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illustrated in Figure S33.
Rf
R
s1 R s2 Cs +
vi (t)
 +
vo(t)
Figure S33 Preemphasis filter The source impedance is 1
1
Rs1 Rs2 +
j 2πf + j 2πfCs j 2πfCsRs1Rs2 + Rs1
RR
CsRs2
= s1 s2
Zs ( f ) =
=
1
1
j 2πfCs ( Rs1 + Rs2 ) + 1 Rs1 + Rs2 j 2πf +
Rs1 + Rs2 +
j 2πfCs
Cs ( Rs1 + Rs2 )
The numerator provides the zero of the source impedance (the pole of the overall gain) and
the denominator provides the pole of the source impedance (the zero of the overall gain). So
we want the gainpole location to be set by
1
= 2π × 12000
CsRs2
and the gainzero location to be set by
1
= 2π × 6000 .
Cs ( Rs1 + Rs2 )
There is no unique solution so let’s arbitrarily set Rs2 = 10 kΩ. Then it follows that
Cs = 1.33 nF and Rs1 = 10 kΩ . The overall gain is Solutions 626 M. J. Roberts  8/16/04 Rf H( f ) = −
Rs1Rs2
Rs1 + Rs2 1
j 2πf +
CsRs2
1
j 2πf +
Cs ( Rs1 + Rs2 ) = −Rf Rs1 + Rs2
Rs1Rs2 1
Cs ( Rs1 + Rs2 )
1
j 2πf +
CsRs2 j 2πf + At low frequencies,
H( f ) = − Rf
.
Rs1 To make the lowfrequency gain one, set R f = Rs1 = 10 kΩ .
34. One problem with causal CT filters is that the response of the filter always lags the
excitation. This problem cannot be eliminated if the filtering is done in real time but if
the signal is recorded for later “offline” filtering one simple way of eliminating the lag
effect is to filter the signal, record the response and then filter that recorded response
with the same filter but playing the signal back through the system backward. Suppose
the filter is a singlepole filter with a transfer function of the form,
H( jω ) = 1 ω
1+ j
ωc , where ω c is the cutoff frequency (halfpower frequency) of the filter.
(a)
What is the effective transfer function of the entire process of filtering the signal
forward, then backward?
(b) What is the effective impulse response? (a)
The impulse response of the filter is h( t) = F −1[H( jω )] . The response of the filter
on the first pass through the filter forward is y1 ( t) = x( t) ∗ h( t) in the time domain or
Y1 ( jω ) = X( jω ) H( jω ) in the frequency domain. The response of the filter on the second
pass through the filter backward is...
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This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.
 Spring '09
 ATOUSA

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