IRW-ANSW.I
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Page 1
Answers to
Selected Problems
Chapter 1
1.1
1.4
1.7
1.10
1.13
1.17
1.20
i=2A
Q=1.5 C
(a) P=–18 W (supplied)
(b) P=–18 W (supplied)
(a) P=–12 W (supplied)
(b) P=24 W (absorbed)
P2=48 W (absorbed)
P36V=–144 W, P1=48 W, P2=48 W, PDS=–8 W, P3=56 W
Vx=18 V
Chapter 2
2.1
2.4
2.7
2.10
2.12
2.14
2.17
2.19
2.22
2.25
2.27
2.29
2.31
2.34
2.36
2.39
2.41
2.44
2.47
I=0.3 mA, P=1.8 mW
Rx=5 k
PS=72 W
I1=6 mA, I2=3 mA
Ix=–8 mA, Iy=±10 mA, Iz=–2 mA
Ix=9 mA, Iy=–10 mA, Iz=–2 mA
Vx=3 V
Vad=7 V
Vad=9 V, Vce=11 V
Vx=10 V
Vx=5 V
P30k=1.2 mW
Io=90 mA
Io=4 mA
IL=0.4 mA
RAB=3 k
RAB=2 k
(a) Min=950 , Max=1050 (b) Min=460.6 , Max=479.4 (c) Min=19.8 k, Max=24.2 k
(a) RNom=3 (b) Positive/Negative
Tolerances=; 8.33%
2.55
2.58
2.60
2.63
2.65
2.68
2.70
2.72
2.75
2.78
2.80
2.83
I1=3 mA
16
Vo = V
3
Io = 2 mA
VS=12 V
VS=48 V
VS=38 V
VS=30 V
IS=4.5 A
Io=6 mA
P=63 mW
Io=4 A
Vo=10 V
Vo=2 V
Io=–20 mA
2.86
Power gain=1.422
2.50
2.52
kW
W
2.87 P10k=10 mW
2FE-1 P=1.2 W
2FE-4 Io =
-4
mA
9
1
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ANSWERS TO SELECTED PROBLEMS
Chapter 3
3.1
3.3
3.5
3.8
3.10
3.13
3.15
3.18
3.21
3.24
3.27
3.30
3.33
3.36
3.39
3.42
3.44
Io=1 mA
V2=22 V
Io=0.6 mA
Io=1.25 mA
Io=2 mA, I1=–6 mA
Io=–1 mA
-5
Vo =
V
6
Io=–1.5 mA
Vo=2 V
Io=–4.8 mA
Vo=4.36 V
Io=1.5 mA
4
Vo = V
3
Io=–0.4 mA
Vo=32.25 V
Vo=4 V
4
Vo = V
3
3.47
Io=7 mA
3.50
Io=5.2 mA
3.53
Io=0.4 mA
3.56
Vo=6 V
3.59
Vo =
3.62
Vo=3 V
3.65
Vo=6 V
3.68
Vo =
3.70
Vo=–5 V
3.73
Vo
= -1
iS
3FE-1 Vo =
8
V
5
-7
V
8
10
V
3
3FE-4 Vo=–3.27 V
3FE-6 Vo=6 V
Chapter 4
4.8
8
mA
7
Vo=0.75 V
-16
mA
Io =
5
Vo=–4.25 V
4.50
4.10
Io=2.4 mA
RAB=1 k
4.52
Vo=–6 V
4.12
Io=0.5 mA
4.55
Io=5.71 mA
4.15
Io=0.4 mA
4.58
Vo=0.43 V
4.18
Vo=10.5 V
4.61
Vo=2.18 V
4.21
Vo=2 V
4.64
Io=1.2 mA
Io =
4.67
4.24
Vo=258 mV
Io=–1 mA
4.70
4.27
RL=2 k, PL = 12.5 mW
4.29
Io=–0.2 mA
4.72
RL=6 k,
4.31
Io=0.67 mA
4.75
Vo=2 V
4.33
Vo=4.8 V
4FE-1 PL=8 mW
4.36
Vo=8 V
4FE-3 RL=12.92 4.1
4.3
4.5
Io =
-7
mA
5
8
V
5
4.39
Vo =
4.42
Io=2 mA
4.44
Io=1.25 mA
4.47
Vo=1.55 V
PL =
25
mW
6
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ANSWERS TO SELECTED PROBLEMS
Chapter 5
5.1
5.3
5.5
5.7
5.9
5.11
5.14
i(t) (A)
5.16
v(t=4)=40 V
C=120 F
i(t)=; 2.92 cos 377t A
(a) i(t)=4.52 cos 377t A
(b) w(t)=360 sin (754t-90°) mJ
v(t)=100t V
0 t 2 ms
=0.2 V
t>2 ms
i(t)=6 mA
0 t 2 ms
=–6 mA
2 t 4 ms
i(t)=0.6 A
0 t 2s
=–2.4 A
2 t 3s
=0.6 A
3 t 5s
=0
t>5 s
i(t)=24 A
0 t 6 s
=–60 A
6 t 10 s
=16 A
10 t 16 s
=0
t>16 s
5.19
30
10
–10
–30
–50
–70
0
2
4
6
8 10 12 14 16 18
t (s)
i(t)=377 cos (2000t) mA
400
i(t) (mA)
IRW-ANSW.I
0
–400
0
0.2
0.4
0.6
t (s)
5.21
5.24
5.27
(a) v(t)=75.4 cos 377t V
(b) w(t)=0.1-0.1 cos 754t J
(a) v(t)=0,
t<0
t>0
=250e–t V,
(b) w(t)=1.25C1-2e–t+e–2t D J
v(t=5 s)=–0.91 V
w(t=5 s)=91.97 J
0.8
1
3
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5.31
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ANSWERS TO SELECTED PROBLEMS
0 t 2 ms
2 t 4 ms
4 t 8 ms
8 t 10 ms
t>10 ms
v(t)=0
=–2.5 V
=2.5 V
=–2.5
=0
v(t) (V)
5.29
3-05-2001
3
2
1
0
–1
–2
–3
0
2
v(t)=6.67 mV
=–10 mV
=0
=25 mV
=0
4
6
t (ms)
0t
3t
6t
9t
t>11 s
8
10
12
3s
6s
9s
11 s
v(t) (mV)
30
15
0
–15
5.33
0
3
6
t (s)
9
12
0 t 2 ms
2 t 3 ms
t 7 3 ms
i(t)=0.5t A
=3*10–3-t A
=0
60
3
40
2
20
0
–2
0
–3
–20
v(t)
i(t) @ CMIN
i(t) @ CMAX
–40
–60
0
1
2
3
4
t (ms)
5
6
7
–5
8
–6
i(t) (mA)
ic(t) (mA)
5.36
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ANSWERS TO SELECTED PROBLEMS
1.5
0.4
1
0.2
0.5
0
0
–0.2
i(t)
v(t) @ LMIN
v(t) @ LMAX
–0.5
–1
–1.5
5.41
0
1
2
3
t (ms)
4
C1=1 F, C2=3 F, and C3=4 F
C1
C2
C3
5.44
5.46
5.49
5.51
5.54
5.56
5.59
5.61
5.64
Vo=12 V
C=3 F
CT=2 F
CT=9 F
(a) CNOM=1.43 F
(b) CMIN=1.254 F, CMAX=1.606 F
(c) %MIN=–12.3%, %MAX=12.3%
L=20 mH
LAB=5 mH
LAB=6 mH
C=1.25 F
5FE-1
2 F
4 F
6 F
Chapter 6
6.1
6.3
6.5
vC(t)=12-8e–t/0.6 V, t>0
vC(t)=6e–t/0.4 V, t>0
2
io(t) = e-10t A, t>0
3
5
–0.4
–0.6
6
–0.8
v(t) (V)
5.38
i(t) (A)
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6.7
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ANSWERS TO SELECTED PROBLEMS
48
A1 - e-11t6 B V, t>0
11
=0,
t>0
vo(t) =
4.5
4
3.5
vo(t) (V)
3
2.5
2
1.5
1
0.5
0
2
4
vC(t)=4 V,
=4e–t/1.2 V,
6
8
10
t (s)
12
14
16
18
20
t<0
t>0
5
4
3
vC(t) (V)
6.9
0
2
1
0
–1
–10
0
10
20
30
40 50
t (s)
60
70
80
90
100
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ANSWERS TO SELECTED PROBLEMS
6.12
vC(t)=6 V,
=6e–15t/4 V,
t<0
t>0
7
6
vC(t) (V)
5
4
3
2
1
0
–1
–2
–1
0
1
2
3
4
5
0.1
0.15
t (s)
6.15
i(t) = 2 mA,
t<0
= A4e-2 * 10 t - 2B mA, t 7 0
6
3
2
1
i(t) (mA)
IRW-ANSW.I
0
–1
–2
–3
–0.2
–0.15
–0.1
–0.05
0
t (s)
0.05
0.2
7
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6.18
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ANSWERS TO SELECTED PROBLEMS
vo(t)=0,
t<0
=–9.6e–9.6t V, t>0
–10
vo(t) (V)
–8
–6
–4
–2
0
–1
0
0.5
t (s)
1
1.5
2
io(t)=1 mA,
t<0
=0.5e–(10/3)t mA, t>0
2
1.5
1
io(t) (mA)
6.21
–0.5
0.5
0
–0.5
–1
–5
–4
–3
–2
–1
0
1
t (s)
2
3
4
5
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ANSWERS TO SELECTED PROBLEMS
6.24
vo(t)=6,
t<0
=12-3e–5t/3 V, t>0
13
12
11
10
vo(t) (V)
IRW-ANSW.I
9
8
7
6
5
4
–2
0
2
4
t (s)
6.27
6.30
io(t)=–3e–10t A, t>0
vo(t)=9e–5t/3 V, t>0
6.33
io(t) = 2.4A1 - e-2.5 * 10 t B mA, t>0
6.36
io(t) = 2 -
5
6.39
6.42
6.45
6.47
e-3t
mA, t>0
4
vo(t)=1.5e–t/0.6 V, t>0
vo(t)=–3.6e–8t V, t>0
io(t)=–0.5e–5t mA, t>0
vo(t)=–6e–4t V, t>0
6.50
vo(t) = -4e-0.889 * 10 t V, t>0
6.52
6.55
6.58
6.61
6
4
2 -9t2
e
A, t>0
3
15
2
(a) s +6s+8=0
(b) s=–2, s=–4
(c) io(t)=k1 e–2t+k2 e–4t
(a) s2+6s+10=0
(b) s=–3+j, s=–3-j
(c) vo(t)=k1 e–3t cos t+k2 e–3t sin t
v(t)=10e–4t cos 2t-40e–4t sin 2t
io(t) =
6
8
10
9
IRW-ANSW.I
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6.64
6.67
6.70
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ANSWERS TO SELECTED PROBLEMS
vo(t) = 0
t<0
5
5
= 16.67Ae-2 * 10 t - e-8 * 10 t B V, t>0
32 -8t 4 -t
i(t) =
e - e A, t>0
7
7
vo(t)=8te–10t V, t>0
vo (t) (V)
0.3
0.2
0.1
0
0
0.2
0.4
t (sec)
0.6
0.8
6.72
1
40
2
150 K
slope=150 kV/s
33.33 V
30
100 K
25 V
20
50 K
slope=37.5 kV/s
10
W
0
0
0
0.5
1 n V(C1:1)
1.0
1.5
2 n
: D(V(C1:1))
t (ms)
6.75 R=2.5 k, C=10 pF, L=333 H
6FE-2 vo(t=1s)=3.79 V
2.0
2.5
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ANSWERS TO SELECTED PROBLEMS
Chapter 7
7.1
7.3
T=0.16 s, f=63.66 Hz
i1(t) leads i2(t) by –85°
i2(t) leads i3(t) by 145°
i1(t) leads i3(t) by 60°
7.6
(a) i(t)=8 cos(377t+68°) A, I = 8 / 68° A
(b) i(t)=4 cos(377t+64°) A, I = 4 / 64° A
7.8
7.10
7.13
7.16
7.18
7.20
7.22
Z=1+j1 Z=1.6+j0.8 Z=5.1+j4.96 Z = 5 / -37° C=431 F
i(t)=4.37 cos(377t+0.75°) A
VR = 8.74 / 0.75° V
VL = 16.47 / 90.75° V
VC = 11.59 / -89.25° V
IMAG
j4
VS
VL+VC
j2
j0
0
7.25
7.28
7.31
7.34
7.37
7.40
7.42
7.44
7.47
7.49
7.52
7.55
VR
2
4
6
8
Real
IR = 9.99 / 27.84° A and IC = 0.38 / 117.84° A
Vo = 10 / -53.1° V
Vo = 1.414 / 15° V
Io = 5.89 / -48.4° A
VS = -8.54 / -20.56° V
IS=8+j4 A
Z = 2 / 83° Io = 4.69 / 78.69° A
Vo = 3.09 / -23.83° V
Vo = 3.58 / 153.43° V
Vo = 5.55 / -86.9° V
Vo = 0.8 + j2.4 V
11
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ANSWERS TO SELECTED PROBLEMS
7.73
Vo = 5.41 / 4.57° V
Io = 2 / -37° A
Vo = 9.03 / 51.3° V
Vo = 1.3 / 12.5° V
Vx = 48.59 / -21.37° V
Vo = 2.53 / -18.43° V
7.76
PROBE results show that the voltage and current phases are equal at 238.9 Hz.
7.58
7.61
7.64
7.67
7.70
40 mA
30 mA
iL
(238.93,18.68 m)
20 mA
iC
10 mA
100 Hz
I(L1)
200 Hz
I(C1)
Frequency
7FE-1 Vo = 5.06 / -71.6° V
7FE-4
Vo
= -133.33
VS
Chapter 8
8.1
di1(t)
di2(t)
- M
dt
dt
di2(t)
di1(t)
vb(t) = -L2
- M
dt
dt
di1(t)
di2(t)
(b) vc(t) = L1
+ M
dt
dt
di2(t)
di1(t)
vd(t) = L2
+ M
dt
dt
(a) va(t) = -L1
300 Hz
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ANSWERS TO SELECTED PROBLEMS
8.7
di1(t)
di2(t)
+ M
dt
dt
di2(t)
di1(t)
- M
vb(t) = -L2
dt
dt
di1(t)
di2(t)
- M
(b) vc(t) = -va(t) = -L1
dt
dt
di2(t)
di1(t)
vd(t) = -vb(t) = L2
+ M
dt
dt
Vo = 2.98 / 26.57° V
8.9
Vo = 2.24 / -153.43° V
8.11
Vo = 20.86 / 4.32° V
8.14
Io = 1.78 / 42° A
8.4
8.17
(a) va(t) = L1
-V1 = I1 AR1 + jL1 B + jM I3
V1 = I2 a jL2 -
j
b - jL2 I3
C1
0 = jM I1 - jL2 I2 + I3 AR2 + jL2 + jL3 B
8.20
Vo = 1.36 / -85.4° V
8.23
Vo = 0.64 / -71.57° V
8.26
Vo = 5.79 / 86.31° V
8.29
Vo = 8.76 / 158.8° V
8.32
Io = 2.78 / -56.31° A
8.35
Zsource = 1.56 / 42.27° 8.38
ZIN = 1.94 / -33.69° L2=3.6 mH
i1(t)=2.46 cos(100t+143.1°) mA
i2(t)=1.54 cos(100t-178.24°) mA
V1 = 8.89 / 30° V, I1 = 1.11 / 30° A
8.41
8.44
8.47
V2 = 4.44 / -150° V, I2 = 2.22 / -150° A
8.49
I1 = 3.16 / -41.56° A, V1 = 4.47 / 3.44° V
8.58
I2 = 1.58 / 138.44° A, V2 = 8.94 / 3.44° V
ZIN=1.5+j0.25 ZS=16-j1 Vo = 44.72 / -153° V
8.60
IS = 2.91 / -75.95° A
8.62
Vo = 15.78 / 189.46° V
8.65
Vo = 1.8 / -139.86° V
8.52
8.55
8FE-1 ZS = 4.88 / 19.75° 8FE-4 P=11.1 W
13
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ANSWERS TO SELECTED PROBLEMS
Chapter 9
9.1
9.3
9.5
9.8
9.11
9.13
9.16
9.18
9.21
9.24
p(t)=11.51+14.4 cos(2t+113.1°) W
P=1.58 W
PS=4.31 W, P2=3.06 W, P4=1.23 W
PABS=35.95 W
P4=10.4 W
PR=4.5 W
PR=2.5 W
PR=32.49 W
ZL=5 , PL=5.28 W
ZL = 0.55 / 33.69° , PMAX=0.42 W
9.26
9.29
ZL=0.9-j0.3 , PMAX=2 W
ZL = 2.83 / 8.13° , PMAX=1.32 W
9.32
9.34
9.37
9.40
9.43
9.46
9.49
9.51
ZL=0.2+j0.4 , PMAX=28.9 W
Vrms=2.31 V
Vrms=1.63 V
Vrms=2.67 V
VL=440 V rms
=36.87°
PF=0.65 Lagging
VS = 281.02 / 8.75° V rms
PFsource=0.756 Lagging
9.54 VS = 320.06 / 9.95° V rms
9.57 C=567.6 F
9.60 C=305 F
9.63 PF=0.88 Lagging
9.65 I=18 A
9.68 Itouch=1.26 A rms, no current near the heart
9FE-1 C=927.6 F
9FE-3 ZL=0.4-j1.2 Chapter 10
Typically, only the a-phase information is listed. The two remaining phases are shifted by –120° and –240°, respectively.
10.1
Vab = 173 / 25° V rms
Vbc = 173 / -45° V rms
Vca = 173 / -165° V rms
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ANSWERS TO SELECTED PROBLEMS
Vab
Van
Vcn
Vca
Vbc
Vbn
10.4
Vab = 208 / 90° V rms
Vbc = 208 / -30° V rms
Vca = 208 / -150° V rms
Vab=208 90° V rms
Van=120 60° V rms
Vcn
Vbc
Vca
Vbn
10.7
Ian = 2.45 / -14° A rms
10.10
Ian = 5.56 / 6.3° A rms
Van = 111.1 / 59.40° V rms
10.13 Vab = 217.4 / 40° V rms
10.16 ZL = 15.62 / 39.8° 10.19 IaA Max = 67.42 A rms
10.22 ZL=19.95+j21.93 10.25 Vab = 242.11 / 40.09° V rms
10.28 IaA = 19.52 / 39.4° A rms
15
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ANSWERS TO SELECTED PROBLEMS
10.31
10.34
ZL=70.48-j25.65 ZL = 32.18 / 25° 10.37
IAN = 9.37 / -4.4° A rms
10.40
Iab = 8.64 / 57.9° A rms
10.43
10.46
10.49
IaA = 37.35 / -1° A rms, PY Load = 7.434 kW
@IL @=10.25 A rms
@IaA @=148.56 A rms, PFLoad=0.74 Lagging
10.52
10.55
10.58
10.61
10FE-1
10FE-4
PFS=0.91 Lagging
SuL=19.94 kVA@0.60 Lagging
PF=0.97 Lagging
C=740.9 F
ST = 2160 / 45° VA
Pp=6.928 kW
Chapter 11
s2LCR + sL + R
s2LC + 1
8s(s + 1)
11.1
Z(s) =
11.4
V0
= 2
Is
2s + 6s + 1
11.7
MAG.
PHASE
0 dB
–20 dB/dec
0°
–45°
0.1
11.10
0.5
0.1
Log MAG.
0.5
Log PHASE
–20 dB/dec
–20 dB/dec
0°
0 dB
–90°
0.1
1
10
Log 10–1
1
10
Log IRW-ANSW.I
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ANSWERS TO SELECTED PROBLEMS
11.13
MAG.
0 dB
–20 dB/dec
±20 dB/dec
–40 dB/dec
1
11.16
MAG.
5
10
50
Log –40 dB/dec
–20 dB/dec
0 dB
–60 dB/dec
0.2
11.19
1
10
100
Log MAG.
0 dB
–20 dB/dec
±40 dB/dec
–60 dB/dec
1
®10
10
100
Log 17
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ANSWERS TO SELECTED PROBLEMS
11.22 MAG
PHASE
=0.1
–20 dB/dec
0 dB
–90°
–40 dB/dec
–180°
1 2
11.25
12
2
Log MAG
0 dB
|H|
=0.5
±20 dB/dec
–40 dB/dec
1
11.28
11.31
11.34
11.37
11.40
11.43
11.46
11.49
11.52
10 a
8
10
Log j
+ 1b
10
H(j) =
2
j
(j) a
+ 1b
20
j
1(j) a
+ 1b
30
H(j) =
2
j
j
(j + 1) a
+ 1b a
+ 1b
100
8
L=12.5 mH, Q=10.42, BW=192 rs
v0=7071 rs, Q=14.14, MAX=7062 rs, @Vo @ max=84.89 V
v0=2 krs, Q=25, BW=80 rs, P=18 W
R=1 k, L=500 H
v0=10 krs, BW=100 rs, Q=100, PLO=PHI=12.5 kW
C=25 nF, L=10 H
Rnew=20 k, Lnew=5 kH, C=12.5 F
12
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ANSWERS TO SELECTED PROBLEMS
11.55 Low-pass filter
Vo
=
Vi
1
j
j 2
a b +
+ 1
10
10
20
0
dB
–20
–40
–60
–80
0.1
1
a1 +
jL
b
R1
11.57 Gv =
L
1 + j a b
R
10
(r/s)
100
103
10
100
R=R1 || R2
A low-pass filter
20
0
–20
dB
IRW-ANSW.I
–40
–60
–80
0.01
0.1
1
(r/s)
19
IRW-ANSW.I
20
11.61
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ANSWERS TO SELECTED PROBLEMS
jCAR1 + R2 B + 1
Vo
=
, a high-pass filter
Vi
jCR1 + 1
MAG.
0 dB
1––
––
CR1
1
––––––––––
C(R1+R2)
Log 11.63 gm=100 S and IABC=5 A
11.66 Leq =
11.68
Vo
=
Vin
0 =
C
Agm 1 gm 2 B
jg1C2
-2 + j £
g1 + g2 + g3 + g3 a
C1
C1
g1 g3
b
C2 § +
C1 C2
g1 g3
2g1 g2 C1 C2
,Q =
A C1 C2
C2 Ag1 + g2 + g3 B + C1 g3
Band-pass filter
11.71
C=100 F, L=101 mH
11FE-2 0=1 krs, R=4 11FE-4 L=100 mH, R=10 Chapter 12
12.1
F(s)=e–(s+a)
12.4
F(s) = e-(s + a) c
12.7
F(s) =
(s + a) sin cos d
2
2 +
(s + a) + (s + a)2 + 2
e-2s
(s + 1)(s + 2)
12.10 F(s) = e-(s + a) c
1
1
d
2 +
s + a
(s + a)
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ANSWERS TO SELECTED PROBLEMS
12.13 (a) f(t) = c
1
1
2
+ e-2t - e-3t d u(t)
6
2
3
1
3
(b) f(t) = c - e-t + e-2t d u(t)
2
2
12.15 (a) f(t) = a
1 -2t
3
e + e-6t b u(t)
4
4
3
1
(b) f(t) = a + e-4t b u(t)
4
4
12.18 (a) f(t)=10e–t cos t u(t)
(b) f(t) = c
1
+ 0.62e-2t cos (t - 108.43°)d u(t)
5
12.21 (a) f(t) = C2e-3t cos 3t - e-3t Du(t)
(b) f(t) = C1 + e-4t sin 4tDu(t)
12.24 (a) f(t) = C2te-2t + e-2t Du(t)
(b) f(t) = C6 - 5te-t - 6e-t Du(t)
12.27 f(t) = c -3 + 3t +
12 -t
2
e + e-2t cos (2t - 26.56°)d u(t)
5
3
1
1 -2(t - 1)
u(t - 1)
12.30 (a) f(t) = u(t - 1) + e
2
2
(b) f(t) = C5e-(t - 2) - 5e-3(t - 2) Du(t - 2)
12.33 (a) f(t) = C-2e-(t - 1) + 4e-3(t - 1) Du(t - 1)
(b) f(t) = c
10 -(t - 2)
20 -4(t - 2)
e
+
e
d u(t - 2)
3
3
1
1
12.36 y(t) = a e-t - e-4t b u(t)
3
3
12.39 f(t)=Ae–t-e–2t Bu(t)
12.42 (a) f(0)=10, f(q)=0
(b) f(0)=0, f(q)=0
(c) f(0)=2, f(q)=0
12.45
9
- t
2 u(t)
i(t) = 4e
A
12.48 iL(t)=A4e -e–t Bu(t) A
12FE-2 vo(t=0.1 s)=0.24 V
–2t
Chapter 13
6s + 8
6s2 + 16s + 11
13.1
Z(s) =
13.3
v(t)=10 u(t) V
21
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22
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ANSWERS TO SELECTED PROBLEMS
13.5
io(t) = a 2 - e-t -
4 - 23 t
e b u(t) A
3
13.7
v(t) = A5e-t - 4500te-t Bu(t) mV
13.10
vo(t) =
13.13
vo(t) = A1 - 5e-4t Bu(t) V
7t
6
A1 - e 4 Bu(t) V
7
13.16
vo(t) = a -
13.19
vo(t) = 212e-t cos (t - 45°)u(t) V
13.22
io(t) = c 1 -
13.25
4
20 -3t
+ 2e-t e b u(t) V
3
3
2 - 4t3
e d u(t) A
3
8
17 - 3t2
vo(t) = a + 4e-2t e b u(t) V
3
3
13.28
vo(t)=5e–3tu(t) V, t>0
13.30
io(t) = -e 2 u(t) A, t>0
13.32
io(t) =
13.35
vo(t) = A4 + 2e
13.37
vo(t) = 1.15Ce
-t
13.40
13.43
13.46
13.49
13.52
13.54
13.56
13.59
13.62
13FE-2
3 - 9t2
e u(t) A, t>0
2
-
5t
12 Bu(t)
-0.42t
- e
V, t>0
-1.58t
Du(t) V, t>0
vo(t) = 2.31 Ce-0.35t - e-5.65t Du(t) V, t>0
vo(t) = A8 - 8e-4t Bu(t) - C8 - 8e-4(t - 1) Du(t - 1) V
Vo
R1 (1 + sCR)
= a1 +
b
, R = R1 || R2
VS
R2 1 + sCR1
-s
Vo
R1 C1
=
C1
C2
R1 R2
VS
s2 + s a
+
b +
C1 C2 R3
C1 C2 R3
C1 C2 R1 R2 R3
Yes
No, overdamped
C=0.5 F
vo(t)=4.7 cos(t-45°) V
io(t) = 1212 cos (2t + 45°) A
Vo
s
= 2
, the network is underdamped.
VS
s + s + 2
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ANSWERS TO SELECTED PROBLEMS
23
Chapter 14
14.1
q
2 jn0t
f(t) = a
e
n = -q jn
nZ0
n odd
14.3
q
6 -jnt
n
f(t) = a
e
sin a
b
n
5
n = -q
14.5
f(t) =
14.8
20
v(t) = a (-1)n + 1
sin nt
n
14.10
1
a0 = , bn = 0
4
q
1
-2
+ a 2 2 ejnt
2 n = -q n nZ0
n odd
q
n=1
an =
14.13
a0 =
an =
bn =
14.16
a0 =
an =
bn =
4
n
2
n
b - 1d +
sin a
b
2 c cos a
n
2
2
n
-
4
1
Acos (n) - 1B
n2
1
(1 - 2 cos n)
n
1
4
1
2 2 Acos(n) - 1B
n
1
Acos(n) - 2B
n
2
q
A
A
2A
+ sin (t) + a
cos n0 t
2
A1
- n2 B
n=2
14.19
f(t) =
14.22
f(t)=–4 sin 20t-5 sin 40t-3 sin 60t-2 sin 80t-sin 100t
n even
14.25
14.28
14.31
(-1)n + 120
∑G(n)∑ cos (nt - 90° + n)
n
jn
G(n) =
, n = / G(n)
1 + 3jn
io(t) =
io(t)=3.18 sin(2t+89.9°)-3.18 sin(4t+89.9°)+3.18 sin(6t+89.9°)-3.18 sin(8t+89.9°) mA
2A
V() =
(1 - cos T)
j
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24
14.34
14.37
14.40
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ANSWERS TO SELECTED PROBLEMS
2a
a 2 + 2
F() =
vo(t) = Ae-3t - e-4t Bu(t) V
2
vo(t) = Ce-t - e-4t Du(t) V
3
14FE-1 a0=0 since the average value is zero
an=0 for all n since this is an odd function
bn=0 for n even because of half-wave symmetry
bn=finite numbers for n odd
Chapter 15
15.1
1
-1
-1
1
,y =
,y =
,y =
ZL 1 2
ZL 2 1
ZL 2 2
ZL
= ZL , z1 2 = ZL , z2 1 = ZL , z2 2 = ZL
(a) y1 1 =
(b) z1 1
1
-1
-1
1
S, y1 2 =
S, y2 1 =
S, y2 2 = S
14
21
21
7
1
1
=
, y1 2 = 0, y2 1 =
, y2 2 =
Z1
Z2
Z2
= 18 , z1 2 = 6 , z2 1 = 6 , z2 2 = 9 15.3
y1 1 =
15.5
y1 1
15.8
z1 1
Vo
= -65.6
V1
15.11
15.13
V2
= -438
V1
15.16
z1 1 = R1 , z1 2 = nR1 , z2 1 = nR1 , z2 2 = n2 AR1 + R2 B
15.19
h1 1 = 6 , h1 2 = 0.5, h2 1 = -0.5, h2 2 =
15.22
A=1, B=–j1 , C=15, D=1-j
R1 R3 + R2 R3 + R1 R2 R1 + R2
A =
,B =
+ R2
+ R2
15.25
C =
1
S
8
R2
R2 + R3
1
,D =
+ R2
+ R2
15.28
y1 1 =
5
-2
-2
3
S, y1 2 =
S, y2 1 =
S, y2 2 =
S
11
11
11
11
15.31
h1 1 =
z1 1 z2 2 - z1 2 z2 1
z1 2
, h1 2 =
z2 2
z2 2
h2 1 =
-z2 1
1
, h2 2 =
z2 2
z2 2
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ANSWERS TO SELECTED PROBLEMS
15.33
15.36
15.39
Y1n=1 S
18 6
ZT = B
R
6 9
B
A B
3
j8
R = B
R
C D
3 - j 3 + j8
15FE-1 V1=36 V
Chapter 16
16.1
16.3
VD (V)
0
0.25
0.50
0.75
(a) 9.4 V,
ID (A)
0
1.7*10–11
2.9*10–7
5.0*10–3
(b) 10 V,
(c) 3.4 V,
(d) –1 V
16.5
Input
Ideal model
Constant voltage
Input and output voltages (V)
IRW-ANSW.I
5
0
–5
5
0
0
0.5
1
1.5
2
2.5
3
t/T
16.8
gm Req
Req
Vo
=
, R0 =
VIN
gm Req + 1
gm Req + 1
Req=RS || rds
For given values Ro=83.3 A good application is a voltage buffer much like the unity gain buffer op-amp.
16.10
vo =
16.12
RON
16.13
RON
gm RD
Av2 - v1 B
2
1
= , R1 = 149.5 2
1.2 25
26
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ANSWERS TO SELECTED PROBLEMS
16FE-1 When VIN is greater than 6 V, D1 is forward biased and D2 is reverse biased. The
circuit reduces to that in Figure A where
VIN=500I+6
and
Vo=300I+6
Solving for Vo yields
Vo=6+0.6AVIN-6B
When VIN is less than –2 V, D2 is forward biased and D1 is reverse biased. Under these
conditions, Vo=–2 V. For VIN between –2 V and ±6 V, both diodes are reverse biased, no
current flows anywhere and Vo=VIN . A plot of Vo versus VIN is shown in Figure B.
I
200 +
300 Vo
VIN
6V
Vo (V)
IRW-ANSW.I
10
8
6
4
2
0
–2
–4
–12 –10 –8 –6 –4 –2
–
Figure A
0
VIN (V)
Figure B
2
4
6
8
10 12