When we looked at filters, we found that the signal $y(n)=\frac{x(n)+x(n-1)}{2}$ is a low pass filter. This is an example of a \emph{Finite Duration Impulse Response}, otherwise known as a finite impulse response, or FIR. FIR filters have no recursion. 

The \emph{Region of Support} is a finite region of support defined as the set of values of $n$ for which $h(n) \neq 0$.

\begin{problem}
Here is an example of recursion. This is a feedback loop represented by the first order difference equation:

$$ y(n) = \alpha y(n-1) + x(n) \mbox{ where } \alpha \in (0,1) \mbox{ and initial conditions } y(-1) = 0$$

Determine the frequency response $H(\omega)$ and the magnitude response $\left| H(\omega) \right|$.

\end{problem}

\begin{solution}

First, lets solve for a few values of $h(n)$:
\begin{align*}
h(n) &= \alpha h(n-1) + \delta(n) \\
h(0) &= \alpha h(-1) + \delta(0) \\
h(0) &= 0 + 1 \\
h(1) &= \alpha h(0) + \delta(1) \\
h(1) &= \alpha + 0 \\
h(2) &= \alpha h(1) + \delta(2) \\
h(2) &= \alpha^2 \\
h(3) &= \alpha^3\\
\end{align*}
This seems to show that $h(n) = \begin{cases} \alpha^n & n \geq 0 \\ 0 & n\lt 0 \end{cases}$

We can simplify this expression by using the \emph{Discrete-Time Unit Step} function, $u(n) \begin{cases} 0 & n\lt 0 \\ 1 & n\geq0 \end{cases}$. Now we can define the impulse response as 

$$ h(n) = \alpha^n \cdot u(n) $$

This is an example of an \emph{Infinite-Duration Impulse Response}, an IIR. Lets write $y(n)$ in terms of its input signal $x$ to change the recursive form by convolving $h$ with an arbitrary input signal.

\begin{enumerate}
\item Convolving $h$ with an arbitrary input signal $x(n)$
\end{enumerate}

\begin{align*}
y(n) &=  (x*h)(n) \\
y(n) &= \sum \limits_{k=-\infty}^{\infty}h(k)x(n-k) \\
y(n) &= \sum \limits_{k=0}^{\infty}\alpha^k x(n-k) \quad \mbox{ change limits since } h(n) \alpha \cdot u(n)\\
y(n) &= x(n) + \alpha x(n-1) + \cdots
\end{align*} 

\begin{enumerate}
\item Finding the frequency response using $H(\omega) = \sum \limits_{n=-\infty}^{\infty} h(n)e^{-i\omega n}$
\end{enumerate}

\begin{align*}
H(\omega) &= \sum \limits_{n=-\infty}^{\infty} h(n) e^{-i\omega n} \\
H(\omega) &= \sum \limits_{n=-\infty}^{\infty} \alpha^n u(n) e^{-i\omega n} \\
H(\omega) &= \sum \limits_{n=0}^{\infty} \alpha^n e^{-i\omega n} \quad \mbox{ change limits since } h(n) \alpha \cdot u(n)\\
H(\omega) &= \sum \limits_{n=0}^{\infty} \left( \alpha e^{-i\omega} \right)^n \quad \mbox{ group as one exponential } \\
H(\omega) &= \frac{1}{1-\alpha e^{-i \omega}} \quad \mbox{ since } \left| \alpha e^{-i \omega} \right| = \left| \alpha \right| \left| e^{-i\omega}  \right| = \left| \alpha \right| \lt  1\\
\end{align*}

Since  $\left| \alpha e^{-i \omega} \right| \lt  1$, the sum  $\sum \limits_{n=-\infty}^{\infty} \left( \alpha e^{-i\omega } \right)^n$ converges to $ \frac{1}{1-\alpha e^{-i \omega}}$.

\begin{enumerate}
\item Find the frequency response using the recursive formula:
\end{enumerate}

Let $x(n) = e^{i\omega n}$ and $y(n) = H(\omega)e^{i\omega n}$, then plug into the recursive formula:
\begin{align*}
y(n) &= \alpha y(n-1) + x(n) \\
H(\omega)e^{i\omega n} &= \alpha H(\omega)e^{i\omega (n-1)} + e^{i\omega n} \\
H(\omega)e^{i\omega n} &= \alpha H(\omega)e^{i\omega n} e^{-i\omega)} + e^{i\omega n} \\
H(\omega) &= \alpha H(\omega) e^{-i\omega)} + 1 \quad \mbox{ divide by } )e^{i\omega n} \\
H(\omega) - \alpha H(\omega) e^{-i\omega)} &=  1 \\
H(\omega) \left(1 - \alpha e^{-i\omega)} \right) &=  1 \\
H(\omega) &= \frac{1}{1 - \alpha e^{-i\omega}} \\
\end{align*}

We implicity used the fact that $\left| \alpha \right| \lt  1$ by the fact that $H(\omega) = \sum \limits_{n\in\Z} h(n)e^{-i\omega n}$ would be divergent otherwise.

Given that $H(\omega) = \frac{1}{1 - \alpha e^{-i\omega}}$, we can find the magnitude response $\left| H(\omega) \right|$.

\begin{align*}
H(\omega) &= \frac{1}{1 - \alpha e^{-i\omega}} \\
H(\omega) &= \frac{e^{i\omega}}{e^{i\omega} - \alpha } \quad \mbox{multiply by } e^{i\omega} \\
\left| H(\omega) \right| &= \left| \frac{e^{i\omega}}{e^{i\omega} - \alpha } \right| \\
\left| H(\omega) \right| &= \frac{\left| e^{i\omega} \right|}{\left| e^{i\omega} - \alpha  \right| } \\
\left| H(\omega) \right| &= \frac{1}{\left| e^{i\omega} - \alpha \right| } \\
\end{align*} 

%\begin{center}
%\includegraphics{images/filters/eminusalpha.ps}
%\end{center}

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 % new vector 
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\psline[linewidth=1.25 pt]{->}(2.121,2.121)

 % new vector 
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\psline[linewidth=1.25 pt]{->}(1.500,0.000)

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\psline[linewidth=1.5 pt]{->}(1.500,0.000)(2.121,2.121)

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\end{pspicture}
\end{center}

By inspection, we can find that this represents a low pass filter. When $\omega = 0$, we have $\left| e^{i\omega} - \alpha \right| = \left| 1 - \alpha \right|$, and when $\omega = \pi$ or $\omega = -\pi$, then  $\left| e^{i\omega} - \alpha \right| = \left| 1 + \alpha \right|$. By looking at this on a unit circle. We have the maximum length when  $\omega = \pm \pi$, so the magnitude response is least since the magnitude of this vector is in the denominator. We have the maximum of the magnitude response when the denominator is the least, when  $\omega = 0$. This is why the filter is a low pass filter. We can also take note of the curvature by acknowledging that when $\omega$ is near $\pm \pi$, the magnitude of $e^{i\omega}-\alpha$ changes much less than when near 0. 

%\begin{center}
%\includegraphics{images/filters/lpf.ps}
%\end{center}

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\begin{center}
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\end{pspicture}
\end{center}


\end{solution}

Design-oriented analysis with an eye toward design allows you to generalize for similar systems. Notice that in the above example, if we wanted to make a high-pass filter, all we would need to do is make $\alpha \in (-1,0)$. Looking at the last figure, we can see that the vector without the negative gets the arrow. Thus, we would have a vector pointing to $e^{i\omega}$ from $\alpha$ on the left side of the unit circle. The magnitude response would still have a value of $\frac{1}{1-\alpha}$ at $\omega = 0$, and $\frac{1}{1+\alpha}$ at $\omega = \pm \pi$. The only difference is that the values at $\omega = \pm \pi$ are the maxima.

To make a LPF or HPF sharper, simply let $\alpha$ approach 1, but not equal 1. This is because we need a safety margin since numerical approximations and noise could send a vector outside of the unit circle and we would have an unstable system. ($\sum \alpha^n$ diverges)

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