s Parameters (1)Nn

8/18/2019 s Parameters (1)Nn

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A


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S-parameters are a useful method for representing a circuit as a “black box”


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.


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.

,

.



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A “black box” or network may have any number of ports.

This diagram shows a simple

network with just 2 ports.


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A “black box” or network may have any number of ports.

This diagram shows a simple

network with just 2 ports.

Note :

A port is a terminal pair of lines.


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S-parameters are measured by sending a single frequency signal into thenetwork or “black box” and detecting what waves exit from each port.

Power, voltage and current

can be considered to be inthe form of waves travellingin both directions.


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Power, voltage and current

can be considered to be inthe form of waves travellingin both directions.

For a wave incident on Port 1,some part of this signal

reflects back out of that portand some portion of the signalexits other ports.

S-parameters are measured by sending a single frequency signal into thenetwork or “black box” and detecting what waves exit from each port.


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I have seen S-parameters described as S11, S21, etc. Can you explain?

First lets look at S11.

S11 refers to the signal

reflected at Port 1 for thesignal incident at Port 1.


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First lets look at S11.


reflected at Port 1 for thesignal incident at Port 1.

Scattering parameter S11

1/1.



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Now lets look at S21.


exiting at Port 2 for thesignal incident at Port 1.


2/1.



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2/1.



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2/1.



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A linear network can be characterised by a set of simultaneous equations

describing the exiting waves from each port in terms of incident waves.

S11 = b1 / a1

S12 = b1 / a2

S21 = b2 / a1

S22 = b2 / a2

Note again how the subscript follows the parameters in the ratio (S11=b1/a1, etc...)



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17/107

S-parameters are complex (i.e. they have magnitude and angle)

because both the magnitude and phase of the input signal arechanged by the network.

(This is why they are sometimes referred to as complex scattering parameters).


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These four S-parameters actually contain eight separate numbers:

the real and imaginary parts (or the modulus and the phase angle)of each of the four complex scattering parameters.


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Quite often we refer to the magnitude only as it is of the most interest.

How much gain (or loss) you get is usually more important than how muchthe signal has been phase shifted.


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S-parameters depend upon the network

and the characteristic impedances of thesource and load used to measure it, andthe frequency measured at.

i.e.

if the network is changed, the S-parameters change.

if the frequency is changed, the S-parameters change.

if the load impedance is changed, the S-parameters change.

if the source impedance is changed, the S-parameters change.

What do S-parameters depend on?


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What do S-parameters depend on?

S-parameters depend upon the network

and the characteristic impedances of thesource and load used to measure it, andthe frequency measured at.

i.e.

if the network is changed, the S-parameters change.

if the frequency is changed, the S-parameters change.

if the load impedance is changed, the S-parameters change.

if the source impedance is changed, the S-parameters change.

In the Si9000e S-parameters arequoted with source and loadimpedances of 50 Ohms


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A little math…

This is the matrix algebraic representation

of 2 port S-parameters:

Some matrices are symmetrical. A symmetrical matrix has symmetry aboutthe leading diagonal.


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A little math…




In the case of a 2-port network, that means that S21 = S12 and interchangingthe input and output ports does not change the transmission properties.


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A little math…




In the case of a symmetrical 2-port network, that means that S21 = S12 andinterchanging the input and output ports does not change the transmission

properties.

A transmission line is an example of a symmetrical 2-port network.


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A little math…

Parameters along the leading diagonal,

S11 & S22, of the S-matrix are referred to asreflection coefficients because they refer tothe reflection occurring at one port only.


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A little math…

Parameters along the leading diagonal,

S11 & S22, of the S-matrix are referred to asreflection coefficients because they refer tothe reflection occurring at one port only.

Off-diagonal S-parameters, S12, S21, are referred to as transmission coefficients because they refer to what happens from one port to another.


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Larger networks:

A Network may have any number of ports.


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Larger networks:


The S-matrix for an n-port network contains n2 coefficients (S-parameters),each one representing a possible input-output path.


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Larger networks:



The number of rows and columns in an S-parameters matrix is equal to thenumber of ports.


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Larger networks:




For the S-parameter subscripts “ij”, “j” is the port that is excited (the input port)

and “i” is the output port.


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Larger networks:




For the S-parameter subscripts “ij”, “j” is the port that is excited (the input port)

and “i” is the output port.


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Sum up…

• S-parameters are a powerful way to describe an electrical network

• S-parameters change with frequency / load impedance / source impedance / network• S11 is the reflection coefficient • S21 describes the forward transmission coefficient (responding port 1

st!)• S-parameters have both magnitude and phase information• Sometimes the gain (or loss) is more important than the phase shift and the phase

information may be ignored• S-parameters may describe large and complex networks

• If you would like to learn more please see next slide:


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Further reading:

//..//51.

//..//154.

//..//.//24321152500542

//..

//.101./.

//..///130.

//..//

140

Online lecture OLL-141 S11 & Smith charts - Eric Bogatinwww.bethesignal.com


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D


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C

36

( ) ( )φ+ω= tcosVtV o

( ) ( ) t j jot j

o eeVReeVRetV ωφφ+ω ==

1 j −=

SinusoidalSource

φ joeV is a complex phasor


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• ,

• C •

37

oV

( )φcosVo

( )φsinVo

φωRe

Im

( ) ( )φ+φ=φ sin jVcosVeV oo j

o

t je

ω


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38

Charge on the inner conductor:

xVCq l∆=∆

where Cl is the capacitance per unit lengthAzimuthal magnetic flux:

xIL l∆=∆Φ

where Ll is the inductance per unit length


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39

i ii ∆+

v vv ∆+

xL l∆

xC l∆

Voltage drop along the inductor:

( )

dt

dixLvvv l∆=∆+−

Current flowing through the capacitor:

dt

dvxCiii l∆−=∆+


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40

Limit as ∆x->0

t

i

Lx

v

l ∂

∂

−=∂

∂

t

B

E ∂

∂

−=×∇

t

vC

x

il∂

∂−=

∂

∂

t

DH

∂

∂=×∇

Solutions are traveling waves

( )

++

−= −+

vel

xtv

vel

xtvx,tv

( )

+−

−=

−+

vel

xt

Z

v

vel

xt

Z

vx,ti

oo

v+ indicates a wave traveling in the +x directionv- indicates a wave traveling in the -x direction


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C

41

vel is the phase velocity of the wave

llCL

1

vel =

For a transverse electromagnetic wave (TEM), the phase velocity isonly a property of the material the wave travels through

µε= 1

CL

1

ll

The characteristic impedance Zo

l

lo

C

LZ =

has units of Ohms and is a function of the material AND thegeometry


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42

Pulse travels down the transmission line as a forward going wave only(v+). However, when the pulse reaches the load resistor:

oo

L

Z

v

Z

v

vvRi

v

−+

−+

−

+==

LR+v

so a reverse wave v-

and i-

must be created to satisfy the boundarycondition imposed by the load resistor


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C

43

The reverse wave can be thought of as the incident wave reflectedfrom the load

Γ =+−=+

−

oL

oLZRZR

vv Reflection coefficient

Three special cases:

RL = ∞ (open) Γ = +1

RL = 0 (short) Γ = -1

RL = Zo Γ = 0

A transmission line terminated with a resistor equal in value to the

characteristic impedance of the transmission line looks the same tothe source as an infinitely long transmission line


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44

( ) t jx j eeVRextcosVv ωβ−+++ =β−ω=

vel=β

ω

phase velocity λ

π=

π=β

2

vel

f 2wave number

By using a single frequency sine wave we can now define compleximpedances such as:

LI jV ω=dt

diLv =

dt

dvCi =

C j

1

Zcap ω=

L jZind ω=

CV jI ω=

Experiment: Send a SINGLE frequency (ω) sine wave into atransmission line and measure how the line responds


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45

LZoZ

0x =

x

d

At x=0

oL

oL

ZZ

ZZVV+−=Γ = +−

Along the transmission line:

( ) ( )xcosV2e1VV

eVeVV

x j

x jx j

βΓ +Γ −=

Γ +=

+β−+

β++β−+

traveling wave standing wave


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()

46

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11.5

1

0.5

0

0.5

1

1.5

Position

( )

( )

( )

( )VSWR

1

1

1V

1V

V

V

min

max

=Γ −

Γ +=

Γ −

Γ +=

+

+

The VSWR is always greater than 1

2

1=Γ

Large voltage

Large current


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()

47

( )

( )

( )( )

VSWR

1

1

1V

1V

V

V

min

max

=

Γ −

Γ +=

Γ −

Γ +=

+

+

The VSWR is always greater than 1

2

1=Γ

Incident wave

Reflected wave

Standing wave


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C A

48

LZoZ

0x =

x

d

oL

oLL

ZZZZ

+−=Γ GΓ

towards load

towards generator

x j

L

x j

eVeVV

β++β−+

Γ +=

d2 jL)d( j

)d( j

L

genreverse

forwardG e

eV

eV

V

V β−−β−+

−β++

Γ =Γ ==Γ

Wave has to traveldown and back


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49

GΓ

LΓ

{ }Γ Re

{ }Γ Im

d2β−=θ

There is a one-to-onecorrespondence between Γ G and

ZL

oG

oG

G ZZ

ZZ

+

−=Γ

G

GoG

1

1ZZ

Γ −

Γ +=

d2 jL

d2 jL

oGe1

e1ZZ

β−

β−

Γ −

Γ +=


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: C

50

For an open circuit ZL= ∞ so Γ L = +1

Impedance at the generator:

( )dtan

jZZ oG

β

−=

For βd


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: C

51

For a short circuit ZL= 0 so Γ L = -1

Impedance at the generator:

( )dtan jZZ oG β=

For βd


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52

LZoZ

0x =

x

d

sP

dx −=

d j

oL

d j

oG

d j

L

d j

G

eZ

Ve

Z

V)d(II

eVeV)d(VV

β−+

β++

β−+β++

Γ −=−=

Γ +=−=

Voltage and Current at the generator (x=-d)

The rate of energy flowing through the plane at x=-d

{ }

o

22

Lo

2

*GG

Z

V

2

1

Z

V

2

1P

IVRe2

1P

++

Γ −=

=

forward power

reflected power


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• ! .

–

• C

–

• :

53

0L =Γ

which implies:

oL ZZ =When ZL = Zo, the load is matched to the transmission line


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54

What if the load cannot be made equal to Zo for some other reasons?

Then, we need to build a matching network so that the sourceeffectively sees a match load.

0=Γ

LZsP 0Z M

Typically we only want to use lossless devices such as capacitors,inductors, transmission lines, in our matching network so that we donot dissipate any power in the network and deliver all the availablepower to the load.


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55

jxrZ

Zz

o

+==

It will be easier if we normalize the load impedance to thecharacteristic impedance of the transmission line attached to the

load.

Γ −

Γ +

= 1

1

z

Since the impedance is a complex number, the reflection coefficientwill be a complex number

jvu +=Γ

( )

22

22

vu1

vu1r

+−

−−=

( ) 22 vu1

v2x

+−=


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B B

56

A dB is defined as a POWER ratio. For example:

( )Γ =

Γ =

=Γ

log20

log10

P

Plog10

2

for

revdB

A dBm is defined as log unit of power referenced to 1mW:

=

mW1

Plog10PdBm


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D


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58

We have only discussed reflection so far. What about transmission?Consider a device that has two ports:

1V 2V

2I1I

[ ] [ ][ ]IZV

IZIZV

IZIZV

2221212

2121111

=

+=

+=

The device can be characterized by a 2x2 matrix:


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()

59

−+

−+

−=

+=

iiio

iii

VVIZ

VVV

Since the voltage and current at each port (i) can be broken downinto forward and reverse waves:

We can characterize the circuit with forward and reverse waves:

[ ] [ ][ ]+−++−

++−

=

+=

+=

VSV

VSVSV

VSVSV

2221212

2121111


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60

[ ] [ ] [ ]( ) [ ] [ ]( )

[ ] [ ] [ ]( ) [ ] [ ]( ) 1o

o1

o

S1S1ZZ

1ZZ1ZZS

−

−

−+=

−+=

Similar to the reflection coefficient, there is a one-to-one

correspondence between the impedance matrix and the scatteringmatrix:


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()

61

The S matrix defined previously is called the un-normalized

scattering matrix. For convenience, define normalized waves:

io

ii

io

ii

Z2

Vb

Z2

Va

−

+

=

=

Where Zoi is the characteristic impedance of the transmission lineconnecting port (i)

|ai|2 is the forward power into port (i)

|bi|2

is the reverse power from port (i)


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()

62

The normalized scattering matrix is:

Where:

[ ] [ ][ ]asbasasb

asasb

22212122121111

=

+=

+=

j,iio

jo

j,i S

Z

Zs =

If the characteristic impedance on both ports is the same then thenormalized and un-normalized S parameters are the same.

Normalized S parameters are the most commonly used.


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63

The s parameters can be drawn pictorially

s11 and s22 can be thought of as reflection coefficientss21 and s12 can be thought of as transmission coefficients

s parameters are complex numbers where the angle corresponds to aphase shift between the forward and reverse waves

s11 s22

s21

s12

a1

a2b1

b2


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64

τ

Zo1 2

[ ]

=

ωτ−

ωτ−

0e

e0s

j

j

21 [ ]

−−=

1001s

1 2

[ ]

=

0G

00s

G

Transmission Line

Short

Amplifier


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65

[ ]

=

010

001

100

s

1

2

[ ]

=

01

00s

Zo

Isolator

1 2

3

Circulator


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66

If the device is made out of linear isotropic materials (resistors,capacitors, inductors, metal, etc..) then:

[ ] [ ]ss T =

j,ii, j ss = ji ≠

or

for

This is equivalent to saying that the transmitting pattern of anantenna is the same as the receiving pattern

reciprocal devices: transmission line

shortdirectional coupler

non-reciprocal devices: amplifier

isolator

circulator

D


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D

67

The s matrix of a lossless device is unitary:

[ ] [ ] [ ]1ss T* =

∑

∑

=

=

j

2 j,i

i

2

j,i

s1

s1for all j

for all i

Lossless devices: transmission line

shortcirculator

Non-lossless devices: amplifier

isolator

A


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68

Network analyzers measure Sparameters as a function of

frequency

At a single frequency, networkanalyzers send out forward waves a1and a2 and measure the phase and

amplitude of the reflected waves b1and b2 with respect to the forward

waves.

a1 a2

b1 b2

02a1

111

a

bs

=

=02a

1

221

a

bs

=

=

01a2

112

a

bs

=

=01a

2

222

a

bs

=

=

A C


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69

To measure the pure S parameters of a device, we need to eliminatethe effects of cables, connectors, etc. attaching the device to the

network analyzer

s11 s22

s21

s12

x11 x22

x21

x12

y11 y22

y21

y12

yx21

yx12

Connector Y Connector X

We want to know the S parameters atthese reference planes

We measure the S parameters at thesereference planes

A C


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A C

• 10

• 10 – D

– D (D)

– B

(), D .

70

A C


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A C•

, .

• :

71

[ ]

−

−=

10

01s

[ ]

=

ωτ−

ωτ−

0e

e0s

j

j

τ

[ ]

=

01

10sThru

Short

Delay**ωτ~90degrees

D


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72

A pure sine wave can be written as:( )zt j

oeVV β−ω=

The phase shift due to a length of cable is:

ph

ph

d

v

d

ωτ=

ω=

β=θ

The phase delay of a device is defined as:

( )ω

−=τ 21phSarg

D


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D

• ,

.• ,

.

•

–

–

73

D


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D• A

–

– •

74

( )( ) ( )tcostcosm1VV o ωω∆+=

( ) ( )( ) ( )( )[ ]tcostcos2

mVtcosVV oo ω∆−ω+ω∆+ω+ω=

The modulation can be de-composed into different frequencycomponents

D


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75

( )

( ) ( )( )

( ) ( )( )ztcos2

mV

ztcos2

mV

ztcosVV

o

o

o

β∆−β−ω∆−ω+

β∆+β−ω∆+ω+

β−ω=

The waves emanating from the source will look like

Which can be re-written as:

( )( ) ( )ztcosztcosm1VV o β−ωβ∆−ω∆+=

D


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76

The information travels at a velocity

ω∂

β∂⇒

ω∆

β∆= 11vgr

The group delay is defined as:

( )( )

ω∂

∂

−=

ω∂

β∂=

=τ

21

grgr

Sarg

d

v

d

D D


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D D

77

Phase Delay:

( )

ω−=τ 21

ph

Sarg

Group Delay:

( )( )ω∂∂−=τ 21gr Sarg


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D

C


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• , , , .

• , () , &

•

, ,, , . ,

79

C


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• C

•

•

80

C


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• ,

• , .

.

• A . (

)

•

81

C


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82

The impedance as a function of reflection coefficient can be re-written in the form:

( ) 22

22

vu1

vu1

r +−

−−

=

( ) 22 vu1

v2x

+−=

( )22

2

r1

1

vr1

r

u +=+

+−

( ) 2

22

x

1

x

1

v1u =

−+−

These are equations for

circles on the (u,v) plane

C C


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83

1 0.5 0 0.5 1

1

0.5

0.5

1

{ }Γ Re

{ }Γ Im

r=0r=1/3

r=1r=2.5

C C


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84

1 0.5 0 0.5 1

1

0.5

0.5

1

{ }Γ Re

{ }Γ Im

x=1/3 x=1 x=2.5

x=-1/3 x=-1 x=-2.5

C


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85

C 1


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86

Given:

Ω= 50Zo

°∠=Γ 455.0L

What is ZL?

( )Ω+Ω=

+Ω=5.67 j5.67

35.1 j35.150ZL

C 2


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87

Given:

Ω= 50Zo

Ω−Ω= 25 j15ZL

What is Γ L?

5.0 j3.050

25 j15z L

−=Ω

Ω−Ω=

°−∠=Γ 124618.0L

C 3

GiΩ+Ω= 50 j50Z

L


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88

Given:Ω= 50Zo

L

What is Zin at 50 MHz?

0.1 j0.1

50

50 j50z L

+=

Ω

Ω+Ω=

°∠=Γ 64445.0L

nS78.6=τ

?Zin

=

ωτ−β− Γ =Γ =Γ 2 jLd2 j

Lin ee

°=ωτ 2442°∠=Γ 180445.0in

( ) Ω=+Ω= 190.0 j38.050ZL

°=ωτ 2442

A

A t hi t k i i t b bi ti f l t


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89

A matching network is going to be a combination of elementsconnected in series AND parallel.

Impedance is NOT well suited when working

with parallel configurations.

21L ZZZ +=

2

Z1

Z

2Z

1Z

21

21L

ZZ

ZZZ

+=

ZIV =

For parallel loads it is better to work withadmittance.

YVI =

2Y1Y

21L YYY +=1

1Z

1Y =

Impedance is well suited when working withseries configurations. For example:

A

jbYZY


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90

jbgYZYYy o

o+===

Γ +

Γ −=

1

1y

( ) 22

22

vu1

vu1g

++

−−=

( ) 22 vu1

v2b

++

−=

( )22

2

g1

1v

g1

gu

+=+

++

( )2

22

b

1

b

1v1u =

+++

These are equations forcircles on the (u,v) plane

A C


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91

1 0.5 0 0.5 1

1

0.5

0.5

1

1 0.5 0 0.5 1

1

0.5

0.5

1

{ }Γ Re

{ }Γ Im

g=1/3

b=-1 b=-1/3

g=1g=2.5 g=0

b=2.5 b=1/3

b=1

b=-2.5

{ }Γ Im

{ }Γ Re

A

C


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C•

,

–

– C

• 180 .

– 180

. 92

A C 1


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93

• Procedure:

• Plot 1+j1 on chart

• vector =• Flip vector 180 degrees

Given:

°∠64445.0

What is Γ

?

1 j1y +=

°−∠=Γ 116445.0

Plot y

Flip 180

degreesRead Γ

A C 2

Given:


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94

• Procedure:

• Plot Γ

• Flip vector by 180 degrees

• Read coordinate

Given:

What is Y?

°+∠=Γ 455.0 Ω= 50Zo

Plot Γ

Flip 180degrees

Read y

36.0 j38.0y −=

( )

( ) mhos10x2.7 j6.7Y

36.0 j38.050

1Y

3−−=

−Ω=

C

Constant ImaginaryImpedance Lines


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95

Constant RealImpedance Circles

Impedance

Z=R+jX

=100+j50

Normalized

z=2+j for

Zo=50

C• (50)

– 1 = 100 + 50


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1 = 100 + 50 – 2 = 75 100 – 3 = 200 – 4 = 150 – 5 = ( )

– 6 = 0 ( ) – 7 = 50 – 8 = 184 900

•, . :

– 1 = 2 + – 2 = 1.5 2

– 3 = 4

– 4 = 3 – 5 = – 6 = 0 – 7 = 1 – 8 = 3.68 18

96

Toward

Generator ConstantReflection


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97

Generator

Away FromGenerator

ReflectionCoefficient Circle

Scale inWavelengths

Full Circle is One HalfWavelength Since

Everything Repeats

C


98/107

• , = 50 + 100

•

• ( )

• A . ,

.

98


99/107

99

0=Γ

Ω100sP Ω= 50Z0 M

Match 100Ω load to a 50Ω system at 100MHz

A 100Ω resistor in parallel would do the trick but ½ of thepower would be dissipated in the matching network. We wantto use only lossless elements such as inductors and capacitors

so we don’t dissipate any power in the matching network

We need to go from


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100

We need to go fromz=2+j0 to z=1+j0 on

the Smith chart

We won’t get anycloser by adding seriesimpedance so we willneed to add somethingin parallel.

We need to flip overto the admittance

chart

ImpedanceChart

y=0.5+j0


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101

y 0.5 j0

Before we add theadmittance, add a

mirror of the r=1circle as a guide.

AdmittanceChart


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102

y=0.5+j0

Before we add the

admittance, add amirror of the r=1circle as a guide

Now add positive

imaginary admittance.

AdmittanceChart

y=0.5+j0


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103

y j Before we add the

admittance, add amirror of the r=1

circle as a guide Now add positive

imaginary admittance jb = j0.5

AdmittanceChart

( )

pF16C

CMHz1002 j50

5.0 j

5.0 j jb

=

π=Ω

=

pF16 Ω100


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104

We will now add seriesimpedance

Flip to the impedanceSmith Chart

We land at on the r=1circle at x=-1

ImpedanceChart

Add positive imaginaryd itt t t t


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admittance to get toz=1+j0

ImpedanceChart

pF16

Ω100

( ) ( )

nH80L

LMHz1002 j500.1 j0.1 j jx

=

π=Ω=

nH 80

This solution wouldh ve ls rked


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have also worked

ImpedanceChart

pF32

Ω100nH160

B

5

0

nH80


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107

50 60 70 80 90 100 110 120 130 140 15040

35

30

25

20

15

10

Frequency (MHz)

R e f l e c t i o n C o e f f i c i e n t ( d B )

50 MHz

100 MHz

Because the inductor and capacitorimpedances change with frequency, thematch works over a narrow frequency range

pF16Ω100

ImpedanceChart

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s Parameters (1)Nn