by Nannapaneni Narayana Rao Edward C. Jordan Professor Emeritus
133
Fundamentals of Electromagnetics Fundamentals of Electromagnetics for Teaching and Learning: for Teaching and Learning: A Two-Week Intensive Course for Faculty in A Two-Week Intensive Course for Faculty in Electrical-, Electronics-, Communication-, Electrical-, Electronics-, Communication-, and Computer- Related Engineering and Computer- Related Engineering Departments in Engineering Colleges in India Departments in Engineering Colleges in India by by Nannapaneni Narayana Rao Nannapaneni Narayana Rao Edward C. Jordan Professor Emeritus Edward C. Jordan Professor Emeritus of Electrical and Computer Engineering of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, USA University of Illinois at Urbana-Champaign, USA Distinguished Amrita Professor of Engineering Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, India Amrita Vishwa Vidyapeetham, India
description
Fundamentals of Electromagnetics for Teaching and Learning: A Two-Week Intensive Course for Faculty in Electrical-, Electronics-, Communication-, and Computer- Related Engineering Departments in Engineering Colleges in India. by Nannapaneni Narayana Rao Edward C. Jordan Professor Emeritus - PowerPoint PPT Presentation
Transcript of by Nannapaneni Narayana Rao Edward C. Jordan Professor Emeritus
Fundamentals of ElectromagneticsFundamentals of Electromagneticsfor Teaching and Learning:for Teaching and Learning:
A Two-Week Intensive Course for Faculty inA Two-Week Intensive Course for Faculty inElectrical-, Electronics-, Communication-, and Electrical-, Electronics-, Communication-, and
Computer- Related Engineering Departments in Computer- Related Engineering Departments in Engineering Colleges in IndiaEngineering Colleges in India
byby
Nannapaneni Narayana RaoNannapaneni Narayana RaoEdward C. Jordan Professor EmeritusEdward C. Jordan Professor Emeritus
of Electrical and Computer Engineeringof Electrical and Computer EngineeringUniversity of Illinois at Urbana-Champaign, USAUniversity of Illinois at Urbana-Champaign, USADistinguished Amrita Professor of EngineeringDistinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, IndiaAmrita Vishwa Vidyapeetham, India
7-2
Program for Hyderabad Area and Andhra Pradesh FacultySponsored by IEEE Hyderabad Section, IETE Hyderabad
Center, and Vasavi College of EngineeringIETE Conference Hall, Osmania University Campus
Hyderabad, Andhra PradeshJune 3 – June 11, 2009
Workshop for Master Trainer Faculty Sponsored byIUCEE (Indo-US Coalition for Engineering Education)
Infosys Campus, Mysore, KarnatakaJune 22 – July 3, 2009
7-3
Module 7Transmission Line Analysis
in Time Domain
7.1 Line terminated by a resistive load7.2 Transmission-line discontinuity7.3 Lines with reactive terminations and discontinuities7.4 Lines with initial conditions7.5 Lines with nonlinear elements
7-4
Instructional Objectives49. Find the voltage and current variations at a location on a lossless transmission line a functions of time and at an instant of time as functions of distance, and the steady state values of the line voltage and current, for a line terminated by a resistive load and excited by turning on a constant voltage source, by using the bounce-diagram technique50. Design a lossless transmission line system by determining its parameters from information specified concerning the voltage and/or current variations on the line51. Design a system of three lines in cascade for achieving a specified unit impulse response
7-5
Instructional Objectives (Continued)52. Compute the reflected power for a wave incident on a junction of multiple lossless transmission lines from one of the lines and the values of power transmitted into each of the other lines, where the junction may consist of lines connected in series, parallel, or series-parallel, and include resistive elements53. Find the solutions for voltage and current along a transmission-line system excited by a constant voltage source and having reactive elements as terminations/discontinuities54. Find the voltage and current variations at a location on a lossless transmission-line system as functions of time and at an instant of time as functions of distance, for specified nonzero initial voltage and/or current distributions along the system
7-6
Instructional Objectives (Continued)55. Analyze a transmission line terminated by a nonlinear element by using the load line technique56. Understand the effect of time delay in interconnections between logic gates
7-7
7.1 Line Terminatedby Resistive Load(EEE, Sec. 6.2; FEME, Sec. 7.4)
7-8
, p pV z t V t z v V t z v
0
1, p pI z t V t z v V t z vZ
Notation
0 0
,
V V V
I I I
V VI IZ Z
7-9
I+, I–
+
–
V+, V– P+, P–
I+, I– z
2
0 0
2
0 0
VVP V I VZ Z
VVP V I VZ Z
7-10
I+ = –0.2 A
P+ = 2 W
0.2 A
–2 W
+
10 V
+
–
V+ = –10 V
0.2 A
Assuming Z0 = 50 Ω,
I+ = 0.1 A
+
–
V+ = 5 VP+ = 0.5 W
0.1 A
+
–
5 V0.5 W
7-11
Assuming Z0 = 50 Ω,
0.08 A
I– = 0.08 A
P– = –0.32 W
0.08 A
–0.32 W
+
4 V
+
–
V– = –4 V
I– = –0.12 A
P– = 0.72 W
0.12 A
+0.72 W
–
6 V
+
–
V– = 6 V0.72WP
7-12
V0t = 0 Z0, vp
z = 0 z
Excitation by Constant Voltage SourceSemi-infinite Line, No Source Resistance
V – 0
0
,
1,
p
p
V z t V t z v
I z t V t z vZ
7-13
00,V t V u t
0V t V u t
0
0
0
0
,
for 0
0 for 0
for z
0 for z
for
0 for
p
p
p
p
p
p
p
p
V z t V t z v
V u t z v
V t z v
t z v
V t v
t v
V z v t
z v t
7-14
V(z)V0
z/vpt
V(t)V0
0
0 vpt z
7-15
z = 150 mI, A
0.50
0.2V, V
10
0 0.5
V0t = 0 Z0, vp
z = 0 z
to
SE7.1
V0 10 V Z0 50 vp 3 108 m s
t, st, s
7-16
V, V10
0 300 z, m
I, A
3000
0.2
z, m
[V ]z150 m,V [V ]t1s ,V
t = 1 s
t, s 3000
1010
0 0.5 z, m
7-17
Effect of Source Resistance
z = 0
–+
+
–
t = 0
Rg
VgV+
I+
0
0g gV I R V
VIZ
B.C.
(+) Wave
7-18
Vg –V
Z0Rg – V 0
Vg VRgZ0
1
V VgZ0
Rg Z0
I V
Z0
VgRg Z0 z = 0
Rg
Vg
V+
I+
+– Z0
7-19
RgV0
t = 0 Z0, vp
z = 0 z = l
S
RL
Line Terminated by Resistance
z = 0
Rg
Vg
I+
Z0 V++–
I V0
Rg Z0
V IZ0
aa
t = 0+
7-207-20
LV V R I I
t l vp B.C:
V V – RLV
Z0– V –
Z0
V – 1 RLZ0
V RL
Z0– 1
0
0
VI Z
VI Z
RLV+ + V–
I+ + I–
z = l
+
7-21
Define Voltage Reflection Coefficient,
Then, Current Reflection Coefficient
V – V RL – Z0RL Z0
V –
V RL – Z0RL Z0
I –
I
– V – Z0V Z0
–V –
V –
7-22
0 gV V V V R I I I
t 2l vp
z = 0
+
–
Rg
V0V+ + V– + V–+
I+ + I– + I–+
0 0 0
, ,V V VI I IZ Z Z
00
RgV V V V V V VZ
7-23
V 1 RgZ0
V – 1
RgZ0
V0 V –RgZ0
– 1
But V + =V0
Rg + Z0Z0
V – 1 RgZ0
V –
RgZ0
– 1
V –
V – Rg – Z0Rg Z0
7-247-24
z = 0
Rg(–)(–+)
t (steady state)
2 20 0
0
2 2
1
1
R S R Sg
R R S R S
V ZR Z
2 2 2 21SS R R S R S R SV V
7-257-25
0 0
0
0
00 0
0 00
0 0
0
11
1
1
R
g R S
L
L
g gL
L g
LL g
V ZR Z
R ZR ZV Z
R Z R ZR ZR Z R Z
VR
R R
7-26
Rg
V0
RL–+
+++++++
– – – – – – –
0 L gV R R
V0 RLRL Rg
2
0
0
0 0
0
1 1
11
SS R R S R S
R S R R Sg
R
g R S L g
I I I
VR Z
V VR Z R R
7-27
Rg
V0
z = l
RL
z = 0
I SS I –
SS I SS I –
SS
V SS V –
SS V SS V –
SS
+
–
+
–
(+)
(–)
For constant voltage source,
Actual Situation in the Steady StateOne (+) Wave and One (–) Wave
ISS V0
RL Rg, VSS
V0 RLRL Rg
7-28
V SS V –
SS V0 – Rg ISS I –SS
B.C. at z 0
V SS V –
SS RL ISS I–SS
B.C. at z l
I SS V
SS
Z0 () wave
I –SS –
V –SS
Z0 (–) wave
Four equations for the four unknownsV
SS , V –SS , I SS , I –
SS
7-29
25
z = lz = 0
I SS I –
SS I SS I –
SS
V SS V –
SS V SS V –
SS
+
–
+
–100 V
Z0 = 50 75
E7.2
V SS V –
SS 100 – 25 ISS I –SS
V SS V –
SS 75 ISS I –SS
7-30
Solving, we obtain
I SS V
SS50
, I –SS –
V –SS
50
V SS 62.5 V , V –
SS 12.5 V
I SS 1.25 A , I –SS – 0.25 A
–+
++++++++
– – – – – – – –
1 A
25
100 V75 V 75
7-317-317-31
Bounce Diagram Technique: Constant Voltage SourceE7.3
V 100 6040 60
60 V , I 6060
1 A
R 120 – 60120 60
13
, S 40 – 6040 60
–15
t = 0 Z0 = 60
z = 0 z =
l
S
T = 1 s
z
40
100 V
120
7-327-32
Voltage
S –15
R 13
0
2
4
1
3
5
t, s
z = 0 z = l
60 V60 20
76 –4/34/15
–4
0
80
22431124
15z
, st
7-337-33
Current
–S 15
13R
0
2
4
1
3
5
t, s
z = 0 z = l
1 A1 –1/3
1/451/225
–1/15
0
141225
915 28
45
23
z
, st
7-347-34
Voltage Current
0
2
4
1
3
5
t, s
z = 0 z = l
60 V60 20
76 –4/34/15
–4
0
80
22431124
15z
aaa
–S15
aaa
R 13
0
2
4
1
3
5
t, s
z = 0 z = l
1 A1 –1/3
1/451/225
–1/15
0
141225
915 28
45
23
z
aaa
S15
–
aaa
R 13
––
, st , st
7-35
V, Vz = 0
76100
60
0 2 4 6 8
112415
I, A1
0 2 4 6 8
141225
915
t, s
t, s
7-36
V, V
z = l
100 80
0 1 3 5 7
2243
t, s
I, A1
0 1 3 5 7
2845
23
t, s
7-37
V, V
100 80 76
60
0 0.5 2.5 4.5 6.5
2243
t, s
z l2
I, A
1
0 0.5 2.5 4.5 6.5
1
23
915
2845
t, s
7-38
[V ]t2.5 s ,V100
0 l/2 l z
76 80
1
0 l/3 l z2l/3
2/3
[I]t1
13 s
,A
7-397-39
Rectangular Pulse SourceUse superposition.
E7.4
V
V0
0 t0t
=
+
V
V0
t0
0
–V0
t0 t
V
100
0 1
Vg, V
t, s
t = 0 Z0 = 60
z = 0 z = l
S
T = 1 s
z
40 Vg
120 1 s
7-407-40
0
2
4
1
3
z = 0 z = l
10
2
4
0
3
80
0
60
0
16
0
1615
– 0
–
–60
–4
4
4/15
20 –20
20
–4/3
4/3
–4
43–
60 V
163
z
S – 15
16 20
0 l/4 l/2 3l/4 lz
[V ]t2
14 s
, V
t, s
13R
0 1 243
5 61613
–
80
t, s
[V]zl,V
163
7-41
Review Questions7.1. Discuss the general solutions for the line voltage and line current and the notation associated with their interpretation in concise form.7.2. What is the fundamental distinction between the occurrence of the response in one branch of a lumped circuit to the application of an excitation in a different branch of the circuit and the occurrence of the response at one location on a transmission line to the application
of an excitation at a different location on the line? 7.3. Describe the phenomenon of the bouncing back and forth of transient waves on a transmission line excited by a constant voltage source in series with internal resistance and terminated by a resistance.
7-42
Review Questions (Continued)7.4. What is the nature of the formula for the voltage reflection coefficient? Discuss its values for some special cases.7.5. What is the steady state equivalent of a line excited by a constant voltage source? What is the actual situation in the steady state?7.6. Discuss the bounce diagram technique of keeping track of the bouncing back and forth of transient waves on a transmission line for a constant voltage source.7.7. Discuss the bounce diagram technique of keeping track of the bouncing back and forth of transient waves on a transmission line for a pulse voltage source.
7-43
Problem S7.1. Plotting line voltage and line current on a transmission-line system involving two lines
7-44
Problem S7.2. Finding several quantities associated with a transmission-line system from given observations
7-45
Problem S7.3. Time-domain analysis of a transmission-line system using the bounce diagram technique
7-46
Problem S7.4. Time-domain analysis of a transmission-line system for a sinusoidal excitation
7-47
7.2 Transmission-LineDiscontinuity
(EEE, Sec. 6.3)
7-48
V + + V
–
I + + I
– I ++
V ++
+
–
+
–
Transmission-Line Discontinuity
(+)
(–)Z01, vp1
(++) Z02, vp2
7-49
V V – V
I I – I
B.C.
I V
Z01, I – – V –
Z01, I
V
Z02
01 01 02
02
01
02 02
01 01
1 1
V V V VZ Z Z
ZV V V VZ
Z ZV VZ Z
7-50
V –
V Z02 – Z0lZ02 Z0l
Z01(+)
Z02
7-517-51
Current Transmission Coefficient,
– –
1
C
I I I II I I
1 – C
1 V
– –
1
V
V V V VV V V
Define Voltage Transmission Coefficient,
7-527-52
Note that
2
2
1 1
1
1
V C
P V I
V I
V I
V I
P
P
7-537-53
Three Lines in CascadeE7.5
–Vo
+
–
50 Z0l = 50 Z02 = 100 Z03 = 50
T1 = 2 s T2 = 2 s T3 = 2 s50
(t)2 s2 s 2 s
0
4
8
12t, s
1/22/3
–2/92/27
–2/812/243
4/9
4/81
4/729
6
10
14
4/9
4/92
4/93
= 1/3 = –1/3 = –1/3 = 0 V = 4/3 V = 2/3 V = 2/3 = 0
, st
7-547-54
2 00
0 1 2 3
4 1 29 9
g
n
on
V t t
V t t nT T
T T T T
0 6 10 14
4/94/92 4/93
and so on
t, s, st
7-55
h(t)
Vg(t)
System
SystemVg (t – ) h( ) d–
(t)
7-567-567-56
For Vg (t) = cos t ,
2 00
0
2 0
2 00
cos
4 1 29 9
4 1 cos 9 9
2
4 1 cos 29 9
o
n
n
n
n
n
n
V t t
nT T d
t
nT T d
t nT T
7-577-57
2 0
0 2
0
2
2
0
2
0
2
4 19 9
4 19 9
49119
nj nT T
on
nj T j T
n
j T
j T
V e
e e
e
e
7-58
V o () 4 9
1 –19
e– j2T2
V o () max 4 9
1 – 1 90.5
V o () min 4 91 1 9
0.4
7-59
0.50.4
0
V o ()
2T2 T2 3 2T2 2 T2
7-60
Junction of Three LinesE7.6
Line 1
Z0 = 50 50 100
0 50Z
Line 1 PLine 3
Line 2
0100
Z
0
50
Z
7-617-61
eff 2
100 3 50 50 1100 3 50 250 5
415615
10050 100
2 2 6 123 3 5 15
V
C
C C
C
7-627-62
eff 3
5050 100
1 1 6 63 3 5 15
C C
C
Pref1 2P 1
25P
eff2trans2 =
4 12 485 15 75
V CP P
P P
7-637-637-63
Note that 125
4875
2475
1
eff3trans3 =
4 6 245 15 75
V CP P
P P
7-64
Review Questions7.8. Discuss the phenomenon of reflection and transmission for a wave incident on a junction between two transmission lines.7.9. How are the voltage and current transmission coefficients at a junction between two lines related to voltage reflection coefficient?7.10. Explain how it is possible for the transmitted voltage or current at a junction between two transmission lines to exceed the incident voltage or current, respectively.7.11. Discuss the determination of the unit impulse response of a system of three lines in cascade.7.12. Outline the procedure for the determination of the frequency response of a system of three lines in cascade from its unit impulse response.
7-65
Review Questions (Continued)7.13. Discuss the determination of the reflection and transmission coefficients at a junction connecting a transmission line to two lines in parallel.7.14. How would you determine the reflection and transmission coefficients at a junction connecting a transmission line to two lines in series?
7-66
Problem S7.5. Finding three parameters for a system of three media from unit impulse response
7-67
Problem S7.5. Finding three parameters for a system of three media from unit impulse response (Continued)
7-68
Problem S7.6. Analysis of a system of three transmission lines excited by a pulse voltage source
7-69
Problem S7.6. Analysis of a system of three transmission lines excited by a pulse voltage source (Continued)
7-70
Problem S7.7. A system of three transmission lines with a resistive network at the junction
7-71
7.3 Lines withReactive Terminations
and Discontinuities(EEE, Sec. 6.4)
7-72
aa
LV02 + V
–
–V02Z0
V –
Z0+
–
aa
t = 0Z0 , T
z = 0 z = l
S
Z0 LV0
IL
Line Terminated by an InductorE7.7
t = T+
0 0LI
7-737-73
0 0
0 0
0 I.C.2 2t T
t T
V VV VZ Z
V02
V – L ddt
V02Z0
– V –
Z0
B.C.
Using I.C.,0
0 0
2 2
ZT
LV V Ae
0
0
0
0
2
2
Zt
L
VL dV VZ dt
VV Ae
7-747-74
t T
t T
0
0Z
TLA V e
00
02– ( – )– , –
Zt T
LVV l t V e( )
0
0
0 0
0 02
––
– ( – )
( , )( , ) –
–
V
Z
t TL
l tI l tZ
V Ve
Z Z
7-75
(+)
z
(–)
V00T
2T
3T
0V0/2–V0/2
(–) (+)
0T
2T VL
–V0/2V0/21
3T
Voltage
= 0
7-76
(+)
z
0T
2T
3T
0V0/2Z0
(–) (–)
0
T
2TIL
l
3T
–V0/2Z0
–V0/2Z0
V0/2Z0
(+)
V0Z0
– = 0
7-77
VV0/2
0 l/2 l z
t = T/2
V0/2V0 t = 3T/2
(+)
0 l/2 (–) l z
V
V
V0/2
0(–)
t = 5T/2(+)
zl
7-78
I
V0/2Z0
0 l/2 l z
t = T/2
I
V0/2Z0
0 l/2 l z
t = 3T/2(+)
(–)
(–)
(+)IV0/2Z0
0l/2 l z
t = 5T/2
7-79
t = 0Z0 , T
z = 0 z = l
S
Z0C
V0
VL
+
–
Line Terminated by a CapacitorE7.8
t = T+
VL(0–) 0
V02 + V
–
–V02Z0
V –
Z0+
–
7-807-80
0 0
0 0
0 0
B.C.2 2
0 I.C.2 2t T
t T
V VV dC VZ Z dt
V VV V
0
00
10
2
2
––
––
t
CZ
VdVCZ VdtV
V Ae
7-817-81
Using I.C.,
t > T
0
0
10 0
1
0
2 2–
–
–
–
TCZ
TCZ
V VAe
A V e
0
10
02– ( – )
– ( , ) –t T
CZVV l t V e
0
0
10 0
0 02
––
– ( – )
( , )( , ) –
–
t T
CZ
V l tI l tZ
V Ve
Z Z t > T
7-82
Review Questions7.16. Discuss the transient analysis of a line driven by a constant voltage source in series with a resistance equal to Z0 of the line and terminated by an inductor.7.17. Why is the concept of reflection coefficient not useful for studying the transient behavior of lines with reactive terminations and discontinuities?7.18. Outline the transient analysis of a line driven by a constant voltage source in series with a resistance equal to Z0 of the line and terminated by a capacitor.
7-83
Problem S7.8. Transient analysis of a transmission line terminated by a capacitive network
7-84
Problem S7.9. Finding the nature of a discontinuity in a transmission line system
7-85
7.4 Lines withInitial Conditions(EEE, Sec. 6.5; FEME, Sec. 7.5)
7-86
aa
++++++++
I(z, 0)
Z0, vp V(z, 0)--------
Line with Initial Conditions
V (z,0) V – (z,0) V(z,0)I (z,0) I – (z,0) I(z,0)
I V
Z0, I – – V –
Z0
V (z,0) – V – (z,0) Z0 I(z,0)
7-87
01,0 ,0 ,02V z V z Z I z
01,0 ,0 ,02V z V z Z I z
7-88
E7.9
aa
++++++++
I(z, 0)
Z0, vp V(z, 0)--------
z = 0 z = l
aa
50
0 l z
V(z, 0), V
1
0 l z
I(z, 0), AZ0 50 z l
7-897-89
A
B
C
CD
0 V, ,V z 0 A, ,I z
0
50
l z
0 V, ,V z 0 A, ,I z
50
0
0
1
-1
l
l
l
z
z
1
z
0
7-907-90t l
2vp
0
50
lz
0
50
lz
0
50
lz
lz
l
z0
1
-1
l
z
DC
B
V,V A,I
V,V A,I
, VV, AI
0
1
B
A
100
1
0
1
7-917-917-91
t l
vp
0
50
lz
0
50
lz
0
50
lz
lz
z0
1
-1
l
DC
C
V,V A,I
V,V A,I
, VV
0
1
B
A
z0
1
-1
l
, AI
7-92
+++++++
I(z, 0)
Z0, vp V(z, 0)-------
z = 0 z = l
RL = Z0 = 50
t = 0
1
0 lz
I(z, 0), A50
0 lz
V(z, 0), V
0If 50 is connected at = 0,LR Z t
7-937-937-93
A
BC
Dt
50
0
V ,VLR
3 2 pl vpl v2 pl v
50
0
B
Al
zC
V ,0 Vz
50
0l
z
CD
V ,0 Vz
7-947-94Uniform DistributionE7.10
+++++++I(z, 0) = 0
Z0, T V(z, 0) = V0-------
z = 0 z = l
– 0
–0 0
0 0
( ,0) ( ,0) 2
( ,0) , ( ,0) –2 2V
VV z V z
VI z I zZ Z
7-95
aa
z
(–)
(+)
V, V
50
500 l z l
(+)
(–)
I, A
1
0
–1
V0 100 V, Z0 50
t = 0Z0 , T
z = 0 z = l
S
RL
+++++++
I(z, 0) = 0
-------V(z, 0) = V0
V0 100 V, Z0 50
150 , 1 mSLR T
7-96
7-97
7-98
aa
z = 0 z = l
RL
0 + I +
V0 + V +
+
–
Bounce Diagram Technique for Uniform Distribution
0
0
0 B.C.LV V R I
VI Z
7-99
V0 V –RLZ0
V
V 1 RLZ0
– V0
V – V0Z0
RL Z0
For V0 100 V, Z0 50 , andRL = 150 ,
V – 100 50150 50
– 25 V
7-100
aa
75
0 2 4 6 t, mS9.37518.7537.5
[V]RL
2
4
0
z = 0
1
3
5
75
37.5
18.75
–25
–12.5
–25
–12.5
–6.25
100
50
25
z = l
100 V
t, mS
z
= 12
= 1
7-101
Energy Storage in Transmission Lines
we, Electric stored energy density =
We, Electric stored energy =
12CV 2
12z0
l CV2 dz
12CV 2
0 l (for uniform distribution)
12CV 2
0vpT 12CV 2
01LC
T
12
V 20
Z0T
7-102
wm, Magnetic stored energy density =
Wm, Magnetic stored energy =
12
LI 2
12z0
l LI2 dz
12
LI 20 l (for uniform distribution)
12
LI 20 vpT
12
LI 20
1LC
T
= 12
I 20 Z0T
7-103
Check of Energy Balance
Initial stored energy
We Wm
12
V 20
Z0T
12
I 20 Z0T
12
(100)2
5010–3 0
0.1 J
7-104
Energy dissipated in RL
3 3
3
2
0
2 22 10 4 10
0 2 10
32
32
75 37.5150 150
2 10 1 175 1150 4 162 10 475150 30.1 J
LR
t L
VdtR
dt dt
7-1057-105
aa
z = 0 z = l– z = l+ z = 2l
100 120 V
100 Z0 = 100
T = 1 s
t = 0
100 T = 1 s
Z0 = 50 S
E7.11
System in steady state at t = 0–.
ss
7-106
t = 0–: steady state
V, V60
0 l 2l z
I, A0.6
0 l 2l z
7-107
aa
z = l+
100 60 + V
– 60 + V +
z = l–
0.6 + I – 0.6 + I
+
+
–
+
–
t = 0+:
60 V – 60 V
0.6 I – = 0.6 I+ 60 + V +
100
B.C.
I – –V –
100, I
V
50
7-108
Solving, we obtain
V – V– 15
I – 0.15
I – 0.3
7-1097-109
Voltage
aa
0
12
3
60–1545
40
–540
z = l+ z = 2l
60 V
t, s
0
12
3z = 0
60
45
40
–1545
–540z = l
60 V = 0
= 0V = 1
13
s
7-1107-110
aa
0
12
3
0.6–0.30.3
0.40.1
0.4
z = l+ z = 2l
0.6 A
t, s
0
12
3z = 0
0.6
0.75
0.8
0.150.75
0.050.8z = l
0.6 A
Current
= 0 = 0, C = 1Ceff = 0.5
s,t
7-1117-111
New steady state
aa
I, A0.8
0.4
0 l 2lz
V, V
40
0 l 2l z
3 s + :t
7-1127-112
aa
100 +
–
+
–
0.8 A 0.4 A
100
120 V
40 V0.4 A
100 40 V
z = 0 z = l– z = l+ z = 2l
3 s + :t
7-113
Review Questions7.19. Discuss the determination of the voltage and current distributions on an initially charged line for any given time from the knowledge of the initial voltage and current distributions.7.20. Discuss with the aid of an example the discharging of
an initially charged line into a resistor.7.21. Discuss the bounce-diagram technique of transient analysis of a line with uniform initial voltage and current distributions.7.22. How do you check the energy balance for the case of a line with initial voltage and/or current distribution(s) and discharged into a resistor?
7-114
Problem S7.10. For the analysis of an initially-charged transmission line discharging into a resistor
7-115
Problem S7.10. For the analysis of an initially-charged transmission line discharging into a resistor (Continued)
7-116
Problem S7.11. Analysis of a system of an initially-charged line connected to another initially-charged line
7-117
Problem S7.12. For the analysis of an initially charged transmission line connected to a capacitor
7-118
7.5 Lines withNonlinear Elements
(EEE, Sec. 6.6; FEME, Sec. 7.6)
7-119
Nonlinear Termination: Load line TechniqueE7.12
7-120
7-121
7-122
7-123
7-124
7-125
Interconnection Between Logic Gates
7-126
7-127
7-128
7-129
7-130
7-131
Review Questions7.23. Discuss the load-line technique of obtaining the time variations of the voltages and currents at the source and load ends of a line from the knowledge of the terminal V-I characteristics.7.24. Discuss the analysis of interconnection between logic gates, using transmission line.
7-132
Problem S7.13. Application of load-line technique for an initially charged line discharging into a nonlinear resistor
The End
