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 + :