13-1

EE334 - Transient Bounce Diagrams

13 Lecture: pp 92-96 2-11

Transients:

Vs

Rs

Vo Zo

+

-

t=0

ZL

L

At t = 0 a switch is thrown, what is the voltage across ZL if L goes to zero?

LS

LS

ZRZV

V +=0 When L is not zero, if we leave the switch closed long enough, this should be the ultimate voltage across the load impedance. What does this mean in a transmission line problem? When we close the switch a voltage will begin to travel toward the load at the phase velocity of the transmission line.

V1+

u

L What is that voltage +1V , the first transient traveling in the positive direction.

13-2

Its magnitude is as calculated from the source voltage and impedance and the line impedance, (it only sees the line impedance, it doesnt know there is a load at the end of it)

Vs

Rs

Vo Zo

+

-

01

0

01 ZR

VI

ZRZV

VS

S

S

S

+=+=++

What happens at the end if the load impedance does not equal the line impedance? (Reflection)

z = L

V 1+

u

ttd

V 1-

V 1+ V 1

-+

Transmittedto the load

Once the step reach the end of the line t equals the delay time some of the energy gets transmitted to the load and some of the energy (V1-) reflects. The reflected voltage adds to the initial step that is already there and travels in the opposite direction towards the source.

+ = 11 VV L when this negative traveling step reaches the source it will reflect if the generator impedance does not match the line impedance

+ = 12 VV S

13-3

At steady-state the switch has been left on for a long time so there has been infinite reflections, then:

[ ]( )[ ]KK

KK

++++=++++=

++++=+++=

+

+

++++

++

221

21

12

111

2211

11

1

SLSLL

SLSLL

SLSLL

VV

VVVVVVVVV

This last series is a binomial series

K++++=321

11 xxxx

( )

=

+

SLLVV 1

111

apply the refection coef. in terms of impedances:

0

0

0

0

ZZZZ

ZZZZ

S

SS

L

LL +

=+=

and simplify

LS

LS

ZZZV

V += which is the equivalent to no transmission line effect (as it should)

similarly LS

S

L ZZV

ZVI +==

We need a method to keep track of the transients as they reflect in the transmission line, the method we will use is the bounce diagram

13-4

VS

RS

z=0 z=L

ZL

L,u

time

position in TL (z)S

L

L/u=Time to cross

line

2

3

4

5

6

Slope of line isthe velocity

V1

+

V1

-= V

1

+ L

z1

t1

t2

t3

t4

t5

t6

to find what is happening at any given position of the transmission line, draw a vertical line at the position, each time a bounce crosses the position line the corresponding reflected magnitude is added to the potential at that position.

13-5

V (z1,t)

t1 t2 t3 t4 t5 t6

V1

+

V1

-V

2

+ V2

- V3

+ V3

-

( ) ( )( )

13-6

(3) Use these values to fill in the bounce diagram:

s = -0.6 = 1L

2

3

4

5

V = 0.8Vo1+

V = 0.8Vo1-

V = -.48Vo2+

V = -.48Vo2-

V = -.288Vo3+

64928.288.48.48.8.8.4212.148.08.08.0

208.00

13-7

1

Vs/Vo

2 3 4 5 6

5364..048.48.8.8.36.18.08.0

000

13-8

( ) ( ) ( ) ( ) ( )==+ tuVtuVtVtVtV 0021 EXAMPLE

1V200ps pulse

Rs = 900

R = 25L

Zo = 100t = 400ps

( )( ) mVVVZR

ZVV

ZZZZ

ZRZR

S

L

LL

S

SS

1001.0100900

1001

6.010025100258.0

100900100900

0

001

0

0

0

0

==+=+=

=+=+

==+=+

=

+

13-9

G = 0.8 G = -0.6

200ps

400ps

600ps

800ps

1000ps

1200ps

1400ps

1600ps

1800ps

V(0,t)V(L,t)

100mV-100mV

100-60

-60mV

60mV

100

-60-48-48mV

48mV-48+28.8

28.8mV

-28.8mV

28.0+23.023.0mV

23.0mV

Junctions in transmission lines or cascaded transmission lines If lines not matched there will be reflections at the junctions

R =100L

Rs = 50

Za = 50 = 500ps

Zb = 25 = 200psVo=1.5V

31

7525

50255025

050505050

==+=+

=

=+=+

=

ab

abab

aS

aSS

ZZZZZRZR

13-10

( )( ) 75.5050505.1

6.012575

2510025100

31

7525

25502550

01 =+=+=

==+=+

=

==+=+

=

+

aS

a

bL

bLL

ba

baba

ZRZVV

ZZZZZZZZ

We now need to calculate the transmission coefficient.

32

3111

34

3111

==+=

=+=+=

abab

baba

or we can remember that the total voltage on the left has to equal the total voltage on the right of the connection between lines.

=0 =-1/3ab

=1/3bas

= 0.6L

V=.75V = .5

V=-.25 V=.3 .7 ns

1.0 ns

1.4 ns

1.8 ns

1.1 ns

1.5 ns

V =.4V =.1

V=.06

V = .02

V = .08

we can always calculate the steady-state potential and compare the load and source voltages. They should converge to the steady-state values. ( )

( )( ) VZRZV

VVLS

LSSS 110050

1005.1 ==+==

13-11

V(0,t)

0.75V0.5V

0.9V.098V

V(L,t)

0.8V 0.96V 0.99V