by Nannapaneni Narayana Rao Edward C. Jordan Professor of Electrical and Computer Engineering
description
Transcript of by Nannapaneni Narayana Rao Edward C. Jordan Professor of Electrical and Computer Engineering
Edward C. Jordan Memorial Offering of the First Edward C. Jordan Memorial Offering of the First Course under the Indo-US Inter-University Course under the Indo-US Inter-University
Collaborative Initiative in Higher Education and Collaborative Initiative in Higher Education and Research: Electromagnetics for Electrical and Research: Electromagnetics for Electrical and
Computer EngineeringComputer Engineering
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-ChampaignUniversity of Illinois at Urbana-ChampaignUrbana, Illinois, USAUrbana, Illinois, USA
Amrita Viswa Vidya Peetham, CoimbatoreAmrita Viswa Vidya Peetham, CoimbatoreJuly 10 – August 11, 2006 July 10 – August 11, 2006
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 ,02
V z V z Z I z
01,0 ,0 ,02
V 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), A
Z0 50 z l
6.5-6
aa
50
0
B
C Al 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
0z
0z
1
I +, A
1
0z
l
I –, A
–1
50
0z
l
V –, V
A
B
1
0 lz
I, A100
50
0 lz
V, V
t l
2vp
6.5-8
aa
50
0
A
BC
lz
V’–, V
DC
l
V’+, V
50
0z 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), A
50
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
6.5-11
Uniform 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
(–)(+)
50
250 zl
I, A
1
0
–1
(–)(+)
aa
zl(–)
I, A
1
0
–1
(+)(+)
V, V
50
25
0 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, mS
9.37518.7537.5
[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
VIZ
6.5-16
V0 V –RLZ0
V
V 1 RLZ0
– V0
V – V0Z0
RL Z0
For V0 100 V, Z0 50 , and
RL = 150 ,
V – 10050
150 50– 25 V
6.5-17
aa
75
0 2 4 6 t, mS
9.37518.7537.5
[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 =
1
2CV 2
1
2z0l CV2 dz
1
2CV 2
0 l (for uniform distribution)
1
2CV 2
0vpT 1
2CV 2
01
LCT
1
2
V 20
Z0T
6.5-19
wm, Magnetic stored energy density =
Wm, Magnetic stored energy =
1
2LI 2
1
2z0l LI2 dz
1
2LI 2
0 l (for uniform distribution)
1
2LI 2
0 vpT 1
2LI 2
01
LCT
=1
2I 2
0 Z0T
6.5-20
Check of Energy Balance
Initial stored energy
We Wm
1
2
V 20
Z0T
1
2I 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 16
2 10 475150 3
0.1 J
LR
tL
Vdt
R
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 2lz
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
1
2
3
60–15
45
40
–540
z = l+ z = 2l
60 V
t, s
0
1
2
3
z = 0
60
45
40
–1545
–540z = l
60 V = 0
= 0V = 1
6.5-27
aa
0
1
2
3
0.6–0.3
0.3
0.4
0.10.4
z = l+ z = 2l
0.6 A
t, s
0
1
2
3
z = 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 2lz
6.5-29
aa
100 +
–
+
–
0.8 A 0.4 A
100
120 V
40 V
0.4 A
100 40 V
z = 0 z = l– z = l+ z = 2l
t = 3 s + :