Network Synthesis Part-1 Lecture Notes

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) = Z1(s) + Z2(s)

Rearranging,

Z2(s) = Z(s)  –  Z1(s)

That is we find Z2(s) by removing Z1(s) from Z(s).

This removal of Z1(s) from Z(s) can be performed in following different ways:

  1. 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:

  1. 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= an-1/bn.

Comparing the polynomial Z(s) and the Z2(s) reveals that:

  • Z(s) and Z2(s) have the same poles, since Z(s) is a prf, all poles of Z2(s) lies on the left half s-plane.
  • If poles on the imaginary axis then the poles are simple
  • Re[Z2(jω) = Re[Z(jω) ]>=0 for all values of ω; thus Z2(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 Y1(s) means the removal of a capacitor.
  1. 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 Z2(s).

= Z1(s)    +        Z2(s)

Z1(s) term has ‘s’ in denominator indicating that removal of a pole at zero means removal of a capacitor (of value k0= a0/b1) from the network. If the original function was an admittance function, then it would mean the removal of an inductor (of value b1/a0 ) 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 (s212) then roots will be complex conjugate and the poles (s+jω) and (s-jω) at the imaginary axis. Since the denominator factor is (s212) we have to find the residues.

 

Z(s) = A/(s-jω1)  + A*/(s+jω1)     (since the residues are complex conjugate A=A* = say K; so, the impedance function:

Z(s)         = 2Ks/(s212) + Z2(s)

= Z1(s)             + Z2(s)

Next, we note that

Re[Z1(s)] =            Re[2Ks/((jω)212)]

=              Re[2Ks/(-ω212)]

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

Now Z(s)        = 2Ks / (s212)

=  1/ {(s2/2Ks) + w12/2ks)}

=  1/(Yas  + Ybs)

Here Yas = s/2K represent a capacitor and Ybs = ω12/2ks represent an inductor connected in parallel to form Z1(s)

Similarly, if we write the Y(s) as the riginal function, then we can prove that Y1(s) = 1/{(Za(s) + Zb(s)}

Za(s)=s/2K representing an inductor of value 1/2K henry and a Zb(s)= ω12/2ks representing a capacitor of value ω12/2K and both Za(s) and Zb(s) connected in series.

 

Example of Network Synthesis

Realize the network whose impedance is given as: Z1(s) = (s4 + 10s2 + 7) / (s3 + 2s)

Solution:

Step-1:

  1. substitute s= jw and check whether Z1(s) has any real parts

Z1(s)           = ((jω)4 + 10(jω)2 + 7) / ((jω)3 + 2(jω))

=  (ω4 -10ω2 + 7) / (-jω3 + 2jω)

Function reveal that for s=jω, Z1(s) is not real, therefore it is inferred that Z1(s) does not have any resistance element.

  1. 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:

s3 + 2s   ) s4 + 10s2 + 7 (   s

-(s4 + 2s2 )

————–

8s2 + 7

Therefore Z1(s) becomes:

Z1(s)    =  s       + (8s2 + 7) / s3 + 2s

=  Z2(s) + Z3(s) ;    where Z2(s) = s representing an inductor of 1 henry

Step-3

Z3(s)        = (8s2 + 7) / s3 + 2s

Inverting we get Y3(s)          = 1/Z3(s)

= (s3 + 2s) / (8s2 + 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 Y3(s) represent a capacitor, its value can be found by long division as:

8s2 + 7 )  s3 + 2s    ( s/8

-( s3 + 7s/8)

—————————

9s/8

Therefore we can write Y3(s) as

Y3(s) = s/8              +  (9s/8) / (8s2 + 7)

=  Y4(s)+  Y5(s)

The admittance Y4(s) = s/8 represent a capacitor of 1/8 Farad

Step-5

Convert the admittance Y5(s) in to impedance Z5(s)  as

Z5(s)     = 1/Y5(s)

= 8(8s2 + 7)/9s

= 64s2/9s   +  56/9s

= 64s/9     +  56/9s

=  Z6(s)   +   Z7(s)

Where  Z6(s) = 64s/9 represent an inductor of value 64/9 Henry

Step-6:

Invert Z7(s) to convert it into admittance Y7(s) as:

Y7(s)    =  1/ Z7(s)

 =9s/56

Y7(s) represent a capacitive impedance of value 9/56 Farad

Updated: June 4, 2020 — 1:38 pm

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