Download No Slide Title

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Fields and Waves
Lesson 1.3
TRANSMISSION LINES - INPUT IMPEDANCE
Darryl Michael/GE CRD
Input Impedance
Do Experiment 1
• After Experiment: Cable+Load can look very different
than Load
• Characterize V/I as Impedance
Input Impedance
From previous class:
V̂ ( z )  V  ( z )  V  ( z )  Vm  e  j  z  ( 1   ( z ))
What about I?
V  ( z ) V  ( z ) Vm  j    z
Î ( z ) 


e
 ( 1   ( z ))
Rc
Rc
Rc
Also,
 ( z )   L  e 2 j ( L z )
Input Impedance
Form the Ratio:
V̂ ( z )
1  ( z )
 Rc 
Î ( z )
1  ( z )
Book calls this Zin(z)
We are primarily interested in z=0 value
• treat connection to rest of circuit as 2 port with,
Z in ( z  0 )  Rc 
1  ( z  0 )
1  ( z  0 )
Input Impedance
After lots of algebra, one can show:
Z L  j  Rc  tan(   L )
Z in ( z  0 )  Rc 
Rc  j  Z L  tan(   L )
Special Case example: ZL=0 (short circuit)
0  j  Rc  tan(   L )
Z in ( z  0 )  Rc 
 j  Rc  tan(   L )
Rc  j  0  tan(   L )
Input Impedance - SHORT CIRCUIT
Z in( z  0 )  j  Rc  tan(   L )
Can change Zin by
changing these two
parameters
• Fix , vary L - different effects
• Vary , fix L - get same effects
Note that L is the
length of the
Transmission Line
Input Impedance - TL
Setup for Problem 1a & 1b
1a. Open Circuit Case
ZL  
small
Z L  j  Rc  tan(   L )
Z L  Rc
Rc  j  Z L  tan(   L )
small
1b. ZL = 93W - lots of complex algebra
Problem 1a and 1b
We know Zin (z=0) - treat as 2-PORT
80 W
1V
Power 
Zin

1
Re Vin  I in
2

Vin
Voltage
Divider
2



V
V

V
1
1 

 Re in  in   Re in 
2  Z in  2 
 Z in 

Do problem 1b - parts 2 &3
Problem 1a and 1b
In a Lossless Transmission Line, Pin flows into the
Transmission Line and it is dissipated at the
2
LOAD
1 VL
Pin  
2 ZL
Do Problem 1b - part 4
V̂ ( z )  Vm  e  j   z  ( 1   ( z ))
V̂ ( z  0 )  Vin  Vm  e  j  z  ( 1   ( z  0 ))
 Vm 
Vin
Can then plug back and get the full phasor
1  (0 )
Do Problem 1b - part 5
Related documents