KAVOSHCOMKAVOSHCOM

RF Communication Circuits

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

ZB

ZC

ZA

Solution

V1 V2

I1 I2

CAI

ZZIVZ +==

=01

111

2

( ) CCB

CCB

CB

C

IZ

ZZZZZ

ZZZ

IV

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

=τi

rVV

=ρ

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

=τ,1

11

i

rVV

=ρ

2

22and

i

rVV

=ρ

1

121

i

tVV

=τ

Then we can rewrite

S-parameters

212111 aab τρ +=

221212 aab ρτ +=Hence

In matrix form

=

2

1

221

121

2

1

aa

bb

ρττρ

=

2

1

2221

1211

2

1

aa

SSSS

bbS-matrix

•S11and S22 are a measureof reflected signal at port1 and port 2 respectively

•S21 is a measure of gain orloss of a signal from port 1to port 2.

•S12 ia a measure of gain orloss of a signal from port 2to port 1.

Logarithmic formS11=20 log(ρ1) S22=20 log(ρ2)S12=20 log(τ12)S21=20 log(τ21)

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 4

Port 5

=

5

4

3

2

1

5554535251

4544434241

3534333231

2524232221

1514131211

5

43

2

1

aaaaa

SSSSSSSSSSSSSSSSSSSSSSSSS

bbbbb

Example8.56 Ω 8.56 Ω

141.8 Ω

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

Solution

Z1=Z2= 8.56 Ω and Z3= 141.8 Ω

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

oin

oin

Vi

rZZZZ

VVS

r+−

====

ρ01

111

2

[ ] Ω=++++= 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 Ω 8.56 Ω

141.8 ΩV1 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

ρ

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

(3)

( ) dB97.3633.0log20log20 =−=− ρReturn 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 =−=

VIID

02

1

2 =−=

VIVB

ABCD matrix for series impedanceZ

I1I2

V1V2

02

1

2 =

=IV

VA02

1

2 =−=

VIVB

02

1

2 =

=IV

IC02

1

2 =−=

VIID

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 =−=

VIVB 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 =−=

VIID for port 2 short circuit I2

Z1

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 γTransmission line

z =0z = - l

For transmission line

ztjb

ztjf eeVeeVzV γωγω += −)(

( )ztjb

ztjf

oeeVeeV

ZzI γωγω −= −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 γωγω +−

continueAt the input z = - l

lll γγ −+ +=−= eVeVVV bf)(1( )lll γγ −−=−= 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

)(l

ll

γγγ

=+

=−

o

oV

eeVA

Note that

2)()cosh(

xx eex−+

=

2)()sinh(

xx eex−−

=

02

1

2 =−=

VIVB for port 2 short circuit

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

)sinh(2

)(l

ll

γγγ

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

)( lll γγγ=

−=

−

02

1

2 =−=

VIID for port 2 short circuit

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

)cosh(2

)(l

ll

γγγ

=−

+−=

−

oo

ooVZ

eeVZD

continue

The complete matrix is therefore

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

ll

ll

γγ

γγ

o

o

Z

Z

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

ll

ll

kZ

kj

kjZk

o

o

When the transmission line is lossless this reduces to

Note that

)cos()cosh( ll kjk =

)sin()sinh( ll kjjk =

jk+=αγ

Whereα= attenuationk=wave propagation

constant

Lossless lineα = 0

Table of ABCD network

Transmission line

Series impedance

Shunt impedance

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

ll

ll

γγ

γγ

o

o

Z

Z

10

1 Z

1101

Z

Z

Z

Table of ABCD network

T-network

π-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(

ll

ll

kZ

kj

kjZkABCD

o

o

tlineLossless transmission line

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

= 11

1

l

l

kZ

j

kjZABCD

o

o

tlineshort

Embedded short transmission line

Z1 Z1Transmission line

= 11

01

111

1101

11 ZkZ

j

kjZ

ZABCD

o

o

embed l

l

+++

+=

12

11

1

12

1

ZkjZ

Zkj

ZkjZ

Z

kjZZ

kjZ

ABCDo

o

o

oo

embed lll

ll

Solving, we have

Comparison with π-network

+++

+=−

1

3

21

3

21

32

3

111

1

ZZ

ZZZ

ZZ

ZZZ

ABCD netπ

+++

+=

12

11

1

12

1

ZkjZ

Zkj

ZkjZ

Z

kjZZ

kjZ

ABCDo

o

o

oo

embed lll

ll

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

l

oZkj l

o

oZk

ZkZ ll

>>21

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

Comparison with series and shunt

Series

If Zo >> Z1 then the series impedance lkjZZ o=

This is an inductance which is given by cZL o l

=

Where c is a velocity of light

Shunt

If Zo << Z1 then the series impedance

cZC

o

l=This is a capacitance which is given by

oZkjZ l

=

Equivalent circuits

Zo ZoZoc

l

Zo ZoZoL

l

cZL o l

=

cZC

o

l=

Zo >> Z1

Zo << Z1

Transmission line parameters

CBZo =

( )

−±== − 1ln1cosh1 21 AAA

llγ

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

( )

+−+−−−−+

+++=

DZCZBAZZBCADZDZCZBAZ

DZCZBAZSSSS

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)( SSSSSSSS

Z

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 Zorespectively. Similarly the propagation coefficients βe and βo 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 β,thus

= )cos()sin(

)sin()cos(

ll

ll

ββ

ββ

Zj

jZABCDtline

[ ] ( )

+−+−−−−+

+++=

DZCZBAZZBCADZDZCZBAZ

DZCZBAZS

oooo

oooo

ooo2

2

2 221

Using conversion

continue

[ ]

−

−

++

=

ZZZjZ

ZZ

ZZj

ZZZjZ

So

o

oo

o22

22

2

22sin2

2sin

sincos2

1

l

l

ll β

β

ββ

Taking 2π

β =l

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 βev = βod = Then

+−

+−==== 222223321441 2 ood

od

oev

evo

ZZZ

ZZZjZSSSS

+−

+

−−=

)()(

)()(

2 2222

2

ood

evod

oev

oodevo

ZZZZ

ZZZZZjZ

l2π

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

++

−=

))((21

2222

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

ZZZZ

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=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

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 π±=

SSArg

This S-parameter must satisfy network characteristic:

continueFor 3 dB coupler

21

2

=

+

−

odev

odev

ZZZZ

21

±=

+

−

odev

odev

ZZZZ

or

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

2<od

evZZ

(Gap size ) also βev and βod are not pure TEM thus not equal

A λ/4 branch line coupler

Z2

Z1

Z2

Z1

1

3

2

4

90o

90o

90o90o

Z2

Z1

Z1

1 2

90o

45o45o

Z2

Z1

Z1

1 2

90o

45o45o

O/C O/C

Symmetrical line

Odd

Even

AnalysisStub odd (short circuit) 11, 4

tan ZZX ods =

=

π

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

−=

π

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 =

−= m 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 Ω then Z2 = 35.4 Ω