**Network Synthesis Lecture Notes**

**Network analysis: **You are given a network then you analyze the behavior (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 behavior.

** **

**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 the degree of P(s) is of one degree higher than the degree of Q(s). It infers that there is an inductance in series.

**The 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 the same poles, since Z(s) is a prf, all poles of Z_{2}(s) lies on the left half s-plane. - If poles on the 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 the original function is considered as impedance function, then removal of Z1(s) means the removal of an inductor, whereas if the original function is given as admittance function the removal of Y
_{1}(s) means the removal of a capacitor.

**Removal of a pole at zero:**It means that the degree of Q(s) is of one degree higher than the degree of P(s). It means there is a capacitor in parallel to 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 the 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 the 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 of Network Synthesis**

**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 indicates 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