\magnification=\magstep1
\centerline{Mathematics 174 -- Applied Mathematics 2}
\centerline{Oops! -- Revenge of the $2\pi$'s}
\centerline{April 11, 2000}
\bigskip
In class today, I got an extra $2\pi$ thrown in with the 
frequency response function for the ideal ``brick wall'' low-pass
filter (D'OH!!)  Here is the way I wanted to say everything,
before I got mixed up in class:

\bigskip
\item{1)}  By the Big Theorem from class April 6, we know any 
filter is characterized by its impulse response:  If we know 
$\gamma = \Gamma(\delta)$, then
$$g(x) = \Gamma(f)(x) = (f * \gamma)(x)$$
where $\gamma(x) = \Gamma(\delta)(x)$ is the {\it impulse response}.
\item{2)}  By the convolution property of Fourier transforms, the
Fourier transform $G$ of the output is related to the Fourier Transform
$F$ of the input like this:
$$\eqalign{
G(\omega) = {\cal F}[g](\omega) &= {\cal F}[f * \gamma](\omega)\cr
                                &= 2\pi F(\omega) {\cal F}[\gamma](\omega)\cr
                                &= H(\omega) F(\omega)\cr}$$
where $H(\omega) = 2\pi {\cal F}[\gamma](\omega)$ is the {\it frequency
response} -- $2\pi$ times the Fourier transform of the impulse response.

\item{3)} For the ideal ``brick wall'' low-pass filter with cut-off $\omega_c$, 
we want to make the frequency response $H(\omega)$
so that the product $H(\omega)F(\omega)$ cuts out all the non-zero
high-frequency values in $F$, but leaves $F$ unchanged in low frequencies.
In other words, we really do want:
$$H(\omega) = \cases{1 & if $|\omega| \le \omega_c$\cr
                     0 & otherwise\cr}$$
(1 on the pass-band, not $2\pi$!)  This means that the Fourier transform
of the impulse response is 
$${\cal F}[\gamma](\omega) = {1\over 2\pi} H(\omega) = 
\cases{{1\over 2\pi} & if $|\omega| \le \omega_c$\cr
                   0 & otherwise\cr}$$
Taking the inverse transform gives
$$\gamma(x) = {1\over 2\pi} {2\sin(\omega_c x)\over x} = 
{\sin(\omega_c x)\over \pi x}.$$

\item{4)} The magnitude of the frequency response $H(\omega)$ for the averaging 
filter {\it was correct} as given in the handout:
$$|H(\omega)| = \sqrt{2 - 2\cos(\omega)\over \omega^2}$$
\bye
