12. Network Synthesis and Readability – Network Analysis and Synthesis

12. Network Synthesis and Readability

CHAPTER OBJECTIVES

After carefully studying this chapter, you should be able to do the following:

Explain the concept ofnetwork synthesis.

Explain the Hurwitz conditions for stability.

Test polynomials for stability applying Hurwitz stability criteria.

State the properties of a positive real function.

Test a network function to find whether it is a positive real function and hence realizable.

Synthesize networks by Foster method.

Synthesize networks by Cauer method.

From the driving point impedance or admittance function realize the network in Foster and Cauer form I and II, respectively.

12.1 INTRODUCTION

Network analysis deals with finding out the output response, using various techniques, when the excitation signal (input signal) and the network are known.

Network synthesis deals with the realisation of the network from the given excitation and output response.

The network synthesis provides alternate solutions. The network synthesis technique is used in filter design where computations, innovations and judgement are required. In the network synthesis, we first determine the driving point function in impedance and admittance form. The driving point admittance/immittance is given by the following form:

where R(s) is the output response and E(s) is the excitation response.

There are two basic considerations to be taken care of. These are: (i) realisability and (ii) stability.

It is determined first whether the function is realisable or not; that is, to find out whether it is possible to obtain a passive network. This is termed as realisability. For stability, there are certain conditions that must be satisfied by the network function.

A network function is tested first for its realisability. It is then tested for its stability. After that, we proceed to realize the network using Foster and Cauer methods.

In this chapter, we will examine the network function for their stability condition. We will also explain the process of realisation of the network, when the driving point network function is given.

12.2 HURWITZ CONDITIONS FOR STABILITY

The driving point impedance or admittance is expressed as the ratio of two polynomials, P(s) and Q(s). These polynomials are tested for Hurwitz stability conditions. If they satisfy the conditions, then we say that they are Hurwitz.

Consider a polynomial:

Then, for the polynomial to be Hurwitz, it has to satisfy the following Hurwitz’s conditions for stability:

Condition1: All the coefficients a0, a1, a2an must be positive.
Condition2: No powers of ‘s’ should be absent between the highest and lowest degree term of ‘s’ unless all the even or all the odd terms are missing. For example, P(s) = s5 + 3s3 + 2s2 + 3s + 1 is not Hurwitz as the term s4 is missing. The polynomial P(s) = s5 + 2s3 + 3s is Hurwitz as all coefficients are positive and all the even terms are missing.
Condition3: The continued fraction expansion of the ratio of the odd to even parts or even to odd parts of Hurwitz polynomial gives all positive quotient terms.

Now, consider a polynomial P(s) = s4 + s3 + 2s2 + 3s + 2.

The even parts of the polynomial P(s), that is, e(s) = s4 + 2s2 + 2,

while the odd parts of the polynomial P(s), that is, e(o) = s3 + 3s.

By continued fraction expansion, if it is seen that all the quotient terms are positive, then the given polynomial P(s) is Hurwitz.

If a polynomial satisfies the conditions of Hurwitz for stability, we say that the polynomial must be Hurwitz. The testing of a polynomial for satisfying the conditions of Hurwitz is further explained through examples.

Example 12.1 Check whether the given polynomial is Hurwitz or not

Solution: Given

All the coefficients (that is, s4, s3, s2 and s are positive). There are no terms missing between the highest and lowest degree terms. Both the conditions, that is, condition 1 and condition 2 are satisfied.

For condition 3 we will perform the continued fraction expansion.

The continued fraction expansion of is given by the following:

Since all the quotient terms are not positive, the given polynomial is not Hurwitz.

Example 12.2 Test if the polynomial s3 + 6s2 + 12s + 8 is Hurwitz.

Solution: Given

We have to test for stability with respect to all the three conditions namely

1.All the coefficients must be positive.

2.No powers of s should be absent and

3.Continued fraction expansion should give all positive quotient terms

The given polynomial satisfies the first two conditions. We will test for the third condition. Now, even part of P(s) = 6s2 + 8 = e(s) and odd part of P(s) = s3 + 12s = o(s)

We will use the continued fraction expansion as follows:

Since all the quotient terms of are positive, the given polynomial is Hurwitz.

Example 12.3 Test if the polynomial s4 + 8s2 + 32 is Hurwitz.

Solution: In this case, only even part is given in the following:

To get odd part, we will differentiate even part e(s) as follows:

o(s) = e′(s) = 4s3 + 16s. We find continued fraction expansion of even to odd parts as

Since all powers of s are not present and all the quotient terms are not positive, and the polynomial is not Hurwitz.

Example 12.4 Test whether the polynomial s5 + 5s3 + 4s is Hurwitz.

Solution: In this case, only odd part is given.

That is,

o(s) = s5 + 5s3 + 4s

To get the even part, we will differentiate the odd part.

That is,

e(s) = o′(s) = 5s4 + 15s2 + 4

The continued fraction expansion is given by the following:

Since all the quotient terms are positive, the polynomial is Hurwitz.

Example 12.5 Find the limits of K so that the polynomial s3 + 14s2 + 56s + K may be Hurwitz.

Solution: By inspection, we see that the first two conditions are satisfied.

Odd part of the given polynomial, that is, o(s) = s3 + 56s

Even part of the given polynomial, that is, e(s) = 14s2 + K

The continued fraction expansion is given by the following:

Now, for the polynomial to be Hurwitz, quotient terms should be positive.

This gives no value for K.

Therefore, the limit of K is 0 < K < 784.

Example 12.6 Test whether the following polynomial is Hurwitz or not.

Solution: Given

Let us first check for P(s):

The continued fraction expansion is given by the following:

Since all the quotient terms of P(s) are positive, P(s) is Hurwitz.

Now, let us check for Q(s):

Even part of Q(s) = s4 + 8s2 + 6 and Odd part of Q(s) = 5s3 + 9s

The continued fraction expansion is carried out as follows:

All the quotient terms of Q(s) are also positive, and therefore, Q(s) is also Hurwitz.

Since P(s) and Q(s) both are Hurwitz, is also Hurwitz.

12.3 PROPERTIES OF POSITIVE REAL FUNCTIONS

Positive real functions (PRF) represent physically realisable networks.

The driving point impedance function Z(s) and driving point admittance function Y(s) of a network can be expressed as the ratio of two polynomials.

If is an immittance function, it is a PRF if the following conditions are satisfied:

1.All the coefficients of P(s) and Q(s) are real and positive.

2.Q(s) is a Hurwitz polynomial.

3.All the terms in the numerator and denominator polynomial are present. There is no missing term.

4.All the poles and zeros of F(s) lie in the left half of the s-plane.

Testing of a function to be positive real or not is illustrated through a few examples.

The necessary and sufficient conditions for a rational network function F(s) to be positive real (that is, physically realisable) are stated as follows:

Condition 1: F(s) must not have any pole in the right-hand side of s-plane; that is, both P(s) and Q(s) of are Hurwitz.

This condition can be checked by continued fraction expansion of the odd to even parts or even to odd parts of P(s) and Q(s).

Condition 2: real part of F(s), that is,

Re[F()] ≥ 0, for all ω

Where M1(s) and N1(s) are even and odd parts of P(s) respectively. M2(s) and N2(s) are the even part and odd part of Q(s) respectively.

Condition 3: This condition is tested when the poles of F(s) lie on the -axis. This is done by using the partial fraction expansion of F(s) to check whether the residues of the poles on the -axis are positive and real.

Example 12.7 Determine whether the function is a positive real functions (PRF) and hence realisable.

Solution: Given

The following necessary conditions are first applied to the given function as

1.All the coefficients of P(s) and Q(s) are real and positive.

2.All the terms in the numerator and denominator polynomial are present. There is no missing term.

Hence, the function F(s) may be a PRF.

To check whether it is truly PRF or not, we will check for the sufficient conditions.

Step 1: firstly, we will check whether F(s) is Hurwitz or not

Let us consider P(s) as follows:

The continued fraction expansion is carried out for P(s) and Q(s) as follows:

Since all the quotient terms of P(s) are positive, P(s) is Hurwitz.

Now, let us consider Q(s) as in the following:

Even part of Q(s) = s2 + 14 and the odd part of Q(s) = 9s

The continued fraction expansion is as follows:

All the quotient terms of Q(s) are positive. Therefore, Q(s) is also Hurwitz.

Since P(s) and Q(s) are both Hurwitz, is also Hurwitz.

Step 2: to check whether Re|F()| ≥ 0, for all ω, that is,

The even part is M1(s) = s2 + 5 and its odd part is N1(s) = 6s

The even part is M2(s) = s2 + 14 and its odd part is N2(s) = 9s

Since A(ω2) ≥ 0 for all ω, the given function is a positive real functions (PRF) and the function is physically realisable network.

Example 12.8 Test whether is a positive real functions (PRF) and represents a physically realisable network.

Solution:

Step 1: check whether F(s) is Hurwitz or not

Now,

Now, consider P(s) as in the following:

The even part and its odd part

Continued fraction expansion is as follows:

Since all the quotient terms are positive, P(s) is Hurwitz.

Now, take

Q(s) = s2 + s + 4

The even part is M2(s) = s2 + 4 and its odd part is N2(s) = s

Continued fraction expansion is given as follows:

Quotient terms are positive ad Q(s) is also Hurwitz.

Since both P(s) and Q(s) are Hurwitz, and therefore, F(s) is also Hurwitz.

Step 2: consider

Since all terms in A(ω2) are not positive, we have to check whether A(ω2) ≥ 0 for all values of ω.

Let us apply Sturm test to check A(ω2) ≥ 0. (It is obtained by considering the derivative of P0(x).)

For this, we substitute ω2 = x, in (i) as

Then, P0(x) = x2 − 4x + 3

Derivative of P0(x), that is, P1(x) = 2x − 4

Now, dividing P0(x) by P1(x), we get the following:

Now, we find the sign changes of P0(x), P1(x), P2(x) as x changes from 0 to ∞

Since there are two sign changes, A(ω2) is always positive for all values of ω. The given function is not PRF.

(overall sign changes = 2 − 0 = 2)

Example 12.9 Test whether the function A(ω2) = 3ω4 − 12ω2 + 4 is positive real.

Solution: Given A(ω2) = 3ω4 − 12ω2 + 4. To check whether A(ω2) ≥ 0 for all values of ω, let us apply Sturm test.

Now, divide P0(x) by P1(x)

We find sign changes of P0(x), P1(x), P2(x) for x = 0 and x = ∞.

Since there are two sign changes, the function is not positive real.

(Overall sign changes = 2 − 0 = 2)

Example 12.10 Test whether is a positive real function.

Solution:

Step 1: let us check whether Y(s) is Hurwitz or not.

Consider P(s) as in the following:

Its odd part N1(s) = 2s3 + 3s and even part M1(s) = 2s2 + 2

Now,

Since all the quotients are positive, P(s) is Hurwitz.

Now, consider Q(s) = s2 + 1

Its even part M2(s) = s2 + 1 and there is no odd part; that is, N2(s) = 0

Therefore, odd part (derivative of M2(s)) = 2s

Since all the quotients are positive, Q(s) is also Hurwitz.

Now, both P(s) and Q(s) are Hurwitz, and therefore, Y(s) is also Hurwitz.

Step 2: let us find

Let us apply Sturm test to check whether A(ω2) ≥ 0 or not

Now, divide P0(x) by P1(x), we get the following:

Therefore, overall sign changes = 1 − 1 = 0.

Since there is no overall sign change, the given function is a PRF, that is, positive real functions.

12.4 SYNTHESIS OF NETWORKS BY FOSTER’S AND CAUER’S METHODS

The Foster’s method of network synthesis uses the partial fraction expansion of the driving point immittance function. Foster is of two types: Foster form-I and Foster form-II. When the driving point function is an impedance, it is referred to as Foster form-I. Foster form-II is used for driving point admittance function.

The Cauer’s methods employ continued expansion approach to synthesise a given immittance function. In Cauer form-I, the terms of both numerator and denominator are arranged in descending degree of s. In Cauer form-II, the terms in the numerator and denominator polynomials of the driving point immittance function are arranged in ascending order.

Foster and Cauer methods of network synthesis are presented in details.

12.5 FOSTER AND CAUER FORMS

12.5.1 Synthesis of R-C Network

The impedance Z(s) of R-C network is of the type as in the following:

Further, for R-C impedance function, Z(0) ≥ Z(∞).

12.5.2 Properties of the R-C Impedance or R-L Admittance Function

1.The poles and zeros of the ZRC(s) or YRL(s) lie on the negative real axis of the complex s-plane.

2.The poles and zeros are interlacing, that is, they alternate along the negative real axis.

3.The poles and zeros are simple. There are no multiple poles and zeros.

4.ZRC(0) > ZRC(∞)

12.5.3 Foster Form-I of R-C Network

It is obtained by using the partial fraction expansion of ZRC(s).

The partial fraction expansion of ZRC(s) is given as follows:

Accordingly, the Foster form-I of R-C network is shown in Figure 12.1.

Figure 12.1 Foster Form-I of R-C Network

12.5.4 Foster Form-II of R-C Network

It is obtained from the following:

Foster from-II of R-C network is shown in Figure 12.2.

Figure 12.2 Foster Form-II of R-C network

12.5.5 Cauer Forms of R-C Network

The R-C network for Cauer form-I and Cauer form-II is shown in Figure 12.3 (a) and (b), respectively. Note that in Cauer form-I resistors are in series and capacitors are in parallel. In Cauer form-II capacitors are in series while the resistors are in parallel. Method of calculation of their values will be illustrated through an example.

Figure 12.3 (a) Cauer Form-I of R-C Network and (b) Cauer Form-II of R-C Network

12.5.6 Synthesis of R-L Network

General driving point impedance of R-L network is given by the following:

In the case of R-L network, ZRL(∞) > ZRL(0).

12.5.7 Properties of R-L Impedance Function/R-C Admittance Function

1.The poles and zeros are simple. There are no multiple poles or zeros.

2.The poles and zero interlace (that is, alternate) each other along the negative real axis.

3.The poles and zeros lie on the negative real axis of the complex s-plane.

4.Z(∞) > Z(0)

12.5.8 Foster Form-I of R-L Network

The partial fraction expansion of ZRL(s) is given as follows:

Figure 12.4 shows the Foster form-I of R-L network.

Figure 12.4 Foster Form-I of R-L Network

12.5.9 Foster Form-II of R-L Network

Consider the foster form-II of R-L network shown in Figure 12.5.

The network is designed by using the partial fraction method. Here, we consider the partial fraction expansion of Y(s).

Figure 12.5 Foster Form-II of R-L Network

12.5.10 Cauer Form-I of R-L Network

Figure 12.6(a) shows the representation of cauer form-I for R-L network.

Figure 12.6(a) Schematic Diagram Shows Cauer Form-I of R-L Network

12.5.11 Cauer Form-II R-L Network

Figure 12.6(b) shows the representation of cauer form-II for R-L network.

Figure 12.6(b) Schematic Diagram Shows Cauer Form-II of R-L Network

12.5.12 Synthesis of L-C Networks

LC immittance is given by

12.5.13 Properties of L-C Immittance

1.The immittance function (that is, Z(s) or Y(s)) is a ratio of odd to even polynomial or even to odd polynomial.

2.The highest order terms in the numerator and denominator differ by one.

3.There is a pole or zero at origin

4.There is a pole or zero at infinity

5.The poles and zeros occur in complex conjugate pairs.

12.5.14 Foster Form-I of L-C Network

The partial fractions of ZLC(s) are given by

The Foster form-I of the L-C network is shown in Figure 12.7.

Figure 12.7 Foster Form-I of L-C Network

12.5.15 Foster Form-II of L-C Network

The Foster form-II for the L-C network is shown in Figure 12.8.

Figure 12.8 Foster Form-II of L-C Network

12.5.16 Cauer Form-I of L-C Network

The Cauer form-I of L-C network is shown in Figure 12.9(a) below.

Figure 12.9(a) Cauer Form-I of L-C Network

12.5.17 Cauer Form-II of L-C Network

Cauer form-II of the L-C network is shown in Figure 12.9(b).

Figure 12.9(b) Cauer Form-II of L-C Network

Example 12.11 Show whether the following driving point functions represent a L-C or a R-C network. Give reasons for your answer.

Solution:

For function 1:

In the above, all the poles and zeros are real, and therefore, it may be R-C or R-L impedance function. We have to check whether it is R-C or R-L. For this, we have to find Z ().

Substituting s = ∞, we get the following:

That is, Z(∞) = 1

Thus, Z(0) > Z(∞). Therefore, the given impedance function is of R-C type. The property of R-C impedance as stated in section 12.5.2, is that ZRC(0) > ZRC(∞).

For function 2:

Since Z(s) has quadratic factors and there is a difference of one between degree of numerator and denominator, it can be written as follows:

is of L-C type. (See properties of L-C impedance section in 12.5.13)

Example 12.12 Synthesise all the four forms of the R-C driving point function represented as

Solution: To synthesis the Foster form-I circuit, let us consider the given driving point function as

The partial fraction expansion of Z(s) will be as follows:

[∵ degree of numerator and denominator is same]

Let us find the values of A and B

To find A, substitute s = −1 in equation (12.2), we get

To find the value of B, substitute s = −3 in equation (12.2)

Now, substituting the values of A and B in equation (12.1), we get the following form:

In this case, term is missing, and therefore, C0 is absent. (For reference, see Figure 12.1) Term represents the parallel combination of R2 and C2.

General form of Foster form-I is represented as follows. The corresponding network is shown in Figure 12.10.

Figure 12.10

Term represents the parallel combination of R4 and C4.

Required Foster form-I is shown in Figure 12.11. Note that in the network, C0 is absent.

Figure 12.11

Foster form-II circuit is developed from Y(s) of the given impedance function as

Since the degree of numerator and denominator is same, the partial fraction will be as follows:

Let us find the values of A and B.

To find A, substitute s = −2 in equation (12.5) [∵ in the denominator of A factor is s + 2]

Here, none of the coefficient can be negative. In such cases, we will use the partial fractions as in the following:

Now, degree of denominator > degree of numerator, and therefore, partial fraction will be used as follows:

Now, substituting the values of A, B and C in equation (12.6), we get the following:

General partial fractions of YRC (s) are

Therefore, comparing equations (12.7) and (12.8), we get the following:

term will represent resistor of value

Term represents series combination of R2 and C2

where

Term represents series combination of R4 and C4.

Figure 12.12

Therefore, required Foster form-II, is shown in Figure 12.12.

In Cauer form-I, we arrange numerator and denominator of Z(s) in descending power of s and then carry out continued fraction expansion to identify the circuit elements from the quotient terms.

Given

Note that in Cauer form-I realization of the circuit element from Z(s), quotient terms with s in the numerator represents the capacitor. The first quotient element will be Z1 since we started with Z(s).

We know, in Cauer form-I of R-C, resistors are in series and capacitors are in shunt, that is, parallel.

Therefore, required Cauer form-I is shown in Figure 12.13.

For Cauer from-II, we arrange the numerator and denominator in the ascending power of ‘s’ so that

Figure 12.13

Since we got a negative quotient as above, we will have to find Cauer form-II circuit by using the function Y(s)

Let us find continued expansion

First element will be . In Cauer form-II, capacitors are in series and resistors are in parallel, as shown in Figure 12.14.

Figure 12.14 Cauer Form-II

Example 12.13 Synthesise all the four forms of the driving point function represented as

Solution: To find Z(∞), we write the following:

Further, to find Z(0), substitute s = 0 in Z(s)

Now, Z(∞) > Z(0)

Therefore, given impedance is of R-L type.

For Foster from-I, we start write Z(s) as

Since degree of numerator is equal to the degree of the denominator, the partial fractions of Z(s) will be given as follows:

Let us find values of A and B

To find A, substitute s = −3 in equation (12.10), we get the following:

Since none of the coefficient should be negative, and we will use the partial fractions of as

Now, the degree of denominator > degree of numerator, and the partial fractions will be used as follows:

Substituting the values of A, B and C in equation (12.11), we get the following form:

Writing Z(s) as, we find the values of Foster from-I circuit parameter as,

The required Foster form-I circuit is shown in Figure 12.15.

Figure 12.15

Foster form-II circuit is realised using the function Y(s)

Here, degree of denominator = degree of numerator. Therefore, partial fractions will be as follows:

Let us find values of A and B

To find A, substitute s = −1 in equation (12.13)

To find the value of B, substitute s = −5 in equation (12.13).

Now, substituting the values of A and B in equation (12.12), we get the following form:

Figure 12.16

The required Foster form-II is shown in Figure 12.16.

For Cauer form-I, we start with,

(Arranging the polynomials in ascending power of ‘s’)

Let us find continued fractions as,

Therefore, we will visualized the Cauer form-I circuit using Y(s).

Now we divide as,

The Cauer form-I circuit is shown in Figure 12.17

Figure 12.17 Cauer Form-I

For Cauer form-II, we arrange the polynomials in descending power of ‘s’.

Now we divide as,

The Cauer form-II circuit is shown in Figure 12.18.

Example 12.14 Find the Foster form-I and Foster form-II of the given expression for Z(s).

Figure 12.18 Cauer Form-II

Solution: First, let us check whether the given impedance is of R-C type or R-L type. Remember, for R-C, Z(0) > Z(∞) and for R-L, Z(∞) > Z(0).

For this, let us find Z(∞)

Z(s) can be written as follows:

Substituting s = ∞, we get the following equation:

Further, to find Z(0),

Clearly, Z(0) > Z(∞), and therefore, given impedance is of R-C type.

For realizing the Foster Form-I network,

Here, degree of denominator = degree of numerator, and therefore, we will find partial fractions as follows:

Let us find the values of A and B

Now, substitute the values of A and B in equation (12.14),

The Foster form-I circuit is shown in Figure 12.19.

For Foster form-II, we write

Let us use the partial fractions as follows:

Let us find values of A and B

Figure 12.19

To find A, substitute s = −1 in equation (12.16),

Since none of the coefficient of Z(s) should be negative, we will obtain the partial fractions of .

(Here, degree of denominator > degree of numerator)

Therefore, partial fractions will be given as follows:

Let us find values of A and B

Substituting the values of A and B in equation (12.18), we get the following:

Therefore, required Foster form-II circuit is drawn as shown in Figure 12.20.

Example 12.15 Obtain the Cauer-I and Cauer-II form of network for

Solution: From the given Z(s), Z(∞) = 1, Z(0) = ∞; Clearly, Z(0) > Z(∞), and therefore, the given impedance function is of R-C type.

Figure 12.20

Cauer-I: To obtain Cauer-I,

We arrange the numerator and denomination in descending power ‘s’ and carry out the division work. The circuit elements are identified from the division work as,

Figure 12.21

Cauer form-I of representation of the circuit has been shown in Figure 12.21. In the circuit the resistors are in series and capacitors are in parallel.

For Cauer Form-II representation of the circuit for the given Z(s), we arrange the numerator and denominator in ascending order of s.

We carry out the division work to identify the circuit elements from the quotient terms.

Required Cauer form-II circuit for Z(s) is shown in Figure 12.22 where the capacitors are in series and the resistors are in parallel.

Example 12.16 Synthesise the impedance function Z(s) in Foster form-II where Z(s) is given as

Figure 12.22

Solution: We find Z(∞) and Z(0) to check the nature of the circuit being represented by Z(s)

and . Since Z(0) > Z(∞), the given impedance function represents an R-C network.

Foster form-II circuit is designed using Y(s) as in the following:

Here, degree of denominator < degree of numerator.

For the partial fractions, degree of denominator ≥ degree of numerator.

Therefore, to make degree of denominator = degree of numerator, we divide both sides of r(s) by s as follows:

Substitute s = 0 in equation (12.20) to get A as

To find B, substitute s = −5 in equation (12.20)

Substitute s = −7 in equation (12.20) to get C as

Now, substitute the values of A, B and C in equation (12.19), we get the following form:

The circuit elements are calculated as

Therefore, the required Foster form-II is shown in Figure 12.23. The students many compare the s circuit with the standard Foster from-II of R-C network shown in Figure 12.2.

Figure 12.23

Example 12.17 A function F(s) has poles and zeros as follows:

Poles at s = 0, −4, −6

Zeros at z = −2, −5

Synthesise F(s) (a) as an impedance in Foster form; (b) as an admittance in Cauer form

Solution: Since the numerator equated to zero given the zeros and the denominator equated to zero given the poles, the impedance function F(s) is written as,

Substituting the given values,

(a) We calculate Z(∞) and Z(0) to identify the nature of the network in Z(s) as synthesis of F(s) as impedance in Foster form. We first

Z(0) > Z(∞), and therefore, Z(s) represents an R-C network.

Using Z(s), we can now proceed to design Foster form-I circuit. We consider the partial fraction of Z(s) as in the following:

Substituting the values of A, B and C in equation (12.21), we get the following:

From the above we find the circuit elements as,

The required Foster form-I is shown in Figure 12.24.

Figure 12.24

(b) We now write F(s) as an admittance and visualize the

Cauer form-I circuit.

We now carry out the division work and determine the circuit elements from the quotient as,

In Cauer from-I of R-C network Z stands for R and Y stands for C.

Therefore, required Cauer form-I circuit is shown in Figure 12.25. Note that in the circuit the first element starts with C1 and not R1 as we started with Y(s) and not Z(s).

Figure 12.25

Example 12.18 An impedance function is given as,

Find Foster form-I, Foster form-II, Cauer form-I and Cauer form-II of circuits, which this impedance function represents.

Solution: Given,

Firstly, we will check whether the given impedance is of R-L or R-C type. Z(∞) can be calculated as follows:

Clearly, Z(∞) > Z(0), and therefore, the given impedance is of R-L type

Foster form-I circuit is designed using Z(s) as follows:

Here, the degree of s in the denominator < degree of s in the numerator. For the partial fractions, the degree of denominator ≥ numerator.

Therefore, to make degree of denominator = degree of numerator, we will divide both sides of Z(s) by s. Accordingly.

Its partial fractions will be as follows:

Let us find values of A and B

To find A, substitute s = −1 in equation (12.23)

To find B, substitute s = −4 in equation (12.23)

Now, putting the values of A and B in equation (12.22), we get the following:

Z(s) can be written in the following form:

From (i) and (ii), then circuit parameters are found as

The Foster form-I circuit is shown in Figure 12.26.

Figure 12.26

Foster form-II circuit is designed using Y(s).

Let us use the partial fractions.

Since degree of denominator > degree of numerator, the partial fractions will be given as follows:

Substituting the values of A, B and C in equation (12.24)

Y(s) is written in the standard from as

Comparing the above two from of Y(s), we find the circuit elements as

With these values, we draw the Foster form-II circuit as shown in Figure 12.27.

Now, we will find the Cauer form-I circuit.

We have,

Now we will carry out the division work and determine the circuit elements from the quotient as

Figure 12.27

Accordingly, the Cauer form-I circuit is drawn as in Figure 12.28.

Now, we will find the Cauer form-II circuit.

Figure 12.28

For Cauer form-II, we arrange the polynomials in ascending powers of ‘s’ as in Z(s) as

We now carry out the division work and determine the circuit elements from the quotient as

The terms are negative, and hence, we cannot design Cauer form-II circuit using Z(s).

Let us find the circuit using Y(s).

The division work is carried out and the circuit elements are found from the quotients as

In Cauer form-II of R-L circuit, Z stands for R and Y stands for L. The Cauer form-II circuit is shown in Figure 12.29

Figure 12.29

12.6 MORE NUMERICALS ON SYNTHESIS OF L-C NETWORK

Example 12.19 Synthesise the function Z(s) using Foster forms

Solution: Foster form-I

Partial fraction expansion of generalised L-C impedance or admittance from results in

From equation (12.25), we get the following form:

and

Required Foster form-I is shown in Figure 12.30. Foster form-II is obtained from Y(s).

Figure 12.30

Now, degree of denominator < degree of numerator

Now, considering equation (12.27), we get the following form:

Required Foster form-II is shown in Figure 12.31.

Example 12.20 Synthesise the impedance function as follows:

Figure 12.31

Solution:

Here the degree of denominator > numerator

Required Foster form-I is shown in Figure 12.32.

Figure 12.32

Example 12.21 The driving point impedance of a one-port LC network is given by the following:

Obtain the first and second Foster forms.

Solution: Foster form-I:

Since degree of denominator < numerator, we write the equation as follows:

The Foster form-I circuit, using the calculated data, has been shown in Figure 12.33.

For Foster form-II,

Figure 12.33

Now, degree of denominator > numerator.

The required Foster form-II circuit is shown in Figure 12.34

Example 12.22 Find Foster forms of the following:

Figure 12.34

Solution: This is a question for practice for the students. Please check your result with the solution provided in Figure 12.35.

Figure 12.35

Example 12.23 Find Cauer form-I and Cauer form-II of the following:

Solution: Cauer form-I can be given as follows:

The required Cauer form-I is shown in Figure 12.36.

For Cauer Form-II,

Figure 12.36

For Cauer form-II, we arrange the terms of ‘s’ in ascending order.

The coefficients should not be negative, and hence, we cannot design Cauer form-II circuit using Z(s). Let us design the circuit using Y(s).

In Cauer form-II, Y corresponds to L and Z corresponds to C.

Accordingly, the circuit is drawn as shown in Figure 12.37.

Figure 12.37

Example 12.24 Find two faster realisations of the given function

Solution: Foster form-I circuit is designed using Z(s)

; clearly, the given impedance is of LC type.

Using the partial fractions, we get the following equation:

Comparing the equation with the general form, we can write it as

We get the following:

Foster form-I circuit is shown in Figure 12.38.

Foster form-II circuit is designed using function Y(s) as follows:

Figure 12.38

Therefore, Foster form-II will be as shown in Figure 12.39.

Figure 12.39

Example 12.25 Find the two Foster realisations of the following function:

Solution: Foster form-I can be obtained as follows:

The required Foster form-I circuit is shown in Figure 12.40.

Figure 12.40

Foster form-II circuit is designed using Y(s).

Therefore, the required Foster form-II circuit is shown in Figure 12.41

Figure 12.41

Example 12.26 Synthesise first and second Foster form of L-C network for the impedance

Solution: Foster form-I is calculated as follows:

Figure 12.42

Foster form-I circuit will be as shown in Figure 12.42.

Foster form-II circuit is designed using Y(s).

Therefore, required Foster Form-II circuit will be as shown in Figure 12.43

Figure 12.43

Example 12.27 Realise in the form of Cauer L-C network

Solution: Cauer form-I can be calculated as follows:

Therefore, the required Cauer form-I circuit is shown in Figure 12.44.

For Cauer form-II

Figure 12.44

Therefore, Cauer form-II circuit will be as shown in Figure 12.45.

Example 12.28 Find two Cauer realisations of the following function:

Figure 12.45

Solution: Cauer form-I can be obtained as follows:

Therefore, the required Cauer form-I circuit will be as shown in Figure 12.46.

Cauer form-II can be obtained as follows:

Figure 12.46

Therefore, required Cauer form-II is shown in Figure 12.47.

Example 12.29 Find the Foster form-I and the Cauer form-II of the function represented as

Figure 12.47

Solution:

Z(0) = 2; now, Z(∞) > Z(0), and therefore, the given impedance is of R-L type.

Foster form-I can be calculated as follows:

Let us find values of A and B.

Substituting s = −2 in equation (12.30), we get the following:

. Since the coefficient should not be negative, we need to use the partial fractions of

Substituting the values of P0, P2 and P4 in equation (12.31), we get the following equation:

Therefore, the required Foster form-I circuit is shown in Figure 12.48.

Figure 12.48

For Cauer form-II, we calculate Z(s) can be calculated as follows:

Therefore, the required Cauer form-II circuit is shown in Figure 12.49.

Example 12.30 Find the circuit in second Cauer form of the following function

Figure 12.49

Solution: Given network function is as follows:

Further, Z(∞) can be calculated as follows:

Therefore, the given impedance is of R-L type.

Now, Cauer form-II can be obtained as in the following:

The Cauer form-II circuit is shown in Figure 12.50.

Example 12.31 For a given function

Figure 12.50

Find the circuit in the following:

(a) Foster form-I and II and

(b) Cauer form-I

Solution: Firstly, let us check whether the given impedance function is of R-C or of R-L type.

In this case, Z(0) > Z(∞)

Therefore, the given impedance is of R-C type.

Foster form-I can be obtained as in the following:

Therefore, the required Foster form-I circuit is drawn as in Figure 12.51.

Figure 12.51

Foster form-II can be calculated as follows:

Let us find values of P2, P4 and P6

Substituting s = −1 in equation (12.34), we get the following:

Substituting s = −3 in equation (12.34), P4 can be calculated as follows:

Substituting s = −5 in equation (12.34), we get the following:

Substituting the values of P2, P4 and P6 in equation P, the following equation can be obtained:

The required Foster form-II circuit is shown in Figure 12.52.

Figure 12.52

Cauer form-I can be calculated as follows:

we will consider Y(s) as follows:

Therefore, Cauer form-I circuit is shown in Figure 12.53.

Example 12.32 Find second Cauer form of the network whose function is

Figure 12.53

Solution: Given

Since Z(0) > Z(∞), the given function is of R-C type.

For finding the circuit in Cauer form-II, the process is as follows

Since, the coefficient is negative, we will now consider Y(s) as

Therefore, required Cauer form-II circuit is shown in Figure 12.54.

Example 12.33 Find two Foster realisation of the network function

Figure 12.54

Solution:

and therefore, Z(0) > Z(∞)

The given impedance is the R-C type.

Foster form-I can be calculated as follows:

Let us find P0 and P2

To find P0, substitute s = 0 in equation (12.36)

To find P2, substitute s = −4 in equation (12.36)

Substituting the values of P0 and P2 in equation (12.35), we get the following:

Therefore, required Foster form-I circuit is shown in Figure 12.55

Foster form-II can be calculated as follows:

Figure 12.55

Let us find P2 and P4.

Substituting s = −3 in equation (12.38), we get the following:

Since, the coefficient cannot be negative, and therefore, we will do the partial fraction of the following:

Now, substituting the values of P2 and P4 in equation (12.39), we get the following form:

Therefore, Foster form-II circuit is shown in Figure 12.56.

Example 12.34 Find Cauer form-I and Cauer Form-II for the network function

Figure 12.56

Solution: Given network function is as follows:

That is, Z(0) > Z(∞), and therefore, given network function is of R-C type

For Cauer form-I, the procedure is followed as follows:

Therefore, required Cauer form-I circuit is shown in Figure 12.57.

For Cauer form-II, the calculation are as follows:

Figure 12.57

Figure 12.58

Therefore, required Cauer form-II circuit is shown in Figure 12.58.

REVIEW QUESTIONS

Short Answer Type

1.Differentiate network analysis and network synthesis.

2.What are Hurwitz conditions for stability?

3.What are the properties of a positive real Function?

4.Explain the process of synthesis of R-C network using Foster method.

5.Explain Cauer’s method to synthesising R-C network.

6.What are the properties of R-C admittance function?

7.What are the properties of R-L impedance function?

8.What are the properties of L-C immittance?

9.Explain how one-port R-L network can be synthesised using Foster method.

10.Explain how one-port R-L network can be synthesised using Cauer method.

11.Draw Foster forms of L-C network.

12.Draw Cauer forms of L-C network

13.Explain the synthesis of L-C network using Foster method.

14.What do you mean by a Hurwitz polynomial?

Numerical Questions

1.Determine the condition for which the function is positive real. It is given that a0, b0, a1 and b1 are real and positive.

2.Express the impedance Z(s) for the network shown in Figure 12.59

Figure 12.59

. Plot its poles and zeros. From the pole-zero plot, what can you infer about the stability of the system?

Z(s) is unstable.

3.Find voltage-transfer function for the network shown in Figure 12.60

Figure 12.60

4.Test whether the polynomial F(s)=s4+s3+2s2 +3s+ 2 is Hurwitz

[Ans. Not Hurwitz]

5.Test whether the function

is positive real, where α and k are positive constants

[Ans. Positive real]

6.Given the network function , plot the zero and poles on s-plane

7.Find the driving point impedance , for the network shown in Figure 12.61

Figure 12.61

Verify that Z(s) is positive real and that the polynomial D(s) + K.N(s) is Hurwitz

[Ans. , Z(s) is PRF, D(s) + K.N(s) is Hurwitz]

8.Design a one-port R-L network to realise the driving point function

[Ans. as in Figure 12.62]

Figure 12.62

9.Draw the pole-zero plot in s-plane for network function

10.Express the driving point admittance Y(s) in the form for the network shown in Figure 12.63 and also verify that Y(s) is PRF.

Figure 12.63

[Ans. is a PRF]

11.Consider the system function . Determine the corresponding

(a)R-C network;

(b)R-L network;

[Ans. as in Figure 12.64]

Figure 12.64

12.Determine if the function is positive real

[Ans. F(s) is a PRF]

13.Realise the impedance in three different ways

[Ans. as in Figure 12.65]

Figure 12.65

14.List out the properties of LC immittance function and then realise the network having the driving point impedance function by continued fraction method.

[Ans. as in Figure 12.66]

Figure 12.66

15.For the network function , synthesise in one Foster and one Cauer form

[Ans. as in Figure 12.67]

Figure 12.67

16.Consider the function . Plot its poles and zeroes

17.Determine whether the function is positive real or not.

[Ans. F(s) is a PRF]

18.Synthesise a one-port LC network in Cauer form-II whose driving point impedance is

[Ans. as in Figure 12.68]

Figure 12.68

19.Given , draw its pole-zero plot. Further, find i(t) from it.

[Ans. i(t) = −2et−6e−3t A]

20.Find the voltage transfer function for the network shown in Figure 12.69.

Figure 12.69

[Ans. ]

21.Determine the range of constant ‘K’ for the polynomial to be Hurwitz

P(s) = s3 + 3s2 + 2s + k

[Ans. 0 < k < 6]

22.Synthesise an R-C ladder and an R-L ladder network to realise the function

as an impedance or an admittance.

[Ans. as in Figure 12.70]

Figure 12.70

23.Calculate the driving point admittance of the network shown in Figure 12.71

Figure 12.71

24.A transform voltage is given by

. Make the pole-zero plot in the s-plane and obtain the time domain response.

[Ans. i(t) = −et + 4e−4t]

25.Check if the given driving point impedance Z(s) represents a passive one-port network

[Ans. No]

MULTIPLE CHOICE QUESTIONS

1.A network function is said to have simple pole or simple zero if

(a)The poles and zeros are on the real axis

(b)The poles and zeros are repetitive

(c)The poles and zeros are complex conjugate to each other

(d)The poles and zeros are not repeated.

2.A function will have a zero at

(a)s = ± j4

(b)Anywhere on the s-plane

(c)On the imaginary axis

(d)On the origin

3.The network shown in Figure 12.72 has zeros at

Figure 12.72

(a)s = 0 and s = ∞;

(b)s = 0 and ;

(c)s = ∞ and ;

(d)s = ∞ and

4.Which of the following is a PRF

(a)

(b)

(c)

(d)

5.A network function can completely be specified by:

(a)Real parts of zeros

(b)Poles and zeros

(c)Real parts of poles

(d)Poles, zeros and a scale factor

6.In the complex frequency s = σ + , while ω has the unit of rad/s and σ has the unit of

(a)Hz

(b)Neper/s

(c)Rad/s

(d)Rad

7.Which of the following property relates to L-C impedance or admittance functions:

(a)The poles and zeros are simple and lie on the -axis

(b)There must be either a zero or a pole at origin and infinity.

(c)The highest (or lowest) powers of numerator or denominator differ by unity.

(d)All of the above.

8.If a network function has zeros only in the left-half of the s-plane, then it is said to be

(a)A stable function

(b)A non-minimum phase function

(c)A minimum phase function

(d)An all-pass function.

9.Zeroes in the right half of the s-plane are possible only for

(a)d.p impedance function.

(b)d.p admittance functions.

(c)d.p impedance as well as admittance functions.

(d)transfer functions.

10.An L-C impedance or admittance function:

(a)has simple poles and zeros in the left half of the s-plane.

(b)has no zero or pole at the origin or infinity.

(c)is an odd rational function.

(d)has all poles on the negative real axis of the s-plane.

11.The Laplace-transformed equivalent of a given network will have capacitor replaced by

(a)

(b)

(c)

(d)

12.If a network function contains only poles whose real-parts are zero or negative, the network is

(a)always stable.

(b)stable, if the -axis poles are simple.

(c)stable, if the -axis poles are at most of multiplicity 2.

(d)always unstable.

13.Which of the following kind of network has the same admittance and impedance properties?

(a)L-C type

(b)R-L type

(c)R-C type

(d)R-L-C type

14.Both odd and even parts of a Hurwitz polynomial P(s) have roots

(a)in the right-half of s-plane

(b)in the left-half of s-plane

(c)on the o-axis only

(d)on the -axis only

15.The impedance of a network is given as . The function is

(a)not a PRF

(b)R-L network

(c)R-C network

(d)L-C network

16.If F1(s) and F2(s) are positive real, then which of the following are positive real?

(a)

(b)F1(s)+F2(s)

(c)

(d)All of these.

17.Which of these is not a positive real Function?

(a)F(s) = Ls

(b)F(s) = R

(c)

(d)

18.A stable system must have

(a)zero or negative real part for poles and zeros.

(b)at least one pole or zero lying in the right-half s-plane

(c)positive real part for any pole or zero

(d)negative real part for all poles and zeros.

ANSWERS

1.d

2.d

3.c

4.c

5.d

6.b

7.d

8.c

9.d

10.a

11.d

12.b

13.a

14.d

15.a

16.d

17.d

18.a