KAVOSHCOM

RF Communication Circuits

Dr. Fotowat-AhmadiSharif University of TechnologyFall-1391Prepared by: Siavash Kananian & Alireza Zabetian

Lecture 3: S- Parameters

Dr. Fotowat-AhmadiSharif University of TechnologyFall-1391Prepared by: Siavash Kananian & Alireza Zabetian

Impedance and Admittance matrices

nnnnn

n

n

n I

II

ZZZ

ZZZZZZ

V

VV

.

.

............

..

..

.

.2

1

21

22212

12111

2

1

nnnnn

n

n

n V

VV

YYY

YYYYYY

I

II

.

.

............

..

..

..

2

1

21

22212

12111

2

1

Impedance matrix Admittance matrix

For n ports network we can relate the voltages and currents by impedance and admittance matrices

1 ZYwhere

Reciprocal and Lossless NetworksReciprocal networks usually contain nonreciprocal media such as ferrites or plasma, or active devices. We can show that the impedance and admittance matrices are symmetrical, so that.

Lossless networks can be shown that Zij or Yij are imaginary

jiijjiij YYor ZZ

Refer to text book Pozar pg193-195

ExampleFind the Z parameters of the two-port T –network as shown below

Z B

Z C

Z A

Solution

V1 V2

I1 I2

CAI

ZZIVZ

01

111

2

CCB

CCB

CB

C

I

ZZZ

ZZZZZ

ZIV

IVZ

2

2

02

112

1

Port 2 open-circuited

Port 1 open-circuited

CBI

ZZIVZ

02

222

1

Similarly we can show that

CI

ZIVZ

01

221

2

This is an example of reciprocal network!!

S-parameters

Microwave device

Port 1 Port 2

Vi1

Vr1

Vt2

Vi2

Vr2

Vt1

Transmission and reflection coefficients

i

tVV

i

rVV

Input signalreflected signal

transmitted signal

S-parametersVoltage of traveling wave away from port 1 is

22

21

1

11 i

i

ti

i

rb V

VVV

VVV

Voltage of Reflected waveFrom port 1

Voltage ofTransmitted waveFrom port 2

22

21

1

12 i

i

ri

i

tb V

VVV

VVV

Voltage of transmitted wave away from port 2 is

Let Vb1= b1 , Vi1=a1 , Vi2=a2 , ,2

212

i

tVV

,1

11

i

rVV

2

22and

i

rVV

1

121

i

tVV

Then we can rewrite

S-parameters

212111 aab

221212 aab Hence

In matrix form

2

1

221

121

2

1

aa

bb

2

1

2221

1211

2

1

aa

SSSS

bbS-matrix

•S11and S22 are a measure of reflected signal at port 1 and port 2 respectively•S21 is a measure of gain or loss of a signal from port 1 to port 2.•S12 ia a measure of gain or loss of a signal from port 2 to port 1.

Logarithmic formS11=20 log(1) S22=20 log(2)S12=20 log(12)S21=20 log(21)

S-parameters

0111

2

1

r

r

Vi

V

VS 02

212

2

rVi

tVVS

01

121

1

rVi

tVVS

02

222

1

rVi

rVVS

Vr2=0 means port 2 is matched

Vr1=0 means port 1 is matched

Multi-port network

network

Port 1

Port 2

Port

3

Port 4

Port 5

5

4

3

2

1

5554535251

4544434241

3534333231

2524232221

1514131211

5

43

2

1

aaaaa

SSSSSSSSSSSSSSSSSSSSSSSSS

bbbbb

Example8.56 8.56

141.8

Below is a matched 3 dB attenuator. Find the S-parameter of the circuit.

Solution

Z1=Z2= 8.56 and Z3= 141.8

By assuming the output port is terminated by Zo = 50 , then

oin

oin

Vi

rZZZZ

VVS

r

01

111

2

50)5056.88.141/()5056.8(8.14156.8

050505050

11

S Because of symmetry , then S22=0

)//( 231 oin ZZZZZ

Continue

02

221

2

rVi

tVVS

From the fact that S11=S22=0 , we know that Vr1=0 when port 2 is matched, and that Vi2=0. Therefore Vi1= V1 and Vt2=V2

11

33132

32122

707.056.850

5056.844.41

44.41

////

VV

ZZZV

ZZZ

ZZZZZVVV

o

oo

o

ot

8.56 8.56

141.8 V1 V2

Therefore S12 = S21 = 0.707

Vo

0707.0707.00

S

Lossless networkFor lossless n-network , total input power = total output power. Thus

n

iiii

n

ii bbaa

1

**

1Where a and b are the amplitude of the signal

Putting in matrix form at a* = bt b*

=at St S* a*

Thus at (I – St S* )a* =0 This implies that St S* =I

Note that bt=atSt and b*=S*a*

In summation form

jifor

jiforSS kj

n

kki 0

1*

1

Called unitary matrix

Conversion of Z to S and S to Z

UZUZS 1

SUSUZ 1

where

1..0.1.....00.01

U

Reciprocal and symmetrical network

For reciprocal network

tUU Since the [U] is diagonal , thus

tZZ

tSS

Since [Z] is symmetry

Thus it can be shown that

ExampleA certain two-port network is measured and the following scattering matrix is obtained:

From the data , determine whether the network is reciprocal or lossless. If a short circuit is placed on port 2, what will be the resulting return loss at port 1?

oo

ooS

02.0908.0908.001.0

Solution

Since [S] is symmetry, the network is reciprocal. To be lossless, the S parameters must satisfy

jifor

jiforSS kj

n

kki 0

1*

1

|S11|2 + |S12|2 = (0.1)2 + (0.8)2 = 0.65

Since the summation is not equal to 1, thus it is not a lossless network.

For i=j

continueReflected power at port 1 when port 2 is shorted can be calculated as follow and the fact that a2= -b2 for port 2 being short circuited, thus

b1=S11a1 + S12a2 = S11a1 - S12b2

b2=S21a1 + S22a2 = S21a1 - S22b2

(1)

(2)

From (2) we have

a2

-a2=b2

Short at port 2

122

212 1

aS

Sb

633.02.01

8.08.01.01 22

211211

1

21211

1

1

jj

SSSS

abSS

ab

Dividing (1) by a1 and substitute the result in (3) ,we have

(3)

dB97.3633.0log20log20 Return loss

ABCD parameters

NetworkV1V2

I1 I2

Voltages and currents in a general circuit

122 VVI 212 IIV

This can be written as

221 IVV 221 IVI

Or

221 BIAVV 221 DICVI

A –ve sign is included in the definition of D

In matrix form

Given V1 and I1, V2 and I2 can be determined if ABDC matrix is known.

2

2

1

1

IV

DCBA

IV

Cascaded network

a b

I1a

V1a

I2a

V2a V1b

I1b I2b

V2b

a

a

aa

aa

a

a

IV

DCBA

IV

2

2

1

1

b

b

bb

bb

b

b

IV

DCBA

IV

2

2

1

1

However V2a=V1b and –I2a=I1b then

b

b

bb

bb

aa

aa

a

a

IV

DCBA

DCBA

IV

2

2

1

1

Or just convert to one matrix

b

b

a

a

IV

DCBA

IV

2

2

1

1

Where

bb

bb

aa

aa

DCBA

DCBA

DCBA

The main use of ABCD matrices are for chaining circuit elements together

Determination of ABCD parameters221 BIAVV 221 DICVI

Because A is independent of B, to determine A put I2 equal to zero and determine the voltage gain V1/V2=A of the circuit. In this case port 2 must be open circuit.

02

1

2

IV

VA for port 2 open circuit for port 2 short circuit

for port 2 open circuit for port 2 short circuit

02

1

2

IV

IC02

1

2

VI

ID

02

1

2

VI

VB

ABCD matrix for series impedanceZ

I1I2

V1V2

02

1

2

IV

VA02

1

2

VI

VB

02

1

2

IV

IC02

1

2

VI

ID

for port 2 open circuit for port 2 short circuit

for port 2 open circuit for port 2 short circuit

V1= V2 hence A=1 V1= - I2 Z hence B= Z

I1 = - I2 = 0 hence C= 0 I1 = - I2 hence D= 1

The full ABCD matrix can be written

10

1 Z

ABCD for T impedance network

Z1 Z2

Z3V1

I1 I2

V2

02

1

2

IV

VA for port 2 open circuit

131

32 V

ZZZV

then

therefore

3

1

3

31

2

1 1ZZ

ZZZ

VVA

Continue

02

1

2

VI

VB for port 2 short circuit

1

32

321

32

32

2V

ZZZZZ

ZZZZ

VZ

Solving for voltage in Z2

But

222ZIVZ

Hence

3

2112

2

1ZZZZZ

IVB

I2Z1

Z3Z2

VZ2

Continue

02

1

2

IV

IC for port 2 open circuitI1 I2

Z1

Z3V2

31322 ZIZIV

12 II

32

1 1ZV

IC

Therefore

Analysis

Continue

02

1

2

VI

ID for port 2 short circuit I2Z1

Z3Z2

VZ2

132

32 I

ZZZI

I1

I1 is divided into Z2 and Z3, thus

Hence

3

2

2

1 1ZZ

IID

Full matrix

3

2

3

3

2121

2

1

11

1

ZZ

Z

ZZZZZ

ZZ

ABCD for transmission line

Input V1

I1

V2

I2

Zo gTransmission line

z =0z = -

For transmission line

ztjb

ztjf eeVeeVzV gg )(

ztjb

ztjf

oeeVeeV

ZzI gg 1)(

b

b

f

fo I

VIV

Z

f and b represent forward and backward propagation voltage and current Amplitudes. The time varying term can be dropped in further analysis.

ztjb

ztjf eeVeeV gg

continueAt the input z = -

gg eVeVVV bf)(1 gg eVeV

ZII bf

o

1)(1

At the output z = 0

bf VVVV )0(2 bf

oVV

ZII

1)0(2

(1) (2)

(3) (4)

Now find A,B,C and D using the above 4 equations

02

1

2

IV

VA for port 2 open circuit

For I2 =0 Eq.( 4 ) gives Vf= Vb=Vo giving

continueFrom Eq. (1) and (3) we have

)cosh(2

)(

g

gg

o

oV

eeVA

Note that

2)()cosh(

xx eex

2)()sinh(

xx eex

02

1

2

VI

VB for port 2 short circuit

For V2 = 0 , Eq. (3) implies –Vf= Vb = Vo . From Eq. (1) and (4) we have

)sinh(2

)(

g

gg

oo

oo ZV

eeVZB

continue

02

1

2

IV

IC for port 2 open circuit

For I2=0 , Eq. (4) implies Vf = Vb = Vo . From Eq.(2) and (3) we have

ooo

oZZV

eeVC )sinh(2

)( ggg

02

1

2

VI

ID for port 2 short circuit

For V2=0 , Eq. (3) implies Vf = -Vb = Vo . From Eq.(2) and (4) we have

)cosh(2

)(

g

gg

oo

ooVZ

eeVZD

continue

The complete matrix is therefore

)cosh()sinh()sinh()cosh(

gggg

o

o

Z

Z

)cos()sin()sin()cos(

kZkj

kjZk

o

o

When the transmission line is lossless this reduces to

Note that

)cos()cosh( kjk

)sin()sinh( kjjk

jkg

Where= attenuationk=wave propagation constant

Lossless line = 0

Table of ABCD network

Transmission line

Series impedance

Shunt impedance

)cosh()sinh()sinh()cosh(

gggg

o

o

Z

Z

10

1 Z

1101

Z

Z

Z

Table of ABCD network

T-network

p-network

3

2

3

3

2121

2

1

11

1

ZZ

Z

ZZZZZ

ZZ

1

3

21

3

21

32

3

111

1

ZZ

ZZZ

ZZ

ZZZ

n

n100

Ideal transformer

n:1

Z1 Z2

Z3

Z3

Z1 Z2

Short transmission line

)cos()sin(

)sin()cos(

kZkj

kjZkABCD

o

o

tlineLossless transmission line

If << l then cos(k ) ~ 1 and sin (k ) ~ k then

11

1

kZ

j

kjZABCD

o

o

tlineshort

Embedded short transmission line

Z1 Z1Transmission line

11

01

111

1101

11 Zk

Zj

kjZ

ZABCD

o

o

embed

12

11

1

12

1

ZkjZ

Zkj

ZkjZ

Z

kjZZkjZ

ABCDo

o

o

oo

embed

Solving, we have

Comparison with p-network

1

3

21

3

21

32

3

111

1

ZZ

ZZZ

ZZ

ZZZ

ABCD netp

12

11

1

12

1

ZkjZ

Zkj

ZkjZ

Z

kjZZkjZ

ABCDo

o

o

oo

embed

It is interesting to note that if we substitute in ABCD matrix in p-network, Z2=Z1 and Z3=jZok we see that the difference is in C element where wehave extra term i.e

oZkj

o

oZk

ZkZ

21

Both are almost same if So the transmission line exhibit a p-network

Comparison with series and shunt

Series

If Zo >> Z1 then the series impedance kjZZ o

This is an inductance which is given by cZL o

Where c is a velocity of light

Shunt

If Zo << Z1 then the series impedance

cZC

o

This is a capacitance which is given by

oZkjZ

Equivalent circuits

Zo ZoZoc

Zo ZoZoL

cZL o

cZC

o

Zo >> Z1

Zo << Z1

Transmission line parameters

CBZo

1ln1cosh1 21 AAA

g

It is interesting that the characteristic impedance and propagation constant of a transmission line can be determined from ABCD matrix as follows

Conversion S to ABCD

DZCZBAZZ

BCADZDZCZBAZDZCZBAZSS

SS

oooo

oooo

ooo2

2

22221

1211

221

For conversion of ABCD to S-parameter

For conversion of S to ABCD-parameter

2112221121122211

2112221121122211

2111111

1111

21

SSSSSSSSZ

SSSSZSSSS

SDCBA

o

o

Zo is a characteristic impedance of the transmission line connected to theABCD network, usually 50 ohm.

MathCAD functions for conversionFor conversion of ABCD to S-parameter

For conversion of S to ABCD-parameter

2,21,22,11,1

1,22,12,21,12,21,22,11,1

2,21.22,11,1 .....2....2....

....1)(

AZAZZAAZZAAAAZAZAZZAAZ

AZAZZAAZAS

1,22,12,21,11,22,12,21,1

1,22,12,21,11,22,12,21,1

1,2 .1.1.1.1.1.1.1..1.1

..21)(

SSSSSSSSZ

SSSSZSSSS

SSA

o

Odd and Even Mode AnalysisUsually use for analyzing a symmetrical four port network

•Equal ,in phase excitation – even mode•Equal ,out of phase excitation – odd mode

(1) Excitation

(2) Draw intersection line for symmetry and apply •short circuit for odd mode•Open circuit for even mode

(3) Also can apply EM analysis of structure•Tangential E field zero – odd mode•Tangential H field zero – even mode

(4) Single excitation at one port= even mode + odd mode

Example 1

Line ofsymmetry

1 2

43

odevodevodevodev

odevodevodevodev

odevodevodevodev

odevodevodevodev

SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS

S

4444434342424141

3434333332323131

2424232322222121

1414131312121111

21

The matrix contains the odd and even parts

Since the network is symmetry, Instead of 4 ports , we can only analyze 2 port

Edge coupled line

continueWe just analyze for 2 transmission lines with characteristic Ze and Zo respectively. Similarly the propagation coefficients be and bo respectively. Treat the odd and even mode lines as uniform lossless lines. Taking ABCD matrix for a line , length l, characteristic impedance Z and propagation constant b,thus

)cos()sin(

)sin()cos(

bbbb

Zj

jZABCD tline

DZCZBAZZBCADZDZCZBAZ

DZCZBAZS

oooo

oooo

ooo2

2

2 221

Using conversion

continue

ZZZjZ

ZZZZj

ZZZjZ

So

o

oo

o22

22

2

22sin2

2sin

sincos2

1

b

b

bb

Taking 2pb

Then

22

22

22 221

oo

oo

o ZZZZjZZjZZ

ZZS

(equivalent to quarter-wavelength transmission line)

continue

S11

S21

S12

S22

2-port network matrix

Convert toS11

S21

S12

S22

S11

S21

S12

S22

S11

S21

S12

S22

S11

S21

S12

S22S33

S44

S34

S43

S13 S14S23

S24

S32

S31

S42S41

4-port network matrix

Odd + even

continue

Assuming bev = bod = Then

222223321441 2 ood

od

oev

evo

ZZZ

ZZZjZSSSS

)()(

)()(

2 2222

2

ood

evod

oev

oodevo

ZZZZ

ZZZZZjZ

2p

For perfect isolation (I.e S41=S14=S32=S23=0 ),we choose Zev and Zod such that Zev Zod=Zo

2.

S11

S21

S12

S22

S13

S23

S14

S24

S31

S41

S32

S42

S33

S43

S34

S44

ev+ od

ev+ od

ev- od

ev- od

Follow symmetrical properties

continue

Similarly we have

22

22

22

22

44332211 21

ood

ood

oev

oev

ZZZZ

ZZZZSSSS

))((2

12222

422

oodoev

oodev

ZZZZZZZ

Equal to zero if Zev Zod=Zo2.

S11

S21

S12

S22

S13

S23

S14

S24

S31

S41

S32

S42

S33

S43

S34

S44

ev+ od

ev+ od

ev- od

ev- od

continue

We have

22

22

22

22

42241331 21

ood

ood

oev

oev

ZZZZ

ZZZZSSSS

))((

)(2222

222

oodoev

oodev

ZZZZZZZ

if Zev Zod=Zo2.

odev

odev

ZZ

ZZ

S11

S21

S12

S22

S13

S23

S14

S24

S31

S41

S32

S42

S33

S43

S34

S44

ev+ od

ev+ od

ev- od

ev- od

continue

222243341221 2 ood

od

oev

evo

ZZZ

ZZZjZSSSS

odevo

ZZjZ 1

if Zev Zod=Zo2.

S11

S21

S12

S22

S13

S23

S14

S24

S31

S41

S32

S42

S33

S43

S34

S44

ev+ od

ev+ od

ev- od

ev- od

continue

1241

231

221

211 SSSS

(1) Power conservation

Reflected power transmitted

power to port 4

transmitted power to

port 3

transmitted power to

port 2

Since S11 and S41=0 , then

1231

221 SS

(2) And quadrature condition 221

11 p

SSArg

This S-parameter must satisfy network characteristic:

continueFor 3 dB coupler

21

2

odev

odev

ZZ

ZZ21

odev

odev

ZZ

ZZor

Rewrite we have

223)2(1)2(1

od

evZZ

In practice Zev > Zod so 83.5223 od

evZZ

However the limitation for coupled edge

2od

evZZ

(Gap size ) also bev and bod are not pure TEM thus not equal

A l/4 branch line coupler

Z 2

Z1

Z 2

Z1

1

3

2

4

90 o

90 o

90 o90 o

Z 2

Z1

Z1

1 2

90 o

45 o45 o

Z 2

Z1

Z1

1 2

90 o

45 o45 o

O/C O/C

Symmetrical line

Odd

Even

AnalysisStub odd (short circuit) 11, 4

tan ZZX ods

p

Stub even (open circuit) 11, 4cot ZZX evs

p

The ABCD matrices for the two networks may then found :

ss

s

ss XZ

XjZ

Zj

jZXZ

jXZj

jZ

jXABCD

222

2

22

2

2

1101

0

0

1101

stub stubTransmission line

continue

DZCZBAZZBCADZDZCZBAZ

DZCZBAZS

oooo

oooo

ooo2

2

2 221

Convert to S

2

2

22

2

2

2

2

22

2

2

2

2

22

2

22 2

2

21

ZZj

XZZjjZZ

ZZZj

XZZjjZ

ZZj

XZZjjZ

XZZ o

s

oo

oo

s

o

o

s

o

s

o

For perfect isolation we require

011111111 odevodev SSSS Thus 01111 odev SS

02

2

22

2

211 ZZj

XZZjjZS o

s

o or 122

22 ZZZ

ZZXo

os

From

previous definition

continueSubstituting into S-parameter gives us

001

222

2 o

o

oodd Z

Z

jZZZS

001

222

2 o

o

oeven Z

Z

jZZZSand

Therefore for full four port

o

odev ZZjSSSSSS 2

212134431221 21

2

22

212123321441 121

oodev

ZZSSSSSS

And

044332211 SSSS

024421331 SSSS

continue

For power conservation and quadrature conditions to be met

Equal split S

212

21 oZ

ZS or22oZZ

And

o

oo

oo

o

os Z

ZZ

ZZ

ZZ

ZZZX

2

222

22

1

2

2

If Zo= 50 then Z2 = 35.4