enable Fourier. After simulation, we compare the output and output waveforms as
shown. The output includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
DC COMPONENT = 7.658051E-01
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 5.000E-01 1.070E+00 1.000E+00 1.004E+01 0.000E+00
2 1.000E+00 3.758E-01 3.512E-01 -3.924E+01 –4.928E+01
Solution 17.71
The schematic is shown below. In the Transient dialog box, we set Print Step = 0.02 s,
Final Step = 12 s, Center Frequency = 0.5, Output Vars = V(1) and V(2), and click enable
Fourier. After simulation, we compare the output and output waveforms as shown. The
output includes the following Fourier components.
FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1)
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED
NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)
1 5.000E-01 1.070E+00 1.000E+00 1.004E+01 0.000E+00
2 1.000E+00 3.758E-01 3.512E-01 -3.924E+01 –4.928E+01
6 3.000E+00 6.139E-02 5.738E-02 -6.563E+01 –7.567E+01
7 3.500E+00 4.743E-02 4.433E-02 -6.520E+01 –7.524E+01
Solution 17.72
T = 5, ωo = 2π/T = 2π/5
bn =
∫ ∫ π=ω
2/T
0
10
0
odt)tn4.0sin(10
5
4
dt)
tnsin()t(f
T
4
Solution 17.73
Solution 17.74
(a) An =
2
n
2
nba +
, φ = tan–1(bn/an)
– –
For the nth harmonic,
o
n
A2
1
πτ
From (1) and (2),
Solution 17.76
vs(t) is the same as f(t) in Figure 16.1 except that the magnitude is multiplied by 10.
Hence
vo(t) =
∑
∞
=
π
π
+
1k
)tnsin(
n
120
5
, n = 2k – 1
The source current Is is
Is =
)R10(n2jR10
n
20
)R10(
R10
V
n2jZ
V
ss
+π+
π
+
=
=
π+
For the DC case, L acts like a short–circuit.
+
π
+
+
−
1
2
)R10
(
5
tancos)R10(
20
1
)R10(25
We want po = (70/100)ps = 0.7ps. Due to the complexity of the terms, we
consider only the DC component as an approximation. In fact the DC component
Solution 17.77
(a) For the first two AC terms, the frequency ratio is 6/4 = 1.5 so that the highest
Solution 17.78
(a) p =
∑ ∑
+=+ R
V
R
V
R
V
2
1
R
V2
rms,n
2
DC
2
n
2
DC
Solution 17.79
From Table 17.3, it is evident that an = 0,
A Fortran program to calculate bn is shown below. The result is also shown.
C FOR PROBLEM 17.79
DIMENSION B(20)
n
bn
1
12.731
2
4.243
3
2.546
4
1.8187
5
1.414
6
1.1573
7
0.9793
8
0.8487
9
0.7498
10
0.6700
Solution 17.80
From Problem 17.55,
This is calculated using the Fortran program shown below. The results are also shown.
C FOR PROBLEM 17.80
COMPLEX X, C(0:20)
n
cn
0
1
0
2
–0.1061 + j0
3
0
4
5
0
6
7
0
8
9
0
Solution 17.81
(a)
2T
The total average power is pavg = Frms2R = Frms2 since R = 1 ohm.
(b) From the Fourier series above
n
ωo
|cn|
|co|2 or 2|cn|2
% power
0
0
81.1%
2A/π
4A2/(π2)
1
2ωo
2A/(3π)
8A2/(9π2)
18.01%
2
4ωo
3
6ωo
2A/(35π)
8A2/(1225π2)
0.13%
4
8ωo
2A/(63π)
8A2/(3969π2)
0.04%
A
Solution 17.82
P = ∑
∞
=
+
1n
2
n
2
DC
R
V
2
1
R
V
Assuming V is an amplitude-phase form of Fourier series. But
|An| = 2|Cn|, co = ao
Alternatively,
P =
R
V
2
rms