\begin{enumerate}
\item Low-Pass Filter
$$ H(\omega) = \frac{1}{1-\alpha e^{i\omega}} \iff h(n) = \alpha^nu(n) $$
for $\alpha \in (0,1)$

\item High-Pass Filter
$$ H(\omega) = \frac{1}{1-\alpha e^{i\omega}} \iff h(n) = \alpha^nu(n) $$
for $\alpha \in (-1,0)$

\item All-Pass Filter
$$ H(\omega) = \frac{\alpha - e^{-i\omega}}{1-\alpha e^{-i\omega}} \iff h(n) = \alpha^{n+1}u(n) - \alpha^{n-1}u(n-1) $$

\begin{claim}
$\abs{H(\omega)} = \abs{\frac{\alpha - e^{-i\omega}}{1-\alpha e^{-i\omega}}} = 1$
\end{claim}
\begin{proof}
\begin{align*}
H(\omega) &= \frac{\alpha - e^{-i\omega}}{1-\alpha e^{-i\omega}} \\
\abs{H(\omega)} &= \abs{\frac{\alpha - e^{-i\omega}}{1-\alpha e^{-i\omega}}} \\
\abs{H(\omega)} &= \frac{\abs{e^{-i\omega}-\alpha}}{\abs{1-\alpha e^{-i\omega}}} \\
\abs{H(\omega)} &= \frac{\abs{e^{-i\omega}-\alpha}}{\abs{e^{-\omega}}\abs{e^{i\omega}-\alpha}} \\
\abs{H(\omega)} &= \frac{\abs{e^{-i\omega}-\alpha}}{\abs{e^{i\omega}-\alpha}} \\
\abs{H(\omega)} &= 1 \quad \mbox{ since complex conjugate have equal magnitude} \\
\end{align*}
\end{proof}

\item Continuous-time High-Pass Filter / Notch Filter
$$ H(\omega) = \frac{i\omega}{i\omega + \alpha} \iff h(t) = 1 - \alpha e^{-\alpha t}u(t)$$
This filter becomes as notch-filter if $\alpha \approx 1$, and high-pass if $\alpha \approx 0$. 

\item Continuous-time Low-Pass Filter / Anti-notch Filter
$$ H(\omega) = \frac{1}{i\omega + \alpha} \iff h(t) = e^{-\alpha t}u(t) $$
This filter becomes an anti-notch filter if $\alpha \approx 0$, and low-pass if $\alpha \approx 1$. 

\item Notch Filter / Low Pass Filter
$$ H(\omega) = \frac{e^{i\omega}+1}{e^{i\omega}+\alpha} \iff h(n) = \alpha^nu(n)+\alpha^{n-1}u(n-1) $$
This filter is a notch if $\alpha \approx 1$, and is low-pass when $0 \lt \alpha \lt\lt 1$. 

\item Comb Filter
$$ H(\omega) = \frac{e^{i\omega N}}{e^{i\omega N}-\alpha} \iff h(n) = \sum \limits_{k=0}^\infty\alpha^k\delta(n-kN) $$
for $\abs{\alpha} \lt 1$
\end{enumerate}

We can generalize a systems effect on a complex exponential:

$$ e^{i\omega_0 n} \to \fbox{H} \to \abs{H(\omega_0)}e^{i\angle H(\omega_0)}e^{i\omega_0 n} =  \abs{H(\omega_0)}e^{i\lr{\omega_0 n + \angle H(\omega_0)}}  $$


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