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 + :