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Foster’s Form 1 and 2 of Network Synthesis with examples

Foster’s Form 1 and 2 of Network Synthesis with examples

Foster’s Form of Synthesis

Foster has given two types of network synthesis for one-port reactive network

  • Series combination of parallel LC networks
  • The parallel combination of series LC networks

Foster 1st Form

When numerator polynomial is of higher degree than the denominator polynomial

  1. The first element capacitor is represented by A0/s
  2. The last element Inductor is represented by Hs corresponds t a pole at infinity
  3. 2Ai/(s2 + ω 2) represents the conjugate poles results in LC resonance.
  4. When no terms in the denominator of Z(s), there will not be any A0/s term indicating the absence of capacitor.

Example: Obtain the first form of the Foster Network for the driving point impedance of LC network given as:

Z(s) = 10(s2+4)(s2+16) / s(s2+9)

Solution:

It is observed from the function Z(s) that the numerator has one higher degree of s than the denominator, hence two poles exist one at ω =0 and another at ω=∞, therefore there will be the presence of the first element as the capacitor and the last element as the inductor

The partial fraction expansion of the network function is:

Z(s) = A0/s + A1/(s+j3)  + A1*/(s-j3)   + Hs

The values of residues is found to be as:

A0 at s=0 is  = 640/9 = 71.11;

so we find C0 = 1/A0

= 1 / 71.11    =   0.0141 Farad

A1 at s= -j3 is= 350/18  =  19.45

Now we know Li and Ci of the parallel network is calculated as

C1    = 1/(2A1)

= 1/(2*19.45)

= 0.0257 Farads

and

L1    = 2A12

= 2*19.45 / 9

= 4.322 Henry

The end element inductor is ‘H’     = 10 henry

Therefore the realization of the impedance function is:

Foster’s 2nd Form

2nd for is for the synthesis of admittance function, and is:

As observed:

Given an admittance function:

i. The inductor is represented by the term B0/s and this corresponds to a pole at origin

ii. The capacitor  C∞   is represented by Hs and corresponds to a pole at infinity.

iii. The series combination of L and C is determined from 2*B1 / (s2 + ω 2)

iv. In case there is no pole at  ω=0 or at ω=∞ or at both, signifies absence of end elements.

Example:

 

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