INDUCTANCES, CAPACITANCES AND RESISTANCES 



A and B would not have been zero and the response of the filter would not 

 have been Butterworth. 



If the R values are altered, how can we make the response return to 

 Butterworth ? By finding new values of L and C such that A and B are zero 

 and D = (IIcoq)^. We take as an example a very important case in valve 

 work, a filter to operate between a generator of resistance R and a load of 

 infinity. We take the low-pass case. 



Referring to Figure 5.35, the first point to notice is that the section must 



nnnpmnr^ 



'•"' vLT^ 



figure 5.35 



be of the pi form. If it were a T, since the load current is zero, the right-hand 

 horizontal element would also pass no current; if it passes no current it 

 can make no contribution to the working of the filter. The filter would 

 degenerate to a two-element one, producing an attenuation slope of only 

 12 dB's/octave. 

 Proceeding as usual we see by inspection that 



1 



jcoCi \yc0C2 



+ 



jojLj 



V, 



out 



>Q 



+ t;-^ + J^L 



jcoCi ytoC2 



in 



1 . J_ /_J_ . 



which simphfies to 



1 1 , . . 



— :^ + '■ — TT + J<J^^ 

 jcoCi J0JC2 



-1 



+ R 



Hence, 



((o^LC - 1) i-jico^C^C^R - coR(Cj_ + Q)] 



1 



{1 + co\R\C^ + C^f - 2LC2] + (o^lL^C^^ - IR^LC^C^iC^ + Q)] 



If this is to be Butterworth, then : 

 (1) The coefficient of co^ must be 



/ 1 \<^ 1 _ 1 



and 



(2) The coefficient of w* must be zero. Therefore 



L^C^^ = IR^LC^C^iC^ + C2) 

 94 



(Dn = 



C-1C2RL 



