Given the following information and pole-zero diagram, find an explicit formula for $\hat{H}(z)$ and the value of $h(0)$.

\begin{enumerate}
    \item $\hat{H}(z)$ is rational
    \item $\hat{H}(z)$ is causal
    \item given $x(n) = 1$, $y(n) = -4/3$
\end{enumerate}

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\rput(1.5,-0.4){ $\frac{3}{2}$ }
\rput(0.75,-0.4){ $\frac{3}{4}$ }
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Looking at the pole-zero diagram, we can write an explicit formula for $\hat{H}(z)$:

\begin{align*}
\hat{H}(z) &= H_0 \frac{ \lr{z - \frac{3}{2} } } { \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} } } \\
\end{align*}

We also can determine the $RoC$ of $H$ because it is causal. This tells us $RoC(h) = \{ z \st \abs{z} \gt \frac{3}{4} \}$. Lets find $H_0$. Recall from the fourier transform days that we found $x(n) = 1 = e^{i 0 n}$, and $y(n) = H(0)e^{i 0 n}$. Here we can't use this exact trick. Recall that when the input of a system is a complex exponential, we get $y(n) = z_0^n\hat{H}(z)$. The trick here is to let $z_0 = 1$ (note that $\hat{H}(1) = H(0)$):

\begin{align*}
y(n) &= z_0^n \hat{H}(z) \\
-\frac{4}{3} &= 1^n \hat{H}(1) \\
-\frac{4}{3} &= \hat{H}(1) \\
\end{align*}

Now we can use the $\hat{H}(1) = -\frac{4}{3}$ to find $H_0$:

\begin{align*}
\hat{H}(z) &= H_0 \frac{ \lr{z - \frac{3}{2} } } { \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} } } \\
\hat{H}(1) &= H_0 \frac{ \lr{ 1- \frac{3}{2} } } { \lr{ 1 + \frac{1}{2}} \lr{ 1 - \frac{3}{4} } } \\
\hat{H}(1) &= H_0 \frac{ \lr{ - \frac{1}{2} } } { \lr{ \frac{3}{2}} \lr{ \frac{1}{4} } } \\
-\frac{4}{3} &= -H_0\frac{4}{3} \\
H_0 &= 1 \\
\end{align*}

This tells the exact formula for $\hat{H}$:
$$
\hat{H}(z) = \frac{ \lr{z - \frac{3}{2} } } { \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} } } \\
$$

How can we find $h(0)$? If $h$ is causal, then $h(n) = 0$ for $n \lt 0$. 

\begin{align*}
\hat{H}(z) &= \sum \limits_{n \in \Z} h(n) z^{-n} \\
\hat{H}(z) &= h(0) + \frac{h(1)}{z} + \frac{h(2)}{z^2} + \cdots  \\
\end{align*}

This brings us to the \emph{Initial Value Theorem}:

$$ 
h(0) = \lim_{z \to \infty} \hat{H}(z)  \\
$$

The initial value theorem applies when the system is causal. There is a corresponding theorem called the final value theorem for anti-caual systems. 

\begin{align*}
h(0) &= \lim_{z \to \infty} \hat{H}(z)  \\
h(0) &= \lim_{z \to \infty}  \frac{ \lr{z - \frac{3}{2} } } { \lr{ z + \frac{1}{2}} \lr{ z - \frac{3}{4} } } \\
h(0) &= 0 \\
\end{align*}
