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

I

I

ZZZ

ZZZ

ZZZ

V

V

V

.

.

..

.....

.....

..

..

.

.

2

1

21

22212

12111

2

1

nnnnn

n

n

n V

V

V

YYY

YYY

YYY

I

I

I

.

.

..

.....

.....

..

..

..

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

ZZI

VZ

01

111

2

CCB

CCB

CB

C

I

ZZZ

ZZZ

ZZ

Z

I

V

I

VZ

2

2

02

112

1

Port 2 open-circuited

Port 1 open-circuited

CBI

ZZI

VZ

02

222

1

Similarly we can show that

CI

ZI

VZ

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

t

V

V

i

r

V

V

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

V

VV

V

VV

Voltage of Reflected waveFrom port 1

Voltage ofTransmitted waveFrom port 2

22

21

1

12 i

i

ri

i

tb V

V

VV

V

VV

Voltage of transmitted wave away from port 2 is

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

212

i

t

V

V,

1

11

i

r

V

V

2

22and

i

r

V

V

1

121

i

t

V

V

Then we can rewrite

S-parameters

212111 aab

221212 aab Hence

In matrix form

2

1

221

121

2

1

a

a

b

b

2

1

2221

1211

2

1

a

a

SS

SS

b

bS-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(r1) S22=20 log(r2)S12=20 log(t12)S21=20 log(t21)

S-parameters

0111

2

1

r

r

Vi

V

VS 02

212

2

rVi

t

V

VS

01

121

1

rVi

t

V

VS

02

222

1

rVi

r

V

VS

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

4

3

2

1

a

a

a

a

a

SSSSS

SSSSS

SSSSS

SSSSS

SSSSS

b

bb

b

b

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 W and Z3= 141.8 W

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

oin

oin

Vi

r

ZZ

ZZ

V

VS

r

01

111

2

50)5056.88.141/()5056.8(8.14156.8

05050

505011

S Because of symmetry , then S22=0

)//( 231 oin ZZZZZ

Continue

02

221

2

rVi

t

V

VS

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

50

56.844.41

44.41

//

//

VV

ZZ

ZV

ZZ

Z

ZZZ

ZZVVV

o

oo

o

ot

8.56 8.56

141.8 V1 V2

Therefore S12 = S21 = 0.707

Vo

0707.0

707.00S

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

n

iiii

n

ii bbaa

1

**

1

Where 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..

...0

0.01

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

oo

S02.0908.0

908.001.0

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

633.0

2.01

8.08.01.0

1 22

211211

1

21211

1

1

jj

S

SSS

a

bSS

a

b

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

I

V

DC

BA

I

V

Cascaded network

a b

I1a

V1a

I2a

V2a V1b

I1b I2b

V2b

a

a

aa

aa

a

a

I

V

DC

BA

I

V

2

2

1

1

b

b

bb

bb

b

b

I

V

DC

BA

I

V

2

2

1

1

However V2a=V1b and –I2a=I1b then

b

b

bb

bb

aa

aa

a

a

I

V

DC

BA

DC

BA

I

V

2

2

1

1

Or just convert to one matrix

b

b

a

a

I

V

DC

BA

I

V

2

2

1

1

Where

bb

bb

aa

aa

DC

BA

DC

BA

DC

BA

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

I

V

VA for port 2 open circuit for port 2 short circuit

for port 2 open circuit for port 2 short circuit

02

1

2

I

V

IC

02

1

2

V

I

ID

02

1

2

V

I

VB

ABCD matrix for series impedanceZ

I1I2

V1V2

02

1

2

I

V

VA

02

1

2

V

I

VB

02

1

2

I

V

IC

02

1

2

V

I

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

I

V

VA for port 2 open circuit

131

32 V

ZZ

ZV

then

therefore

3

1

3

31

2

1 1Z

Z

Z

ZZ

V

VA

Continue

02

1

2

V

I

VB for port 2 short circuit

1

32

321

32

32

2V

ZZZZ

Z

ZZ

ZZ

VZ

Solving for voltage in Z2

But

222ZIVZ

Hence

3

2112

2

1

Z

ZZZZ

I

VB

I2Z1

Z3Z2

VZ2

Continue

02

1

2

I

V

IC for port 2 open circuit

I1 I2

Z1

Z3V2

31322 ZIZIV

12 II

32

1 1

ZV

IC

Therefore

Analysis

Continue

02

1

2

V

I

ID for port 2 short circuit I2

Z1

Z3Z2

VZ2

132

32 I

ZZ

ZI

I1

I1 is divided into Z2 and Z3, thus

Hence

3

2

2

1 1Z

Z

I

ID

Full matrix

3

2

3

3

2121

2

1

11

1

Z

Z

Z

Z

ZZZZ

Z

Z

ABCD for transmission line

Input V1

I1

V2

I2

Zo gTransmission line

z =0z = -

For transmission line

ztjb

ztjf eeVeeVzV )(

ztjb

ztjf

oeeVeeV

ZzI 1)(

b

b

f

fo I

V

I

VZ

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

continueAt the input z = -

eVeVVV bf)(1 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

I

V

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

)(

o

o

V

eeVA

Note that

2

)()cosh(

xx eex

2

)()sinh(

xx eex

02

1

2

V

I

VB for port 2 short circuit

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

)sinh(2

)(

oo

oo ZV

eeVZB

continue

02

1

2

I

V

IC for port 2 open circuit

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

ooo

o

ZZV

eeVC

)sinh(

2

)(

02

1

2

V

I

ID for port 2 short circuit

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

)cosh(2

)(

oo

oo

VZ

eeVZD

continue

The complete matrix is therefore

)cosh()sinh(

)sinh()cosh(

o

o

Z

Z

)cos()sin(

)sin()cos(

kZ

kj

kjZk

o

o

When the transmission line is lossless this reduces to

Note that

)cos()cosh( kjk

)sin()sinh( kjjk

jk

Wherea= attenuationk=wave propagation constant

Lossless linea = 0

Table of ABCD network

Transmission line

Series impedance

Shunt impedance

)cosh()sinh(

)sinh()cosh(

o

o

Z

Z

10

1 Z

1

101

Z

Z

Z

Table of ABCD network

T-network

p-network

3

2

3

3

2121

2

1

11

1

Z

Z

Z

Z

ZZZZ

Z

Z

1

3

21

3

21

32

3

111

1

Z

Z

ZZ

Z

ZZ

ZZ

Z

n

n1

0

0Ideal transformer

n:1

Z1 Z2

Z3

Z3

Z1 Z2

Short transmission line

)cos(

)sin()sin()cos(

kZ

kj

kjZkABCD

o

o

tlineLossless transmission line

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

1

11

kZ

j

kjZABCD

o

o

tlineshort

Embedded short transmission line

Z1 Z1Transmission line

1

101

111

11

01

11 Zk

Zj

kjZ

ZABCD

o

o

embed

12

11

1

12

1

Z

kjZ

Z

kj

Z

kjZ

Z

kjZZ

kjZ

ABCDo

o

o

oo

embed

Solving, we have

Comparison with p-network

1

3

21

3

21

32

3

111

1

Z

Z

ZZ

Z

ZZ

ZZ

Z

ABCD net

12

11

1

12

1

Z

kjZ

Z

kj

Z

kjZ

Z

kjZZ

kjZ

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

oZ

kj

o

o

Z

k

Z

kZ

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 byc

ZL 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

oZ

kjZ

Equivalent circuits

Zo ZoZoc

Zo ZoZoL

c

ZL o

cZC

o

Zo >> Z1

Zo << Z1

Transmission line parameters

C

BZo

1ln

1cosh

1 21 AAA

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

BCADZDZCZBAZ

DZCZBAZSS

SS

oooo

oooo

ooo2

2

22221

1211

2

21

For conversion of ABCD to S-parameter

For conversion of S to ABCD-parameter

2112221121122211

2112221121122211

211111

11111

2

1SSSSSSSS

Z

SSSSZSSSS

SDC

BA

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)(

AZAZZAAZZ

AAAAZAZAZZAAZ

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

..2

1)(

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

SSSSSSSS

SSSSSSSS

SSSSSSSS

SSSSSSSS

S

4444434342424141

3434333332323131

2424232322222121

1414131312121111

2

1

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(

Zj

jZABCD

tline

DZCZBAZZ

BCADZDZCZBAZ

DZCZBAZS

oooo

oooo

ooo2

2

2 2

21

Using conversion

continue

Z

ZZjZ

ZZ

ZZj

Z

ZZjZ

So

o

oo

o22

22

2

22

sin2

2sin

sincos2

1

Taking 2

Then

22

22

22 2

21

oo

oo

o ZZZZj

ZZjZZ

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

S22

S33

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

ZZ

Z

ZZ

ZjZSSSS

)(

)(

)(

)(

2 2222

2

ood

evod

oev

oodevo

ZZ

ZZ

ZZ

ZZZjZ

2

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 2

1

ood

ood

oev

oev

ZZ

ZZ

ZZ

ZZSSSS

))((2

12222

422

oodoev

oodev

ZZZZ

ZZZ

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 2

1

ood

ood

oev

oev

ZZ

ZZ

ZZ

ZZSSSS

))((

)(2222

222

oodoev

oodev

ZZZZ

ZZZ

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

ZZ

Z

ZZ

ZjZSSSS

odevo

ZZjZ

1if 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

continue

12

412

312

212

11 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

12

312

21 SS

(2) And quadrature condition221

11

S

SArg

This S-parameter must satisfy network characteristic:

continueFor 3 dB coupler

2

12

odev

odev

ZZ

ZZ2

1

odev

odev

ZZ

ZZor

Rewrite we have

223)2(1

)2(1

od

ev

Z

Z

In practice Zev > Zod so 83.5223 od

ev

Z

Z

However the limitation for coupled edge

2od

ev

Z

Z(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

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

The ABCD matrices for the two networks may then found :

ss

s

ss X

Z

X

jZ

Z

j

jZX

Z

jXZ

jjZ

jXABCD

222

2

22

2

2

11

01

0

0

11

01

stub stubTransmission line

continue

DZCZBAZZ

BCADZDZCZBAZ

DZCZBAZS

oooo

oooo

ooo2

2

2 2

21

Convert to S

2

2

22

2

2

2

2

22

2

2

2

2

22

2

22 2

2

2

1

Z

Zj

X

ZZjjZZ

ZZ

Zj

X

ZZjjZ

Z

Zj

X

ZZjjZ

X

ZZ 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 Z

Zj

X

ZZjjZS o

s

o or 122

2

2 ZZZ

ZZX

o

os

From

previous definition

continueSubstituting into S-parameter gives us

0

01

222

2 o

o

o

odd Z

Z

jZZZS

0

01

222

2 o

o

o

even Z

Z

jZZZSand

Therefore for full four port

o

odev Z

ZjSSSSSS 2

212134431221 2

1

2

22

212123321441 12

1

oodev

Z

ZSSSSSS

And

044332211 SSSS

024421331 SSSS

continue

For power conservation and quadrature conditions to be met

Equal split S

2

1221

oZ

ZS or

22

oZZ

And

o

oo

oo

o

os Z

ZZ

ZZ

ZZ

ZZZX

2

222

2

21

2

2

If Zo= 50 W then Z2 = 35.4 W