Section 6 5 TRƯỜNG ĐIỆN TỪ

Slide Presentations for ECE 329,
Introduction to Electromagnetic Fields,
to supplement “Elements of Engineering
Electromagnetics, Sixth Edition”
by

Nannapaneni Narayana Rao

Edward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, Urbana, Illinois, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India


6.5
Lines with Initial Conditions


6.5-3

Line with Initial Conditions

I(z, 0)
++++++++

Z 0, v p
--------

V(z, 0)

V  (z, 0)  V – (z, 0) V(z, 0)

I  (z, 0)  I – (z, 0)  I(z, 0)

–
V
V
I 
, I – –
Z0
Z0

V  (z, 0) – V – (z, 0) Z0 I(z, 0)


6.5-4

1
V  z , 0    V  z , 0   Z0 I  z , 0 
2


1
V  z , 0    V  z , 0   Z0 I  z , 0 
2



6.5-5

Example:

V(z, 0), V

I(z, 0)
++++++++

Z0, vp
--------

V(z, 0)

z = 0 Z 50 
0

zz =ll

50
0

l z
I(z, 0), A

1
0

l

z


6.5-6

V +(z, 0), V
I +(z, 0), A
B
50
1
A
C
0
l z 0
l z
I –(z, 0), A

V –(z, 0), V
50 C
0

D

1
0
l z
–1

l z


6.5-7

l

t
2vp

V +, V
50
0

V –, V

50

B

C

D

l

1
z

B

A

0

I +, A

l

0

I –, A

z

1
z

V, V

0
–1

l

100

z

I, A

50
0

l

1

l

z

0

l

z


6.5-8

t  vl
p
V +, V

I +, A

’

’

50
D

0

1

C
l

z

0

V –, V

I –, A

’

50

z

l
’

B

1

C

A

0

l

z

0

l

z

–1

V, V

I, A
1

50
0

l

z

0
–1

l

z

6.5-9

I(z, 0)

t=0

+++++++

Z0, vp
-------

RL = Z0 = 50 

V(z, 0)

z=0

z=l

V(z, 0), V

I(z, 0), A

50
0

1
l

z

0

l

z


6.5-10

V +(z, 0) V
V –(z, 0) V
B
50
50 C
D
A
C
0
l z 0

l z

[V]RL, V
50
0

A

B

l/2vp

C
l/vp

D
3l/2vp

t


Uniform Distribution

6.5-11

I(z, 0) = 0
+++++++

Z0, T V(z, 0) = V0
------z=0
z=l
V0

–
V (z, 0) V (z, 0) 
2

V0
V0

–
I (z, 0) 
, I (z, 0) –
2Z0
2Z 0


6.5-12

V0 100 V, Z 0 50 
V, V
50

I, A
(–)

50

1

(+)

0

l

z

0
–1

S

(+)

l z

(–)

I(z, 0) = 0
+++++++

RL

t=0
Z0 , T V(z, 0) = V0
------z=0

V0 100 V, Z 0 50 
RL 150 , T 1 mS

z =l


6.5-13

t = 0.5 mS

V, V
50

I, A
(–)

25

(+)

0

l

(–)

0

z

(+)

0

(–)

–1

t = 1.5 mS

V, V
50
25

1

l z

I, A

1
(+)

l

z

0
–1

(+)
(–)

l z


6.5-14

t = 2.5 mS
V, V

50
12.5

I, A
(–)

0.5
0
l z
–0.5

(+)

0

75
0

(+)
(–)

l z

[V]RL, V
37.5
2

18.75
4

9.375
6

t, mS


6.5-15

Bounce Diagram Technique
for
Uniform Distribution

0 + I+

RL

+
V0 + V +
–
z=0
V0  V   RL 0  I  

V
I 
Z0


z=l
B.C.

6.5-16

RL 

V0  V –
V
Z0
 RL 

V 1 
– V0
 Z0 
V  – V0

Z0
R L  Z0

For V0 100 V, Z0 50 , and
RL = 150 ,
50

V – 100
– 25 V
150  50


6.5-17

=

0
75

1
2

100 V
–25

–12.5

t, mS 37.5
4

1

50
3

–12.5
–6.25

18.75
z=0

0

100

–25

2

75

=1

25
5
z=l

z

[V]RL
37.5
2

18.75
4

9.375
6

t, mS


6.5-18

Energy Storage in Transmission Lines
1 2

we, Electric stored energy density = CV
2
l 1
We, Electric stored energy = z0 CV 2 dz
2
1 2
 CV 0 l (for uniform distribution)
2
1 2
1 2 1
 CV 0 v p T  CV 0
T
2
2
LC
2
1V0

T
2 Z0


6.5-19

1 2
wm, Magnetic stored energy density = LI
2
l 1 2
Wm, Magnetic stored energy = z0 LI dz
2

1 2
 LI 0 l (for uniform distribution)
2
1 2
1 2 1
 LI 0 v p T  LI 0
T
2
2
LC
1 2
= I 0 Z0 T
2


6.5-20

Check of Energy Balance
Initial stored energy

We  Wm
2
1V0
1 2

T  I 0 Z0 T
2 Z0
2
1 (100)2


10 –3  0
2 50
0.1 J