Section 6 3 TRƯỜNG ĐIỆN TỪ

No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]

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

Discontinuity

V+ + V–

I+ + I– I++

V++

+

–

+

–

(+) (–)

Z01, v p1 (++) Z02, v p2

V  V V

I  I –  I





B.C.

I  V 

Z01 , I –  – V

–

Z01 , I V



Z02

02 01

V V V V

Z

Z



 V–

V  

Z02 – Z0l

Z02  Z0l

Current Transmission Coefficient,

C  I 

I   I

  I –

I 1  I

–

I 

C 1 – 

V 1  

V V

  V –

V  1  V

–

V

Define Voltage Transmission Coefficient,

Note that

2 2

1 1

V I P P

  

 









  

  







–V o

+ –

50  Z0l = 50  Z02 = 100  Z03 = 50 

T1 = 2 s T2 = 2 s T3 = 2 s50 

0 4 8 12

t, s

1/2

2/3 –2/9 2/27 –2/81

2/243

4/9 4/81 4/729

6 10 14

4/9 4/92 4/93

 = 1/3  = –1/3  = –1/3

 = 0 V = 4/3 V = 2/3 V = 2/3  = 0

Three Lines in Cascade

(t)

0 6 10 14

4/9

4/92 4/93

and so on

t, s

   

0

g

n

n

T T T T











 

 

  



V g (t)

System System – V g (t –  ) h( ) d

(t)

For V g (t) = cos t ,

0

2 0 0

0

2 0

2 0 0

cos

n

n

n n

n n

nT T d

t

nT T d

t nT T

 

 





 





 







 

 

 

 

 

 











   2 0 

0

2

2 0

0

2 0

2

4 9 1 1

9

n

j nT T

n

n

n

j T

j T

e e

















 

 









V o()  4 9

1 – 1

9 e – j2T2

V o() max  4 9

1 – 1 9 0.5

V o() min  4 9

1  1 9 0.4

0.4

0

V o()

 Line 1

Z0 = 50  50  100 



Z  

Lin e 3

Li ne 2

0

100

Z





0

Z 

 eff 2

4 1

5 6 1

5 100

50 100

V

C

C











   

   









 eff 3 50 100 50

C











Pref1 2P  1

25 P

eff2

trans2 =

V C

 

Note that 1

25  4875  2475 1

eff3

trans3=

4 12 24

5 15 75

 

  