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Slide Presentations for ECE 329,Slide Presentations for ECE 329,Introduction to Electromagnetic Fields,Introduction to Electromagnetic Fields,
to supplement “Elements of Engineering to supplement “Elements of Engineering Electromagnetics, Sixth Edition”Electromagnetics, Sixth Edition”
byby
Nannapaneni Narayana RaoNannapaneni Narayana RaoEdward C. Jordan Professor of Electrical and Computer EngineeringEdward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, Urbana, Illinois, USAUniversity of Illinois at Urbana-Champaign, Urbana, Illinois, USADistinguished Amrita Professor of EngineeringDistinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, IndiaAmrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India
6.5
Lines with Initial Conditions
6.5-3
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)
6.5-4
01,0 ,0 ,02V z V z Z I z
01,0 ,0 ,02V z V z Z I z
6.5-5
Example:
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
6.5-6
aa
50
0
BC A
l z
V +(z, 0), V I
+(z, 0), A
1
0 l z
50
0 l z
V –(z, 0), V
0 l z
I –(z, 0), A
1
–1
C D
6.5-7
l l
50 BD
C
V +, V
0 z 0 z1
I +, A
1
0 zl
I –, A
–1
50
0z
l
V –, V
AB
1
0 lz
I, A100
50
0 lz
V, V
t l
2vp
6.5-8
aa
50
0A
BC
lz
V’–, V
DC
l
V’+, V
50
0 z z
1
0 lz
I’+, A
1
0 lz
I’–, A
–1
V, V
50
0 lz
I, A
1
0 lz
–1
t lvp
6.5-9
+++++++
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
6.5-10
aa
50
0 l z
V –(z, 0) V
C D50
0
BAl z
V +(z, 0) V
C
AB
CD
t
[V]RL, V
50
0 l/2vp l/vp 3l/2vp
6.5-11Uniform Distribution
+++++++I(z, 0) = 0
Z0, T V(z, 0) = V0-------
z = 0 z = l
V (z,0) V – (z,0) V02
I (z,0) V0
2Z0, I – (z,0) –
V02Z0
6.5-12
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
6.5-13
aa
V, V
l z
(–)(+)
5025
0 zl
I, A
1
0
–1(–)
(+)
aa
zl(–)
I, A
1
0
–1
(+)(+)
V, V
50
250 l z
(–)
t = 0.5 mS
t = 1.5 mS
6.5-14
aa
I, A
0.50
–0.5
V, V
l z
5012.5
0(+) (+)
(–)
(–) l z
t = 2.5 mS
75
0 2 4 6 t, mS9.37518.75
37.5
[V]RL, V
6.5-15
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
6.5-16
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
6.5-17
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
6.5-18
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
6.5-19
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
6.5-20
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
6.5-21
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
6.5-22
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
Another Example:
System in steady state at t = 0–.
6.5-23
t = 0–: steady state
V, V60
0 l 2l z
I, A0.6
0 l 2l z
6.5-24
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
6.5-25
Solving, we obtain
V – V – 15
I – 0.15
I – 0.3
6.5-26
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
6.5-27
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
6.5-28
t = 3 s + : New steady state
aa
I, A0.8
0.4
0 l 2lz
V, V
40
0 l 2l z
6.5-29
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
t = 3 s + :