cse 464 transmission line equations (dally/poulton notation) january 2009

CSE 464 Transmission Line

Equations(Dally/Poulton Notation)

January 2009Based on Notes by Fred U.

RosenbergerDavid M. Zar

Washington University in St. [email protected]

T-Line Equations 2 - David M. Zar - 04/19/23

Transmission Line Equations

)( and )(v

xtV

v

xtV rf

rf VxV anddirection positive in the gpropagatin with waves traveling twoare

rf VVV

.city with velomoveBoth direction. negative in the gpropagatin vx

C

LZ 0 LC

v1

rf III v

ZL 0

vZC

0

1

ff IZV 00

0

ZZZZK

T

TrT

rr IZV 00

0

ZZZZK

S

SrS

2

1

vCL

Superposition (if linear)


General Equations

cmpsinpsftns

cmECLv

CLZ

mCEmCmLvCL

rrr

r

r

rr

/)33( /)83( /)1(Or

sec/)/103(/1

1) (s/w squareper /377/

)//(17112.1)//(/

/1

0

0

2

For Parallel Plates:


Transmission Line Equations

)()( 0 v

ltVtV ffl

)()(0 v

ltVtV rlr

0

00

0

00 )()()(

ZZ

ZZtV

ZZ

ZtVtV

S

Sr

SSf

0

0

0

0 )()()(ZZ

ZZtV

ZZ

ZtVtV

T

Tfl

TTrl

TT

T

Tfll ZZ

ZV

ZZ

ZVV

0

0

0

2

00

0

00 2

ZZ

ZV

ZZ

ZVV

S

Sr

SS

Z0

Zs

ZT

V0+-

+

-Vs

L0

+

-Vl

(ZS-Z0)/(ZS + Z0)= KrS

(ZT-Z0)/(ZT + Z0)= KrT


General Equations (VT = 0)Use Superposition for Multiple Sources

SSrSrf ZZ

ZtVKtVtV

0

000 )()()( 1)

rTfr Kv

ltVtV )

2()( 00 2)

2) into 1),S

SrSrTff ZZ

ZtVKK

v

ltVtV

0

000 )()

2()(

4)S

SrSrTf

ffx

ZZ

Z

v

xtVKK

v

x

v

ltV

v

xtVtV

0

00

0

)()2

(

)()(

SrTS

rSrTfrx

ZZ

ZK

v

x

v

ltV

KKv

x

v

ltVtV

0

0

20

)2

(

)4

()(

3)

5)

At arbitrary x,

From 2) & 3),

2l since we have to go “there and back” to see it at the start of the line


Equations Continued

Repeat 3) into 4), 3) into 5),

0

1

0

0 ))1(2

()2

()(n

nrS

nrTS

nrT

nrSS

Sx KK

v

x

v

lntVKK

v

nl

v

xtV

ZZ

ZtV

fVrV 100

At x = 0,

1

1

0

00 )()

2()()(

n

nrS

nrS

nrTSS

S

KKKv

nltVtV

ZZ

ZtV

fVrV fV

00

0 )2

()1()(n

nrS

nrTSrT

Sl KK

v

l

v

nltVK

ZZ

ZtV

fVrV

Now can find V0, Vl at any time.

At x = l,


Example Driving T-Line

Apply to problem,

01

00)(

t

ttVS

00

0 )2

()1()(n

nrS

nrTSrT

Sl KK

v

l

v

nltVK

ZZ

ZtV

0)12(

0

0)12(

1

v

lnt

v

lnt

v

lNt

v

lNKKK

ZZ

Z N

n

nrS

nrTrT

S

)32()12(for )1(

00

0

Z0

Zs

ZT

This is the time to go from l0 l for the last (N) time.


Steady State Value and Error Term, Example

With algebra,

))(1( 1

NrSrT

ST

T KKZZ

Z

Attenuation in steady state

Error 0 as N .

For 1rSK,0SZ We get

9.0 ,1K ,0 ,1900 ,100 :Example rS0 rTST KZZZ

))9.0(1()( 1

N

ST

Tl ZZ

ZtV

(l/v).N 642 t,8.21)9.0ln(/)1.0ln(1 :10% error For

v

lNt

v

lN )32()12(for


Example Continued For 2 foot line, this could be on the order of

85ns (or even 3.55 ns for a 1 inch line!) What to do?

»Wait a long time»Real wire will have some resistance (Zs≠0)»CS solution: use 1900 Ohm transmission Line»Reduce resistance of receiver to match t-line

– Increased power– May be a problem with driver resistance (e.g. 10 to 50

Ohms)»Increase resistance of driver to match t-line

– 100 Ohms in series with driver gives 2.5% error after only l/v,or after 2ns for one foot line (this is the minimum possible)


Transmission Line Equations for Wire

)/ln()/60(

)/ln(/)/6.55()/ln(/2Coax

120

1212

rrZ

rrmpFrrC

r

r

)/2ln()/60(

)/2ln(/)/(6.55)/2ln(/2/

0 rsZ

rsmpFrsmC

r

r

r

r2

r1

s

Wire over ground plane(warning, not accurate for r close to s)

Between 2 Wires: 1/2 capacitance of wire over ground plane


PC Board Equations

w

3770 w

sZ

r s

Parallel Plates(ignoring fringing)

0

2

2

0rw

sws

swsw

C

LZ


PC Board Equations

67.0475.0017.1)/(

)8.0

98.5ln(

41.1

870

rpd

r

ftnst

tw

hZ

w

t

h dielectric

air

wt

h2=h1rpd

r

ftnst

bttbw

wtw

bZ

017.1)/(

25.0/ ,35.0)/(for

))/8.0(67.0

4ln(

600

b

Stripline

Microstrip

h1

(Careful here, approximations grossly wrong for some values: large w/h)

cse 464 transmission line equations (dally/poulton notation) january 2009

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