Ch13DP

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Design Problems 
 
DP13-1 
Pick the appropriate circuit from Table 13.4-2. 
 
 
We require 
 
2 1
1 1 2 2 21
1 12 1000 , 2 10000 , 2 and 5 
R p Cz p k
C R C R z CR
π π× < = × > = = = = =k 
 
1Try 2 2000. Pick 0.05 F.z Cπ μ= × = Then 
 
1 1
1 2 1 2
1
1 1.592 k , 2 3.183 k and 0.01 F
2
C CR R R C pC z k
z
μ= = Ω = = Ω = = = 
2 2
1Check: 31.42 k rad s 2 10,000 rad s.p
C R
π= = < ⋅ 
 
13-1 
DP13-2 
 
 
o
2s
1 1||
( ) 1( ) 1 1( ) 1 || 1
RR
j C j C R LC
Rj L jj L R j C R RC LCj C
ω ω ωω ω ω ω ωω ωω
+= = = =⎛ ⎞ + − ++⎜ ⎟ +⎝ ⎠
H
V
V
+
 
 
Pick 30
1 2 (100 10 ) rad s
LC
ω π= = ⋅ . When 0ω ω= 
0
1
( ) 1 1 1
LC
j
LC RC LCLC
ω =
− + +
H 1 
So 0( ) 
CR
L
ω =H . We require 
03 dB 0.707 ( ) 1000
C CR
L L
ω− = = = =H 
 
Finally 
31 2 (10010 )
1.13 nF
2.26 mH
0.707 1000
LC C
LC
L
π ⎫= ⋅ ⎪ =⎪⇒⎬ =⎪= ⎪⎭
 
 
 
 
 
 
 
 
 
 
 
 
 
13-2 
DP13-3 
 
 
1
 2
 3
 4
 5
 6
 1
 2
10 k
 866 k
 8.06 k
 1 M
 2.37 M
 499 k
 0.47 
 0.1 
R
R
R
R
R
R
C F
C F
μ
μ
= Ω
= Ω
= Ω
= Ω
= Ω
= Ω
=
=
 
 
 
Circuit A 
3 3
a c s 1 c
2 1
R R
R R
= − − = − −V V V H V H 2 sV 
 
 
 
Circuit B 
5
4
o a
1 51
R
R
j C Rω= − = −+V V 3 aH V 
 
 
Circuit C c o
2 6
1
j C Rω= − = −V V 4 oH V
a
 
Then 
c 3 4=V H H V 
2
a 2 s 1 3 4 a a
1 3 41
s
−= − − ⇒ = +
HV H V H H H V V
H H H
V 
2 3
o 3 a
1 3 41
= − = +
H HV H V
H H H s
V 
After some algebra 
3
1 4 1
o s
3 2
2 4 6 1 2 5 1
R
j
R R C
R
j
R R R C C R C
ω
ωω
=
− +
V V 
 
This MATLAB program plots the Bode plot: 
 
R1=10; % units: kOhms and mF so RC has units of sec 
R2=866; 
R3=8.060; 
13-3 
R4=1000; 
R5=2370; 
R6=449; 
C1=0.00047; 
C2=0.0001; 
 
pi=3.14159; 
fmin=5*10^5; 
fmax=2*10^6; 
f=logspace(log10(fmin),log10(fmax),200); 
w=2*pi*f; 
 
b1=R3/R1/R4/C1; 
a0=R3/R2/R4/R6/C1/C2; 
a1=R5/C1; 
 
for k=1:length(w) 
 H(k)=(j*w(k)*b1)/(a0-w(k)*w(k)+j+w(k)*a1); 
 gain(k)=abs(H(k)); 
 phase(k)=angle(H(k)); 
end 
 
subplot(2,1,1), semilogx(f, 20*log10(gain)) 
xlabel('Frequency, Hz'), ylabel('Gain, dB') 
title('Bode Plot') 
subplot(2,1,2), semilogx(f, phase*180/pi) 
xlabel('Frequency, Hz'), ylabel('Phase, deg') 
 
 
 
13-4 
 
DP13-4 
Pick the appropriate circuits from Table 13.4-2. 
 
 
We require 
4
1 2 2 1 1 2
3 1 1
1 110 , 200 and 500 
R
k k R C p p
2 4R R C C R
= − = = = = = 
1 1
1 1
1Pick 1 F. Then 5 k .C R
p C
μ= = = Ω 2 4
2 2
1Pick 0.1 μF. Then 20 k .C R
p C
= = = Ω 
Next 
2 26 3
3 3
10 (10 )(20 10 ) 500
R R
R R
−= ⋅ ⇒ =
Ω
 
2 3Let 500 k and 1 k .R R= Ω = 
 
 
 
 
 
 
 
 
 
13-5 
 
 
DP13-5 
Pick the appropriate circuits from Table 13.4-2. 
 
 
We require 
4
1 2 2 1 1 2
3 1 1
1 120 dB 10 , 0.1 and 100 
R
k k R C p p
2 4R R C C R
= = − = = = = = 
1 1
1 1
1Pick 20 F. Then 500 k .C R
p C
μ= = = Ω 2 4
2 2
1Pick 1 μF. Then 10 k .C R
p C
= = = Ω 
Next 
2 26 3
3 3
10 (20 10 )(10 10 ) 50
R R
R R
−= ⋅ ⋅ ⇒ = 
2 3Let 200 k and 4 k .R R= Ω = Ω 
 
 
 
 
 
 
 
 
13-6 
 
 
 
DP13-6 
The network function of this circuit is 
2
3
1
1
( )
1
R
R
j R C
ω ω
+
= +H 
The phase shift of this network function is 1 1tan R Cθ ω−= − 
The gain of this network function is ( ) ( )
3 3
2 2
2 2
1
1 1
1 ( ) 1 tan
R R
R R
G
R C
ω ω θ
+ +
= = =+ +
H 
 
Design of this circuit proceeds as follows. Since the frequency and capacitance are known, R1 is 
calculated from 1
tan( )R
C
θ
ω
−= . Next pick R2 = 10kΩ (a convenient value) and calculated R3 using 
2
3 2( 1 (tan ) 1)R G Rθ= ⋅ + − ⋅ . Finally 
 
1 2 345 deg, 2, 1000 rad s 10 k , 10 k , 18.284 k , 0.1 μFG R R R Cθ ω= − = = ⇒ = Ω = Ω = Ω = 
 
 
DP13-7 
From Table 13.4-2 and the Bode plot: 
16
1
1800 2.5 k
(0.5 10 )
z R
R −
= = ⇒ = Ω× 
2
 2
1
32 dB 40 100 k
R
R
R
= = ⇒ = Ω 
3
2
1 1200 0.05 F
(200)(100 10 )
p C
R C
μ= = ⇒ = =× 
(Check: 
6 6
6
0.5 10 0.5 1020 dB 10 
0.05 10
pk
z C
− −
−
× ×= = = = × ) 
 
DP13-8 
2 2
1
( ) 1 11
R j C R
j C R
j C
ωω ω
ω
−= = − ++
H 
13-7 
1
1 1 6
tan(270 195 )195 180 90 tan 37.3 k
(1000)(0.1 10 )
C R Rω− −°− °° = + − ⇒ = = Ω×
2
2 1
1
10 ( ) 10 373 klim
R
R R
Rω
ω
→∞
= = ⇒ = = ΩH 
 
13-8 
	Design Problems 
	DP13-1 
	DP13-2 
	DP13-3 
	DP13-4 
	DP13-5 
	DP13-6 
	DP13-7 
	DP13-8