transmission lines

Winter 2005

ECE

ECE 766Computer Interfacing and Protocols

108 -

Transmission LinesTransmission Lines

• Transmission line effects must be considered when length is comparable to ¼ wavelength

• We will ignore the energy loss on transmission lines

• Concentrate on time-domain description rather than frequency domain

eg

+

-

zg

zl

i

i

e

x dl

Winter 2005

ECE


208 -

Transmission LinesTransmission LinesxR xL

xG xC

),( txxi ),( txi

),( txe

-

+

),( txxe

-

+

x

C: Capacitance / Unit Length [F/m]L: Inductance / Unit Length [H/m]R: Conductor Resistance / Unit Length [Ω/m]G: Insulation Conductance /Unit Length [ /m]

Note G ≠ 1/R !

Ω

Winter 2005

ECE


308 -

Transmission LinesTransmission Lines

xR xL

),( txi),( txe ),( txxe

By KVL:

t

txiLtxiR

x

txetxxe

txxet

txixLtxixRtxe

),(),(

),(),(

0),(),(

),(,

In the limit as :

t

txiLtxiR

x

txe

),(),(

),(

0x

Winter 2005

ECE


408 -

Transmission LinesTransmission Lines),( txi

By KCL:

t

txxeCtxxeG

x

txitxxi

txxit

txxexCtxxexGtxi

),(),(

),(),(

0),(),(

),(,

In the limit as :

t

txeCtxeG

x

txi

),(),(

),(

0x

xC

),( txxi

xG

Winter 2005

ECE


508 -

Lossless CaseLossless Case

(2)

(1)

t

eC

x

it

iL

x

e

0

0

G

R

Take partial derivative w.r.t. x in (1) and partial derivative w.r.t. t in (2),then substitute.

2

2

2

2

2

2

2

2 1or

1

t

i

x

i

LCt

e

x

e

LC

Recognize as wave equations

Winter 2005

ECE


608 -

Lossless CaseLossless Case

Show that the solution is in the form . txLCfe

Q.E.D.

obtain weequation, wavethe into ngSubstituti

Similarly,

Let

2

2

2

2

2

2

2

2

2

2

2

2

2

2

1

.

.

ds

fd

ds

fdLC

LC

ds

fd

dt

ed

ds

fdLC

x

s

ds

fdLC

ds

dfLC

xx

e

ds

dfLC

x

s

ds

df

x

e

txLCs

Winter 2005

ECE


708 -

InterpretationInterpretation

• is a wave traveling to right with velocity

• is a wave traveling to left with velocity

• Solving for i, we obtain

• is characteristic of line

txLCf

txLCfCL

i /

1

LC/1

txLCf LC/1

CL /

line lossless the of impedancestic characteri is CLz /0

Winter 2005

ECE


808 -

InterpretationInterpretation

txLCftxLCf

zi

txLCftxLCfe

210

21

1

Note:Each traveling wave direction, e and i are related by z0.

Winter 2005

ECE


908 -

ReflectionsReflections

• Look at terminations with real impedances frequency independent

• Wave of voltage and current traveling to right

• At termination

zl

z0

e-

+el

+

-

il

0zi

e

lzi

e

Winter 2005

ECE


1008 -


• Hence, there must be reflected waves e- and i- such that

• In terms of voltage

l

ll

ll zii

ee

z

ei

and

0

l

ll

ll z

ze

ze

ee

00

kzz

zz

e

e

l

l

l

l

0

0 Reflection coefficient

Winter 2005

ECE


1108 -


• Special cases:– zl = z0 k=0

Matched, no reflections. Line looks infinite.

– zl = 0, short circuit k = -z0/z0 = -1

– zl = , open circuit k = 1

8

Winter 2005

ECE


1208 -

Multiple ReflectionsMultiple Reflections

10

0

00

z

zk

5.033

00

00

zz

zzkl

R=3z0z0

t=0

x=0 x=l

E

Hitting load, a wave of E/2 is produced

Et=.3T

Et=1.3T 3E/2

Et=2.3T 3E/2

Arriving wave of E/2 isreflected toward load

T 2T 3T 4T 5T

Voltage at load1.5E

.75E

1.125E.9375E

6T 7T

E

Winter 2005

ECE


1308 -

Time-Space (Bounce) DiagramTime-Space (Bounce) Diagram

• Mark reflection coefficients

• Write initial voltages• Write wave amplitudes• Update

– Wave amplitudes when reflected

– Voltages as waves cross

xk0=-1 kl=.5

e=E

e=0

e=E

e=E

E

E/2

-E/2

-E/4

E/4

E/8

T

2T

3T

4T

5T

e=3E/2

e=3E/4

e=9E/8

Tim

e

Winter 2005

ECE


1408 -

ExampleExample

3z0

E z0

Magnitude of the first wave:

43 00

00

E

zz

zEE

xk0=.5 kl=1

e=E/4

e=0

e=5E/8

e=13E/16

E/4

E/4

E/8

E/8

E/16

E/16

T

2T

3T

4T

5T

e=E/2

e=3E/4

e=7E/8

Tim

e

z0

t=0

k0=.5

E

R=3z0

lkl=1

E

t

At x=0

T 2T 3T 4T 5T

Winter 2005

ECE


1508 -

Reflections in Digital LinesReflections in Digital Lines

• Consider one source and one load

• Option 1:Do not terminate either end. Ringing will stop eventually.– Pro: Simple, no additional power loss– Con: Limited speed

Winter 2005

ECE


1608 -


• Option 2:Matched termination at the end

– Pro: No reflections– Con: Excessive power consumption

For z0=150Ω, power consumption 135mWReduce by duty factor

(0.5 for regular lines, 0.05 for floppy drives)Multiply by number of lines

z0 Open-collectordriver

Winter 2005

ECE


1708 -


• Option 3:Matched termination at the source end

– Pro: Can be run at the same speed as loadtermination If receiver has very high input impedance, fullvoltage appears at the receiverNo power dissipated at constant voltage level

– Con: Special, high input impedance line receiversrequired (not suited for standard TTL)Look at multiple terminations

z0

Winter 2005

ECE


1808 -

Reflection in Digital LinesReflection in Digital Lines

• Problem for multiple receivers

– Assume the ideal case, where taps are infinitely short and have infinite impedances

– Even for this case, intermediate taps do not get full signal immediately

z0

E

xk0=0 kl=1

e=.5E

e=0E/2

E/2

T

2T

e=E

E

t

At x=.5L

.5T T 1.5T 2T

Winter 2005

ECE


1908 -

Reflection and Transmission at Reflection and Transmission at JunctionsJunctions

e1+, i1+

e1-, i1-

e2+ , i2

+

e3

+, i3 +

0

33

0

22

0

11

0

11

3211

32

211

, , ,z

ei

z

ei

z

ei

z

ei

iiii

ee

eee

320

3

0

2

0

1

0

1

211

, eez

e

z

e

z

e

z

e

eee

121

121

2 eee

eee

3

21

11

21

11

111

eee

1122 3

2

3

11

11

eeee

Winter 2005

ECE


2008 -

Reflection and Transmission at Reflection and Transmission at JunctionsJunctions

3/2 L

2/3 L

Lz0 t=0T

2T

3T

4T

k=0

k=1

k=1E/2

-E/6

2E/9

E/3E/3

2E/9

E/3

E/3

-E/9

Many multiple reflections,eventually come to rest

transmission lines

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