1

Prof. David R. JacksonDept. of ECE

Notes 14

ECE 5317-6351 Microwave Engineering

Fall 2011

Network AnalysisMultiport Networks

2

Multiport NetworksA general circuit can be represented by a multi-port network, where the “ports” are defined as access terminals at which we can define voltages and currents.

Examples:

1) One-port network

2) Two-port network

Note: Equal and opposite currents are assumed on the two wires of a port.

R

1I

1V

+

-

+

-

1I

1V

+

-

+

-

+

-

+

-

1I 1I2I2I

1V 1V2V

2V

3

3) N-port Network

To represent multi-port networks we use:

- Z (impedance) parameters- Y (admittance) parameters- h (hybrid) parameters- ABCD parameters

- S (scattering) parameters

Not easily measurable at high frequency

Measurable at high frequency

Multiport Networks (cont.)

+

+

+

+

+

1I

2I

3I

mI

NI

1V

2V

3V

mV

NV

-

-

-

-

-

4

Poynting Theorem (Phasor Domain)

2s f d m eP P j P W W

*

2 2

2 2

1ˆ

2

1 1

2 2

1 12

4 4

i

V S

c

V

c

V

E J dV S n dS

E H dV

j H E dV

The last term is the VARS consumed by the region.

The notation < > denotes time-average.

5

Consider a general one-port network

Complex power delivered to network:

Self Impedance

*1 1

1

2

1

22ˆi S m en dPP E H n j Wd W

V I

s

Average power dissipated in [W]

Average magnetic energy (in [J]) stored inside V

Average electric energy (in [J]) stored inside V

,E H n

S

V1V

1I

+

-

d dP P

m mW W

e eW W

6

** 1 1

1 12

2 21

2

1

2

2

1 1

1

1

1

1

121 12

4 (

221

)

22

di

in

in in

m

in

e

n

i

m e

n

dR jX

PR

I

V I

W WX

V I P

I I I

P j W W

VZ

I

I

I

Define Self Impedance (Zin)

,E H n

S

V1V

1I

+

-

7

We can show that for physically realizable networks the following apply:

*1 1

*1

*

1

in

in in

in

Z Z

V V

I

X

I

R

is an even function of

is an odd function of

Self Impedance (cont.)

Please see the Pozar book for a proof.

in in inZ R jX

Note: Frequency is usually defined as a positive quantity. However, we consider the analytic continuation of the functions into the complex frequency plane.

8

1 11 1 12 2

2 21 1 2

1 11 12 1

2 21 22 2

2 2

V Z I Z I

V Z Z I

V Z Z IV Z I

V Z I Z I

Consider a general 2-port linear network:

In terms of Z-parameters, we have (from superposition)

Impedance (Z) matrix

Two-Port Networks

1 2+

-

+

-1V

1I 2I

2V

9

Elements of Z-Matrix: Z-Parameters(open-circuit parameters )

0k

iij

j I k j

VZ

I

1

1

112

2 0

222

2 0

I

I

VZ

I

VZ

I

Port 2 open circuited Port 1 open circuited

2

2

111

1 0

221

1 0

I

I

VZ

I

VZ

I

1 2+

-

+

-1V

1I 2I

2V

1 11 1 12 2

2 21 1 22 2

V Z I Z I

V Z I Z I

10

Z-Parameters (cont.)

0k

iij

j I k j

VZ

I

+

jI

iV -

We inject a current into port j and measure the voltage (with an ideal voltmeter) at port i. All ports are open-circuited except j.

N-port network

11

Z-parameters are convenient for series connected networks.

1 1 1

2 2 2

A B

A B

I I I

I I I

Series

Z-Parameters (cont.)

A

B

1AI

1BI 2

BI

2AI

+

+

+

+

2AV

2BV

1AV

1BV

1V 2V

2I1I++

--

1 2

1 2

- -

--

1 1 1

2 2 2

1

2

1

2

A B

A B

A A B B

A B

A B

A B

V V V

V V V

Z I Z I

Z Z I

IZ Z

I

IZ Z

I

1 111 11 12 12

2 221 21 22 22

A B A B

A B A B

V IZ Z Z Z

V IZ Z Z Z

12

1 11 12 1

2 21 2

0

2 2

k

iij

j V k j

I Y Y VI Y V

IY

Y Y V

V

I

Admittance (Y) ParametersConsider a 2-port network:

Admittance matrix

Short-circuit parameters

1 21V

1I

+

-

2I

2V+

-

1 11 1 12 2

2 21 1 22 2

I Y V Y V

I Y V Y V

or

13

Y-Parameters (cont.)

0k

iij

j V k j

IY

V

jV

iI

+-

We apply a voltage across port j and measure the current (with an ideal current meter) at port i. All ports are short-circuited except j.

N-port network

14

1 1 1

2 2 2

111 11 12 12

221 21 22 22

A B

A B

A B A B

A B A B

I I I

I I I

VY Y Y Y

VY Y Y Y

1 1 1

2 2 2

A B

A B

V V V

V V V

Parallel

1AI

1AV

1I

1V A

B1BV

2V

2I

2BI

2AI

2BV

2AV

+ +

+ +

+ +1

1

2

2

1BI

- -

--

- -

Admittance (Y) Parameters

Y-parameters are convenient for parallel connected networks.

15

1Y Z

Admittance (Y) Parameters

Relation between [Z] and [Y] matrices:

V Z I

I Y V

V Z Y V

Z Y V

Z Y U Identity Matrix

Hence

Therefore

16

Reciprocal Networks

If a network does not contain non-reciprocal devices or materials* (i.e. ferrites, or active devices), then the network is “reciprocal.”

ij ji ij jiZ Z Y Y

Z Y

and are symmetric

* A reciprocal material is one that has reciprocal permittivity and permeability tensors. A reciprocal device is one that is made from reciprocal materials

Note: The inverse of a symmetric matrix is symmetric.

Example of a nonreciprocal material: a biased ferrite

(This is very useful for making isolators and circulators.)

17

Reciprocal MaterialsD E

B Η

x xx xy xz x

y yx yy yz y

z zx zy zz z

D E

D E

D E

x xx xy xz x

y yx yy yz y

z zx zy zz z

B

B

B

Ferrite: 0

0

0

0 0 1

j

j

is not symmetric!

Reciprocal: ,ij ji ij ji

18

We can show that the equivalent circuits for reciprocal 2-port networks are:

T-equivalent

Pi-equivalent

11 21Z Z 22 21Z Z

21Z

21Y

11 21Y Y22 21Y Y

Reciprocal Networks (cont.)

19

ABCD-ParametersThere are defined only for 2-port networks.

1 2'

1 2

V VA B

I IC D

' '2 2

2 2

1 1

2 20 0

1 1' '2 20 0

I I

V V

V IA C

V V

V IB D

I I

1 2 1I

1V2V

'2I

'2 2I I

20

Cascaded Networks

'

'

21 1

1 1 2

1

1

2

2

AAA

A A

BA

B

B

A B

B

VV VABCD

I I I

VABCD

I

VABCD ABCD

I

1 2'

1 2

ABV VABCD

I I

1 2 1 2

1AI '

2AI

1AV

2AV

1BV 2

BV

1I'2I

1V 2V

1BI

'

2BI

A B

A nice property of the ABCD matrix is that it is easy to use with cascaded networks: you simply multiply the ABCD matrices together.

21

At high frequencies, Z, Y, h & ABCD parameters are difficult (if not impossible) to measure.

o V and I are not uniquely definedo Even if defined, V and I are extremely difficult

to measure (particularly I).o Required open and short-circuit conditions are

often difficult to achieve.

Scattering (S) parameters are often the best representation for multi-port networks at high frequency.

Scattering Parameters

22

S-parameters are definedassuming transmission lines are connected to each port.

On each transmission line:

0 0

0 0

i i i iz zi i i i

i i i ii

i

i i i

ii

iV zV z V e V e

V z V zI z

Z Z

V z

1, 2i

1 2

1z 2z

1a

1b 2b

2a01 1,Z 02 2,Z

Scattering Parameters (cont.)

Local coordinates

0

0

i i i i i

i i i i i

a z V z Z

b z V z Z

Incoming wave function

Outgoing wave function

23

For a One-Port Network

01

01

1

1

1

1

11

0

0

0

0

L

Z

Z

V

V

S

b

a

1l

1a

1b01Z

L

1 1

11 1

0 0

0

Lb a

S a

0

0

i i i i i

i i i i i

a z V z Z

b z V z Z

Incoming wave function

Outgoing wave function

For a one-port network, S11 is defined to be the same as L.

24

Scattering matrix

1 11 1 12 2

2 21 1 22

1 11 12 1

2 21 22 2

2

0 0 0

0 0

0

0

0

0

0

b S a S a

b S a S

b S S ab S a

b S S

a

a

For a Two-Port Network

1 2

1z 2z

1a

1b 2b

2a01 1,Z 02 2,Z

25

Scattering Parameters

Output is matched

Input is matched

input reflection coef. w/ output matched

reverse transmission coef. w/ input matched

forward transmission coef. w/ output matched

output reflection coef. w/ input matched

1 11 1 12 2

2 21 1 22 2

0 0 0

0 0 0

b S a S a

b S a S a

2

1

2

1

111

1 0

112

2 0

221

1 0

222

2 0

0

0

0

0

0

0

0

0

a

a

a

a

bS

a

bS

a

bS

a

bS

a

Output is matched

Input is matched

26

Scattering Parameters (cont.)

ja

ib

For a general multiport network:

All ports except j are semi-infinite (or matched)

0

0

0k

iij

j a k j

bS

a

Port j

Port i

Semi-infinite

N-port network

27

Scattering Parameters (cont.)

1

2

3

2a

2b

3a

3b

1a

1b

Illustration of a three-port network

28

For reciprocal networks, the S-matrix is symmetric.

ij jiS S i j

Scattering Parameters (cont.)

Note: If all lines entering the network have the same characteristic impedance, then

0

0

0k

iij

j V k j

VS

V

29

Why are the wave functions (a and b) defined as they are?

Note:

*

2

0

01 10 Re 0 0

2 2i

i i ii

VP V I

Z

0

2

0 0

10 0

2

i i i

i i

a V Z

P a

Scattering Parameters (cont.)

1 2

1z 2z

1a

1b 2b

2a01 1,Z 02 2,Z

(assuming lossless lines)

30

Similarly,

2

0

20100

2

1

2i

i ii

VP b

Z

Also,

Scattering Parameters (cont.)

2 2 2

2 2 2

0

1 10

21 1

2

0

2

02

i i

i

i i

i

i

i

li i i i i

li i i i i

li i i

li i i

P

V

l a l a e

l V e

V l

l b

e

P l b e

V

31

Example

1a

1b

2a

2b0Z 0Z

Z2V1V

1z 2z

Find the S parameters for a series impedance Z.

Note that two different coordinate systems are being used here!

32

Example (cont.)

02 2

1 1 011

1 1 0

0

00

0

0

0 0

0 0in

ina a

b V Z ZS

a V Z Z

Z Z Z

Z Z Z

1102

ZS

Z Z

By symmetry:

S11 Calculation:

1a

1b 2b0Z 0Z

Z2V1V

inZ

1z 2z

Semi-infinite

22 11S S

33

2

2

221

1 0

2

1 0

0

0

0

0

a

a

bS

a

V

V

2 2 2

02 1

0

1 1 0 11

02 2 1 0 11

0

0 0 0

0 0

0 1

0 0 1

a V V

ZV V

Z Z

V a Z S

ZV V a Z S

Z Z

Example (cont.)S21 Calculation:

1a

1b 2b0Z 0Z

Z2V1V

inZ

1z 2z

Semi-infinite

1 1 00 0V a Z

34

01 0 11

021

1 0

0 0 0 011

0 0 0 0 0

0 1

0

2 21 1

2 2

Za Z S

Z ZS

a Z

Z Z Z Z Z ZS

Z Z Z Z Z Z Z Z Z Z

021

0

2

2

ZS

Z Z

Hence

Example (cont.)

1a

1b 2b0Z 0Z

Z2V1V

inZ

1z 2z

Semi-infinite

12 21S S

35

Example

0Z 0Z0 ,s sZ

L

z 2z1z

Find the S parameters for a length L of transmission line.

Note that three different coordinate systems are being used here!

36

Example (cont.)

2

2

2

0022

0

111

0 01

( )in a

in aa

Z ZS

b

Z ZS

a

by symmetry

2

2

0 00 0 20

0 0

1tan

tan 1

s

s

j LLs s

in s s j Las s L

eZ jZ LZ Z Z

Z jZ L e

0 0

0 0

sL

s

Z Z

Z Z

0Z 0Z0 ,s sZ

L

z 2z1z

1 0V 2 0V+ +

- -

L

+-

V z

Semi-infinite

S11 Calculation:

37

Example (cont.)

0 00 0

0 011 22

0 00 0

0 0

tan

tan

tan

tan

s ss

s s

s ss

s s

Z jZ LZ Z

Z jZ LS S

Z jZ LZ Z

Z jZ L

20 0 0 11 220

0s in aZ Z Z Z S S

Note: If

0Z 0Z0 ,s sZ

L

z 2z1z

1 0V 2 0V+ +

- -

L

+-

V z

Hence

38

2 2

2221

1

0

1 0 00

0

0a a

bS

a

V Z

V Z

1 1 110 0 1V V S

0 0

0 0

sL

s

Z Z

Z Z

Example (cont.)

0Z 0Z0 ,s sZ

L

z 2z1z

1 0V 2 0V+ +

- -

L

+-

V z

11

11

00

1

VV

S

We now try to put the numerator of the S21 equation in terms of V1 (0).

Hence, for the denominator of the S21 equation we have

S21 Calculation:

Semi-infinite

39

21

1

2

2

0 0 1

00

1

0 1s

s s

s s

sj z j z

j L j LL

j L j LL

L

V

V z V e e

V L V e e

VV

e e

Example (cont.)

0Z 0Z0 ,s sZ

L

z 2z1z

1 0V 2 0V+ +

- -

L

+-

V z

2 20 0 0 0 1 LV V V V

Next, use

1

2 2

00 1

1s sLj L j L

L

VV

e e

Hence, we have

40

2

2 1121 2

1 0

0 1 1

0 1

s

s

j LL

j LLa

V S eS

V e

Example (cont.)

0Z 0Z0 ,s sZ

L

z 2z1z

1 0V 2 0V+ +

- -

L

+-

V z

Therefore, we have

1121 122

1 1

1

s

s

j LL

j LL

S eS S

e

by symmetry

11

11

00

1

VV

S

1

2 2

00 1

1s sLj L j L

L

VV

e e

so

41

Special cases:

0 0 11 22

21 12

) 0, 0s

s L

j L

a Z Z S S

S S e

0

0

s

s

j L

j L

eS

e

20 11 220

21

2)

2 2

0

1 1s

g gs

g

in a

j L

b L L

Z Z S S

e S

0 1

1 0S

0 ,s sZ

L

Example (cont.)

42

02 0111

02 01

01 0222 11

02 01

Z ZS

Z Z

Z ZS S

Z Z

2

2

2

221

1

02

1

01 0

0

0

0

0 0

a

a

b

a

V

Z

VS

Z

Example

Find the S parameters for a step-impedance discontinuity.

01Z 02Z

43

22 2

2 2 1 1 1100 00 0 0 0 1

aa aV V V V S

01 0221 12

01 02

2Z Z

S SZ Z

Example (cont.)

1V2V

++

01Z 02Z --Because of continuity of the voltage

across the junction, we have:

2 2

2

02 02

1 1

01 010

1

0

2

11

1

0 10

0 0

a a

V

Z Z

V V

Z Z

V

S

S

02 0111

02 01

02

02 01

1 1

2

Z ZS

Z Z

Z

Z Z

0121 11

02

1Z

S SZ

Hence

so

Semi-infiniteS21 Calculation:

44

For reciprocal networks, the S-matrix is symmetric.

For lossless networks, the S-matrix is unitary.

Identity matrix

TS S

* *T TS S S S U

* *

1 1

( , )N N

Tik kj ki kj ij

k k

i j S S S S

Take element1 ;

0;ij

i j

i j

Properties of the S Matrix

Equivalently,

* 1TS S

Notation: † *H T

S S S

† 1S S

so

A B U

B A U

Note :

If

then

N-port network

45

Example:

11 12 13

21 22 23

31 32 33

S S S

S S S S

S S S

* * *11 11 21 21 31 31

* * *12 12 22 22 32 32

* * *13 13 23 23 33 33

* * *11 12 21 22 31 32

* * *11 13 21 23 31 33

* * *12 13 22 23 32 33

1

1

1

0

0

0

S S S S S S

S S S S S S

S S S S S S

S S S S S S

S S S S S S

S S S S S S

Unitary

Properties of the S Matrix (cont.)

The column vectors form an orthogonal set.

The rows also form orthogonal sets (see the note on the previous slide).

46

Example

50

02 2

0 02

0 02

j j

jS

j

12

350

S

Not unitary Not lossless

1) Find the input impedance looking into port 1 when ports 2 and 3 are terminated in 50 [] loads.

2) Find the input impedance looking into port 1 when port 2 is terminated in a 75 [] load and port 3 is terminated in a 50 [] load.

(For example, column 2 doted with the conjugate of column three is not zero.

47

11 11 1

1

0 50[ ]in in

bS Z

a

1 If ports 2 and 3 are terminated in 50 [Ω]:

Example (cont.)

(a2 = a3 = 0)

2 0 1

2

3

2a

2b

3a

3b3 0

50S 1a

1b1in

1 11 1 12 2 13 3b S a S a S a

48

22

2

75 50 1

75 50 5

a

b

2) If port 2 is terminated in 75 [Ω] and port 3 in 50 [Ω]:

Example (cont.)

2

1

5

1

2

3

2a

2b

3a

3b3 0

50S 1a

1b1in

49

31 21 11 12 13

1 1 1

212 2 12 2 21

1

1 1

5 102 2

in

b j jS

ab aS S S

a a a

S Sa

2 2 2a b

1

1

1 44.55[1

501

]inin

in

Z

Example (cont.)

11

2 2

3 3

02 2

0 02

0 02

ab

b a

j j

j

jb a

322 1 21 22 23

1 1

/aa

b a S S Sa a

2

1

5

1

2

3

2a

2b

3a

3b3 0

50S 1a

1b1in

50

Transfer (T) Matrix

For cascaded 2-port networks:

2

1 11 12 2

1 21 22 2

2

a T T b

b T

b

T

T

a

a

22

21 21

11 11 2212

21 21

1 S

S ST

S S SS

S S

A1Aa 2

Aa2Ba

1Ba

1Bb

2Bb

2Ab1

Ab B1 12 2

T Matrix:

21

22 22

212 12

1122 22

1T

T TS

T TT

T T

(Derivation omitted)

51

1 2

1 2

1

1

2 1

2 1

21

1

1

1 2

A AA

A A

AA

A

A

A B

A

A

A B

B

B

BB

B

a bT

b a

aT

b

b a

a

a

b

a

b

bT

b aT

But

Transfer (T) Matrix (cont.)

Hence

The T matrix of a cascaded set of networks is the product of the T matrices.

ABT

52

Conversion Between Parameters

53

ExampleDerive Sij from the Z parameters.

(The result is given inside row 1, column 2, of the previous table.)

11 21Z Z 22 21Z Z

21Z

Semi infinite

0Z 0Z11S

1 2

011 1

0

inin

in

Z ZS

Z Z

11 21 21 22 21 0||inZ Z Z Z Z Z Z

S11 Calculation:

54

Example (cont.)

11 21 21 22 21 0

21 22 0 2111 21

22 0

11 21 22 0 21 22 0 21

22 0

211 22 11 0 21 22 21 0 21 22 21 0 21

22 0

211 22 11 0 21

22 0

||inZ Z Z Z Z Z Z

Z Z Z ZZ Z

Z Z

Z Z Z Z Z Z Z Z

Z Z

Z Z Z Z Z Z Z Z Z Z Z Z Z

Z Z

Z Z Z Z Z

Z Z

11 21Z Z 22 21Z Z

21Z

Semi infinite

0Z 0Z11S

1 2

55

Example (cont.)

211 0 22 21

22 0in

Z Z Z ZZ

Z Z

11 21Z Z 22 21Z Z

21Z

Semi infinite

0Z 0Z11S

1 2

011

0

in

in

Z ZS

Z Z

211 0 22 21 0 0 22

11 211 0 22 21 0 0 22

Z Z Z Z Z Z ZS

Z Z Z Z Z Z Z

so

56

Example (cont.)11 21Z Z 22 21Z Z

21Z

Semi infinite

0Z 0Z11S

1 2

211 0 22 21 0 0 22

11 211 0 22 21 0 0 22

2 211 0 11 22 21 0 0 22

2 211 0 11 22 21 0 0 22

20 22 11 0 21

20 22 11 0 21

Z Z Z Z Z Z ZS

Z Z Z Z Z Z Z

Z Z Z Z Z Z Z Z

Z Z Z Z Z Z Z Z

Z Z Z Z Z

Z Z Z Z Z

57

Example (cont.)11 21Z Z 22 21Z Z

21Z

Semi infinite

0Z 0Z11S

1 2

20 22 11 0 21

11 20 22 11 0 21

Z Z Z Z ZS

Z Z Z Z Z

Note: to get S22, simply let Z11 Z22 in the previous result.

20 11 22 0 21

22 20 11 22 0 21

Z Z Z Z ZS

Z Z Z Z Z

58

Example (cont.)

11 21Z Z 22 21Z Z

21Z

Semi infinite

0Z 0Z11S

1 2cV 2 0V 1 0V

1 110 1V S

Assume 1 0 1 [V]V

21 22 21 01

11 21 21 22 21 0

||0

||c

Z Z Z ZV V

Z Z Z Z Z Z

Use voltage divider equation twice:

0

222 21 0

0 c

ZV V

Z Z Z

21 2 20 0S V V

S21 Calculation:

59

Example (cont.)

Hence

21 22 21 0 021 11

22 21 011 21 21 22 21 0

||1

||

Z Z Z Z ZS S

Z Z ZZ Z Z Z Z Z

11 21Z Z 22 21Z Z

21Z

Semi infinite

0Z 0Z11S

1 2cV 2 0V 1 0V

After simplifying, we should get the result in the table.

(You are welcome to check it!)