Notes 3.1 Reflection from a load / bounce diagram PDF

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Notes 3.1 Reflection from a load / bounce diagram /
Reflection coefficient/ Transmission coefficient / Solving transmission line circuits / special cases...

Description

ELEC 351 Part 3 Reflection and Transmission in the Time Domain 1. 2. 3. 4. 5. 6. 7.

Reflection from a Load The Bounce Diagram Response to a Pulse Function Final Values Junctions between Transmission Lines Time Domain Reflectometry Time Domain Demonstration

(c) Copyright C.W. Trueman 2020

Transmission Line Terminated with a Resistor 



Transmission line m/s,



ohms



The Reflection Coefficient at the Load 



z V  ( z , t )  f  (t  ) u z V  ( z, t )  f  (t  ) u

Simplify the notation

 



Reflection from the Load  

Define “reflection coefficient”



Important Special Cases  



(1) Matched load:   so The incident wave is completely absorbed by the load. (2) Load resistance greater than line characteristic resistance:     so   



Extreme case: open circuit load so .  The reflected voltage is the same as the incident voltage,





(3) Load resistance less than line characteristic resistance:     so   



Extreme case: short circuit load so .  The reflected voltage is the negative of the incident voltage,





Solving Transmission-Line Circuits

  ohms,

cm/ns



cm

• What is the voltage across the generator terminals at z=0? • What is the voltage across the load, at z=L?

Solving Transmission-Line Circuits 



5 volts 

t=0

time

ohms,

cm/ns

cm

Find the voltage across the load resistor as a function of time.



Values of the coefficients: 



5 volts 

t=0

ohms,

cm/ns

cm

time



  







 





Initial Step Launched onto the Transmission Line • In general the voltage and current on the transmission line are 











• But there is no reflected voltage at the generator until enough time has passed for the generator voltage to travel to the load, be reflected, and travel back to the generator. for so at • At the generator    

• The input resistance of the transmission line is  

 

Initial Step Launched onto the Transmission Line • The input resistance of the transmission line is

.

• We can represent the transmission line with a resistor of value

 

   



=5 volt step

 

 ,  ,



volt step

• The generator launches a step function onto the transmission line of height 

volts

After one transit time:  =4.901

+

  









 =0.9047

volts The voltage at the load after one transit time is the incident plus the reflected voltage: volts After t  Td the load voltage is

=4.901+4.434= 9.335 volts.









  

After two transit times:

 =-4.260

+

 

After three transit times:















 =0.9047

The voltage at the load after one transit time is the incident plus the reflected   voltage: After t  Td the voltage is 4.901+4.434= 9.335 volts.

After three transit times, a new incident wave arrives at the load and generates a new reflected wave. The voltage is the previous value plus the new incident voltage plus the new reflected voltage. At t  3Td the voltage is 9.335 -4.260-3.854=1.221 volts. Homework: show that at



𝑉

the voltage is 1.221 + 7.035 = 8.256 volts

9.335

8.256 8.114

7.053

1.221

time 1

2

3

4

5 ns

Removing the Reflections: 



5 volts 

t=0

time

,

cm/ns cm



Better match at the source: Rs  10

Rs  25

Rs  50

Perfect match!

Better match at the load:

Better match at the generator and at the load:

BOUNCE Demonstration 



5 volts 

t=0

time

ohms,

cm/ns



cm

On the course web site you will find a demonstration of the circuit solved in this set of notes: Notes_3_1.mp4 You can watch the step function bouncing back and forth better the load and the source, and check the values we calculated in these notes.

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