The magnitude and phase plots are shown below... Therefore, to increase the bandwidth, we merely add another 20 µF in series with the first one... a This is an RLC series circuit... a Co
Trang 1Chapter 14, Solution 1
RCj1
RCjCj1R
R)
(
i
o
ω+
ω
=ω+
=
=ω
V
V H
=ω)(
H
0
0
j 1
j
ωω+
ωω
, where
RC
1
0 =ω
2 0
0)(1)(H
ωω+
ωω
=ω
ω
−
π
=ω
∠
=φ
0
1 -tan2)(
Trang 2Chapter 14, Solution 2
=ω+
=ω+
=ω
RLj1
1L
jR
R)
(
H
0
j 1
1
ωω+ , where L
R
0 =ω
2
0)(1
1)
(H
ωω+
=ω
ω
=ω
∠
=φ
0
1 -tan-)(
H
The frequency response is identical to the response in Example 14.1 except that
LR
RR
sC1
||
RR
Z
sRC1sC
1R
Trang 3sRC1(sC
1
sC1
Th
i Th
Th
V V
Z
V
++
=
⋅+
=
)sRC1(sRCsRC1)(
sRC1(
1)
sRC1)(
sC1(
1)
s
Th i
o
++
++
=+
+
=
=
Z V
V H(
=)s(
H
1 sRC 3 C R s
1
2 2
0.383-
=
=RC
2.617-
Chapter 14, Solution 4
(a)
RCj1
RC
=ω
)RCj1(LjR
RRC
j1
RL
j
RCj1
R)
(
i
o
ω+ω+
=ω++ω
ω+
=
=ω
V
V H
=ω)(
H
L j R RLC -
)LjR(CjCj1LjR
LjR)
(
ω+ω+
ω+ω
=ω+ω+
ω+
=ω
H
=ω)(
H
RC j LC 1
RC j LC -
2
2
ω+ω
−
ω+ω
Trang 4Chapter 14, Solution 5
(a)
Cj1LjR
Cj1)
(
i
o
ω+ω+
ω
=
=ω
V
V H
=ω)(
H
LC RC
j 1
1
2
ω
−ω+
(b)
RCj1
RC
=ω
)RCj1(LjR
)RCj1(Lj)RCj1(RLj
Lj)
(
i
o
ω+ω+
ω+ω
=ω++ω
ω
=
=ω
V
V H
=ω)(
H
RLC L
j R
RLC L
ω
−ω
Chapter 14, Solution 6
(a) Using current division,
Cj1LjR
R)
(
i
o
ω+ω+
=
=ω
I
I H
)25.0)(
10()25.0)(
20(j1
)25.0)(
20(jLC
RCj1
RCj)
ω
−ω
+
ω
=ω
−ω+
ω
=ω
H
=ω)(
5 2 5 j 1
5 j
ω
−ω+ω
(b) We apply nodal analysis to the circuit below
Trang 55.0R
x x
x
−+
I
Cj1Lj
5.0
o x
x
ω+ω
I
Cj1Lj
5.0R
1x
s
ω+ω+
=
V I
)Cj1Lj(2
1R
1)Cj1Lj(
2 o
s
ω+ω+
=ω+ω
I I
1R
)Cj1Lj(2o
s = ω + ω +
I I
)LC1
(2RCj
RCjR
)Cj1Lj(21
1)
ω
=ω
+ω+
=
=ω
I
I H
)25.01(2j
j)
ω
−+ω
ω
=ω
H
=ω)(
5 0 j 2
j
ω
−ω+ω
Chapter 14, Solution 7
(a) 0.05=20log10H
Hlog10
5
5 = 10
=
=105 235
H 1 718×10 5
Trang 610j)1(H
= 3.9 j1.7 4.254 -23.55
j2
6j1
3)1(H
j1(
1)
(
ω+ω+
=ω
H
10/j1log20j
1log20-
HdB = 10 + ω − 10 + ω
)10/(tan)(tan
- -1 ω − -1 ω
=φ
The magnitude and phase plots are shown below
HdB
0.1
-40
10 / j 1
1 log
20 10
ω +
ω + j 1
1 log
1 arg
φ
-135°
10 / j 1 arg ω +
ω +-90°
-180°
-45°
Trang 7=ω
5
j1j1
10)
j5(j
50)
j(H
ω j
1 log 20
j 1
1 log 20
1 arg ω +
ω j
1 arg-90°
)10j1(5)(
ω+ω
ω+
=ω
H
2j1log20j
log2010j1log205log20
HdB = 10 + 10 + ω − 10 ω − 10 + ω
2tan10tan90
- °+ -1ω − -1ω
=φ
Trang 8The magnitude and phase plots are shown below
1 10 0.1
-90°
90°
Trang 9Chapter 14, Solution 12
201.0log20,)10/1
(
)1
ω
j j
j w
-90o
Trang 10
Chapter 14, Solution 13
)10j1()j(
)j1)(
101()j10()j(
j1)
ω+ω
ω+
=ω+ω
ω+
=ω
G
10j1log20j
log40j
1log2020-
GdB = + 10 + ω − 10 ω − 10 + ω
10tantan
-180°+ -1ω− -1ω
=φ
The magnitude and phase plots are shown below
φ 90°
-90°
-180°
Trang 11ω+
=
5
j25
10j1j
j125
50)(
H
ω
−ω++
=20log 2 20log 1 j 20log j
2
10 1 j 2 5 (j 5)log
−
ω
−ω+
°
=φ
51
2510tantan
Trang 12Chapter 14, Solution 15
)10j1)(
2j1(
)j1(2)
j10)(
j2(
)j1(40)
(
ω+ω+
ω+
=ω+ω+
ω+
=ω
H
10j1log202j1log20j
1log202log20
HdB = 10 + 10 + ω − 10 + ω − 10 + ω
10tan2tantan-1ω− -1ω − -1ω
=φ
The magnitude and phase plots are shown below
j)
(G
+
ω
=ω
Trang 13-20
ωjlog20
10
jlog
1arg
ω+ j1
1argarg(jω)
-180°
ω90°
20 log(1/100) -60
j1(
j)41()
(
ω+ω+
ω
=ω
G
2j1log40j
1log20j
log204-20log
GdB = 10 + 10 ω − 10 + ω − 10 + ω
2tan2tan 90° -1ω− -1ω
=φ
The magnitude and phase plots are shown below
GdB
-20
20
ω-12
100
1 10 0.1
-40
Trang 14)2j1(4)
(
2ω+ω+ω
ω+
=ω
G
ω
−ω++
=20log 4 50 40log 1 j 2 20log j
10j1log205j1log
20 10 + ω − 10 + ω
− where 20log104 50=-21.94
10tan5tan2tan290
- °+ -1ω − -1ω − -1ω
=φ
The magnitude and phase plots are shown below
20
Trang 15Chapter 14, Solution 19
)10010
j1(100
j)
ω
−ω+
ω
=ω
H
10010
j1log20100log20j
log20
10 10
1001
10tan
Trang 16Chapter 14, Solution 20
)10j1)(
j1(
)j
1(10)(
2ω+ω+
ω
−ω+
=ω
N
2 10
10 10
dB 20 20log 1 j 20log 1 j 10 20log 1 j
10tantan
1
2 1
−
ω
=φ
The magnitude and phase plots are shown below
j1)(
10j1(100
)j1(j)
ω
−ω+ω
+
ω+ω
=ω
T
100log20j
1log20j
log20
TdB = 10 ω + 10 + ω − 10
Trang 17j1log2010j1log
−
ω
−ω
−ω+
°
=φ
1001
10tan
10tantan
log20
A zero of slope +20dB/dec at ω=2 → 1+jω 2
A pole of slope -20dB/dec at
20j1
120
ω+
→
=ω
Trang 18A pole of slope -20dB/dec at
100j1
1100
ω+
→
=ωHence,
)100j1)(
20j1(
)2j1(10)
(
ω+ω
+
ω+
j 20 (
) j 2 (
10 4
ω+ω
+
ω+
Chapter 14, Solution 23
A zero of slope +20dB/dec at the origin → j ω
A pole of slope -20dB/dec at
1j1
11
ω+
→
=ω
A pole of slope -40dB/dec at 2
)10j1(
110
ω+
→
=ωHence,
2)10j1)(
j1(
j)
(
ω+ω+
ω
=ω
H
=ω)(
) j 10 )(
j 1 (
j 100
ω+ω+
1 arg
) 10 / j 1 ( arg + ω
) j ( arg ω
10090°
-90°
45°
Trang 19)j10)(
10(k)100j1(j
)10j1(k)(
ω+ω
ω+
′
=ω
+ω
ω+
′
=ω
G
where k′ is a constant since argk′=0
Hence, G(ω)=
) j 100 ( j
) j 10 ( k
ω+ω
ω+
, where k=10 k′ is constant
Chapter 14, Solution 25
s/krad5)101)(
1040(
1LC
1
6 - 3
=ω
C
4L4jR)4(
0
0 0
=ω
)101)(
105(
410
404
105j2000)
4
3 0
Z
)5400050
(j2000)
4(ω0 = + −
Z
=
ω 4)( 0
=ω
C
2L2jR)2(
0
0 0
=ω
)101)(
105(
2)
1040(2
)105(j2000)
2
3 0
Z
)52000100
(j2000)
4(ω0 = + −
Z
=
ω 2)( 0
=ω
C2
1L2jR)2(
0 0
0
Z
Trang 20)101)(
105)(
2(
1)
1040)(
105)(
2(j2000)
=ω
C4
1L4jR)4(
0 0
=ω
)101)(
105)(
4(
1)
1040)(
105)(
4(j2000)
Z 2+j 0 75 kΩ
Chapter 14, Solution 26
10101052
12
LC
f o
ππ
1010
R B
101.0
101050
101
3
3 6
L LC R
)80)(
10(RQL0
=
=ω
=
Trang 211L
=
s/rad625.016
10L
R
B= = =Therefore,
10B
R
L= = =
F2)5.0()1000(
1L
=
5020
1000B
ω+ω+ω
=
j1
jj
1j
Z
Trang 2221
j1
j
ω+
ω+ω+
−ω
=
Z
Since and v(t) i(t) are in phase,
21
10
)Im(
ω+
ω+ω
−ω
=
=
Z
012
4 +ω − =ω
618.02
411-
2 = ± + =ω
10(
10Q
RL0
=
=
=ω
=
F2.0)05.0)(
100(
1L
1
C 20
=
=ω
=
s/rad5.0)2.0)(
10(
1RC
LL
X
rad/s10x796.810
x40
10x.510x10x2X
RL
Trang 23Chapter 14, Solution 32
Since Q>10,
2
B0
1 =ω −
2
B0
2 =ω +ω
=
ω2 6 0.025 6 025×10 6 rad / s
Chapter 14, Solution 33
pF 84.5610x40x10x6.5x2
80R
f2
QC
=
→
ω
=
H 21.1480x10x6.5x2
10x40Q
f2
RL
L
RQ
6
3 o
o
µ
=π
=π
=
→
ω
=
Chapter 14, Solution 34
10x60x10x
1LC
1
6 3
1RC
1B
Trang 241RR
=
→
ω
=
)40)(
10200(
80R
QCRC
=
→
=ω
)1010)(
104(
1C
1LLC
1
6 - 10
2 0
=
ω 200 2.5
2
B0
=+
=+ω
=
ω 200 2.5
2
B0
Chapter 14, Solution 36
s/rad5000LC
kS75.18j5.0L
4C4
jR
1)4(
=ω
Y
=
−
=ω
01875.0j0005.0
1)
4( 0
kS5.7j5.0L
2C2
jR
1)2(
=ω
Y
Trang 25−
=ω
0075.0j0005.0
1)
2( 0
kS5.7j5.0C2
1L2jR
1)2(
0 0
=ω
Y
=
ω )2( 0
Z 8 85−j 132 74Ω
kS75.18j5.0C4
1L4jR
1)4(
0 0
=ω
Y
=
ω )4( 0
)C
1L(jRLRjCLLjCj
1R
)Cj
1R(Lj)Cj
1R//(
Lj
Z
ω
−ω+
ω+ω
=ω+ω
=
1)LCC
R(0
)C
1L(R
C
1LC
LLR)
Z
2 2
2
=+ω
→
=ω
−ω+
=
Thus,
2
2CRLC
1+
=ω
Chapter 14, Solution 38
2 2
R
LjRCjCjLjR
1Y
ω+
ω
−+ω
=ω+ω+
At resonance, Im(Y)=0, i.e
Trang 26L
0 2
0
ω+
ω
−ω
C
LL
R 2 2
0
2 +ω =
2 3 - 6
3
2
-2
50)
1010)(
1040(
1L
RLC
=ω+ω
=
ω ( ) 2 (88)x10 1762
2 1 o
nF89.1910x2x10x8
1BR
1CRC
1C
1LLC
1
9 2
o 2
π
=ω
−
0
10x20x10x15
1LC
1
1.8257 k rad/sec
Trang 27=ω
1RC
1B
C
2 1
2
+
C = and C1 = 20 µF, we then obtain C2 = 20 µF
Therefore, to increase the bandwidth, we merely add another 20 µF in series with the first one
Chapter 14, Solution 41
(a) This is a series RLC circuit
Ω
=+
4.0
1LC
L
=
=L
R
B 8 rad / s
(b) This is a parallel RLC circuit
F263
)6)(
3(F
6andF
+
→
µµ
F2
C= µ , R= k2 Ω, L=20mH
Trang 28102(
1LC
1
3 - 6
=
)1020)(
105(
102L
R
3 0
102(
1RC
LjRC
j
1LjR
Cj
LjR
2
ω+
=ω+ω+ω
ω+
=
Z
2 2 2 2 2
2
in (1 LC) R C
)RCjLC1
)(
LjR(
ω+ω
−
ω
−ω
−ω+
=
Z
At resonance, Im(Zin)=0, i.e
CR)LC1
(L
0=ω −ω2 −ω 2
CRL
2 = −ω
=
−
=ω
LC
CR
L 2
R C
−(b) Zin = jωL||(R+1 jωC)
RCj)LC1
(
)RCj1(LjCj1LjR
)Cj1R(Lj
2
ω+ω
=ω+ω+
ω+ω
=
Z
2 2 2 2 2
2 2
]RCj)LC1
[(
)LjRLC-
ω+ω
−
ω
−ω
−ω+ω
=
Z
At resonance, Im(Zin)=0, i.e
LCR)LC1
(L
0=ω −ω2 +ω3 2 2
Trang 291
−
(c) Zin =R||(jωL+1 jωC)
RCj)LC1
(
)LC1
(RC
j1LjR
)Cj1Lj(R
ω+ω
=
Z
2 2 2 2 2
2 2
in (1 LC) R C
]RCj)LC1
)[(
LC1
(R
ω+ω
−
ω
−ω
−ω
(R
0= −ω2 ω
0LC
1−ω2 =
=
ω0
LC 1
ω
=
Cj
1R
||
LjR
LjR
2 1
1 in
Z
Trang 30LjRCj
1R
Cj
1RLjR
LRj
1
1 2
2 1
1 in
ω+
ω+
ω+
⋅ω+
1
2 1
in (R j L)(1 j R C) LCR
)CRj1(LRj
ω
−ω
+ω+
ω+ω
=
Z
)CRRL(jLCRLCR
R
LRjLCRR-
2 1 2
2 1
2 1
1 2
1 2
ω+ω
=
Z
2 2 1 2
2 2
2 1
2 1
2 1 2
2 1
2 1 1 2
1 2
)]CRRL(jLCRLCR
R)[
LRjLCRR-
+ω+ω
−ω
−
+ω
−ω
−ω
−ω
+ω
=
Z
At resonance, Im(Zin)=0, i.e
)LCRLCR
R(LR)CRRL(LCRR
1
2 1 1 2
1 2
1 3 2
1 2 2 2
2 1
ω
=
LC1
CR
2
2 + −ωω
=
1)CRLC
0
CRLC
1
−
=ω
2 6 - 2
6 - 0
)109()1.0()109)(
02.0(
=
ω0 2 357 krad / s
(b) At , ω=ω0 =2.357krad/s
14.47j)1020)(
10357.2(jL
jω = × 3 × -3 =
0212.0j9996.014.47j1
14.47jLj
||
+
=ω
14.47j1.0)109)(
10357.2(j
11
.0Cj
1
×
×+
=ω+
)Cj1R(
||
)Lj
||
R()
Z
Trang 31)14.47j1.0)(
0212.0j9996.0()( 0
−+
=ω
Z
=
ω )( 0in
1j
2j
31
ω++ω+ω
=
2
2C1
jCC-j0.5jC
1
jCj-j1.5
1
+
++
=+++
1
C-0.5
++
2
1C1
C1
2
2
Z Z
20)10)(
2
=
= I Z V
V)tsin(
20)t(
v = , i.e Vo = 20 V
Trang 32Chapter 14, Solution 45
(a)
ω+
ω
=ω
j1
jj
||
ω+
=ω+
ω
=
1j
11
j1j
1
||
1Transform the current source gives the circuit below
ω
j 1
I j 1
j
ω+
ω+
ω
⋅ω+
ω+ω++
ω+
=
j1jj1
jj1
11
j11o
=
=ω
I
V
) j 1 ( 2
j
ω+ω
)j1(2
1)
1(
1)
1(
Chapter 14, Solution 46
(a) This is an RLC series circuit
nF26.1110
x10x)10x15x2(
1L
1CLC
1
3 2
3 o
2
π
=ω
Trang 33Chapter 14, Solution 47
RLj1
1L
jR
R)
(
i
o
ω+
=ω+
1)0(
H = and H(∞)=0 showing that this circuit is a lowpass filter
At the corner frequency,
2
1)(
H ωc = , i.e
R
L1R
L1
12
2 c
L
R
c =ω
=
=ω
=
×
×
⋅π
=
⋅π
10102
1L
R2
1
Chapter 14, Solution 48
Cj
1
||
RLj
Cj
1
||
R)
(
ω+
CjRLj
Cj1R
CjR)
(
ω+
ω+
ω
ω+
j R
R
2
ω
−ω+1)
0
(
H = and H(∞)=0 showing that this circuit is a lowpass filter
Trang 34Chapter 14, Solution 49
2
4)0
1)(
2 c1004
42
2
ω+
=
2.08
100
c = → ω =ω
+
10j1
220j2
4)2(H
+
=+
=
199.0101
2)2(
In dB, 20log10 H(2) = - 14.023
=
=-tan 10)
2(Harg -1 - 84.3°
Chapter 14, Solution 50
LjR
Lj)
(
i
o
ω+
0)
R1
12
1)
(
c 2
=
=ω
H
or c 2 fc
L
Rπ
=
=ω
=
⋅π
=
⋅π
=
1.0
2002
1L
R2
1
Trang 35Chapter 14, Solution 51
RC1j
jRC
j1
RCj)
(
+ω
ω
=ω+
ω
=ω
′
This has a unity passband gain, i.e H(∞)=1
50RC
1
c =ω
=
ω+
ω
=ω
′
=ω
j50
10j)(10)
ω
j 50
10 j
=π
1 =2πf =20π×10
ω
3 2
2 =2πf =22π×10
ω
3 1
B=ω −ω = π×
3 1
2
2ω = π×
+ω
=
ω
Trang 36C
1LLC
1
2 0
=
)1080()1021
(
1
L 3 2 -12 2 872 H
BLRL
R
B= → =
=
×π
=π
=
→
=π
=
10x300x10xx2
1C
f2
1RRC
1f2
12 3
c c
c
Chapter 14, Solution 56
s/krad10)104.0)(
1025(
1LC
1
6 3
10L
10
s/krad8.92.0102Bo
1 =ω − = − =
2
8.9
π
=f
s/krad2.102.0102Bo
2 =ω + = + =
2
2.10
1 < <
Trang 37Chapter 14, Solution 57
(a) From Eq 14.54,
LC
1L
RssL
RsLC
ssRC1sRC
sC
1sLR
R)
s(
2 2
++
=+
+
=++
0
2 sB s
sB
ω++
(b) From Eq 14.56,
LC
1L
Rss
LC
1s
sC
1sLR
sC
1sL)
s(
2
2++
+
=++
+
=
H
=)s(
0 2
2 0 2
sB s
s
ω++
ω+
Chapter 14, Solution 58
(a) Consider the circuit below
R 1/sC
1RsC
1RsC
1R
||
sC
1R)s(
=
Z
Trang 38sRC1R)
s
(
+
++
=
Z
)sRC2(sC
CRssRC31)
s
(
2 2 2+
++
sC2
sC
V I
I
2 2 2
s 1
)sRC2(sCsRC
2
RR
++
+
⋅+
=
V
2 2 2 s
o
CRssRC31
sRC)
s
(
++
=
=
V
V H
=
2 2
2
CR
1sRC
3s
sRC33
1)
3
B 3 rad / s
(b) Similarly,
sLR2
)sLR(RsL)sLR(
||
RsL)
s
(
+
++
=++
=
Z
sLR2
LssRL3R)
s
(
2 2 2
+
++
RsL
R2
V I
I
2 2 2
s 1
sLR2sL
R2
sLRsL
++
+
⋅+
=
⋅
V
Trang 392 2
2 2 2
s o
L
RsL
R3s
sL
R331
LssRL3R
sRL)
s(
++
R3
)1040)(
1.0(
1LC
1
12 -
3
1021.0
102L
R
25010
2
105.0B
=+ω
Trang 401sL
||
R)s(
++
=
Z
LCssRC1
)LCs1(R)s
2++
+
=
Z
LCRsRLCRsCsRRR
)LCs1(R
o
2 o o
2 o
i
o
+++
+
+
=+
=
=
Z
Z V
V H
LCssRC1
)LCs1(RR
2 o
o
++
=+
Z
LCssRC1
LCRsRLCRsCsRRR
2
2 o
2 o o
+++
RCjLC1
LCRR
LCRC
RRjR
2
2 o
2 o o
ω
−+ω
−ω
+
=
Z
2 2
2
2 o
2 o
2 o
)RCjLC1
)(
CRRjLCRLCR
RR(
ω+ω
−
ω
−ω
−ω
+ω
−ω
−+
=
Z
0)Im(Zin = implies that
0)LC1
(CRR]
LCRLCR
RR[RC
o
2 o
2
ω
Trang 41LCRLCR
R
o
2 o
101(
1LC
1
6 - 3
LCRLCR
RCRR
j
R
)LC1
(R
2 o
2 o
o
2
ω
−ω
−+ω
R)
or
o o
o 2
o
2 max
RR
R)
RR(LCCRRjRR
LC
1Rlim
)(HH
+
=+
−ω
+ω+
R
)LC1
(R)
2 o
2 o
2 o
2 o
))RR(LCR
R()CRR
(
)LC1
)(
RR(2
1
+ω
−++ω
ω
−+
=
2 8 - 2
2 6 -
-9 2
)10410
()10
(96
)1041
(102
1
×
⋅ω
−+ω
×
×
⋅ω
−
=
2
1)10410
()10
(96
)1041
(10
0
2 8 - 2
2 6 -
-9 2
−
×
⋅ω
−+ω
×
×
⋅ω
−
=
0)10410
()1096()2)(
2 -6 2
()
10
96
( × -6ω 2 − −ω2 ⋅ × -8 2 =
010010
092.810
6
1 × -15ω4− × -7ω2 + =
Trang 42101471.2
109109.2
Hence,
s/krad653.14
1 =ω
s/krad061.17
2 =ω
=
−
=ω
−ω
Cj1
V V
ω+
1
V
V =ω
+
=
=ω
i
o)(
V
V H
RC j 1
1
ω+
Cj1R
R
V V
ω+
=
Since V+ = V−,
o iRCj1
RCj
V
V =ω
+ω
=
=ω
i
o)(
V
V H
RC j 1
RC j
ω+ω
Trang 43Chapter 14, Solution 62
This is a highpass filter
RCj1
1RC
j1
RCj)(
ω
−
=ω+
ω
=ω
H
ωω
−
=ω
cj1
1)
f1000j1
1f
fj1
1)
(
c = −
−
=ω
H
(a)
i
o5j1
1)Hz200f
mV120o
(b)
i
o5.0j1
1)
kHz2f
mV120o
(c)
i
o1.0j1
1)kHz10f
mV120o
Chapter 14, Solution 63
For an active highpass filter,
i i
f i
RsC1
RsC)
s(H
+
−
Trang 44But
10/s1
s10)
s(H
RC
i f f
.0RC
i i i
i
Chapter 14, Solution 64
f f
f f
f
RC
j
1
||
RZ
ω+
=ω
=
i
i i i
i
CRj1Cj
1RZ
ω
ω+
=ω+
i
f i
o )
-(
Z
Z V
V H
) C R j 1 )(
C R j 1 (
C R j -
i i f
f
i f
ω+ω
+
ω
This is a bandpass filter H(ω) is similar to the product of the transfer function
of a lowpass filter and a highpass filter
Chapter 14, Solution 65
i
i 1 j RC
RCjC
j1R
R
V V
V
ω+
ω
=ω+
=+
o f i
iRR
R
V V
+
=
−
Since V+ = V−,
Trang 45i o
f i
i
RCj1
RCjR
R
R
V V
ω+
ω
=+
=
=ω
i
o)(
RC j 1
RC j R
R 1
i f
It is evident that as ω→∞, the gain is
sR
R
R)
s(
2 4
3
4+
⋅+
=
H
(c) When R 3 →∞,
CR1s
CR1-)s(
2
1+
=
H
Chapter 14, Solution 67
f i i
4
1R
Rgain
s/rad)500(2CR
1frequency
Corner
f f
500)(
2(
1
×π
=Therefore, if Rf = 20 kΩ, then Ri = 80 kΩ and C=15 915 nF
Trang 46Chapter 14, Solution 68
i f i
f
R5RR
R5gainfrequency
s/rad)200(2CR
1frequency
Corner
i i
200)(
2(
1
×π
=Therefore, if Ri = 20 kΩ, then Rf =100 kΩ and C=39 8 nF
Chapter 14, Solution 69
This is a highpass filter with fc = kHz 2
RC
1f
2 c
c = π =ω
3
c 4 10
1f
2
1RC
×π
=π
1040001
C 7 96 nF
Trang 47Chapter 14, Solution 70
(a)
)YYY(YYY
YY)
s(
)s()s(
3 2 1 4 2 1
2 1 i
o
+++
=
=
V
V H
H
) sC G G ( sC G G
G G
1 2 1 2 2 1
2 1
+++
GG
GG)0(H
2 1
K1L
f m
-6 f
101K
K
C
m f
6
K
1K
Substituting (1) into (2),
f f
6
K25
1K
10 =
=f
Trang 48Chapter 14, Solution 72
CL
LCK
K
LCC
f 2
-3 2
)2)(
1(
)1020)(
104(
LKK
6 2
)104)(
2(
)1020)(
1(
800(
10300K
K
CC
-9 f
m
pF 375 0
10LK
=
nF110101K
K
C'
C
8 f
m
=
=
=
Trang 491K
K
C'
C
5 f
x50RK'
R
3
3 3
m
mH1010
10x10x10LH
10LK
K
CF
p40
'
f m
10B
RLL
R
B= → = = =
F5.312)2)(
1600(
1L
1CLC
1
2 0
CC
-4 m
F 5208
Trang 50(b) ′= = 3 =
f 10
2K
400(LK
400(
10125.3KK
C
-4 f
m
pF 81 7
10LK
K
3 f
=
′
F1.0)10)(
10(
1K
K
C
f m
Trang 51Also,
sC1sLsLsC
1sL
2
V V
V
Combining (2) and (3)
o o
1
sC1sL
2
o 1 4s LC
LCs
V V
+
Substituting (3) and (4) into (1) gives
1 2 o
1
LCs41
sCsL
R
1
V V
2 1
2
sRCLCs41LCs41
sRC
+
++
=+
+
=
sRCLCs41
LCs412
Trang 52sRCR
o in
++
=
=
I Z
sC
1RsL4
in = + +
When R =5, L=2, C=0.1,
=)s(in
Z
s
10 5 s
At resonance,
C
1L40)Im( in
ω
−ω
1.0(2
1LC
10LK
KL
m
10)100)(
10(
1.0K
Z
s
10 50 s 8 0
4
++
=
=
=ω
)10)(
2.0(2
1LC
2
1
4 -
Trang 53Chapter 14, Solution 80
(a) R′=KmR =(200)(2)=400Ω
mH2010
)1)(
200(K
LK
200(
5.0K
K
C
f m
.0sL
2 x 2
I V
V
=+
−
But, Ix =sC V1
Trang 545.0sL
2 1 2
V V
V
=+
−
(2) Solving (1) and (2),
1sCR5.0LCs
RsL2
RsL
1025.0)(
10j(5.0)1025.0)(
1020()10j(
400)1020)(
10j(
6 - 4
6 - 3
2
-4
3 - 4
.0j5.0
200j400
Z 632 5∠- 18 435°ohms
Chapter 14, Solution 81
(a)
1GR)LGRC(jLC
RLjZ
toleadswhich
LjR
1)LjR)(
CjG(LjR
1C
jGZ1
ω
−
+ω
=
ω+
+ω+ω
+
=ω++ω+
=
LC
1GRC
GL
Rj
LC
RC
j)
(Z
j2
)1j(1000)
(
++ω+ω
−
+ω
=
Trang 55Comparing (1) and (2) shows that
LR1
R/LmF,
1C1000
C
1CG2
C
GL
R + = → = = mS
L4.0RR
10
1R10LC
1GR2501
4.0K
L'L
3 f
C'C
3
3 f
C
C′=
2001
110
1K
1C
Trang 56Chapter 14, Solution 83
pF1.010x100
10CKK
1'CF
f m
5µ → =
Ω
=Ω
Trang 57Chapter 14, Solution 85
We let Is =1∠0o A so that Vo/Is =Vo The schematic is shown below The circuit
is simulated for 100 < f < 10 kHz
Trang 58Chapter 14, Solution 86
The schematic is shown below A current marker is inserted to measure I We set
Total Points = 101, start Frequency = 1, and End Frequency = 10 kHz in the
AC sweep box After simulation, the magnitude and phase plots are obtained in the Probe menu as shown below