15.tex

Notes for video 11.mp4
\subsection*{Butterworth Filter}

Choose $C(\omega)$ for maximum flatness at $\infty$.

We want the taylor expansion about infinity to be maximum flat. 

Work with $P(1/\omega)$ and go for maximum flatness at "new DC". \\
You want a maximally flat approximation to zero. 
\begin{align*}
C(\omega) &= 1 \\
C(\omega) &= \frac{1}{1 + C_{2N}\omega^{2N}}
\end{align*}

\paulhint{See frame 15a taken at 3:24}\\
\includegraphics[scale=0.5]{frames/15a}


It's already a taylor series approximation starting at $2M + 2$.

Cutoff frequency gets changed by changing the C coefficient. \\

The more zeroes you have at inifity, the steeper the rolloff.  \\

\subsection*{Good for phase response}


\paulhint{See frame 15b taken at top chart.}\\
\includegraphics[scale=0.5]{frames/15b}

\begin{itemize}
\item{Maximally flat pass band, and a maximally slow transition band. }
\item{There isn't really a stop band, it's all transition band.}
\item{Stop band is at $\infty$, the only place you make it to zero. 
    Maximally gentle lowpass filter.} 

The impulse response is going to be short:

\paulhint{See frame 15c. Short imp} \\
\includegraphics[scale=0.5]{frames/15c}

\item{Short means, there isn't much phase distortion.}

\item{Suppose you put an impulse into the LPF filter: 

\paulhint{See frame 15d. Little diagram with LPF box} \\
\includegraphics[scale=0.5]{frames/15d}
}

\item{Intuitively, the filter will take the broadband click and output a narrow
    band click. It should not ring. If you hear a ring, it is phase distortion. }

\item{The ideal lowpass filter has a "corner", and the sync function rings like
crazy at the cutoff. Think of that as delaying those components. 
Group delay has a huge spike at cutoff frequency and polezero diagram.}

%\paulhint{See frame 15e. Little diagram with sync box, pole zero 
%diagrams, and group delay} \\
%\includegraphics[scale=0.5]{frames/15d}

\item{In the class of filters you often use, butterworth is the most friendly and
    gentle to use. It's a nice smooth warm thing. Good thing to try first.}

\item{
$P(\omega) = \frac{1}{1 + \omega^{2N}}$ \\
$H(s)H(-s) = P(\omega) = P(\omega) \vert_{\omega = \frac{s}{j}}
= \frac{1}{1 + (\frac{s}{j})^{2N}}
$ \\
}

\item{You might recognize this as roots of unity. A bunch of poles at the roots
of unity. Unit circle in the S-plane, which is where the poles of the butterworth
filter lie.
\paulhint{See frame 15e. chart} \\
\includegraphics[scale=0.5]{frames/15e}
}

\item{Poles are at 2nth roots of unity for odd N, else rot($pi/wN$).}

\item{ Take the spectral factorization:  \\
LHP: \\
$H(s) = \frac{1}{(s - p_1) \cdots (s- p_n)}$
\paulhint{See frame 15g. for H(s) eqn} \\
\includegraphics[scale=0.5]{frames/15g}
}

\item{Intuitively, $N =1$, real pole, and minus s is mirrored at roots of unity.}

%\paulhint{graph is being written for N = 1 15h}\\
\item{
\includegraphics[scale=0.9]{frames/15h}
}
\item{
When $N = 2$, you get into trouble when you get roots of unity would be like
the DFT case, so you rotate the poles:
}
  
\item{\paulhint{graph is being written for N = 2 15i}\\
\includegraphics[scale=0.9]{frames/15i}
}

\end{itemize}

\subsection*{Series Biquad Realization}

The poles are in closed form, so we can easily take a series biquad realization.

\paulhint{Stonehenge like graphics. screenshot taken at 23:38 15j}

Elliptic function filter design are hard. Will NOT be covered in class. 


\paulhint{graph is being written for N = 2 15j}\\
\includegraphics[scale=0.9]{frames/15j}\\