Foster’s Form 1 and 2 of Network Synthesis with examples Network Analysis by Ravinder Nath Rajotiya - April 22, 2020June 4, 20200 Share on Facebook Share Send email Mail Print Print Table of Contents Toggle Foster’s Form of SynthesisFoster 1st FormFoster’s 2nd Form 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 The first element capacitor is represented by A0/s The last element Inductor is represented by Hs corresponds t a pole at infinity 2Ai/(s2 + ω 2) represents the conjugate poles results in LC resonance. 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 = 2A1/ω2 = 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: Share on Facebook Share Send email Mail Print Print