Dany Silatcha Woussah
Professor Viviana
TCET 2220
Homework Chap 3
1)
𝑖 = 8 cos(2𝜋 ∗ 106 𝑡 − 0.025𝑥)
a. Direction of propagation
Since the argument of the function has the difference between the time and displacement terms, the wave is
traveling in the positive x-direction
b. Peak value
Ip = 8 A
c. Angular frequency
2π ∗ 106 rad/sec
d. Phase constant
β = 0.025 rad/m
e. Cyclic frequency
F=
f.
ω
2π ∗ 106
=
= 1MHz
2π
2π
Period
T=
1
1
=
= 1 us
F 1MHz
g. Wavelength
λ=
2π
2π
=
= 251 meter
β
0.025
h. Velocity of propagation
V = F ∗ λ = 1MHz ∗ 251.2 m = 251.2 ∗ 106 m/s
2)
𝑣 = 15 cos(108 𝑡 + 0.35𝑥)
a. Direction of propagation
Since the argument of the function has the difference between the time and displacement terms, the wave is
traveling in the negative x-direction
b. Peak value
Vp = 15 V
c. Angular frequency
108 rad/sec
d. Phase constant
β = 0.35 rad/m
e. Cyclic frequency
ω 108
F=
=
= 15.9 MHz
2π 2π
f.
Period
T=
1
1
=
= 6.3 ∗ 10−8 𝑠𝑒𝑐
F 15.9 MHz
g. Wavelength
λ=
2π
2π
=
= 17.9 meter
β
0.35
h. Velocity of propagation
V = F ∗ λ = 15.9 MHz ∗ 17.9 m = 2.8 ∗ 108 m/s
3)
a. Period
1
1
=
= 2 ∗ 10−8 𝑠𝑒𝑐
F 50 MHz
T=
b. Angular Frequency
2π ∗ 50MHz = 314 ∗ 106 rad/se
c. Phase constant
β=
ω 314 ∗ 106
=
= 1.57 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑉
2 ∗ 108
d. Wavelength
λ=
2π
2π
=
=4m
β
1.57
e. An equation for the current
𝑖 = 2 cos(314 ∗ 106 𝑡 − 1.57𝑥)
4)
a. Cyclic frequency
F=
ω 20 MHz
=
= 3.18 MHz
2π
2π
b. Period
T=
1
1
=
= 3.14 ∗ 10−7
F 3.18 MHz
β=
ω 20 ∗ 106
=
= 0.06 𝑟𝑎𝑑/𝑚
𝑉
3 ∗ 108
c. Phase constant
d. Wavelength
λ=
2π
2π
=
= 106.67 m
β
0.06
e. An equation for the current
V = 25 cos(20 ∗ 106 t + 0.06x)
5)
a.
̅̅̅
𝐼 + = (Ipk ) ∗ (𝑒 𝑗𝑜 ) = 8∠0 𝐴
b.
I(̅ x) = I̅+ ∗ (e−jβx ) = 8ej0 ∗ e−j0.025 = 8∠ − 0.025x
c. The wave of the distances-varying phasor at x = 100 m.
I(̅ 100) = 8∠ − 0.025(100) = 8∠ − 2.5
6)
+ or 𝑉
−
̅̅̅̅
̅̅̅̅
a. A fixed phasor representation in peak units as either 𝑉
̅̅̅̅
𝑉 + = (Vp ) ∗ (𝑒 𝑗𝑜 ) = 15∠0 𝑉
b. The corresponding distance-varying phasor 𝑉̅ (𝑥) in peak units.
̅(x) = 15∠0.35𝑥
V
c. The wave of the distances-varying phasor at x = 4 m
̅(4) = 15∠0.35 ∗ 4 = 15∠1.40 𝑉
V
7)
𝑖 = 8 cos(2𝜋 ∗ 106 𝑡 − 0.025𝑥 + 1.5)
a. A fixed phasor representation in peak units as either ̅̅̅
𝐼 + or ̅̅̅
𝐼−
̅̅̅
𝐼 + = (Ip ) ∗ (𝑒 𝑗𝑜 ) = 8∠1.5 𝐴
b. The corresponding distance-varying phasor 𝐼 (̅ 𝑥) in peak units
I(̅ x) = I̅+ ∗ (e−jβx ) = (8∠1.5) ∗ (1∠ − 0.025x) = 8∠1.5 − 0.025x
c. The wave of the distances-varying phasor at x = 100 m.
I(̅ 100) = 8∠1.5 − 0.025(100) = 8∠ − 1.0 𝐴
8)
𝑣 = 15 cos(108 𝑡 + 0.35𝑥 + 𝜋/3)
+ or 𝑉
−
̅̅̅̅
̅̅̅̅
a. A fixed phasor representation in peak units as either 𝑉
̅̅̅̅
𝑉 − = (Vp ) ∗ (𝑒 𝑗𝑜 ) = 15∠
𝜋
𝑉
3
b. The corresponding distance-varying phasor 𝑉̅ (𝑥) in peak units.
𝜋
𝜋
̅(x) = (15∠ ) ∗ (1∠0.35𝑥) = 15∠ + 0.35𝑥
V
3
3
c. The wave of the distances-varying phasor at x = 4 m
𝜋
̅(4) = 15∠ + 1.40 𝑉
V
3
9)
(IRMS ) = (Ipk ) ∗ 0.707 = 8 A ∗ 0.707 = 5.66 A
P = IRMS 2 ∗ R = (5.662 A) ∗ (50Ω) = 1,601.78 W
10)
(VRMS ) = (Vp ) ∗ 0.707 = 15 v ∗ 0.707 = 10.61 V
P=
VRMS 2 10.612 V
=
= 1.50 W
𝑅
75Ω
11)
Yes, the power would dissipate
(IRMS ) = (Ip ) ∗ 0.707 = 8 A ∗ 0.707 = 𝟓. 𝟔𝟔 ∠𝟏. 𝟓 𝐀
P = IRMS 2 ∗ R = (5.662 A) ∗ (50Ω) = 𝟏, 𝟔𝟎𝟏. 𝟕𝟖 𝐖
12) Redefine the fixed phasor of Problem 3-8 so that the phasor magnitude is expressed in RMS units. Would the power
dissipated in a 75-Ω resistance be the same as in Problem 3-10?
Yes, the power would dissipate
(VRMS ) = (Vpk ) ∗ 0.707 = 15 v ∗ 0.707 = 10.61 V
P=
VRMS 2 10.612 V
=
= 1.50 W
𝑅
75Ω
13) Under steady-state ac conditions, the forward current wave on a certain lossless 50-Ω line is ̅̅̅
𝐼 + = 2∠0 𝐴. Determine
the voltage forward wave.
+ = 2∠0 𝐴 ∗ (50Ω) = 100∠0 𝑉
̅̅̅̅
𝑉
14) Under steady-state ac conditions, the forward current wave in 300Ω lossless line is ̅̅̅̅
𝑉 + = 15∠3. Determine the current
forward wave.
I̅+ =
̅̅̅̅
𝑉 + 15∠3
=
= 300∠3 mA
𝑅𝑜
50
− = 200∠0 V. Determine the
̅̅̅̅
15) Under steady-state ac conditions, the reverse voltage wave on a lossless 50Ω line is 𝑉
reverse current wave.
I̅− = −
−
̅̅̅̅
𝑉
200∠0
=−
= −4∠0 A
𝑅𝑜
50
16) Under steady-state ac conditions, the reverse current wave on a lossless 75-Ω line is ̅̅̅
𝐼 − = 0.5∠2. Determine the
reverse voltage wave.
− = −(𝑅 ) ∗ (I̅
− ) = (−0.5∠2 𝐴 ∗ (75Ω) = −37.5∠0 𝑉
̅̅̅̅
𝑉
𝑜
17) A table of specifications for one version of RG - 8/U 50-Ω coaxial cable indicates that the attenuation per 100 ft. at 50
MHz is 1.2 dB. At this frequency, determine the following:
a. Attenuation factor in decibel per ft
1.2 dB
= 0.012 dB/Ft
100
b. Attenuation factor in np per ft
0.012 dB
= 1.382 ∗ 10−3 Np/ft
8.686
For a length of 300 ft. determine the following:
c. Total attenuation in decibel
LdB = αdB ∗ d = 0.012 ∗ 300 = 3.6 dB
d. Total attenuation in np
L = α ∗ d = (1.3815 ∗ 10−3 ) ∗ 300 = 0.4145
e. 𝑉2 ⁄𝑉1 Ratio using both decibels and np for a single wave
V2
= e−L = e−0.4145 = 0.6607
V1
In the RMS of decibels, this ratio may be expressed as
−𝐿𝑑𝐵
−3.6
V2
= 10 20 = 10 20 = 10−0.18 = 0.6607
V1
18) A transmission line has an attenuation of 0.05 dB/m. Determine the following:
a. Attenuation factor in np/m
𝛼=
𝛼𝑑𝐵
0.05
=
= 5.756 ∗ 10−3 Np/m
8.686 8.686
For a length of 400 m, determine the following:
b. Total attenuation in decibel
LdB = αdB ∗ d = 0.05 ∗ 400 = 20 dB
c. Total attenuation in np
L = α ∗ d = (5.756 ∗ 10−3 )(400) = 2.303 Np
d. 𝑉2 ⁄𝑉1 Ratio using both decibels and np for a single wave
V2
= 𝑒 −𝐿 = 𝑒 −2.303 = 0.1
V1
In term of decibles:
−𝐿𝑑𝐵
−20
V2
= 10 20 = 10 20 = 10−1 = 0.1
V1
19) A single-Frequency wave is propagating in one direction on a transmission line of length 200m. with an input RMS
voltage of 50 V, the output RMS voltage is measured as 20 V. Determine the following:
a. Total attenuation in decibel
V1
50
LdB = 20log10 ( ) = 20log10 ( ) = 7.959 dB
V2
20
b. Total attenuation in np
L=
LdB
7.959
=
= 0.9163 Np
8.686 8.686
c. Attenuation factor in decibel/meter
α𝑑𝐵 =
LdB 7.959
=
= 0.03980 dB/m
d
200
d. Attenuation factor in np/meter
α=
𝐿 0.9163
=
= 4.581 ∗ 10−3 𝑁𝑝/𝑚
𝑑
200
20) A single-Frequency wave is propagating in one direction on a transmission line of length 400m. The input power to
the line is 40 W, and the output power is 12 W. Determine the following:
a. Total attenuation in decibel
V1
40
LdB = 20log10 ( ) = 20log10 ( ) = 5.229 dB
V2
12
b. Total attenuation in np
L=
LdB
5.229
=
= 0.6020 Np
8.686 8.686
c. Attenuation factor in decibel/meter
α𝑑𝐵 =
LdB 5.229
=
= 0.01307 dB/m
d
400
d. Attenuation factor in np/meter
α=
L 0.6020
=
= 1.505 ∗ 10−3 Np/m
d
400
21) A transmission line has the following parameters at 50 MHz: 𝐿 = 1.2uH/m, 𝑅 = 15Ω/m, 𝐶 = 10pF/m, and 𝐺 = 4uS/
m. Determine the following:
a. Z
𝑧 = 𝑅 + 𝑗ωL = 0.15 + 𝑗(2π ∗ 50 MHz ∗ 1.2 µ𝐻)
0.15 + 𝑗376.99
Ω
𝑚
377.29 ∠1.5310
Ω
𝑚
b. Y
Y = G + 𝑗ωC = 4 ∗ 10−6 + j(2π ∗ 50 MHz ∗ 10pC)
4 ∗ 10−6 + j3.1416 x 10−3 𝑆/𝑚
3.1416 ∗ 10−3 ∠1.5695 𝑆/𝑚
c. γ, α and β
ɣ = √ZY = √(377.29 ∠1.5310)(3.1416 ∗ 10−3 ∠1.5695)
= √1.1853∠3.1005 = 1.0887 ∠1.5503
= 22.36 * 10-3 + j1.088
α = 22.36 x 10-3 Np/m
β = j1.008 rad/ft
d. v
e. 𝑍𝑜
22) A lossy audio-frequency line has the following parameters at 2 kHz: 𝐿 = 0.1 uH/ft, 𝑅 = 0.2Ω/ft, 𝐶 = 2pF/ft, and
𝐺 𝑖𝑠 𝑛𝑒𝑔𝑙𝑖𝑔𝑖𝑏𝑙𝑒. Determine the following:
a. Z
𝑧 = 𝑅 + 𝑗ωL = 0.20 + 𝑗(2π ∗ 2 KHz x 0.1µ𝐻)
0.20 + 𝑗1.2566𝑚
Ω
𝑓𝑡
0.200 ∠6.283𝑚
Ω
𝑓𝑡
b. Y
Y = G + 𝑗ωC = 0 + j(2π ∗ 2 KHz x 2pC)
0 + j2.5133 x 10−8 𝑆/𝑓𝑡
251.33 x 10−10 ∠1.571 𝑆/𝑓𝑡
c. γ, α and β
ɣ = √ZY = √(0.200∠6.283𝑚)(251.33 𝑥 10−10 ∠1.5701
= √50.27 𝑥 10−10 ∠1.58 = 70.9 𝑥 10−6 ∠0.79
= 5 x 10-5 + j5.029 x 10-5
α = 5 x 10-5 Np/ft
β = j5.029 x 10-5 rad/ft
d. Attenuation in dB/ft
αdB = 8.69 𝑥 5 𝑥10−5 = 4.34 𝑥 10−4
𝑑𝐵
𝑓𝑡
e. v
𝑣=
f.
𝜔
β
2π x 2 kHz
= 5.029 𝑥 10−5 25 𝑥 107 𝑓𝑡/𝑠
𝑍𝑜
0.200 ∠6.28𝑚
𝑍𝑜 = √
= √8 𝑥 106 ∠ − 1.57
251.33𝑥10−10 ∠1.571
= 2821∠ − 0.783 Ω = 2001 − 𝑗1989 Ω
23) A coaxial cable has the following parameters at a frequency of 1 MHz:
Z = 0.3 + j2 Ω/m = 2.022∠1.419 Ω/𝑚
Y = 0.5µ + j0.6m S/m = 0.6m ∠1.57
𝑆
𝑚
a. α and β
ɣ = √ZY = √(2.02∠1.422𝑚)( 0.6𝑚 ∠1.570
= √1.213𝑚 ∠3 = 0.0348 ∠ 1.5 = 2.6m + j0.035
α = 2.6m Np/m
𝛽 = 0.0348 rad/m
b. Attenuation in dB/ft
αdB = 8.69 𝑥 2.6m = 0.02263 dB/m
c. V
v=
ω 2π x 1MHz
=
= 18.1 x 107 m/s
β
0.0348
d. 𝑍𝑜
𝑍𝑜 =
𝑍
2.02 ∠1.4219𝑚
= √
= √3.37𝑘∠ − 0.148
𝑌
0.6𝑚 ∠1.57
58.06 ∠ − 0.074 Ω = 57.9 − 𝑗4.3 Ω
24) For the coaxial cable of problem 3-23, repeat the analysis at 100 MHz if the series resistance increases to 1 Ω/m, but
the shunt conductance remains essentially the same.
L=
X
2
=
= 0.32 µ𝐻/𝑚
ω 2π x 1MHz
𝐶=
ϐ
0.6𝑚
=
= 95.5 𝑝𝐹/𝑚
𝜔 2π x 1MHz
𝑍 = 𝑅 + 𝑗 𝜔𝐿 = 1 + 𝑗2πx108 𝑥 0.312 𝑥10−6 Ω/𝑚
1 + j200 = 200 ∠1.57 Ω/m
Y = G + jB = 0.5 x 10-6 + 𝑗2πx108 𝑥 95.5 𝑥 10−12
= 0.5 x 10-6 + j0.06 = 0.06 ∠1.571 S/m
a. ɣ = √ZY = √(200∠1.57)( 0.06∠1.571
= √12∠3.137 = 3.46 ∠1.57
= 8.3m + j3.46
α = 8.3m Np/m
ϐ = 3.46 rad/m
b. αdB = 8.69 x 8.3 x 10-3 = 0.072 dB/m
c. 𝑣 =
𝜔
ϐ
d. 𝑍𝑜 =
=
𝑍
𝑌
2π x 100MHz
3.46
= 1.814 𝑥 108 m/s
200 ∠1.57𝑚
0.06𝑚 ∠1.571
= √
= √3.33𝑘 ∠ − 0.0048
= 57.74 ∠ − 0.0024 Ω = 57.74 – j0.139
25) For the circuit of fig. P3-25, determine the following:
a. Input current 𝐼1
I =
E
50∠0
50∠0
=
=
Z1 + Zo 300 + (290 − 𝑗60) 590 − 𝑗60
=
50∠0
= 84.31 ∠0.1013 mA
593 ∠ − 0.1011
b. Input Voltage 𝑉1
𝑣1 = 𝑍𝑜 ∗ 𝐼1 = (290 − 𝑗60) ∗ (0.84∠0.101
= (296.14 ∠ − 0.204) 0.084 ∠0.101) = 24.97∠ − 0.103 V
c. Input power 𝑃1
𝑃1 = 𝐼1 2 ∗ 𝑅𝑜 = (0.0842 ) ∗ 290 = 2.06 𝑊
d. Load current 𝐼2
𝐼2 = 0.084 ∠0.101 ∗ (𝑒 −1.5 )1∠ − 0.6 = 0.019 ∠ − 0.5
e. Load Voltage 𝑉2
𝑉2 = (24.97 ∠(𝑒 −1.5 )(1∠ − 0.6) = 5.57 ∠ − 0.703 V
f.
Load Power 𝑃2
𝑃2 = 𝐼2 2 ∗ 𝑅𝑜 = (0.0192 ) ∗ 290 = 0.103 𝑊
g. Line loss in dB
𝐿𝑑𝐵 = 10 log10
𝑃1
2.026
= 10𝑙𝑜𝑔10
= 13.03 dB
𝑃2
0.103
26) For the circuit of fig. P3-26, determine the following:
a. Input current 𝐼1
𝐼 =
𝐸
80∠0
80∠0
=
=
𝑍1 + 𝑍𝑜 (600 + 𝑗100) + (600 + 𝑗100) 1200 − 𝑗200
b. Input Voltage 𝑉1
Zo = Ro + jXo = 600 + j100 = 608.28 ∠0.165 Ω
𝑉1 = Zo 𝐼1 = (608.28∠0.1652) (0.0658∠ − 0.1652) = 40 ∠0 V
c. Input power 𝑃1
𝑃1 = 𝐼1 2 ∗ Ro = 0.0662 x 600 = 2.59 W
d. Load current 𝐼2
𝐿 =
𝐿𝑑𝐵
24
=
= 2.76 Np
8.69 8.69
𝐼1 = (0.066∠ − 0.165)(𝑒 −2.763 )(1∠ − 3) = 4.150 𝑥 10^ − 3 ∠ − 3.165 𝐴
e. Load Voltage 𝑉2
V2 = (40∠0)(𝑒 −2.763 )(1∠ − 3) = 2.52 ∠ − 3
f.
Load Power 𝑃2
P2 = 𝐼2 2 ∗ Ro = (4.15m2 ) ∗ 600 = 0.0103 W
g. Line loss in np
L=
V1
40
= ln (
) = 2.76 Np
V2
2.52