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GUC (Dr. Hany Hammad) 1/31/2016

COMM (603) Lecture #1 1

Radio Frequency Engineering

Associate Prof. Hany Hammad

Why Use Network Analysis?

• Objectives:

– Equivalent circuit that is open to all the tools of the circuit analysis.

• Reasons:

– Maxwell’s equations are much more difficult and provide more information than we need.

– We are only interested signal flow and the voltage and current at a set of terminals.

© Dr. Hany Hammad, German University in Cairo



When to apply microwave network anlysis?

• Circuit dimensions << wavelength

– Lumped passive and active components.

– Negligible phase change throughout the circuit.

– Circuit theory (Kirchoff’s and Ohms laws)

• Circuit dimensions ~ wavelength

– Distributed passive and active components.

– Phase depends on position. Components are characterized by their dimensions, propagation constant and characteristics impedance.



Impedance parameters

2121111 IZIZV

2221212 IZIZV

2

1

2221

1211

2

1

I

I

ZZ

ZZ

V

V

01

111

2

II

VZ

IZV

01

221

2

II

VZ

02

112

1

II

VZ

02

222

1

II

VZ

Open Circuit

Linear

Network

1I

1V+

_ Port-1

2I

2V+

_ Port-2

Linear

Network

1I

1V Port-1

02 I

2V+

_ Port-2

+

Linear

Network

01 I

1V+

_ Port-1

2I

2VPort-2 +




Example 1

Find the impedance parameters for the two port network shown in Figure.

6

01

111

2

II

VZ

6

01

221

2

II

VZ

6

02

112

1

II

VZ

6

02

222

1

II

VZ

Ans.:

66

66

2221

1211

ZZ

ZZ

+

V1

+

V2

I1 I2

6

V1

+

V2

I1

6+

+

V1

V2

I2

6 +


Example 2


Ans.:

12

01

111

2

II

VZ

0

01

221

2

II

VZ

0

02

112

1

II

VZ

3

02

222

1

II

VZ

30

012

2221

1211

ZZ

ZZ

21 3I1 I2

+

V1

+

V2

21 3I1

V1

+

V2

+

21 3

V2 V1 +

+

I2




Example 3


18

01

111

2

II

VZ

6

01

221

2

II

VZ

6

02

112

1

II

VZ

9

02

222

1

II

VZ

96

618

2221

1211

ZZ

ZZ

Ω12 3I1 I2

+

V1

+

V2

6

Ans.:

Ω12 3I1

+

V2

6V1 +

+

V1

Ω12 3

6 V2

+


Example 3 (another technique)

2121111 IZIZV 2221212 IZIZV

2121111 IZIZV 2221212 IZIZV

211111111121211121211111 IZZIZZIZIZIZIZVV

222221212122212122212122 IZZIZZIZIZIZIZVV

2

1

22222112

12121111

22

11

I

I

ZZZZ

ZZZZ

VV

VV


Ans.:

96

618

2221

1211

ZZ

ZZ

Ω12 3I1 I2

+

V1

6

+

V2

Ω12 3I1 I2

+

V1

6

+

V2

+

+

+

1V

1V

2V

2V





Ans.:

ljZ

ljZ

Z

lZ

Zj

ZljZZ

ljZZZ

I

VZ o

L

o

L

o

o

Lo

oLoin

cot

tan

tan1

tan

tan

1

1

01

111

2

II

VZ LZ

or

zj

o

zj

o eVeVV

zj

o

zj

o

o

eVeVZ

I 1

z = 0

ljlj

o

lj

o

lj

o eeVeVeVV 1

ljlj

o

olj

o

lj

o

o

eeZ

VeVeV

ZI

11

oo VVLZ

oZ

l

+

V1

+

V2

I1 I2

+

1

o

o

V

V

= 0

= 0

Note: it is symmetrical

Open Circuit

(Open Circuit)

lz

lz

o

o

o

oo

I

V

I

VZ

Example 4


Example 4

ljlj

o

lj

o

lj

o eeVeVeVV 1

ljlj

o

olj

o

lj

o

o

eeZ

VeVeV

ZI

11

11

1

1 cotsin2

cos2ZljZ

lj

lZ

ee

eeZ

I

Vooljlj

ljlj

o

01

221

2

II

VZ

ooo VVVV 22

lj

o

lj

o

o

eVeVZ

I 1

1

oo VV

l

jZ

lj

Z

eeV

ZVZ oo

ljlj

o

oo

sinsin2

2221

1221 ZZ 2211 ZZ

1

ljZl

jZl

jZljZ

oo

oo

cotsin

sincot

Since




Admittance Parameters

2121111 VYVYI

2221212 VYVYI

2

1

2221

1211

2

1

V

V

YY

YY

I

I

01

111

2

VV

IY

VYI

01

221

2

VV

IY

Linear

Network

1I

1V+

_ Port-1

2I

2V+

_ Port-2

Linear

Network

1I

1V Port-1

2I

02 V+

_ Port-2

+

02

112

1

VV

IY

02

222

1

VV

IY Linear

Network

1I

01 V Port-1

2I

2V+

_ Port-2

+


Example 5

Find the admittance parameters for the two port network shown in Figure.

Ans.:

05.0

01

111

2

VV

IY

05.0

01

221

2

VV

IY

05.0

02

112

1

VV

IY

05.0

02

222

1

VV

IY

21 II

05.005.0

05.005.0

S05.0I1 I2

+

V1

+

V2

S05.0I1 I2

V1

V2=0 +

S05.0I1 I2

+

V1=0 V2




Example 6


Ans.:

0692.01.0025.02.0

1.0)025.02.0(

01

111

2

VV

IY

0615.0)889.0(0692.01

2

1

1

01

221

2

I

I

V

I

V

IY

V

112 889.0540

40III

S1.0 S2.0

S025.0

I1

+

V1

I2

+

V2

Current divider

S1.0 S2.0

S025.0

I1

V1

I2

+

V2=0

+

2.01025.01


Example 6

121 8.01040

40III

0615.08.00769.02

1

2

2

02

112

1

I

I

V

I

V

IY

V

0769.02.0025.01.0

2.0)025.01.0(

02

222

1

VV

IY

0769.00615.0

0615.00692.0

2221

1211

YY

YY

S1.0 S2.0

S025.0

I1

V1=0

I2

+

V2

+




Example 7


ans.:

1269.01115.0

1115.01192.0

2221

1211

YY

YY

S1.0 S2.0

S025.0

I1

+

V1

I2

+

V2

S05.0

S1.0 S2.0

S025.0

I1

+

V1

I2

+

V2

S05.01I

1I

2I

2I

2

1

22222121

12121111

2

1

2221

1211

2

1

V

V

YYYY

YYYY

V

V

YY

YY

I

I


Admittance Parameters

2121111 VYVYI

2221212 VYVYI

2121111 VYVYI

2221212 VYVYI

2121211111111 VYYVYYIII

2222212121222 VYYVYYIII

2

1

22222121

12121111

2

1

2221

1211

2

1

V

V

YYYY

YYYY

V

V

YY

YY

I

I




Hybrid Parameters

2121111 VhIhV

2221212 VhIhI

2

1

2221

1211

2

1

V

I

hh

hh

I

V

01

111

2

VI

Vh

01

221

2

VI

Ih

02

112

1

IV

Vh

02

222

1

IV

Ih

Linear

Network

1I

1V+

_ Port-1

2I

2V+

_ Port-2

Are especially important in transistor circuit analysis.

(dimensionless)

(dimensionless)

()

(Siemens)

(Input impedance) (forward current gain)

(reverse voltage gain)

(Output admittance)


Example 8

Find the hybrid parameters for the two port network shown in Figure.

ans.:

14129

63

01

111

2VI

Vh

3

2

36

6

01

221

2

VI

Ih

3

2

36

6

02

112

1

IV

Vh

SV

Ih

I9

1

02

222

1

2

1

2

1

9

1

3

23

214

V

I

I

V

S

12 3

6

I1

+

V1

I2

+

V2

12 3

6

I1

V1

I2

+

V2

+

12 3

6

I1 I2

V2 +




Transmission Parameters

221 BIAVV

221 DICVI

221 BIAVV

221 DICVI

2

2

1

1

I

V

DC

BA

I

V

2

2

1

1

I

V

DC

BA

I

V

Linear

Network

1I

1V+

_ Port-1

2I

2V+

_ Port-2

Linear

Network

1I

1V+

_ Port-1

2I

2V+

_ Port-2



02

1

2

IV

VA

221 BIAVV

221 DICVI

02

1

2

VI

VB

02

1

2

IV

IC

02

1

2

VI

ID

Linear

Network

1I

1V+

_ Port-1

2I

2V+

_ Port-2

Linear

Network

1I

1V Port-1

02 I

2V+

_ Port-2

+

Open Circuit

Linear

Network

1I

1V Port-1

2I

2VPort-2 +




Example 9

Find the transmission parameters for the two port network shown in Figure.

Ans.:

1

02

1

2

IV

VA

1

02

1

2VI

VB

SV

IC

I

0

02

1

2

1

02

1

2

VI

ID

10

11

DC

BA

1I1 I2

+

V1

+

V2

1I1 I2

+

V2

1V +

1I1 I2

+

V2=0

1V +


Example 10

Find the transmission parameters for the two port network shown in Figure.

Ans.:

1

02

1

2

IV

VA

0

02

1

2

VI

VB

SjV

IC

I

02

1

2

1

02

1

2

VI

ID

1

01

jDC

BA

+

V1

+

V2

Sj

I1 I2

V1

+

V2

Sj

I1 I2

+

V1

+

V2=0

Sj

I1 I2

+




Example 11

Find the transmission parameters for the two port network shown in Figure. Ans.:

2

2

1

1

10

11

I

V

I

V

2

2

1

1

1

01

I

V

jI

V

2

2

1

1

10

11

I

V

I

V

2

2

1

1

10

11

1

01

10

11

I

V

jI

V

2

2

1

1

1

21

I

V

jj

jj

I

V

1I1 I2”

+

V1

+

V2”

1

j

1I 2I

1 V

2 V

1I

j

2 I

2 V

1 V

1I 2I

1 V

2 V

2

2

1

11

10

11

I

V

jj



Microwave Engineering, 3rd Edition by David M. Pozar

Copyright © 2004 John Wiley & Sons




Z, Y & ABCD

Z

Z

Y

Y

DABC DABC


Conversion between different parameters

2

1

2221

1211

2

1

V

V

YY

YY

I

I

1121

1222

21122211

1

2221

1211

2221

1211 1

ZZ

ZZ

ZZZZZZ

ZZ

YY

YY

YZ

One type of network parameter can be converted into another via the respective equations. Conversion between the impedance and admittance parameters.

2

1

1

2221

1211

V

V

ZZ

ZZ





YABCD

221 BIAVV

221 DICVI

212

1V

B

AV

BI

212221

1V

B

AV

BDCVDICVI

211 VB

BCADV

B

DI

B

A

B

B

BCAD

B

D

YY

YY

12221

1211

Conversion between the transmission and admittance parameters


Reciprocity

• A network us reciprocal if a zero impedance source and a zero impedance ammeter can be placed at any locations in a network and their positions interchanged without changing the ammeter reading.

• Accordingly, as a consequence of reciprocity, for the Z matrix description of a reciprocal network.

2112 ZZ 2112 YY

2221

1211

YY

YY

2221

1211

ZZ

ZZ




Multiple Port Networks

NNNN

N

N V

V

V

YY

Y

YYY

I

I

I

2

1

1

21

11211

2

1

NNNN

N

N I

I

I

ZZ

Z

ZZZ

V

V

V

2

1

1

21

11211

2

1

jkforIj

iij

k

I

VZ

0

jkforVj

iij

k

V

IY

0Microwave Engineering, 3rd Edition

by David M. Pozar Copyright © 2004 John Wiley & Sons


Scattering parameters (S-parameters)

• Z, Y, and ABCD parameters are very difficult to determine at radio and microwave frequencies.

• The scattering matrix (S-parameters) is to be employed at these frequencies.

ia is an incident wave at port i.

ib is a reflected wave at port i.

2121111 aSaSb

2221212 aSaSb

2

1

2221

1211

2

1

a

a

SS

SS

b

b aSb

Scattering Matrix

ijS is the scattering parameters

Two-port

Network Port-1

1aPort-2

1b

2a

2b





01

111

2

aa

bS 0( 2 a Port 2 is matched)

01

221

2

aa

bS

0( 1 a Port 1 is matched) 02

112

1

aa

bS

02

222

1

aa

bS

iiS is the reflection coefficient i at the ith port when the

other port is matched terminated.

ijSis the forward coefficient i of the jth port if i is greater

than j, whereas it represents the reverse transmission coefficient if i is less than j with the other port terminated by a matched port.



• The steady state total voltage and current at the ith port is given by:

oioii VVV oioi

oi

oioii VVZ

III1

Incident

Reflected

iiioi IZVV 02

1 iiioi IZVV 0

2

1

2*

*

2

1Re

2

1Re

2

1

oi

oioi

oioioioii V

ZZ

VVIVP

2*

*

2

1Re

2

1Re

2

1

oi

oioi

oioioioii V

ZZ

VVIVP

Incident Reflected

oi

oi

oi

oioi

I

V

I

VZ

Note:

Objective: Try to find an expression for ai and bi in term of the incident and reflected voltage and currents

Solving the last two equations for reflected

and incident voltages

Incident power at port i

Reflected power at port i





ioi

oi

i

oi

ioii

oi

oii IZ

Z

V

Z

IZV

Z

Va

22

1

22

1

2

ioi

oi

i

oi

ioii

oi

oii IZ

Z

V

Z

IZV

Z

Vb

22

1

22

1

2

ohmampereohm

voltwattba ii .&

2

11

aPP avsi

i

2

11

bPP refi

i

(Power available from the source)

(Power reflected from port1)

2

1

2

1 baPPP refavsdel (Power delivered to the port)


Evaluating the values of S-parameters

01

111

2

aa

bS

01

221

2

aa

bS

1

1111

22

1

o

o

Z

IZVa

2

2222

22

1

o

o

Z

IZVb

022

1

2

2222

o

o

Z

IZVa 222 IZV o

2

222

2

222

222

2

2

1

o

o

o

o

Z

VZI

Z

IZb

1

11

22 o

S

Z

Va

11

01

1

11

111

2

Sa

b

ZZ

ZZ

ao

o

2

1

1

2

1

1

2

2

01

221

222

22

o

o

SS

o

oaZ

Z

V

V

V

Z

Z

V

a

bS

1SV

11 os ZZ

2oZ

1V

inputZZ 1

Two Port Network Device

2V

1I 2I

2oZ1oZ

?

From Previous Lecture




Evaluating the values of S-parameters

1

2

2

1

02

112

2

1o

o

SaZ

Z

V

V

a

bS

22

02

2

22

222

1

Sa

b

ZZ

ZZ

ao

o

2SV

22 os ZZ

1V

ouputZZ 2

Two Port Network Device

2V

1I 2I

2oZ1oZ1oZ

Same analysis as previous slide


o

o

S ZZ

ZZ

V

V

21

1

o

o

ZZ

Z

V

V

1

2

Example

Find the S-parameters of a series impedance Z connected between the two ports.

Answer:

ooo

oo

ZZ

Z

ZZZ

ZZZS

211

Z

Zo Zo

Z

Zo Zo Zo V2 V1

Zo

VS

?11 S

?21 S1

2

1

221

2

o

o

S Z

Z

V

VS

01112

a

S

11

11

012

o

o

a ZZ

ZZ

1

1

1

2

1

2

SS V

V

V

V

V

V

12 oo ZZ 1

o

o

ZZ

ZS

2

221




Example

o

o

ZZ

ZS

2

212

11

11

2221

1211

1

1

SS

SS

ooo

oo

o

o

a

ZZ

Z

ZZZ

ZZZS

ZZ

ZZS

222

22

22

02221

Z

Zo Zo V2 V1

Zo

VS Zo

Same analysis as previous slide (Symmetry)

oZZ

ZS

222

1

2

2

1

02

112

2

1o

o

SaZ

Z

V

V

a

bS

211212

21 SS

ZZ

Z

o

o

Note:


Reciprocal Networks and Lossless Networks

The impedance and admittance matrices are: • Symmetric for reciprocal networks. • Purely imaginary for lossless networks.

For Scattering parameters the S-parameters are symmetrical if: And Lossless if the matrix is unitary: In terms of Scattering matrix this can be written as:

jiij SS Symmetrical Matrix

1 AAT

The conjugate Transpose

The Inverse

otherwise

kjifSS

N

i

ikij

0

1

1

* Unitary Matrix “lossless”

1AAT




Verification

oZjX

jX

21

If in the previous example if show that the matrix is Unitary

22

2

21

2

112

12 oo ZjX

jX

ZjX

jXSS

022

2

2

2

2

*

2221

*

1211

oo

o

o

o

o ZjX

jX

ZjX

Z

ZjX

Z

ZjX

jXSSSS

12

2

222

2

22

2

o

o

o ZX

Z

ZX

X

02

2

222

2*

2122

*

1112

o

o

ooo

o

ZjX

Z

ZjX

jX

ZjX

jX

ZjX

ZSSSS

otherwise

kjifSS

N

i

ikij

0

1

1

*

11

11

2221

1211

1

1

SS

SS

12

22

2

12 SSalso

jXZ

Unitary Matrix “lossless”


Verification

• It can be done in a simpler way:

– The dot product of any column of [S] with the conjugate of that column gives unity.

– The dot product of any column with the conjugate of a different column gives zero (orthogonal).

2221

1211

SS

SS

0*

2221

*

1211 SSSS

12

21

2

11 SS




Multiport Networks

NNNN

N

N a

a

a

SS

S

SSS

b

b

b

2

1

1

21

11211

2

1

jkforaj

iij

k

a

bS

0










Network Analyzer

R&S® ZVB Vector Network Analyzer

(20 GHz & 4 ports)

http://www2.rohde-schwarz.com/en

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