1 z^{-1}}} \quad \operatorname{ROC}:|z|\gt1\\$

(b) From Euler's identity.

Thus,

$x(z)=\frac{1}{2 j}\left(\frac{1}{1-e^{\sin z^{-1}}-1}-\frac{1}{1-e^{-\tan z^{-1}}}\right) \quad \text { ROC: }|z|\gt1\\$

Time shifting. If,

$x(n) \stackrel{z}{\longleftrightarrow} X(z)\\$

then,

$x(n-k) \stackrel{\sim}{\longleftrightarrow} z^{-k} X(z)\\$

The ROC of $z^{-k} X(z)$ is the same as that of X(z) except for $z=0$ if $k\gt0$ and $z=\infty$ if $k\lt0$. The proof of this property follows immediately from the definition of the z-transform.

The properties of linearity and time shifting are the key features that make the z-transform extremely useful for the analysis of discrete-time LTI systems.