If we look at the low-pass filter $H(\omega) = \frac{1}{1-\alpha e^{-i\omega}}$, we find that we have a filter whose magnitude response looks like

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

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\begin{center}
\begin{pspicture}(-3.5,-1.25)(3.5,3)
\psplot[algebraic,linewidth=1.5pt,plotpoints=1000]{-3.14}{3.14}{cos(x)+1.3}
\psaxes[showorigin=false,labels=none, Dx=3.14](0,0)(-3.25,0)(3.25,2.5)

\psline[linestyle=dashed](-3.14,0.3)(3.14,0.3)
\psline[linestyle=dashed](-3.14,2.3)(3.14,2.3)
\rput(3.6,2.3){$\frac{1}{1-\alpha}$}
\rput(3.6,0.3){$\frac{1}{1+\alpha}$}

\rput(3.14, -0.35){$\pi$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}

Lets define a new frequency response to be $H_N(\omega) = \frac{1}{1-\alpha e^{-i\omega N}}$. Notice that this corresponds to the difference equation $y(n) - \alpha y(n-N) = x(n)$:

\begin{align*}
H(\omega) &= \frac{1}{1-\alpha e^{-i\omega N}} \\
H(\omega) - \alpha H(\omega) e^{-i\omega N} &= 1 \\
H(\omega) e^{i\omega n} - \alpha H(\omega)e^{i\omega n} e^{-i\omega N} &= e^{i\omega n} \\
y(n) - \alpha y(n-N) &= x(n) \\
\end{align*}

To graph the magnitude response for a particular value of $N$, we can analyze the vectors representing the numerator and denominator. But, we can also perform a simple trick, since we know the graph of $H_1(\omega)$:

%\begin{center}
%\includegraphics{images/comb/comb1.ps}
%\end{center}

\psset{unit=1cm, algebraic, arrowscale=1.5}
\begin{center}
\begin{pspicture}(-3.5,-1.25)(3.5,3)
\psplot[algebraic,linewidth=1.5pt,plotpoints=1000]{-3.14}{3.14}{cos(x)+1.3}
\psaxes[showorigin=false,labels=none, Dx=1.62](0,0)(-3.25,0)(3.25,2.5)

\psline[linestyle=dashed](-3.14,0.3)(3.14,0.3)
\psline[linestyle=dashed](-3.14,2.3)(3.14,2.3)
\rput(3.6,2.3){$\frac{1}{1-\alpha}$}
\rput(3.6,0.3){$\frac{1}{1+\alpha}$}

\rput(0,3){$N=1$}

\rput(3.14, -0.35){$\pi$}
\rput(1.62, -0.35){$\pi/2$}
\rput(-1.62, -0.35){$-\pi/2$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}

Now let $H_2(\omega) = H_1(2\omega)$, then we need to divide by 2 on our axes:

%\begin{center}
%\includegraphics{images/comb/comb2.ps}
%\end{center}

\psset{unit=1cm, algebraic, arrowscale=1.5}
\begin{center}
\begin{pspicture}(-3.5,-1.25)(3.5,3)
\psplot[algebraic,linewidth=1.5pt,plotpoints=1000]{-3.14}{3.14}{cos(4*x/2)+1.3}
\psaxes[showorigin=false,labels=none, Dx=1.62](0,0)(-3.25,0)(3.25,2.5)

\psline[linestyle=dashed](-3.14,0.3)(3.14,0.3)
\psline[linestyle=dashed](-3.14,2.3)(3.14,2.3)
\rput(3.6,2.3){$\frac{1}{1-\alpha}$}
\rput(3.6,0.3){$\frac{1}{1+\alpha}$}

\rput(0,3){$N=2$}

\rput(3.14, -0.35){$\pi$}
\rput(1.62, -0.35){$\pi/2$}
\rput(-1.62, -0.35){$-\pi/2$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}

If you notice there will always be $N$ peaks and $N$ valleys. In general, we can use the first graph to find the graph of the magnitude response for all value of $N$. The higher the $N$, the more oscillations.

%\begin{center}
%\includegraphics{images/comb/comb4.ps}
%\end{center}

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\begin{center}
\begin{pspicture}(-3.5,-1.25)(3.5,3)
\psplot[algebraic,linewidth=1.5pt,plotpoints=1000]{-3.14}{3.14}{cos(8*x/2)+1.3}
\psaxes[showorigin=false,labels=none, Dx=1.62](0,0)(-3.25,0)(3.25,2.5)

\psline[linestyle=dashed](-3.14,0.3)(3.14,0.3)
\psline[linestyle=dashed](-3.14,2.3)(3.14,2.3)
\rput(3.6,2.3){$\frac{1}{1-\alpha}$}
\rput(3.6,0.3){$\frac{1}{1+\alpha}$}

\rput(0,3){$N=4$}

\rput(3.14, -0.35){$\pi$}
\rput(1.62, -0.35){$\pi/2$}
\rput(-1.62, -0.35){$-\pi/2$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}


\psset{unit=1cm}

\begin{center}
\begin{pspicture}(-3.5,-1)(3.75,3.5)

\slider{1}{8}{n}{$2N$}{4}
%\rput(0,3){$N=4$}

\psplot[algebraic,linewidth=1.5pt,plotpoints=1000]{-3.14}{3.14}{cos(n*x/2)+1.3}
\psaxes[showorigin=false,labels=none, Dx=1.62](0,0)(-3.25,0)(3.25,2.5)

\psline[linestyle=dashed](-3.14,0.3)(3.14,0.3)
\psline[linestyle=dashed](-3.14,2.3)(3.14,2.3)
\rput(3.6,2.3){$\frac{1}{1-\alpha}$}
\rput(3.6,0.3){$\frac{1}{1+\alpha}$}


\rput(3.14, -0.35){$\pi$}
\rput(1.62, -0.35){$\pi/2$}
\rput(-1.62, -0.35){$-\pi/2$}
\rput(-3.14, -0.35){$-\pi$}
\rput(0, -0.35){$0$}

\end{pspicture}
\end{center}


Comb Filter