TRANSIENT SOLUTIONS OF ELECTRICAL NETWORKS 121 

 II. Solution for M Sections of All- Pass Lattice Network 



The process for solving any type of problem is to replace any 

 resistance and inductance in series by i?Oi(l — e~^P')/2, and any 

 capacity and leakance in parallel by (1 — e~~P^)/2Ro,, where Rq^ -> oo ; 

 i?o, -> ; Ro,Pi = R+ jc^L ■ R0JP2 = 1/(G+ jcoC) / 



The next problem considered here is the solution for any number of 

 sections of all-pass lattice type network^ as shown' on Figure 4. 



■^mwy—^mwy 



R=Y^ 



R=V^ 



Fig. 4 — Sections of all pass lattice network. 



These networks have the property of passing all frequencies without 

 attenuation, and they are much used as phase equalizers. 



The steady state equation for the current at the end of the first 

 section, when this section is terminated at each end by a resistance 

 R = VZ^, is _ _ 



E r -Vz2 - Vzi i 

 2^LVz; + Vz;J' 



^l 



The current flowing out of the wth section of such a structure takes 

 the form 



• -A 



'™ ~ 2R 



iZo - VZi 



_ (35) 



VZ2 + -VZi 



In the structure considered Zi = juiL; Zi = l/j'coC and yLjC = R. 

 In accordance with section I-B, we replace the inductance by a short 

 circuited line, and the capacity by an open circuited line. For the 

 first line, in the limiting case, we have by equation (15), 



Ro -^ <x> ; RqP = jojL. 



For the second line, by equation (18), we have 



i?o->0; —=-^7^ 



Ro 

 P 



jwC 



There is no loss of generality if we take the propagation constants for 

 3 See for example B. S. T. J., July 1928, page 510. 

 9 



