Lecture 5 Transmission Line Impedance Matching

8/2/2019 Lecture 5 Transmission Line Impedance Matching

1/20

EECS 117

Lecture 5: Transmission Line Impedance Matching

Prof. Niknejad

University of California, Berkeley

University of California, Berkeley EECS 117 Lecture 5 p. 1/


2/20

Open Line I/V

The open transmission line has infinite VSWR andL = 1. At any given point along the transmission line

v(z) = V+

(ejz + ejz) = 2V+

cos(z)

whereas the current is given by

i(z) = V+

Z0(ejz ejz)

or

i(z) = 2jV+Z0

sin(z)



3/20

Open Line Impedance (I)

The impedance at any point along the line takes on asimple form

Zin() = v()i() = jZ0 cot()

This is a special case of the more general transmission

line equation with ZL = .Note that the impedance is purely imaginary since anopen lossless transmission line cannot dissipate any

power.

We have learned, though, that the line stores reactiveenergy in a distributed fashion.



4/20

Open Line Impedance (II)

A plot of the input impedance as a function of z isshown below

-1 -0.8 -0.6 -0.4 -0.2 0

2

4

6

8

10

Zin(/4)

Zin(/2)

z

Zin(z)

Z0

The cotangent function takes on zero values when approaches /2 modulo 2



5/20

Open Line Impedance (III)

Open transmission line can have zero input impedance!

This is particularly surprising since the open load is ineffect transformed from an open

A plot of the voltage/current as a function of z is shownbelow

-1 -0.8 -0.6 -0.4 -0.2 00

0. 5

1

1. 5

2

v(z) i(z)Z0

z/

v/v+

v(/4)

i(

/4)



6/20

Open Line Reactance

/4 capacitor < /4 capacitivereactance

= /4 short (actslike resonant seriesLC circuit)

> /4 but < /2 inductive reactance

And the process re-peats ...

.25 .5 .75 1.0 1.25

-7.5

-5

-2.5

0

2. 5

5

7. 5

10

jX(z)

z



7/20

/2 Transmission Line

Plug into the general T-line equation for any multiple of/2

Zin(m/2) = Z0ZL +jZ0 tan(

/2)

Z0 +jZL tan(/2)m/2 = 2

m2

= m

tan m = 0 if m ZZin(m/2) = Z0ZLZ0 = ZLImpedance does not change ... its periodic about /2

(not )



8/20

/4 Transmission Line

Plug into the general T-line equation for any multiple of/4

m/4 =

2

m

4 =

2mtan m

2= if m is an odd integer

Zin(

m/4) = Z

2

0

ZL

/4 line transforms or inverts the impedance of theload



9/20

Effect of Source Impedance

Zs

Vs ZLZ0

Up to now we have considered only a terminatedsemi-infinite line (or matched source)

Consider the effect of the source impedance Zs

The voltage at the input of the line is given by

vin = v() = v+ej(1 + Le2j)



10/20

Effect of Source Impedance

By voltage division, the voltage can also be expressedas

vin =Zin

Zin + ZsVs

Equating the two forms we arrive at

v+ = ZinVs(Zin + Zs)ej(1 + Le2j)

In a matched system, we desire the input impedance

seen into the T-line to be the conjugate of the sourceimpedance (maximum power transfer)

Impedance matching is required to acheive this goal



11/20

/4 Impedance Match

Rs

Vs RL

/4

Z0 =RLRs

If the source and load are real resistors, then a

quarter-wave line can be used to match the source andload impedances

Recall that the impedance looking into the quarter-wave

line is the inverse of the load impedance

Zin(z = /4) = Z20

ZL



12/20

SWR on /4 Line

In this case, therefore, we equate this to the desired

source impedance Zin =Z20RL

= Rs

The quarter-wave line should therefore have acharacteristic impedance that is the geometric mean

Z0 =

RsRL

Since Z0 = RL, the line has a non-zero reflectioncoefficientSW R =

RL

RLRs

RL +

RLRs

It also therefore has standing waves on the T-line

The non-unity SWR is given by1+|L|1|L|


/


13/20

Interpretation of SWR on /4 Line

Consider a generic lossless transformer (RL > Rs)

Thus to make the load look smaller to match to thesource, the voltage of the source should be increased inmagnitude

But since the transformer is lossless, the current willlikewise decrease in magnitude by the same factor

With the /4 transformer, the location of the voltageminimum to maximum is /4 from load (since the load isreal)

Voltage/current is thus increased/decreased by a factorof 1 + |L| at the loadHence the impedance decreased by a factor of

(1 + |L|)2University of California, Berkeley EECS 117 Lecture 5 p. 13/


14/20

Matching with Lumped Elements (I)

Rs

Vs RLZ0

1

Y = Y0 jB

jB

Y = Y0

Recall the input impedance looking into a T-line variesperiodically

Zin(

) = Z0ZL +jZ0 tan()

Z0 +jZL tan()

Move a distance 1 away from the load such that thereal part of Zin has the desired value


M hi i h L d El (II)


15/20

Matching with Lumped Elements (II)

Then place a shunt or series impedance on the T-line toobtain desired reactive part of the input impedance (e.g.zero reactance for a real match)

For instance, for a shunt match, the input admittancelooking into the line is

y(z) = Y(z)/Y0 = 1 Lej2z

1 + Lej2z

At a distance 1 we desire the normalized admittance to

be y1 = 1jbSubstitute L = e

j and solve for 1 and let = 2z +

1 ej

1 + ej= 1

2

j2 sin 1 + 2 cos + 2


M hi i h L d El (III)


16/20

Matching with Lumped Elements (III)

Solve for (and then 1) from (y) = 1

= 2 = cos1()

1 =

2=

4

cos1()

At 1, the imaginary part of the input admittance is

b = (y1) = 2

12

Placing a reactance of value b in shunt providedimpedance match at this particular frequency

If the location of 1 is not convenient, we can achievethe same result by move back a multiple of /2


M t hi ith St b (I)


17/20

Matching with Stubs (I)

Rs

Vs RLZ0

1

Y = Y0 jB

Y = Y0

jB

At high frequencies the matching technique discussedabove is difficult due to the lack of lumped passive

elements (inductors and capacitors)

But short/open pieces of transmission lines simulatefixed reactance over a narrow band

A shorted stub with < /4 looks like an inductorUniversity of California, Berkeley EECS 117 Lecture 5 p. 17/

M t hi ith St b (II)


18/20

Matching with Stubs (II)

Rs

Vs RLZ0

1

Y = Y0 jB

Y = Y0

jB

open stub

An open stub with < /4 looks like a capacitor

The procedure is identical to the case with lumpedelements but instead of using a capacitor or inductor,we use shorted or open transmission lines

Shunt stubs are easier to fabricate than series stubs


L T i i Li


19/20

Lossy Transmission Lines

Lossy lines are analyzed in the same way as losslesslines

Low-loss lines are often approximated as lossless lines

Recall the general voltage and current on the line

v(z) = v+

ez

+ v

ez

i(z) =

v+

Z0 ez

v

Z0 ez

Where = + j is the complex propagation constant.On an infinite line, represents an exponential decay in

the wave amplitude

v(z) = ez

v+ejz


T i i Li Di i


20/20

Transmission Line Dispersion

What about dispersion? Is the amplitude attenuation afunction of frequency? If so, the wave will distort.Moreover, how does the speed of propagation vary with

frequency?For a dispersionless line, the output should be a linearlyscaled delayed version of the input vout(t) = Kvin(t ),or in the frequency domain

Vout(j) = KVin(j)ej

The transfer function has constant magnitude |H(j)|and linear phase H(j) = The propagation constant j should therefore be a

linear function of frequency and should be a constantIn general, a lossy transmission line has dispersion


Lecture 5 Transmission Line Impedance Matching

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