ECE 391 supplemental notes - #11 - Oregon State …web.engr.oregonstate.edu/~traylor/ece391/Andreas_slides/...ECE 391 supplemental notes - #11 168 Oregon State University ECE391–

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ECE 391 supplemental notes - #11

168 Oregon State University ECE391– Transmission Lines Spring Term 2014

Adding a Lumped Series Element Consider the following T-line circuit:

ZL

Z0,1! Z0,2!R

zL =ZL

Z0,2zin,2 = rin,2 + jxin,2

zL,1 = rin,2Z0,2Z0,1

+ RZ0,1

+ jxin,2Z0,2Z0,1

zin,1 = rin,1 + jxin,1Zin,1 = zin,1Z0,1

è All calculations can be done on Smith chart

renormalization

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Smith Chart as Admittance Chart

For circuits involving parallel (shunt) connections of circuit elements, it is more convenient to work with admittances.

How can the Smith chart be used with admittances?

è How is Y=1/Z = G + jB related to reflection coefficient Γ ?

Consider normalized admittance y

y = YY0

= G + jBY0

= g + jb = Y Z0 =Z0Z

= 1z= 1r + jx


Reflection coefficient Γ

Smith chart can directly be used with y=g+jb after rotating reflection coefficient by 180°.

Smith Chart as Admittance Chart

Γ = z −1z +1

= 1 y −11 y +1

= − y −1y +1

y −1y +1

= −Γ = Γ e jπ g = const

b = const circles

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Impedance vs. Admittance Coordinates

r + jx

r - jx SC OC

Re(Γ)

Im(Γ)

inductive

capacitive

0

g + jb

g - jb OC SC

Re(-Γ)

Im(-Γ)

inductive

capacitive

0

Impedance Chart Admittance Chart


Example 4 Given the normalized load admittance

yL = 0.5 + j2.0

Determine the normalized admittance at distance

d = λ/16 = 0.0625λ

Solution: use Smith chart with admittance coordinates è see web applet (example 4)

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Example 5 Given a 50Ω T-line that is terminated in ZL = ZT= 25Ω and has a shunt resistance Rsh = 50Ω connected at distance d = 1m from the termination. The wavelength on the line is λ = 3m.

Determine the input impedance at the location of the shunt resistance.

Solution: see web applet, example 5


Additional Example

( )Ω+= 2525 jZLΩ= 500Z

λ1025.0=l

Results

6.2SWR ≈

45.0≈ΓL5.116≈Lθ

1.15.1zin j+≈j−=1yL

5.05.0z jL +=

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Impedance Matching

Implementation Issues §  Types of Networks

-  lumped elements -  distributed elements -  mixed -  reactive vs. resistive

§  Design Criteria -  bandwidth -  lossless/lossy -  complexity -  parasitics


Impedance Matching

è Matching network needs to have at least

two degrees of freedom (two knobs)

to match a complex load impedance

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Transmission-Line Matching Idea: Use impedance transformation property of transmission line

transformation circle

2βz’ load zL (or yL)


Transmission-Line Matching Transformation to real z or y


2βz’

real z (or y) real z (or y)

load zL (or yL)

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Transmission-Line Matching Transformation to r=1 (or g=1) circle


2βz’

r=1 or g=1

load zL (or yL) z=1+jx (y=1+jb)

z=1-jx (y=1-jb)


Single Stub Matching Stub tuners are useful for matching complex load impedances. The two design parameters in single stub matching networks are

§  distance from load at which stub is connected

§  length of stub

Other design considerations are

§  open or shorted stub

§  series or shunt connected stub

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Basic Design Principle

Objective: Match load admittance by 1.  transforming along T-line to Yin = Y0+jB 2.  tuning out remaining jB with a parallel shunt element of

admittance –jB 3.  shunt element can be realized with a T-line stub

Yin = Y0


Design Steps

1.  Determine distance d/λ from the load YL (or ZL) at which the §  input admittance is Y0 +jB (for shunt stub matching) §  input impedance is Z0+jX (for series stub matching)

2.  Realize the stub with a length l/λ of open- or short-circuited transmission line (stub).

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Example 6

Using a short-circuited stub, match the load impedance ZL=(80 + j70)Ω to a transmission line with Z0 = 50Ω.

Z0 ZL

d

Solution: use Smith chart with admittance coordinates è see web applet (example 6)


Additional Example

Ω=125LRpF54.2=LC

GHz10 =fΩ= 500Z

jL −= 5.0z

58.11yin_1,2 j±≈

mS6.31/58.1 0sh_1,2 jZjY ==

Need shunt element with

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Quarter-Wave Transformer

LZZZ 01 =

increased mismatch

NOTE: for complex load, insert λ/4 transformer where Zin=R

4/λ at design frequency f0


Example: Quarter-Wave Transformer

Ω=125LRpF54.2=LC

GHz10 =fΩ= 500Z

jL −= 5.0z

Ω≈ 5.24Z T0,

24.0zin,1 ≈ Ω≈12Zin,1

in,1

2,0

0in,2ZZZ

Z T==From

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Bandwidth Performance

GHz1at 0 =f


Lumped Element Matching Reactive L-Section Matching Networks

0ZRL <0ZRL >

220

220

LL

LLLLL

XRZRXRZRX

B+

−+±=

LL

L

RBZ

RZX

BX 001 −+=

( )0

0

ZRRZ

B LL−±=

( ) LLL XRZRX −−±= 0

different solutions may result in different bandwidths

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Example

Design criteria §  bandwidth §  realizability of components §  parameter sensitivity §  … cost … ?


Lumped-Distributed and Distributed Matching Networks

Series Reactance Matching Shunt Reactance Matching

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Examples


Double-Stub Tuner •  Double-stub (and multi-stub) tuners use fixed

stub locations (typical spacing: λ/8)

•  Advantages: -  easier to accommodate different loads -  better bandwidth performance

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Smith Chart with Lossy T-Lines Recall for a lossless T-line

For low-loss T-lines, Z0 ≈ real and γ = α + jβ

Then

zin (z ') =Zin (z ')Z0

= 1+ ΓLe−2γ z '

1− ΓLe−2γ z ' =

1+ ΓL e−2αz 'e jφ z '( )

1− ΓL e−2αz 'e jφ z '( )

zin (z ') =Zin (z ')Z0

=1+ ΓL e

jφ z '( )

1− ΓL ejφ z '( ) φ(z ') = θL − 2βz '


Smith Chart with Lossy T-Lines Γ z '( ) = ΓL e

−2αz '

Γre

Γim As z '→∞zin →1(Zin → Z0 )

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Approach

Γre

Γim

zL

Γ inlossy = Γ in

losslesse−2αz '

Γ inlossless = ΓL e

jφ

1.  Assume lossless line and rotate by -2βz’ 2. Change |Γ| by factor

e−2αz '


Example 7

Given a lossy T-line with characteristic impedance Z0 = 100Ω, length l = 1.5m (l < λ/2), and ZSC(z’=l) = 40 – j280 Ω.

(a) Determine α and β.

(b) Determine Zin for ZL = 50 + j50 Ω an l = 1.5m.

Solution using Smith chart

è see web applet (example 7)

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ECE 391 supplemental notes - #11 - Oregon State …web.engr.oregonstate.edu/~traylor/ece391/Andreas_slides/...ECE 391 supplemental notes - #11 168 Oregon State University ECE391–