Network Analysis and Synthesis

Chapter 6Synthesis of two port networks

5.1 Introduction

• In this chapter we will discuss 2 of the most widely used two port synthesis methods– Coefficient matching and – Darlington method (insertion loss method)

5.2 Coefficient matching• It is one of the most simple and effective method

for two port synthesis.• In this method,

– We compare the transfer function we want to synthesize with the transfer function of a circuit we already know.

– If they have the same form, we try to match the coefficients of the two functions. (Remember that the coefficients of the second transfer function is a function of the circuit elements.)

– Once we match the coefficients, we know the values of the elements in the circuit.

Example 1

• The voltage transfer function of the following circuit is

R1-source impedanceR2-load impedance

• If we are required to synthesize the following voltage transfer function.

• Comparing the two equations and assuming equal termination and normalizing the resistors to 1Ω, R1=R2=1Ω

12

1)(

2

ssKsT

2 and 2 ,1R

arefunction transfer normalized theof valueselemental theHence,

2 and 2

211

2

11

21

CLR

CL

LCLC

Example 2

• Consider the following network

• The voltage transfer function is

21

21

21

2

212121

21

1122

23

21

and

11

11

111

)(

RR

RRR

RR

RK

RCLCRRCC

CCL

sCRCR

ss

RCLCK

sT

p

p

p

• If we were to synthesize the following voltage transfer function assuming equal termination with 1Ω resistor.

hLfCC

Solving

CCCCL

CCRCLCK

ssssT

p

2,1

21111

211

,11

,2

1,

2

1R

on terminatiequal assuming and equations two thecompairingBy 122

1)(

21

2121

2121p

23

Pros and Cons of Coefficient matching

• Pros– Simple– Effective

• Cons– When the order of the transfer function increases,

the number of simultaneous equations we have to solve for increases.

– Doesn’t demonstrate sophisticated network design methods.

5.3 Insertion Loss (Darlington Method)

• A low pass characteristics can be obtained by using RC, RL or LC networks.

• Low cost, low sensitivity to component variations and simplicity of design make LC two port networks the most widely used filter networks.

• Here the network is assumed to be doubly terminated. (A valid assumption in almost all cases.)

• A powerful method for designing doubly terminated LC two port networks is the Darlington method.

• It is one of the most effective method of realizing a two port network: Insertion loss method (Darlington method).

• In Darlington method of filter design, – The specifications of the insertion loss of the filter

is converted to the reflection coefficients (related to the maximum power that can be delivered by the source vs. the actual delivered power to the load) of the filter.

– From the reflection coefficients the driving point impedance of the terminated networks is obtained.

– Then this driving-point impedance is developed into resistively terminated LC ladder network.

Procedures of Darlington synthesis

• The derivation of Darlington method is complicated, hence, we will just discuss the procedure for using the Darlington method.

• Procedure1. From F(s) obtain the reflection coefficient p(s)

)(

)()(

14

)(

)()(

41)()(

1

1

221

21

221

21

sq

sps

RR

RR

sq

spsF

RR

RRss

o The zeros of q1(s) are the left plane zeroes of q(s).o The zeros of p(s) are equally distributed between the

zeros of ρ (s) and ρ (-s), with restrictions that conjugate zeros must be together.

2. From ρ(s) determine the normalized Z(s).

3. Expand Z(s) into continued fraction expansion about infinity and obtain the ladder. The two impedances defined above lead to 2 (dual) ladders, one terminated with R2 and the other 1/R2.

)(1

)(1)(or

)(1

)(1)(

s

ssZ

s

ssZ

Example 3• Synthesize the following voltage transfer function using

Insertion loss method

• Solution:– To find the reflection coefficient

– The zeros of q(s) are z1=1, z2=-1, z3=-0.5+j0.866, z4=-0.5-j0.866, z5=0.5+j0.866, z6=0.5-j0.866

– Hence, poles of ρ(s) are z2=-1, z3=-0.5+j0.866, z4=-0.5-j0.866

1 ,1

1)( 216

RRs

sF

1)(

41)()(

6

6

221

21

s

ssF

RR

RRss

– The zeros of p(s) are 6 multiple poles at s=0.– Hence, the zeros of ρ(s) are 3 multiple poles at

s=0.– Hence

– The driving point impedance is

122)(

866.05.0866.05.01)(

23

3

3

sss

ss

jsjss

ss

122

1222

)(1

)(1)(

2

23

ss

sss

s

ssZ

– Using continued fraction expansion

Example 4

• Synthesize the following voltage transfer function using Insertion loss method with R1=R2=1Ω

• Solution:– To find the reflection coefficient

– The zeros of p(s) are 8 multiple poles at s=0, hence the zeros of ρ(s) are 4 multiple poles at s=0.

81

1)(

ssF

1)(

41)()(

8

8

221

21

s

ssF

RR

RRss

(Refer at the end for a more detailed explanation on how to get roots of polynomials of the form sn+a)

– The zeros of q(s) are evenly distributed on the unit circle on the s plane.

– The angle between the two zeros is

– Since no zero on real axis or jw planeand because the zeros of q(s) have to be conjugate complex,the angle of one of the roots from the real axis should be equalfor two conjugate roots.

00

458

360

• The zeros of q(s) are then

• The roots of ρ(s) are the left hand zeros of q(s)

00000000

00000000

5.22sin5.22cos,5.67sin5.67cos,5.67sin5.67cos,5.22sin5.22cos

5.22sin5.22cos,5.67sin5.67cos,5.67sin5.67cos,5.22sin5.22cos

jjjj

jjjj

924.0383.0 and 383.0924.0

67.5 and 22.5 of sine and cosine theComputing

5.22sin5.22cos,5.67sin5.67cos

5.22sin5.22cos,5.67sin5.67cos

4,32,1

00

0000

0000

jpjp

jj

jj

• ρ(s) becomes

• The driving point impedance becomes:

1613.2414.3163.2)(

924.0383.0924.0383.0383.0924.0383.0924.0)(

234

4

4

ssss

ss

jsjsjsjs

ss

1613.2414.3163.2

1613.2414.3163.22)(

)(1

)(1)(

23

234

sss

sssssZ

s

ssZ

• Taking the continuous fraction expansion

• Hence, the network becomes

Finding roots of polynomials of the form sn+a

• The roots will be located on the circle with radius of a1/n on the s plane.

• Separation between them is

• Check if the polynomial has root at s=a1/n or s=-a1/n or s=ja1/n or s=-ja1/n.

• Case 1: root at one of these– Start at that root and plot

the roots with separation of θ.

n

0360

• Case 2: no root at those locations– Because the roots have to be conjugate complex,

the angle one of the roots makes with the real axis will be equal to the negative angle of the conjugate root.

– Hence, one of them will be

– Start from this root and plot the restwith separation angle of θ.

2sin

2cos1

jr

Example

• What are the roots of s8+1• Separation between the roots is

• Root on one of the axis? No• Hence one of the root is

• Plot the rest starting from this one.

00

458

360

2

45sin

2

45cos

00

1 jr

Example• s6-1• The separation angle is

• Root at one of the axis? Yes at r1=1.• Hence, plot the rest with separation of 600

starting from r1=1.• The roots are then

00

606

360

006,5

004,32,1 60sin60cos,60sin60cos,1 jrjrr