**Network Synthesis**

**Network analysis: **You are given a network then you analyse the behaviour (output) by applying certain excitation.

**Network Synthesis: **You know the output response for certain excitation, then you are required to find appropriate Z(s) or Y(s) that will give that desired behaviour.

** **

Let us consider the network equations in terms of impedance and admittance:

Process of network synthesis: let us first consider figure-1(a)

Z(s) = Z_{1}(s) + Z_{2}(s)

Rearranging,

Z2(s) = Z(s) – Z_{1}(s)

That is we find Z_{2}(s) by removing Z_{1}(s) from Z(s).

**This removal of Z _{1}(s) from Z(s) can be performed in following different ways:**

**Removal of a pole at infinity**: It means that degree of P(s) is of one degree higher than the degree of Q(s). It infers that there is an inductance in series.

Procedure of removal of a pole:

- Divide the numerator P(s) polynomial by denominator Q(s) polynomial. We get

Z(s) = Hs + A(s) /B(s) where H is the quotient i.e. H= a_{n-1}/b_{n}.

Comparing the polynomial Z(s) and the Z_{2}(s) reveals that:

- Z(s) and Z
_{2}(s) have same poles, since Z(s) is a prf, all poles of Z_{2}(s) lies on left half s-plane. - If poles on imaginary axis then the poles are simple
- Re[Z
_{2}(jω) = Re[Z(jω) ]>=0 for all values of ω; thus Z_{2}(s) is also a prf

- So if original function is considered as impedance function, then removal of Z1(s) means removal of an inductor, whereas if the original function is given as admittance function the removal of Y
_{1}(s) means removal of a capacitor.

**Removal of a pole at zero:**It means that degree of Q(s) is of one degree higher than the degree of P(s). It means there is a capacitor in parallel of the network Z_{2}(s).

= Z_{1}(s) + Z_{2}(s)

Z_{1}(s) term has ‘s’ in denominator indicating that removal of a pole at zero means removal of a capacitor (of value k0= a_{0}/b_{1}) from the network. If the original function was an admittance function, then it would mean removal of an inductor (of value b_{1}/a_{0} ) from the circuit.

**Removal of conjugate poles:**Factors of Q(s) are conjugate pairs i.e poles are on imaginary (jw) axis

If the Q(s) has roots of the form (s^{2}+ω_{1}^{2}) then roots will be complex conjugate and the poles (s+jω) and (s-jω) at the imaginary axis. Since denominator factor is (s^{2}+ω_{1}^{2}) we have to find the residues.

Z(s) = A/(s-jω_{1}) + A*/(s+jω_{1}) (since the residues are complex conjugate A=A* = say K; so, the impedance function:

Z(s) = 2Ks/(s^{2}+ω_{1}^{2}) + Z_{2}(s)

= Z_{1}(s) + Z_{2}(s)

Next we note that

Re[Z1(s)] = Re[2Ks/((jω)^{2}+ω_{1}^{2})]

= Re[2Ks/(-ω^{2}+ω_{1}^{2})]

Thus, Re[Z2(jω)] becomes zero or +ve for any value of w since Z(s) is a PRF.

Now Z(s) = 2Ks / (s^{2}+ω_{1}^{2})

= 1/ {(s^{2}/2Ks) + w_{1}^{2}/2ks)}

= 1/(Y_{as} + Y_{bs})

Here Yas = s/2K represent a capacitor and Ybs = ω_{1}^{2}/2ks represent an inductor connected in parallel to form Z_{1}(s)

Similarly, if we write the Y(s) as the riginal function, then we can prove that Y_{1}(s) = 1/{(Z_{a}(s) + Z_{b}(s)}

Z_{a}(s)=s/2K representing an inductor of value 1/2K henry and a Z_{b}(s)= ω_{1}^{2}/2ks representing a capacitor of value ω_{1}^{2}/2K and both Za(s) and Zb(s) connected in series.

**Example: Realize the network whose impedance is given as: Z _{1}(s) = (s^{4} + 10s^{2} + 7) / (s^{3} + 2s) **

Solution:

Step-1:

- substitute s= jw and check whether Z
_{1}(s) has any real parts

Z_{1}(s) = ((jω)^{4} + 10(jω)^{2} + 7) / ((jω)^{3} + 2(jω))

= (ω^{4} -10ω^{2} + 7) / (-jω^{3} + 2jω)

Function reveal that for s=jω, Z_{1}(s) is not real, therefore it is inferred that Z_{1}(s) does not have any resistance element.

- Since the degree of numerator polynomial is one higher than denominator polynomial, therefore Z1(s) has a pole at infinity which indicate that a series inductor is present.

Step-2: Since an inductor is present, find its value by long division as:

s^{3} + 2s ) s^{4} + 10s^{2} + 7 ( s

-(s^{4} + 2s^{2} )

————–

8s^{2} + 7

Therefore Z1(s) becomes:

Z1(s) = s + (8s^{2} + 7) / s^{3} + 2s

= Z_{2}(s) + Z_{3}(s) ; where Z_{2}(s) = s representing an inductor of 1 henry

Step-3

Z_{3}(s) = (8s^{2} + 7) / s^{3} + 2s

Inverting we get Y_{3}(s) = 1/Z_{3}(s)

= (s^{3} + 2s) / (8s^{2} + 7)

Again as we that in this admittance function, degree of numerator is one higher than the degree of the numerator, therefore it represents a capacitor in parallel

Step-4:

Since the impedance Y_{3}(s) represent a capacitor, its value can be found by long division as:

8s^{2} + 7 ) s^{3} + 2s ( s/8

-( s^{3} + 7s/8)

—————————

9s/8

Therefore we can write Y_{3}(s) as

Y_{3}(s) = s/8 + (9s/8) / (8s^{2} + 7)

= Y4(s)+ Y5(s)

The admittance Y_{4}(s) = s/8 represent a capacitor of 1/8 Farad

Step-5

Convert the admittance Y_{5}(s) in to impedance Z_{5}(s) as

Z_{5}(s) = 1/Y_{5}(s)

= 8(8s^{2} + 7)/9s

= 64s^{2}/9s + 56/9s

= 64s/9 + 56/9s

= Z_{6}(s) + Z_{7}(s)

Where Z_{6}(s) = 64s/9 represent an inductor of value 64/9 Henry

Step-6:

Invert Z_{7}(s) to convert it into admittance Y_{7}(s) as:

Y_{7}(s) = 1/ Z_{7}(s)

=9s/56

Y_{7}(s) represent a capacitive impedance of value 9/56 Farad