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Unformatted text preview: 9.12 With C2 set to an open circuit,
VOUT = .Rz/[Rz + R] + (l/jmcln
= JwRZCI/[l + ju>C1(Rl + R2)] =: A low frequency pole occurs at
W = molar] + 122) = [(0.5 pF)(5.2 icon“ = 385 rad/s.
Note that the lowfrequency response includes a solitary
factor of jto in the numerator =: the Bode plot begins with
a slope of +20 dB/decade. With C1 set to a short circuit, VOUT = lell(1/JwC2)]/[R2tIGmC2) + R1] Simplifying this expression yields a high frequency pole at (Dr: = l/C2(R1“Rﬂ
= [(200 pF)(l92 (hr1 = 2.6 x 107 rad/s. Between (9L and (DH.
Vout/vin = RZ/(Rl + R2)
= (5 kQ)/(5 k9 + 0.2 kg) = 0.96 Expressed in dB, this ratio becomes 20 loglo (0.96) =
0.34 dB.‘ Here are magnitude and angle Bode plots of the transfer function: wad/Val (dc) O
70' _ _ _._ _ T
!— ' “(Yul/sec) 3,; 240x107 9.16 At 60 Hz, the circuit must attenuate the input by —30
dB, for a maximum noise signal of —40 dB relative to the voice signal. The filter must also pass signals at 200 Hz.
The frequency span from 200 to 60 Hz constitutes log '200
— log 60 z 0.5 decades, hence the slope of the ﬁlter must
exceed 60 .dB/dec. i.e., it must have three poles. Here is
one strategy for quickly designing such a ﬁlter: Cascade
three highpass stages; Make the input impedance to each
one much larger than the output impedance of its
predecessor. The circuit shown below represents one
possibility. For each stage, 1/2an equals approximately
60 Hz. The values of Rh and 1/Cn are increased by factors
. of ten for each stage going from right to left. The tape
"_ recorders are assumed to have abOut 10 an input
{ resistances and 100 9 output resistances. The circuit
l shown results in input and output loading factors on the i order of 1/2  both acceptable values. c,=27,.r C.=2.1,.F 63:01?» V‘m \ Vac(t
. [2 Rs. :3 _
l .
loan. Um. JO kn.  _.... ....u nab purcs a! £0 25, 7 135103523; 312d 2 x 105 lad/s. .At low frequencies well
res use west pole frequency, the magnitude of th
 po approaches lHOco)l = 10 s 20 dB. Aseach pol: w" /o‘ Ia‘ wLmJ/s) to loo m3
9.18 This system function has an initial zero at (u = 0, a
second zero at (.0 = 90 rad/s, a pole at m = 5000 ml] , and
a pole at 105 rad/s. At high frequencies, well above the
highest pole, the magnitude of the response approaches the
value , mean = 900 x (105)(5000)/(90) = 5 x 109 E 194 dB
Just below the highest pole, the magnitude falls off at 20
dB/decade; below the lower pole, the slope becomes 40
dB/decade. Below the highest zero, the slope becomes 20
dB/decade, and retains that value down to do (to = 0).
Here are the magnitude and angle Bode plots of the system function: 9.23 The system function needed to describe this circuit
response must have a solitary m2 = 0, an initial pole at 93,1
= 211: x 10 Hz = 62.8 rad/s and a second pole at mp2 = 1:
x 20 kHz = 125 krad/s. The system function will have the
form
How) = A (tempo/[<1 + tampon + jm/m 2)].
The given information can be used to find the ’midband"
gain A. Speciﬁcally, well above mm and well below mp2,
H003) = A (jco/mPlem/wpl) = A
= A = 500. The overall system function becomes
OLD/62.8)
Hem) = 500' _
(1 + jw/62.8)(l + jw/1.25 x 105) 9.29 a. The system function can be described by
1 H00 = ——————— [1 + (if/50000)]4 The four coincident poles at f = 50 kHz lead to a Bode plot
that is ﬂat (0 dB) at dc and has a rolloff of —80 dB/decade '
at high frequencies (—20 dB/decade per pole): The exact value of the 3dB endpoint can be found by
seuing lHl to 1/1/2 = 0.707, i.e. 1 I
lHl= = ~————— = 0.707
{ 11 +<f/50000121” 14 n + (050000212
; Solving this equation for f results in
t 1 + (f/50000)2 = 1/(0.707)* = 1.19, or
' f = (50 kHz)(1.l9  1)” = 21.7 kHz
b. .The actual magnitude of the response at 50 kHz can
be found by substituting f = 50 kHz into the above
expression for [HI:
1 l
Hl=——5— =———=0.25
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 Spring '06
 Audiffred

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