ANTENNAS and MICROWAVES ENGINEERING (650427) · 2018-04-02 · a linear two port network by...

Philadelphia University

Faculty of Engineering Communication and Electronics Engineering

Part 4 Dr. Omar R Daoud 1

ANTENNAS and MICROWAVES

ENGINEERING

(650427)

4/1/2018 Dr. Omar R Daoud 2

Microwave Network Analysis

General Considerations The relationship between the voltage (V) and

current (I) at the terminals/ports of a complex circuit is the main issue. Thus, it can be considered as a black box.

For a linear circuit, the I-V relationship is linear and can be written in the form of matrix equations.

A simple example of linear 2-port circuit is shown below. Each port is associated with 2 parameters, the V and I.

021

111

VV

Iy

0121

12

V

V

Iy

021

221

VV

Iy

0122

22

V

V

Iy



General Considerations In RF and microwave systems, the ABCD

parameters is the main of interest.

They are the most useful for representing TL and other linear microwave components in general, even for the cascaded ones.

221

221

2

2

1

1

DICVI

BIAVV

I

V

DC

BA

I

V

02

1

2

IV

VA

02

1

2

VI

VB

02

1

2

VI

ID

02

1

2

IV

IC

3

3

33

33

1

1

3

3

22

22

11

11

1

1

I

V

DC

BA

I

V

I

V

DC

BA

DC

BA

I

V



General Considerations The analysis using Y, Z, H or ABCD parameters is considered to describe

a linear two port network by open/short the network ports.

At radio frequencies:

it is difficult to have a proper short or open circuit, there are parasitic inductance and capacitance in most instances.

Open/short condition leads to standing wave, can cause oscillation and destruction of device.

For non-TEM propagation mode, it is not possible to measure voltage and current. We can only measure power from E and H fields.

Hence a new set of parameters (S) is needed which

Do not need open/short condition.

Do not cause standing wave.

Relates to incident and reflected power waves, instead of

voltage and current.



Scattering Matrix As oppose to V and I, S-parameters

relate the reflected and incident

voltage waves.

S-parameters have the following

advantages:

Relates to familiar measurement such

as reflection coefficient, gain, loss etc.

Can cascade S-parameters of multiple

devices to predict system performance

(similar to ABCD parameters).

Can compute Z, Y or H parameters from

S-parameters if needed

222

22

0 VVVV

eVeVzV zjzj

222

22

0 IIII

eIeIzI zjzj



Scattering Matrix If the n–port network is linear, there is

a linear relationship between the

normalized waves.

Considering that we can send energy

into all ports, this can be generalized to

An arbitrary N-port microwave network

2222

VsV

22

VsVnn

2121

VsVConstant that

depends on the

network construction

nnnnnnn

VsVsVsVsV 332211

nn

VsVsVsVsV13132121111

nn

VsVsVsVsV23232221212

VSV



Scattering Matrix Considering that we can send energy

into all ports, this can be generalized to

Vn+ is the amplitude of the voltage wave

incident on port n.

Vn- is the amplitude of the voltage wave

reflected from port n.

An arbitrary N-port microwave network

2222

VsV

22

VsVnn

2121

VsVConstant that

depends on the

network construction

nNNN

N

n V

V

V

SS

S

SSS

V

V

V

.

.

.

....

......

......

......

.....

...

.

.

.

2

1

1

21

11211

2

1



Scattering Matrix A specific element of the [S] matrix can be

determined as:

Sij is the transmission coefficient from port j to

port i when all other ports are terminated in

matched loads.

It can be found by driving port j with an incident

wave Vj+, and measuring the reflected wave

amplitude, Vi-, coming out of port i. The incident

waves on all ports except j-th port are set to zero

(which means that all ports should be terminated

in matched load to avoid reflections).

Sii is the reflection coefficient seen looking into

port i when all other ports are terminated in

matched loads.

jkforVj

iij

k

V

VS

,0

2

1

2

1

2221

1211

2

1

V

VS

V

V

ss

ss

V

V

01

01

02

02

2

1

12

2

2

22

1

2

21

1

1

11

VVVV

V

Vs

V

Vs

V

Vs

V

Vs

Measurement of s11 and s21: 0

20

2

1

2

21

1

1

11

VV

V

Vs

V

Vs

Measurement of s22 and s12:

01

01

2

1

12

2

2

22

VV

V

Vs

V

Vs



Scattering Matrix Reciprocal and Lossless networks

Reciprocal Network means:

Impedance and admittance matrices are symmetric

Lossless Network means:

It is purely imaginary (no real

power can be delivered to the

network) i.e a unitary matrix

2

1

2

1

2221

1211

2

1

V

VS

V

V

ss

ss

V

V

01

01

02

02

2

1

12

2

2

22

1

2

21

1

1

11

VVVV

V

Vs

V

Vs

V

Vs

V

Vs


20

2

1

2

21

1

1

11

VV

V

Vs

V

Vs


01

01

2

1

12

2

2

22

VV

V

Vs

V

Vs

tss ][][

1* ][][ tss



Scattering Matrix Important points to note:

Reflection coefficient looking into port n is not equal to Snn, unless all other ports are matched

Transmission coefficient from port m to port n is not equal to Snm, unless all other ports are matched

S parameters of a network are properties only of the network itself (assuming the network is linear)

Changing the termination or excitation of a network does not change its S parameters, but may change the reflection coefficient seen at a given port, or transmission coefficient between two ports

2

1

2

1

2221

1211

2

1

V

VS

V

V

ss

ss

V

V

01

01

02

02

2

1

12

2

2

22

1

2

21

1

1

11

VVVV

V

Vs

V

Vs

V

Vs

V

Vs


20

2

1

2

21

1

1

11

VV

V

Vs

V

Vs


01

01

2

1

12

2

2

22

VV

V

Vs

V

Vs



Scattering Matrix Example

Find the S parameters of the 3 dB

attenuator circuit with 50 Ω characteristic

impedance. S11 can be found as the reflection coefficient seen at

port 1 when port 2 is terminated with a matched load

(Z0 =50 Ω)

Thus S11 = 0. Because of the symmetry of the circuit, S22 = 0.

022

0

)1(

0

)1(

0

)1(

0

1

1

11 Z

in

in

VV ZZ

ZZ

V

VS

50)5056.8(8.141

)5056.8(8.14156.8

)1(

inZ




Find the S parameters of the 3 dB

attenuator circuit with 50 Ω characteristic

impedance. S21 can be found by applying an incident wave at port

1, V1+, and measuring the outcome at port 2, V2

-. This

is equivalent to the transmission coefficient from port

1 to port 2:

From the fact that S11 = S22 = 0, we know that V1

- = 0

when port 2 is terminated in Z0 = 50 Ω, and that V2+ = 0.

In this case we have V1+ = V1 and V2

- = V2.

Thus, S21 = S12 = 0.707

0

1

2

212

VV

VS

11227071.0

56.850

50

56.844.41

44.41VVVV

Where 41.44 = (141.8//58.56) is the

combined resistance of 50 Ω and 8.56 Ω

paralled with the 141.8 Ω resistor.




A two port network is known to have the following

scattering matrix:

Determine if the network is reciprocal and lossless.

If port 2 is terminated with a matched load, what is the

return loss seen at port 1?

If port 2 is terminated with a short circuit, what is the return

loss seen at port 1?

02.04585.0

4585.0015.0S





scattering matrix:

Determine if the network is reciprocal and lossless.

Since [S] is not symmetric, the network is not reciprocal.

Taking the 1st column, (i = 1) gives;

So the network is not lossless.

02.04585.0

4585.0015.0S

1745.0)85.0()15.0(|||| 222

21

2

11 SS





scattering matrix:

If port 2 is terminated with a matched load, what is the

return loss seen at port 1?

When port 2 is terminated with a matched load, the reflection coefficient

seen at port 1 is Γ = S11 = 0.15. So the return loss is;

02.04585.0

4585.0015.0S

dBRL 5.16)15.0log(20||log20





scattering matrix:

If port 2 is terminated with a short circuit, what is the return

loss seen at port 1?

V2+ = - V2

- (for a short circuit at port 2), we can write:

02.04585.0

4585.0015.0S

2121112121111

VsVsVsVsV

2221212221212

VsVsVsVsV

22

2112

11

1

2

1211

1

1

1 S

SSS

V

VSS

V

V

452.02.01

)4585.0)(4585.0(15.0

00

dBRL 9.6)452.0log(20||log20



Scattering Matrix Reciprocal and Lossless networks

If all the components of the network are passive and it does not contain any active component, then its S parameter matrix must be reciprocal.

The [S] matrix will also be symmetric ( ).

Usually to avoid power loss, we would like to have a network that is matched at all ports and is lossless.

However, it is impossible to construct a three port lossless reciprocal network that is matched at all ports.

jiij SS

0

0

0

2313

2312

1312

SS

SS

SS

S

If all the three ports are matched the [S] matrix can be written as:

If the three port network is not reciprocal then and its [S] matrix will not be symmetric.

jiij SS

jiSSN

k

kjki

11

jiSSN

k

kjki

01



Microwave Filters They:

are linear 2-port network

control the frequency response at a certain point in

a microwave system

provide perfect transmission of signal for

frequencies in a certain passband region

have infinite attenuation for frequencies in the

stopband region

have linear phase response in the passband (to

reduce signal distortion).

Whay?

RF signals consist of:

Desired signals – at desired frequencies

Unwanted Signals (Noise) – at unwanted

frequencies

That is why filters have two very important

bands/regions:

Pass Band – frequency range of filter where it

passes all signals

Stop Band – frequency range of filter where it

rejects all signals

Active filter: there can be amplification of the of the signal power in the passband region. Passive filter: do not provide power amplification in the passband.

Filter used in electronics can be constructed from

resistors, inductors, capacitors, transmission line sections and resonating structures (e.g. piezoelectric crystal, Surface Acoustic Wave (SAW) devices, and also mechanical resonators etc.).

Active filter may contain transistor, FET and Op-amp.



Microwave Filters Parameters:

Pass bandwidth;

BW(3dB) = fu(3dB) – fl(3dB)

Stop band attenuation and frequencies,

Ripple difference between max and min

of amplitude response in passband

Input and output impedances

Return loss

Insertion loss

Group Delay, quality factor



Microwave Filters Frequency Response:

Maximally flat (Butterworth)

Called the binomial or Butterworth

response,

It is optimum in the sense that it

provides the flattest possible passband

response for a given filter complexity.

no ripple is permitted in its attenuation

profile

N

c

LR kP

21

PLR- Power Loss Ratio

– frequency of filter

c – cutoff frequency of filter

N – order of filter

Attenuation versus normalized frequency for

maximally flat filter prototypes.




Equal Ripple (Chebyshev)

also known as Chebyshev.

sharper cutoff

the passband response will have ripples

of amplitude 1 +k2

Where 1 + k2 is the ripple level in the passband.

Since the Chebyshev polynomials have the

property that

It shows that the filter will have a unity power

loss ratio at ω = 0 for N odd, but the power loss

ratio of 1 + k2 at ω = 0 for N even.

c

NLR TkP

221

Attenuation versus normalized frequency for equal-ripple filter prototypes.

(0.5 dB ripple level)

1

0NT





also known as Chebyshev.

sharper cutoff

the passband response will have ripples

of amplitude 1 +k2

Where 1 + k2 is the ripple level in the passband.

Since the Chebyshev polynomials have the

property that

It shows that the filter will have a unity power

loss ratio at ω = 0 for N odd, but the power loss

ratio of 1 + k2 at ω = 0 for N even.

c

NLR TkP

221

Attenuation versus normalized frequency for equal-ripple filter prototypes.

(3 dB ripple level)

1

0NT




Elliptic Function

have equal ripple responses in the

passband and stopband.

maximum attenuation in the passband.

minimum attenuation in the stopband.

Linear Phase

linear phase characteristic in the

passband

to avoid signal distortion

maximally flat function for the group

delay.



Microwave Filters LPF design:

As a matter of practical design procedure, it will

be necessary to determine the size, or order of

the filter.

This is usually dictated by a specification on the

insertion loss at some frequency in the stopband of

the filter.

Low pass filter prototype, N = 2

Ladder circuit for low pass filter prototypes and their

element definitions. (a) begin with shunt element. (b)

begin with series element.

g0 = generator resistance, generator conductance.

gk = inductance for series inductors, capacitance

for shunt capacitors.

(k=1 to N)

gN+1 = load resistance if gN is a shunt capacitor,

load conductance if gN is a series inductor.

LL

s

RRR

RR

R

CC

LRL

0

'

0

'

0

'

0

'

'

'

kk

c

k

kk

c

k

CjCjjB

LjLjjX

c

kkk

c

kkk

R

CCC

LRLL

0

'

0'

c

Impedance Scaling Frequency scaling

new element values of

the prototype filter

new element values



Microwave Filters From LPF design to HPF Transformation:

Frequency scaling new element values

c

kc

k

kc

k

C

RL

LRC

0'

0

' 1



Microwave Filters From LPF design to BPF Transformation:

Frequency scaling

new element values

0

0

0

012

0 1

0

12

210

The center frequency is:

The series inductor, Lk, is

transformed to a parallel LC

circuit with element values:

k

k

kk

LC

LL

0

'

0

'

The shunt capacitor, Ck, is

transformed to a shunt LC circuit

with element values:

0

'

0

'

kk

k

k

CC

CL



Microwave Filters From LPF design to BSF Transformation:

Frequency scaling

new element values

0

12

210

The center frequency is:

The series inductor, Lk, is

transformed to a series LC circuit


The shunt capacitor, Ck, is

transformed to a series LC circuit


1

0

0

k

k

kk

LC

LL

0

'

0

'

1

0

'

0

' 1

kk

k

k

CC

CL



Microwave Filters Example:


Design a maximally flat low pass filter with a

cutoff freq of 2 GHz, impedance of 50 Ω, and

at least 15 dB insertion loss at 3 GHz.

Compute and compare with an equal-ripple

(3.0 dB ripple) having the same order.

First find the order of the maximally flat filter to

satisfy the insertion loss specification at 3 GHz.

Attenuation versus normalized frequency for

maximally flat filter prototypes.

5.012

31

c

Using the normalized freq = 0.5

and min IL = 15dB; It is found

out that N = 5

618.0

618.1

0.2

618.1

618.0

5

4

3

2

1

g

g

g

g

g












5.012

31

c

Using the normalized freq = 0.5

and min IL = 15dB; It is found

out that N = 5

618.0

618.1

0.2

618.1

618.0

5

4

3

2

1

g

g

g

g

g

C3 C5

L2 L4

C1

LL

s

RRR

RR

R

CC

LRL

0

'

0

'

0

'

0

'

cR

gC

0

33

c

gRL

40

4

cR

gC

0

11

c

gRL

20

2

cR

gC

0

55

984.0

102250

618.09

0

11

cR

gC

438.6

1022

618.1509

202

c

gRL

183.3

102250

00.29

0

3

3

cR

gC

438.6

1022

618.1509

404

c

gRL

984.0

102250

618.09

0

55

cR

gC

pF

nH

pF

nH

pF












4817.3

7618.0

5381.4

7618.0

4817.3

5

4

3

2

1

g

g

g

g

g 541.5

102250

4817.39

0

11

cR

gC

031.3

1022

7618.0509

202

c

gRL

223.7

102250

5381.49

0

33

cR

gC

031.3

1022

7618.0509

404

c

gRL

541.5

102250

4817.39

0

55

cR

gC

pF

nH

pF

nH

pF





Design a band pass filter having a 0.5 dB

equal-ripple response, with N = 3. The center

frequency is 1 GHz, the bandwidth is 10%, and

the impedance is 50 Ω.

= 0.1, N = 3, = 1 GHz

LRg

Lg

Cg

Lg

000.1

5963.1

0967.1

5963.1

4

33

2

1

2

1

Transforming the LPF prototype to the BPF prototype





Design a band pass filter having a 0.5 dB equal-ripple response, with N = 3. The

center frequency is 1 GHz, the bandwidth is 10%, and the impedance is 50 Ω.

nH

ZLL 0.127

1.01012

505963.19

0

1

1

0

pF

LZC 199.0

5963.1101250

1.09

100

1

nH

C

ZL 726.0

0967.11012

501.09

20

02

pF

Z

CC 91.34

50)1.0(1012

0967.19

00

22

nH

ZLL 0.127

1.01012

505963.19

0

3 0

3

pF

LZC 199.0

5963.1102250

1.09

300

3





Design a five-section high pass lumped element filter with 3

dB equal-ripple response, a cutoff frequency of 1 GHz, and

an impedance of 50 Ω. What is the resulting attenuation at

0.6 GHz?

N = 5, = 1 GHz. At c = 0.6 GHz, The attenuation for N = 5, is

about 41 dB.

LRg

Lg

Cg

Lg

Cg

Lg

000.1

4817.3

7618.0

5381.4

7618.0

4817.3

6

55

44

33

2

1

2

1

pF

CZC

c

18.47618.0101250

11'

920

2

nH

L

ZL

c

754.15381.41012

50'

93

03

nH

L

ZL

c

28.24817.31012

50'

91

1

0

nH

L

ZL

c

754.15381.41012

50'

95

05

pF

CZC

c

18.47618.0101250

11'

940

4



Microwave Filters Filters Realizations Using Distributed Circuit Elements:

Lumped-element filter realization using surface mounted inductors

and capacitors generally works well at lower frequency (at UHF,

say < 3 GHz).

At higher frequencies, the practical inductors and capacitors loses

their intrinsic characteristics.

Also a limited range of component values are available from

manufacturer.

Therefore for microwave frequencies (> 3 GHz), passive filter is usually

realized using distributed circuit elements such as transmission line

sections. The focus will be on stripline microwave circuits.




Richard’s Transformation

jLLjljZZ cin tan LZ

l

c

tan

jCCjljYY cin tan

CY

l

cZc

1

tan

For LPP design, a further requirement ( regarding wavelength at cut-off frequency )

is that:

1tan cl 8

2 1tan c

cll




Kuroda’s Identities

1

22 1Z

Zn The inductor represents shorted TL while the capacitor represents open-circuit TL.





Design a 3rd order Butterworth Low-Pass Filter. Rs = RL= 50Ohm, fc = 1.5GHz.

500.0000.21

Length = c/8 for all TLs at = 1 rad/s

Convert to TLs using Richard’s Transformation Add extra TL on the series connection






apply Kuroda’s 1st Identity. apply Kuroda’s 2nd Identity. After applying Kuroda’s Identity.






Impedance and frequency denormalization.

Length = c/8

for all TLs at

f = fc = 1.5GHz

Zc/Ω /8 @ 1.5GHz /mm W /mm

50 13.45 2.85

25 12.77 8.00

100 14.23 0.61

Microstrip line using double-sided FR4 PCB (r = 4.6, H=1.57mm)

The layout (top view)





Design a low pass filter for fabrication using microstrip lines. The specifications are: cutoff

freq of 4 GHz, third order, impedance of 50 ohms and a 3dB equal ripple characteristics

Length = c/8 for all TLs at = 1 rad/s

Convert to TLs using Richard’s Transformation

Add extra TL on the series connection

g1 = 3.3487 = L1

g2 = 0.7117 = C2

g3 = 3.3487 = L3

g4 = 1.0000 = RL





Design a low pass filter for fabrication using microstrip lines. The specifications are: cutoff

freq of 4 GHz, third order, impedance of 50 ohms and a 3dB equal ripple characteristics

apply Kuroda’s 1st Identity. apply Kuroda’s 2nd Identity.

After applying Kuroda’s Identity.



Microwave Power Divider & Couplers

Needed if the power from a single microwave power amplifier may be insufficient to power up a device (e.g a radar transmitter)

In such case, several power amplifiers may be used with their power added using power combiners

Microwave couplers allows construction of balanced amplifiers that feature constant gain over wide bandwidth

Passive 3- and 4-port passive components are used to solve such problem.

Power dividers and couplers are passive microwave components used for power division or power combining.

Power division: an input signal is divided by the coupler into two (or more) signal or lesser power.

Power divider types:

equal division (3 dB) type (T-junctions & circulators)

unequal power division type (Resistive power divider)

Directional coupler – an arbitrary power division.

Hybrid junctions – equal power division and they have either:

a 90º (quadrature) or

a 180º (magic-T) phase shift between the output ports.

The simplest type of power divider is a T-junction, which is a three-port network with two inputs and one output.




Both dividers and combiners can be multi-port networks. The most common value for (division ratio) in splitter is –3 dB (when P2 = P3 ). The power ratio in splitter can range up to –40 dB for one path.

1 to 4 power divider




Directional Coupler It is a four port device that samples the

power flowing into port 1 coupled in to port 3 (the coupled port) with the remainder of the power delivered to port 2 (the through port) and no power delivered to the isolated port.

Directional couplers are described by three specifications: Coupling (C) - The ratio of input power

to the couple power. If all the ports are terminated in matched loads, the coupling coefficient becomes

Directivity (D)- The ratio of coupled

power to the power at the isolated port. When all ports are matched

Isolation (I) – The ratio of input power

to power out of the isolated port. When all ports are matched

3

1log10P

PC

4

3log10P

PD

4

1log10P

PI

||log2031

SC

||

||log20

41

31

S

SD

||log2041

SI

DCI

P

P

P

P

P

P

P

P

P

PI

)log10()log10(log10log104

3

3

1

4

3

3

1

4

1

Insertion Loss (IL). It is the level of loss that is occurred by the coupler between the input (P1) and through (P2) ports.

2

1log10P

PIL

For an perfectly matched network

||log2021

SIL




Hybrid Coupler It is a special cases of directional couplers,

where the coupling factor is 3 dB.

There are two types of hybrids:

The quadrature hybrid has a 90 degree phase shift between port 2 and 3 when fed from port 1, with the following [S] matrix:

The magic-T hybrid or rat-race hybrid has a 180 degree phase shift between port 2 and 3 when fed from port 4, with the following [S] matrix:

010

100

001

010

2

1

j

j

j

j

S

0110

1001

1001

0110

2

1S




Directional Coupler Example

An 8 dB directional coupler has a directivity of 35 dB. If the input power is P1 is 40mW, what are the output powers at P2, P3 and P4? Assume that the coupler is (a) lossless, (b) has an insertion loss of 0.1 dB

For the lossless case:

C (dB) = 8dB = = P1(dB) – P3(dB)

P1 = 40mW = 16.02 dBm

P3 = P1 – C = 16.02dBm – 8 dB = 8.02dBm = 6.34 mW

D (dB) = 35 dB = = P3 (dB) – P4 (dB)

P4 = P3 (dB) – D (dB) = 8.02dBm – 35 dB = -26.98 dBm

= 0.002mW

P2 (dBm) = P1 – P3 – P4 = 40 mW – 6.34 mW – 0.002 mW

= 33.66 mW = 15.27 dBm

3

1log10P

P

4

3log10P

P

For insertion loss of 0.1 dB case: (assuming IL at all 3 ports are equal) IL = 0.1 dB P3 = 8.02 dBm – 0.1 dB = 7.92 dBm = 6.19 mW P4 = -35 dBm – 0.1 dB = -35.1 dBm = 0.000309 mW P2 = 15.27 dBm – 0.1dB = 15.17 dBm = 32.89 mW




Power Dividers T-junction

It is the simplest type of power divider. It can be virtually implemented using any type of

transmission line. It is very simple to implement, it must be treated

with care because it does not offer any isolation between its ports.

In general, the fringing fields and higher order modes associated with the discontinuity at such a junction, leading to store energy that can be accounted for by a lumped susceptance, B. In order to match the divider to the input

line of characteristic impedance, Z0, it must have:

If the transmission lines are assumed to be

lossless (or low loss), then the characteristic impedances are real. If assume that B = 0, then :

Various T-junction power dividers. (a) E plane waveguide T. (b) H plane waveguide T. (c) Microstrip T-junction.

021

111

ZZZjBYin

021

111

ZZZ

In order for the input port to be matched, the output lines must be matched (terminated in their characteristic impedance).





The power dividing ratio can be selected by using different values of characteristic impedance for ports 2 and 3.

The input to the T junction can be matched through the correct choice of impedances in port 2 and 3.

For the lossless T junction, it cannot be matched at all three ports simultaneously

In order for the input port to be matched, the output lines must be matched (terminated in their characteristic impedance).

11321 PPPPP

2

1

1

2

1

2

2

22

1

2

1

Z

Z

Z

VP

Z

VP oo

3

1

1

2

1

3

2

32

1

2

1

Z

Z

Z

VP

Z

VP oo

1

0||

||

312132

312132

132

132

1

11

ZZZZZZ

ZZZZZZ

ZZZ

ZZZ

ZZ

ZZ

L

L

32

32132132 )(

ZZ

ZZZZZZZZ

The resistive power divider for an equal power split.

If the T junction contains lossy components

then it is possible to match all the three ports.

In this case the signal power will be reduced

due to loss in the junction.





Assuming that all the lumped-element resistors are terminated in the characteristic impedance Zo, the input impedance looking into any port is:

The voltage at the center of the junction is:

The output voltages V2 and V3 are equal to:

The resistive power divider for an equal power split.

ooo

inoo

ooo

in ZZZ

ZZZ

ZZZ

Z

3

2

3333

The network is symmetric from all three ports,

the output ports are also matched.

S11=S22=S33=0

The network is reciprocal,

S21=S31=S23=1/2.

Thus, the output power is –6 dB below the input

power level (lossy). The power delivered to the

input and outputs of the divider are:

113

2

32

3

32

VZZ

Z

VVoo

o

10

0

322

1

4

3

3

VVZ

Z

ZVVV o

011

101

110

2

1S

o

inZ

VP

2

1

2

1

in

oo

PZ

V

Z

VPP

4

1

8

121

2

1 2

1

2

1

32 Half of the supplied power is

dissipated in the resistors.





Example A lossless T junction power divider has a source impedance of 50 Ohms. Find the output characteristics impedance so that the input power is divided in a 2:1 ratio. Compute the reflection coefficients seen looking into the output ports.

If the voltage at the junction is V0, as shown in the

figure, the input power to the divider is

The results yields the characteristic impedance as:

The input impedance to the junction is:

Looking into the 150 Ω output line, we see an

impedance of 50||75 = 30 Ω, while at the 75 Ω output

line, we see an impedance of 150||50 = 37.5 Ω.

Thus the reflection coefficient looking into these ports

are:

2

0

0 )(2

1

Z

VP

in Then the output powers are: in

PZ

VP

3

1)(

2

1

1

2

0

1

inP

Z

VP

3

2)(

2

1

2

2

0

2

OhmsZZ 150301 OhmsZZ 75

2

302

OhmsZin

50150||75

666.015030

150301

333.0

755.37

755.372




Power Amplifiers Amplifier can be categorized as

According to signal level: Small-signal Amplifier.

Power/Large-signal Amplifier.

According to D.C. biasing scheme of the active component: Class A.

Class B.

Class AB.

Class C.

Class D (D stands for digital),

Class E and

Class F.

Most RF and microwave amplifiers today used transistor devices such as Si or SiGe BJTs, GaAs HBTs, GaAs or InP FETs, or GaAs HEMTs. They are: Rugged,

Low cost,

Reliable and

can be easily integrated in both hybrid an monolithic integrated circuitry.



Microwave Power Divider &

Couplers

Power Amplifiers Characteristics Power Gain

it is preferred for high frequency amplifiers as the impedance encountered is usually low.

The ratio of output power over input power is called the Power Gain (G), usually expressed in dB.

Instead on focusing on voltage or current gain, RF engineers focus on power gain.

By working with power gain, the RF designer is free from the constraint of system impedance.

3 types of power gain can be defined as:

As

LT

As

AoA

in

Lp

P

PG

P

PG

P

PG

powerInput Available

load todeliveredPower Gain Transducer

powerInput Available

Power load AvailableGain Power Available

Amp. power toInput

load todeliveredPower Gain Power

The effective power gain

GP, GA and GT can be expressed as

the S-parameters of the amplifier and

the reflection coefficients of the

source and load networks.




Couplers


Power Gain = G = PL / Pin is the ratio of power dissipated in the load ZL to the power delivered to the input of the two-port network. This gain is independent of ZS although some active circuits are strongly dependent on ZS.

Available Gain = GA = Pavn / Pavs is the ratio of the power available from the two-port network to the power available from the source. This assumes conjugate matching in both the source and the load, and depends on ZS but not ZL.

Transducer Power Gain = GT = PL / Pavs is the ratio of the power delivered to the load to the power available from the source. This depends on both ZS and ZL.

If the input and output are both conjugately matched to the two-port, then the gain is maximized and G = GA = GT

0

0

22

211211

1

1

1 ZZ

ZZ

S

SSS

V

V

in

in

L

Lin

0

0

11

211222

2

2

1 ZZ

ZZ

S

SSS

V

V

out

out

S

Sout

22

11

2

21

2

11

1

outs

s

A

s

sG

22

22

22

21

2

11

11

sinL

sL

T

s

sG

22

22

22

21

11

1

inL

L

P

s

sG

Note:

All GT, GP, GA, 1 and 2

depends on the S- parameters.




Couplers


A special case of the transducer power gain occurs when both input and output are matched for zero reflection (in contrast to conjugate matching).

Another special case is the unilateral transducer power gain, GTU where S12=0 (or is negligibly small). This nonreciprocal characteristic is common to many practical amplifier circuits. Γin = S11 when S12 = 0, so the unilateral transducer gain is:

2

21SGT

2

22

2

11

222

21

11

11

LS

LS

TU

SS

SG

The general transistor amplifier circuit.

The separate effective gain factors:

2

22

2

2

210

2

2

1

1

1

1

L

L

L

Sin

S

S

SG

SG

G

If the transistor is unilateral, the unilateral transducer gain

reduces to GTU = GSG0GL




Couplers


Example

An RF amplifier has the following S-parameters at fo: s11=0.3<-70o,

s21=3.5<85o, s12=0.2<-10o, s22=0.4<-45o. The system is shown

below. Assuming reference impedance (used for measuring the

S-parameters) Zo=50, find:

(a) GT, GA, GP.

(b) PL, PA, Pinc.

Step 1 - Find s and L

111.0

os

os

ZZ

ZZs

187.0

oL

oL

ZZ

ZZL

Step 2 - Find 1 and 2

358.0265.011 11

122122

11

22 js

sss

s

s

s

s

s

sout

151.0146.011 22

211211

22

11 js

sss

s

s

L

L

L

Lin

Step 3 - Find GT, GA, GP

742.1311

122

22

22

21

inL

L

P

s

sGG

739.1411

122

11

2

21

2

outs

s

A

s

sG

562.12

11

1122

22

22

21

2

sinL

sL

T

s

sG




Couplers


Example

An RF amplifier has the following S-parameters at fo: s11=0.3<-70o,

s21=3.5<85o, s12=0.2<-10o, s22=0.4<-45o. The system is shown

below. Assuming reference impedance (used for measuring the

S-parameters) Zo=50, find:

(a) GT, GA, GP.

(b) PL, PA, Pinc.

WPs

s

Z

VA 078.0

Re8

2

WZPP osZZsZZ

Ain 0714.012

1

1

WPGP inPL 9814.0




Couplers

Power Amplifiers Characteristics Harmonic Distortion




Couplers

Power Amplifiers Characteristics Bandwidth




Couplers

Power Amplifiers Characteristics Noise Figure




Power Amplifiers Stability Oscillation is possible if either the input or output port impedance has

the negative real part; this would imply that |Γin|>1 or |Γout|>1. Γin and Γout depends on the source and load matching networks,

The stability of the amplifier depends on ΓS and ΓL as presented by matching networks. Unconditionally stable:

|Γin| < 1 and |Γout| < 1 for all passive source and load impedance.

Conditionally stable:

|Γin| < 1 and |Γout| < 1 only for a certain range of passive source and load impedance (referred as potentially unstable) .

The stability condition of an amplifier circuit is usually frequency dependent.




Power Amplifiers Stability Oscillation is possible if either the input or

output port impedance has the negative real part; this would imply that |Γin|>1 or |Γout|>1. Γin and Γout depends on the source and load

matching networks,

The stability of the amplifier depends on ΓS and ΓL as presented by matching networks. Unconditionally stable:

|Γin| < 1 and |Γout| < 1 for all passive source and load impedance.

Conditionally stable:

|Γin| < 1 and |Γout| < 1 only for a certain range of passive source and load impedance (referred as potentially unstable) .

The stability condition of an amplifier circuit is usually frequency dependent.

The condition that must be satisfied by ΓS and ΓL if

the amplifier is to be unconditionally stable:

11 22

211211

L

Lin

S

SSS 1

111

2112

22

S

S

out

S

SSS

The determinant of the scattering matrix:

21122211 SSSS

The output stability circles:

22

22

2112

22

22

1122

S

SSR

S

SSC

L

L

The input stability circles:

22

11

2112

22

11

2211

S

SSR

S

SSC

S

S

Stability Test

Rollet’s condition: 12

1

2112

22

22

2

11

SS

SSK

The Auxiliary condition: 121122211 SSSS

The μ test: 11

21121122

2

11

SSSS

S




Power Amplifiers Stability

The condition that must be satisfied by ΓS and ΓL if

the amplifier is to be unconditionally stable:

11 22

211211

L

Lin

S

SSS 1

111

2112

22

S

S

out

S

SSS

The determinant of the scattering matrix:

21122211 SSSS

Output stability circles for conditionally stable device. (a) |S11| < 1 (b) |S11| > 1

If the device is unconditionally stable, the stability circles must be

completely outside (or totally enclose) the Smith chart.

11

11

22

11

SRC

SRC

SS

LL

The output stability circles:

22

22

2112

22

22

1122

S

SSR

S

SSC

L

L

The input stability circles:

22

11

2112

22

11

2211

S

SSR

S

SSC

S

S

Stability Test

Rollet’s condition: 12

1

2112

22

22

2

11

SS

SSK

The Auxiliary condition: 121122211 SSSS

The μ test: 11

21121122

2

11

SSSS

S


Power Amplifiers Stability Example

The S parameters for the HP HFET-102 GaAs FET at 2 GHz with a bias

voltage of Vgs = 0 are given as follow (Z0 = 50 Ohm):

S11 = 0.894 < -60.6, S21 = 3.122 < 123.6, S12 = 0.020 < 62.4, S22 = 0.781 < -27.6

Determine the stability of this transistor using the K- test and the μ test,

and plot the stability circles on the Smith Chart



1696.021122211

SSSS

1607.02

1

2112

22

22

2

11

SS

SSK

For the K- test:

For the μ test: 186.01

21121122

2

11

SSSS

S

which indicates potential instability

Calculation for the input and output stability circles:

50.0

47361.1

22

22

2112

22

22

1122

S

SSR

S

SSC

L

L

Input stability circle and radius

199.0

68132.1

22

11

2112

22

11

2211

S

SSR

S

SSC

S

S




Power Amplifiers Design Single Stage Transistor

Maximum power transfer from the input matching network to the transistor and the maximum power transfer from the transistor to the output matching network will occur when:

Then, assuming lossless matching sections, these conditions will maximize the overall transducer gain:

Bilateral transistor case:

Γin is affected by Γout, and vice versa, so that the input and output sections must be matched simultaneously

Lout

Sin

2

22

2

2

2121

1

1

1max

L

L

S

T

SSG

S

SL

L

LS

S

SSS

S

SSS

11

211222

22

211211

1

1

The solution is:

2

2

2

2

22

1

2

1

2

11

2

4

2

4

C

CBB

C

CBB

L

S

11222

22111

22

11

2

222

22

22

2

111

1

1

SSC

SSC

SSB

SSB

When S12 = 0, it shows that ΓS = S11* and

ΓL = S22*, and the maximum transducer

gain for unilateral case:

When the transistor is unconditionally

stable, K > 1, and the max transducer

power gain can be simply re-written as:

12

12

21

max KK

S

SGT

The maximum stable gain with K = 1:

12

21

S

SGmsg




Power Amplifiers Design Single Stage Transistor

Unilateral transistor case:

|S12| is small enough to be ignored.

Error in the transducer gain caused by approximating |S12| as zero is given by the ratio GT/GTU

Where U is defined as the unilateral figure of merit

22 )1(

1

)1(

1

UG

G

UTU

T

)1)(1(2

22

2

11

22112112

SS

SSSSU


Power Amplifiers Design Example

Design an amplifier for a maximum gain at 4.0 GHz. Calculate the overall transducer gain, GT, and the

maximum overall transducer gain GTmax. The S parameters for the GaAs FET at 4 GHz given as follow (Z0

= 50 Ohm): S11 = 0.72 < -116o, S21 = 2.60 < 76o, S12 = 0.03 < 57o, S22 = 0.73 < -68o



Determine the stability of this transistor using the K- test

162488.021122211

SSSS

195.12

1

2112

22

22

2

11

SS

SSK

Since || < 1 and K > 1, the transistor is unconditionally stable

at 4.0 GHz.

For the maximum gain, we should design the matching sections for a conjugate match to the

transistor. Thus, ΓS = Γin* and ΓL = Γout*, ΓS and ΓL can be determined

from:

61876.02

4

123872.02

4

2

2

2

2

12

1

2

1

2

21

C

CBB

C

CBB

L

S

The effective gain factors can calculated as:

dBSG 30.876.62

210

dBS

GS

20.617.41

12

11

dBS

GL

L

L22.267.1

1

12

22

2

So the overall maximum transducer gain will be;

dBGT 72.1622.230.820.6max


Power Amplifiers Design Example

An FET is biased for minimum noise figure, and has the following S parameters at 4 GHz: S11 = 0.60 < -60o,

S21 = 1.90 < 81o, S12 = 0.05 < 26o, S22 = 0.50 < -60o. For design purposes, assume the device is unilateral

and calculate the max error in GT resulting from this assumption.

Compute the unilateral figure of merit: Then the ratio of GT/GTU is bounded as: In dB, this is:

Thus we should expect less than about ± 0.5 dB error in gain.



059.0)1)(1(

2

22

2

11

22112112

SS

SSSSU

22 )1(

1

)1(

1

UG

G

UTU

T

130.1891.0 TU

T

G

G

dBGGTUT

53.050.0




Power Amplifiers Design Constant Gain Circles

It is desirable to design For less than the max obtainable gain (mismatches are

purposely introduced to reduce the overall gain),

To improve bandwidth or/and

To obtain a specific value for an amplifier gain.

Procedure is facilitated by plotting constant gain circles on the Smith Chart (represents loci of ΓS and ΓL,

that give fixed values of GS and GL). Unilateral device (for simplicity sake)

Input constant gain circles:

2

11

2

11

2

11

11

11

11

11

Sg

SgR

Sg

SgC

S

S

S

S

SS

Output constant gain circles:

2

22

2

22

2

22

22

11

11

11

Sg

SgR

Sg

SgC

L

L

L

L

LL

Normalized gain factors gS and gL

)1(1

1 2

112

11

2

max

SSG

Gg

S

S

S

S

S

)1(1

1 2

222

22

2

max

SSG

Gg

L

L

L

L

L

0 ≤ gS ≤ 1, and 0 ≤ gL ≤ 1. A fixed value of gS and gL

represents circles in the ΓS and ΓL planes.

The expression for the GS and GL

2

22

2

1

1

L

L

L

SG

2

11

2

1

1

S

S

S

SG

These gains are maximized when

ΓS = S11* and ΓL = S22*

2

111

1max

SG

S

2

221

1max

SG

L



Design an amplifier to have a gain of 11 dB at 4 GHz. Plot constant gain

circles for GS = 2 dB and 3 dB; and GL = 1 dB and 0 dB. The FET has the

following S parameters (Z0 = 50 Ω):

S11 = 0.75 < -120o S21 = 2.50 < 80o S12 = 0.00 < 0o S22 = 0.60 < -85o

Since S12 = 0 and |S11| < 1 and |S22| < 1,

the transistor is unilateral and unconditionally stable

The gain of the mismatched transistor is:

So the max unilateral transducer gain is:

Thus we have 2.5 dB more available gain than required by specs, since the design

only requires 11 dB gain.



dBS

GS

6.329.21

12

11

max

dB

SG

L9.156.1

1

12

22

max

dBSG 0.825.62

210

dBGUT

5.130.89.16.3max

166.011

11

120706.011

2

11

2

11

2

11

11

Sg

SgR

Sg

SgC

S

S

S

S

S

S875.0max

S

S

S

G

Gg 640.0

max

L

L

L

G

Gg

440.011

11

70440.011

2

22

2

22

2

22

22

Sg

SgR

Sg

SgC

L

L

L

L

L

L

For condition 1 (input side), when GS = 3 dB: For condition 1 (output side), when GL = 0 dB:

Condition 1: GS = 3 dB and GL = 0 dB Condition 2: GS = 2 dB and GL = 1 dB

(Note that these conditions must happens at the same time in order to keep the gain at 11 dB.)




Power Amplifiers Design Low Noise

It is often required to have a preamplifier with as low a noise figure as possible.

Generally it is not possible to obtain both minimum noise figure and maximum gain for an amplifier, so some sort of compromise must be made.

This can be done by using constant gain circles and circles of constant noise figure to select a usable trade of between noise figure and gain.

2

min optS

S

N YYG

RFF

For a fixed noise figure, F, the noise figure parameter, N, is given as:

2

0

min 14

opt

N ZR

FFN

The circles of constant noise figure:

1

1

1

2

N

NNR

NC

opt

F

opt

F



An GaAs FET amplifier is biased for minimum noise figure and has the

following S-parameters (Z0 = 50 Ω):

S11 = 0.75 < -120 S21 = 2.50 < 80 S12 = 0.00 < 0 S22 = 0.60 < -85 RN = 20 Ω

Γopt = 0.62 < 100 Fmin = 1.6 dB. For design purposes assume the unilateral.

Then design an amplifier having 2.0 dB noise figure with the max gain that

is compatible with this noise figure.

Compute the center and radius of the 2 dB noise figure circle:



0986.014

2

0

min

opt

NZR

FFN

24.0

1

1

10056.01

2

N

NNR

NC

opt

F

opt

F

The noise figure circle is plotted in the figure.

Minimum noise figure (Fmin = 1.6 dB) occurs for ΓS

= Γopt = 0.62<100o

It can be seen that GS = 1.7 dB gain circle just

intersects the F = 2.0 dB noise figure circle, and any

higher gain will result in a worse noise figure.



An GaAs FET amplifier is biased for minimum noise figure and has the

following S-parameters (Z0 = 50 Ω):

S11 = 0.75 < -120 S21 = 2.50 < 80 S12 = 0.00 < 0 S22 = 0.60 < -85 RN = 20 Ω

Γopt = 0.62 < 100 Fmin = 1.6 dB. For design purposes assume the unilateral.

Then design an amplifier having 2.0 dB noise figure with the max gain that

is compatible with this noise figure.

For the output section we choose ΓL = S22* = 0.5<60o for a max GL of:



The noise figure circle is plotted in the figure.

Minimum noise figure (Fmin = 1.6 dB) occurs for ΓS

= Γopt = 0.62<100o

It can be seen that GS = 1.7 dB gain circle just

intersects the F = 2.0 dB noise figure circle, and any

higher gain will result in a worse noise figure.

dBS

GL

25.133.11

12

22

dBSG 58.561.32

210

dBGGGGLSTU

53.80max

ANTENNAS and MICROWAVES ENGINEERING (650427) · 2018-04-02 · a linear two port network by...

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Transcript of ANTENNAS and MICROWAVES ENGINEERING (650427) · 2018-04-02 · a linear two port network by...