Network Analysis and Synthesis

Chapter 4Synthesis of deriving point functions

(one port networks)

Elementary Synthesis procedures

• The basic philosophy behind the synthesis of driving-point functions is to break up a positive real (p.r.) function Z(s) into a sum of simpler p.r. functions Z1(s), Z2(s) . . . Zn(s).

• Then to synthesize these individual Zi(s) as elements of the overall network whose dp impedance is

)(...)()()( 21 sZsZsZsZ n

Breaking up process

• One important restriction is that all Zi(s) must be positive real.

• If we were given all the Zi(s), we could synthesize a network whose driving point impendance is Z(s) by simply connecting the Zi(s) in series.

• However, if we were to start from Z(s) alone, how do we decompose Z(s) into Zi(s)?

)(

)(

...

...)(

011

1

011

1

sQ

sP

bsbsbsb

asasasasZ

mm

mm

nn

nn

Removing a pole at s=0

• If there is a pole at s=0, we can write Q(s) as

• Hence, Z(s) becomes

• Z1(s) is a capacitor.

• We know Z1(s) is positive real, is Z2(s) positive real?

)()( ssGsQ

)()(

)()(

21 sZsZ

sRs

DsZ

Is Z2(s) positive real?

• The poles of Z2(s) are also poles of Z1(s), hence, Z2(s) doesn’t’ have poles on the right hand side of the s plane and no multiple poles on the jw axis.– Satisfies the first 2 properties of p.r. functions.

• What about Re(Z2(jw))?

• Since Z(s) is p.r. Re(Z2(jw))=Re(Z(jw))>0.

• Hence, Z2(s) is p.r.

)(Re

)(Re)(Re)()(Re)(Re

2

2121

jwZ

jwZjwZjwZjwZjwZ

Removing a pole at s=∞

• If Z(s) has a pole at s=∞, we can write Z(s) as

• Using a similar argument as previous we can show that Z2(s) is p.r.

• Z1(s) is an inductor.

)()(

)()(

21 sZsZ

sRLssZ

Removing complex conjugate poles on the jw axis.

• If Z(s) has complex conjugate poles on the jw axis, Z(s) can be expanded into

• Note that

• Hence, • Z2(s) is p.r.

)(2

)( 221

2sZ

s

KssZ

02

Re 21

2

s

Kjw

0))(Re()(Re 2 sZsZ

Removing a constant K

• If Re(Z(jw)) is minimum at some point wi and if Re(Z(jw)) = Ki as shown in the figure

• We can remove that Ki as

• Z2(s) is p.r.• This is essentially removing

a resistor.

)()( 2 sZKsZ i

Constructing

• Assume that using one of the removal processes discussed we expanded Z(s) into Z1(s) and Z2(s).

• We connect Z1(s) and Z2(s) in series as shown on the figure.

Example 1

• Synthesize the following p.r. function

• Solution:– Note that we have a pole at s=0. Lets remove it

– Note that 2/s is a capacitor, while s/(s+3) is a parallel connection of a resistor and an inductor.

)3(

62)(

2

ss

sssZ

3

2)(

0,1,23

)(

s

s

ssZ

CBAs

CBs

s

AsZ

• 2/s is a capacitor with C=1/2.• While s/(s+3) is a R=1 connected in parallel

with an inductor L=1/3.

Example 2

• Synthesis the following p.r. function

• Solution– Note that there are no poles on s=0 or s=∞ or jw

axis.– Lets find the minimum of Re(Y(jw))

42

27)(

s

ssY

2

2

2

2

2

88

144

1616

288

1616

4472Re

44

27Re)(Re

w

w

w

w

w

wjwj

jw

jwjwY

• Note that minimum of Re(Y(jw))=1/2.• Lets remove it

• ½ is a conductance in parallel with Y2(s)=

• Note that Y2(s) is a conductance 1/3 in series with an inductor 3/2.

2

3

2

1)(

s

ssY

2

3

ss

Exercise

• Synthesize the following p.r. function.

ss

ssssZ

36

1336)(

3

23

Synthesis of one port networks with two kinds of elements

• In this section we will focus on the synthesis of networks with only L-C, R-C or R-L elements.

• The deriving point impedance/admittance of these kinds of networks have special properties that makes them easy to synthesize.

1. L-C imittance functions

• These networks have only inductors and capacitors.

• Hence, the average power consumed in these kind of networks is zero. (Because an inductor and a capacitor don’t dissipate energy.)

• If we have an L-C deriving point impedance Z(s)

)()(

)()()(

22

11

sNsM

sNsMsZ

M1 and M2 even partsN1 and N2 odd parts

• The average power dissipated by the network is

even

oddsZ

odd

evensZ

sN

sMsZ

sM

sNsZ

sNsMsNsM

sNsNsMsM

sNsM

sNsNsMsM

jwZ

jwIjwZPowerAverage

)(or )(

)(

)()(or

)(

)()(

)(0)(or )(0)(

0)()()()(

)()(

)()()()(

0)(Re

0)()(Re2

1

2

1

2

1

1221

2121

22

22

2121

2

Properties of L-C function

1. The driving point impedance/admittance of an L-C network is even/odd or odd/even.

2. Both are Hurwitz, hence only simple imaginary zeros and poles on the jw axis.

3. Poles and zeros interlace on the jw axis.4. Highest power of the numerator and

denominator may only differ by 1.5. Either a zero or a pole at origin or infinity.

Synthesis of L-C networks

• There are two kinds of network realization types for two element only networks.– Foster and– Cauer

Foster synthesis

• Uses decomposition of the given F(s) into simpler two element impedances/admittances.

• For an L-C network with system function F(s), it can be written as

• This is because F(s) has poles on the jw axis only.

...2

)( 220

i

i

s

sKsK

s

KsF

• Using the above decomposition, we can realize F(s) as

For a driving point impedance

For a driving point admittance

Example

• Synthesize as driving point impedance and admittance.

Solution: – Decompose F(s) into simpler forms

4

912)(

2

22

ss

sssF

2

15,

2

9,2

4

2)(

10

210

KKK

s

sKsK

s

KsF

• For driving point impedance

• For driving point admittance

Cauer synthesis

• Uses partial fraction expansion method.• It is based on removing pole at s=∞.

• Since the degree of the numerator and denominator differ by only 1, there is either a pole at s=∞ or a zero at s=∞.– If a pole at s=∞, then we remove it.– If a zero at s=∞, first we inverse it and remove the

pole at s=∞.

)(

)()(or

)(

)()(

2

1

2

1

sN

sMsZ

sM

sNsZ

• Case 1: pole at s=∞– In this case, F(s) can be written as

...1

1

)()(

1)(

Hence,

1)( ofOrder )(M ofOrder

,)(

)()(

21

3

2

32

2

3

sKsK

sK

sNsM

sKsF

sNs

sM

sNsKsF

• This expansion can easily be realized as

• Case 2: zero at s=∞– In this case will have a pole at s=∞.

– We synthesize G(s) using the procedure in the previous step.

– Remember that if F(s) is an impedance function, G(s) will be an admittance function and vice versa.

)(

1)(

sFsG

Example

• Using Cauer realization synthesize

Solution:– This is an impedance function.– We have a pole at s=∞, hence, we should remove

it.

34

16122)(

24

35

ss

ssssZ

ss

ssssZ

4

1042)(

4

3

• If we were given Y(s) instead our realization would be

R-C driving point impedance/ R-L admittance

• R-C impedance and R-L admittance driving point functions have the same properties.

• By replacing the inductor in LC by a resistor an R-C driving point impedance or R-L driving point admittance, it can be written as

• Where ...)( 10

is

KK

s

KsF

resistorsRepresent ,...,K

admittance L-Rfor inductor and impedance C-Rfor Capaictors ,...1

,1

0

i

i

i

K

KK

Properties of R-C impedance or R-L admittance functions

1. Poles and zeros lie on the negative real axis.2. The singularity nearest origin must be a pole

and a zero near infinity.3. The residues of the poles must be positive

and real.4. Poles and zeros must alternate on the

negative real axis.

Synthesis of R-C impedance or R-L admittance

• Foster– In foster realization we decompose the function

into simple imittances according to the poles. That is we write F(s) as

– For R-C impedance

...)( 10

is

KK

s

KsF

• For R-L admittance

Example

• Synthesize as R-C impedance and R-L admittance in foster realization.

Solution:– Note that the singularity near origin is a pole and a

zero near infinity.– The poles and zeros alternate– We can expand F(s) as– R-C impedance

)3(

)4)(2(3)(

ss

sssF

33

18)(

ss

sF

• R-L admittance

• Cauer realization– Cauer realization uses continued fraction expansion.– For R-C impedance and R-L admittance we remove a

resistor first.– Then invert and remove a capacitor– Then invert and remove a resistor . . .

Example

• Synthesize using Cauer realization as R-C impedance and R-L admittance.

Solution:– Note that the singularity near origin is a pole.– The singularity near infinity is a zero.– The zeros and the poles alternate.

– Note that the power of the numerator and denominator is equal, hence, we remove the resistor first.

)3(

)4)(2(3)(

ss

sssF

F(s) is R-L impedance or R-C admittance

For R-C impedance For R-L admittance

R-L impedance/R-C admittance

• R-L impedance deriving point function and R-C admittance deriving point function have the same property.

• If F(s) is R-L impedance or R-C admittance, it can be written as

...)( 0

i

i

s

sKKsKsF

resistorsRepresent ,...,K

admittance C-Rfor Capacitors and impedance L-Rfor Inductors ,...1

,1

0i

i

i

K

KK

Properties of R-L impedance/R-C admittance

1. Poles and zeros are located on the negative real axis and they alternate.

2. The nearest singularity near origin is zero. The singularity near infinity is a pole.

3. The residues of the poles must be real and negative.• Because the residues are negative, we can’t use

standard decomposition method to synthesize.

Synthesis of R-L impedance and R-C admittance

• Foster– If F(s) is R-L impedance d.p or R-C admittance d.p

function. We can write it as

– Because of the third property of R-L impedance/R-C admittance d.p. functions, we can’t decompose F(s) into synthesizable components with the way we were using till now.

– We have to find a new way where the residues wont be negative.

...)( 0

i

i

s

sKKsKsF

• If we divide F(s) by s, we get

• Note that this is a standard R-C impedance d.p. function, hence, the residues of the poles of F(s)/s will be positive.

• Once we find Ki and σi we multiply by s and draw the foster realization.

...)( 0

i

i

s

KK

s

K

s

sF

Example

• Synthesize as R-L impedance and R-C admittance using Foster realization.

Solution:– Note that the singularity near origin is a zero.– The singularity near infinity is a pole.– The zeros and the poles alternate.

)6)(2(

)3)(1(2)(

ss

sssF

F(s) is R-L impedance or R-C admittance

• We divide F(s) by s.

64

5

24

1

21)(

sby gmultiplyin Then

64

5

24

12

1

)6)(2(

)3)(1()(

s

s

s

ssF

sss

sss

ss

s

sF

• R-L impedance

• R-C admittance

• Cauer realization– Using continued fractional expansion – We first remove R0. To do this we use fractional

expansion method by focusing on removing the lowest s term first.

– We write N(s) and M(s) starting with the lowest term first.

Example

• Synthesize as R-L impedance and R-C admittance using Cauer realization.

• Solution:

– We write P(s) and M(s) as

)6)(2(

)3)(1(2)(

ss

sssF

2

2

812)(

286)(

sssQ

sssP

)(

)()(

sQ

sPsF

• R-L impedance

• R-C admittance