Determine the Z-transform and $RoC$ for $h(n) = \cos(\omega_0 n) u(n)$.

\begin{align*}
h(n) = \cos(\omega_0 n) u(n) &\ztp \hat{H}(z) \\
h(n) = \frac{1}{2} e^{i\omega_0 n}u(n)  + \frac{1}{2} e^{-i\omega_0 n} u(n) &\ztp \frac{1}{2} \hat{U}\lr{\frac{z}{e^{-i\omega_0}}} + \frac{1}{2} \hat{U}\lr{\frac{z}{e^{i\omega_0}}} \\
\end{align*}

We can use the fact that $\hat{U}(z) = \frac{z}{z - 1}$.
\begin{align*}
\hat{H}(z) &= \frac{1}{2} \lr{\frac{z}{z - e^{i\omega_0}}} + \frac{1}{2} \lr{\frac{z}{z - e^{-i\omega_0}}} \\
\hat{H}(z) &= \frac{z}{2}\lr{\frac{2z - e^{-i\omega_0} - e^{i\omega_0}}{\lr{z-e^{i\omega_0}}\lr{z-e^{-i\omega_0}}}} \\
\hat{H}(z) &= \frac{z\lr{z - \cos\omega_0}}{\lr{z-e^{i\omega_0}}\lr{z-e^{-i\omega_0}}} \\
\end{align*}

Lets use $\omega_0 = \frac{\pi}{4}$.

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We have zeros at 0, and $z = \cos\omega_0 = \frac{1}{\sqrt{2}}$. We have poles at $e^{\pm i \omega_0}$. There no fourier transform since the unit circle is not included.



Trigonometric Example