Introduction to
Frequency-Selective Circuits
Assessment Problems
AP 14.1
fc= 8 kHz,ω
c=2πfc= 16πkrad/s;
AP 14.2 [a] ωc=2πfc=2π(2000) = 4πkrad/s;
[b] H(jω)= ωc
ωc+jω=4000π
4000π+jω.
14–1
14
14–2 CHAPTER 14. Introduction to Frequency-Selective Circuits
AP 14.5 Let Zrepresent the parallel combination of (1/SC) and RL. Then
Z=RL
(RLCs +1).
AP 14.6
AP 14.7
ωo=2π(2000) = 4000πrad/s;
AP 14.8
AP 14.9
ω2
AP 14.10
ωo= 8000πrad/s;
C= 500 nF;
AP 14.11
ωo=2πfo=2π(20,000) = 40πkrad/s; R= 100 Ω;Q= 5;
14–4 CHAPTER 14. Introduction to Frequency-Selective Circuits
Problems
P 14.1 [a] ωc=R
L=127
10 ⇥103= 12.7 krad/s;
[c] vo(t)|ωc=7.07 cos(12,700t45) V;
R
L
[c] ωc(UL)=R
L;ωc(L)=R
L✓RL
R+RL◆so the cutofffrequencies are different.
P 14.3 [a] ωo=R
L= 2000πrad/s;
[b] Re= 31.42k270 = 28.14 Ω;
⇣R+Rl
[e] ωc=127 + 75
0.01 = 20,200 rad/s;
14–6 CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.6 [a] ωc=0.9(8000) = 7200 = Req
0.03125 so Req = 7200(0.03125) = 225 Ω.
Then,
P 14.7 [a] ωc=1
RC =1
(103)(100 ⇥109)= 10 krad/s;
[b] H(jω)= ωc
s+ωc
=10,000
s+10,000;
[c] vo(t)|ωc=0.2(0.7071) cos(10,000t45)
P 14.8 [a] Let Z=RL(1/SC)
RL+1/SC =RL
RLCs +1;
p2(R+RL)=(1/RC)
qω2
14–8 CHAPTER 14. Introduction to Frequency-Selective Circuits
[c] With a load resistor added in parallel with the capacitor the transfer
function becomes
This transfer function is in the form of a low-pass filter, with a cutoff
frequency equal to the quantity added to sin the denominator.
Therefore,
P 14.10 [a] ωc=2π(100) = 628.32 rad/s.
[c]
P 14.11 [a] H(s)= sL
R+sL =s
s+R/L =s
s+15,000.
Problems 14–9
P 14.12 [a] H(s)=Vo
Vi
=RLksL
R+RLksL =
s✓RL
R+RL◆
s+R
L✓RL
R+RL◆
P 14.13 [a] ωc=R
Lso R=ωcL= (25 ⇥103)(5 ⇥103) = 125 Ω.
[b] ωc(loaded) = R
L·RL
R+RL
= 24,000;
[c]
14–10 CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.15 [a] 1
RC =1
(50 ⇥103)(5 ⇥109)= 4000 rad/s;
[b] H(s)= s
s+ωc
·
.. H(jω)= jω
4000 + jω;
[c] vo(t)|ωc= (0.7071)(0.5) cos(4000t+45
)
P 14.16 [a] H(s)=Vo
Vi
=R
R+Rc+ (1/sC)
[b] H(jω)= R
R+Rc
·jω
jω+ (1/(R+Rc)C);
Problems 14–11
[e] ωc=1
(62.5⇥103)(5 ⇥109)= 3200 rad/s;
P 14.17 [a] ωc=1
RC =2π(300) = 600πrad/s;
[b] Re= 5305.16k47,000 = 4767.08 Ω;
P 14.18 [a] β=ωc2ωc1and ω2
o=ωc1ωc2.
Then,
14–12 CHAPTER 14. Introduction to Frequency-Selective Circuits
Similarly,
β
2+v
u
u
t β
2!2
+ω2
o=ωc2ωc1
2+s✓ωc2ωc1
2◆2
+ωc1ωc2
[b] ωo2
6
41
2Q+v
u
u
t1+ 1
2Q!23
7
5=ωo
2Q+v
u
u
tω2
o+ ωo
2Q!2
.
But Q=ωo/β. Thus
Similarly,
P 14.19 β=ωo
Q=50,000
4= 12.5 krad/s; 12,500
2π=1.99 kHz
2π=7.02 kHz.
P 14.20 ωo=pωc1ωc2=q(121)(100) = 110 krad/s;
P 14.21 [a] ωo=q1/LC so L=1
ω2
oC=1
[8000(2π)]2(5 ⇥109)= 79.16 mH.
Q=ωo
βso β=ωo
Q=8000
2= 4 kHz.
[b] From part (a), β= 4 kHz. Then,
P 14.22 H(jω)= jω(8000π)
(16,000π)2ω2+jω(8000π).
[a] H(j16,000π)= (j16,000π)(8000π)
(16,000π)2(16,000π)2+(j16,000π)(8000π)= 1;
[c] H(j64,378.8) = j64,378.8(8000π)
(16,000π)264,378.82+j64,378.8(8000π)=1
p2/45;
P 14.23 H(s)=1(R/L)s
s2+(R/L)s+ (1/LC)=s2+ (1/LC)
s2+(R/L)s+ (1/LC);
Problems 14–15
P 14.24 [a] ω2
o=1
LC =1
(40 ⇥103)(40 ⇥109)= 625 ⇥106;
[c] ωc1=ωo2
6
41
2Q+v
u
u
t1+ 1
2Q!23
7
5= 25,000 2
41
10 +s1+ 1
100 3
5
[e] β=ωc2ωc1= 27.62 22.62 = 5 krad/s
14–16 CHAPTER 14. Introduction to Frequency-Selective Circuits
·
[b] From the solution to Problem 14.24,
[c] From the solution to Problem 14.24,
P 14.26 [a] L=1
ω2
oC=1
(50 ⇥109)(20 ⇥103)2= 50 mH;
[b] ωc2=ωo2
6
4
1
2Q+v
u
u
t1+ 1
2Q!23
7
5= 20,000 2
4
1
10 +s1+ 1
100 3
5
Problems 14–17
P 14.27 [a] We need ωc= 20,000 rad/s. There are several possible approaches – this
one starts by choosing L= 1 mH. Then,
[b] % error in ωc=21,320 20,000
20,000 (100) = 6.6%;
P 14.28 [a] ω2
o=1
LC so L=1
[8000(2π)]2(5 ⇥109)= 79.16 mH;
P 14.29 [a] We need ωcclose to 2π(8000) = 50,265.48 rad/s. There are several
possible approaches – this one starts by choosing L= 10 mH. Then,