What will a system do in response to a pure frequency? In other words, when we pass a complex exponential through a system, what does its frequency response tell us about that system?

\begin{problem}
Determine the frequency response when the impulse response is defined $h(n) = \delta(n-N)$, $N \in \Z$, where  the system takes an arbitrary signal $x(n)$:
$$ x(n) \to \fbox{$h(n)$} \to y(n) $$
\end{problem}

\begin{solution}
\begin{align*}
y(n) &= \sum \limits_{\ell \in \Z} = 1 \cdot x(n-N) \\
H(\omega_0) &= \sum \limits_{\ell \in \Z} h(\ell)e^{-i\omega_0\ell} = 1 \cdot e^{-i\omega_0N}
\end{align*}

If $N=0$, $h(n)=\delta(n)$, nothing happens. $\left| H(\omega_0) \right| = 1$, doesn't alter frequency content. The frequency response passes the signal through with the same amplitude and frequency --- it only alters \emph{phase}.

$\angle H(\omega) = -\omega N$
\end{solution}

\begin{problem}
Given a \emph{two-point moving average filter} $y(n) = \frac{x(n) + x(n-1)}{2}$, determine the impulse response $h(n)$ and frequency response $H(\omega)$. Also sketch a graph of the magnitude response.
\end{problem}

\begin{solution}
To find the impulse response, simply let $x(n) = \delta(n)$ to get $h(n) = \frac{1}{2}\delta(n) + \frac{1}{2}\delta(n-1)$.

What about the frequency response? Well, we know that $H(\omega) = \sum \limits_{\ell \in \Z} h(\ell)e^{-i\omega\ell}$ First thing to think about --- how many non-zero terms?

\begin{align*}
H(\omega) &= \sum \limits_{\ell \in \Z} h(\ell)e^{-i\omega\ell} \\
H(\omega) &= \sum \limits_{\ell = 0}^1 h(\ell)e^{-i\omega\ell} \quad \mbox{only looking at } \ell \in \{0,1\}\\
H(\omega) &= h(0)e^{0} + h(1)e^{-i\omega}\\
H(\omega) &= \frac{1}{2} + \frac{1}{2}e^{-i\omega}\\
H(\omega) &= e^{-i\omega/2} \left[ \frac{1}{2}e^{i\omega/2} + \frac{1}{2}e^{-i\omega/2} \right] \quad \mbox{factor out } e^{-i\omega/2} \\
H(\omega) &= e^{-i\omega/2} \frac{\left[ e^{i\omega/2} + e^{-i\omega/2} \right]}{2} \\
H(\omega) &= e^{-i\omega/2} \cos(\omega/2) \\
H(\omega) &= \cos(\omega/2) e^{-i\omega/2}
\end{align*}

Since $\cos(\omega/2)$ is not always positive, we need to take a slight detour. $H(\omega + 2\pi) = H(\omega)$ in \emph{discrete time}.
\begin{proof}
\begin{align*}
H(\omega + 2\pi) &= \sum \limits_{\ell \in \Z} h(\ell)e^{-i(\omega + 2\pi)\ell} \\
H(\omega + 2\pi) &= \sum \limits_{\ell \in \Z} h(\ell)e^{-i\omega} e^{-i 2\pi\ell} \\
H(\omega + 2\pi) &= \sum \limits_{\ell \in \Z} h(\ell)e^{-i\omega} \quad \mbox{since } \ell \in \Z \to e^{-i 2\pi\ell} = 1 \\
H(\omega + 2\pi) &= H(\omega)
\end{align*}
\end{proof}

So we only plot in some contiguous interval of length $2\pi$. So we have $H(\omega) = \cos(\omega/2)e^{-i\omega/2}$ This gives us $\left| H(\omega) \right| = \left| \cos(\omega/2) \right| = \cos(\omega/2) $ for $\left| \omega \right| \leq \pi$.

%\begin{figure}[htp]
%\centering
%\includegraphics{images/magres.ps}
%\caption{ Magnitude response of a low pass filter }
%\label{ fig:magres }
%\end{figure}

%\psset{xunit=1, yunit=1, algebraic, arrowscale=1.5}
\begin{center}
\begin{pspicture}(-3.5,-3.25)(3.5,2)
  \psplot[algebraic,linewidth=1.5pt]{-3.14}{3.14}{cos(x/2)}
  \psaxes[showorigin=false,labels=none, Dx=3.14](0,0)(-3.25,0)(3.25,0)
\rput(3.14, -0.35){$\pi$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}
\rput(1.25,1.25 ){$\left| H(\omega) \right| =\cos(\omega/2)$}

\end{pspicture}
\end{center}

If we look at the graph of the magnitude response $\left| H(\omega) \right|$, we find that it favors lower frequencies. Therefore, we can determine that this system is a \emph{Low Pass Filter}. This is because it zeros out any frequencies near $\pi$ or $-\pi$, and is roughly 1 for values near 0. In general, values near $0$ represent lower frequencies, values near $\pi$ and $-\pi$ represent higher frequencies, and values near $\pi/2$ and $-\pi/2$ represent the fequencies in the range between. The highest frequency discrete-time signal = $(-1)^n = e^{i\pi n}$
\end{solution}

\begin{problem}
Consider the \emph{High Pass Filter} defined by $y(n) = \frac{x(n)-x(n-1)}{2}$. Determine its impulse response $g(n)$ and frequency response $G(\omega)$. Graph its magnitude response $\left| G(\omega) \right|$. Also find the phase response $\angle G(\omega)$
\end{problem}

\begin{solution}
Plugging in $x(n) = \delta(n)$, we get $g(n) = \frac{1}{2}\delta(n) - \frac{1}{2}\delta(n-1)$. Solving for the frequency response we get:
\begin{align*}
G(\omega) &= \sum \limits_{\ell \in \Z} g(\ell)e^{-i\omega\ell} \\
G(\omega) &= \frac{1}{2} - \frac{1}{2}e^{-i\omega} \\
G(\omega) &= e^{-i\omega/2} \left[ \frac{1}{2}e^{i\omega/2} - \frac{1}{2}e^{-i\omega/2} \right] \\
G(\omega) &= e^{-i\omega/2} i\sin(\omega/2) \\
G(\omega) &= e^{-i\omega/2} e^{i\pi/2}\sin(\omega/2) \\
G(\omega) &= \sin(\omega/2)e^{-i(\pi/2 -\omega/2)}
\end{align*}

We can then come up with the phase response, subtracting $\pi$ from $\omega \in (-\pi,0) $ to make sure we have positive values:

$$\angle G(\omega) = \begin{cases} \frac{\pi}{2}-\frac{\omega}{2} & 0 \leq \omega \lt \pi \\ (\frac{\pi}{2}-\frac{\omega}{2})-\pi & -\pi \lt \omega \lt 0 \\ \end{cases}$$


%\begin{figure}[htp]
%\centering
%\includegraphics{images/magres2.ps}
%\caption{ Magnitude response of a high pass filter }
%\label{ fig:magres2 }
%\end{figure}

%\psset{xunit=1, yunit=1, algebraic, arrowscale=1.5}
\begin{center}
\begin{pspicture}(-3.5,-1.25)(3.5,2)
  \psplot[algebraic,linewidth=1.5pt]{-3.14}{3.14}{abs(sin(x/2))}
  \psaxes[showorigin=false,labels=none, Dx=3.14](0,0)(-3.25,0)(3.25,0)
\rput(3.14, -0.35){$\pi$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}
\rput(1,1.25 ){$\left| G(\omega) \right| =\left| \sin(\omega/2) \right|$}

\end{pspicture}
\end{center}


\end{solution}

Filters