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\centerline{Mathematics 174 -- Applied Mathematics 2}
\centerline{Problem Set 6 -- An Analog Filter}
\centerline{April 6, 2000}
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\noindent
In this problem set we will study a rudimentary analog
filter -- the RC circuit we discussed in class.  Recall that
for any input signal $f(x)$, where $x$ represents {\it time}
in this example, the voltage response of the 
system represented by the circuit
is given by a solution $v(x) = \Gamma(f)(x)$ 
of the first order ODE:
$$ {dv\over dx} + {1\over RC} v = {f \over RC}\leqno(1)$$
(where $R,C > 0$ are constants -- the resistance of the resistor
and the capacitance of the capacitor in the circuit).  
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\item{A.} We will assume that the response to a zero input $f \equiv 0$ is 
zero.  (In physical terms, this means that there is no initial energy in the
system before the input is supplied.)  Show that in 
that case the response to a general input $f$ can be written as 
$$v(x) = \Gamma(f)(x) = 
{1\over RC} e^{-x\over RC} \int_{-\infty}^x f(s) e^{s\over RC}\ ds\leqno(2)$$
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\item{B.} Explain why this system is {\it linear, time-invariant, and causal}.
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\item{C.} Rewriting integral on the right hand side in (2) as a
convolution, identify the {\it impulse response} $\gamma = \Gamma(\delta)$ 
of the system (the response to a delta function input at $x = 0$).   
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\item{D.}  What is the response of this system to the Heaviside step
function input:
$$f(x) = \cases{1 & if $x \ge 0$\cr
                0 & if $x < 0$\cr}$$
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\item{E.}  Show that if $f(x)$ is any input signal, and 
$F(\omega) = {\cal F}[f](\omega)$
is its Fourier transform, then the Fourier transform $G$ of the corresponding
output $g = \Gamma(f)$ is given by
$$G(\omega) = {\cal F}[\Gamma(f)](\omega) = 2\pi F(\omega) H(\omega),$$
where $H$ is the Fourier transform of $\gamma = \Gamma(\delta)$.
The function $2\pi H(\omega)$ is called the {\it frequency response}
of the system.
\item{F.}  Show that the {\it frequency response} $H(\omega)$ of the 
RC circuit is:
$$H(\omega) = {1\over 1 + iRC \omega}$$
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\item{G.}  The function $|H(\omega)|$ is called the {\it magnitude response}
of the system.  Plot this as a function of $\omega$, taking $R = 1, C = 5$.  
Also, determine the interval of $\omega$ values for which 
$|H(\omega)| \ge {1\over \sqrt{2}}$.
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\noindent
The system here is a realizable {\it low-pass filter},
whose ``pass-band'' is taken (by convention) to be your interval from 
part G. 
Of course, a good low-pass filter should have a sharper pass-band
cut-off, and should have $|H(\omega)|$ closer to 1 on more of its
pass-band.  One sub-area of signal processing studies methods for
constructing such filters using more complicated circuits.  The
main tool is the form of the frequency response $H(\omega)$. 
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\item{H.}  Plot the functions 
$$|H(\omega)| = {1\over \sqrt{1 + (5\omega)^{2N}}}$$
for $N = 2,3,4$.  These magnitude response functions correspond to a 
family of realizable low-pass filters called {\it Butterworth
filters\/}.  Each can be realized by an electrical circuit like the RC
circuit above, but with more components.
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\noindent
{\it Assignment}
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Individual write-ups due Thursday, April 13.
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