CONSTANT RESISTANCE NETWORKS 181 



With a voltage £o applied to the common terminals the power 

 absorbed by the first network is E^^Gi and by the second is E^^Gi. 

 Since in both cases the power delivered to the network must be 

 absorbed in the single resistance, the two insertion losses are given by 



e-2«2 = G2 = ^- • (4) 



1 H — 



Since F{\) must be an even function of X, the poles of (3) may be 

 written ± c„i ± idn and (3) is 



^ ^ m=(n— 1)/2 A A 



e-2a, = _ 2)2 _^_^ n , , ' • , T —r^ (5) 



A + To A — fo 1 A -|- Tm ± iCm A — f m ± Mm 



if the degree of F{\) is 2n. If w is even, the terms X + Co and \ — Cq 

 are omitted and the product taken from w = 1 to w = njl. The 

 quantity D"^ is the denominator of F{\), a polynomial in X^. 



Let jSi be the phase angle between Eo and the voltage Ei across the 

 resistance in the first network. The left side of equation (3) may 

 then be factored in the form e~'^"^ — e-(ai+'^i)e-(«i-j^i\ Similarly half 

 of the factors on the right belong with e-(«i+'^i) and half with e~(«i~*''*i^ 

 Now the terms with poles having a negative real part^ must belong 

 with e-^^i+'^i^ so that: 



e-(«i+i^i) = D-A — n ^ 



X + f X + r^ ± idm 



" ^X + c^cj + dj + X2 + 2c„X * ^^^ 



Since X is an imaginary function of frequency, say X = ix, and Z) is 

 real if X or x is real, the phase /3i is given by 



jn=(n— 1)/2 ^ 



<- 7«=1 '" 1^ "• 



If w is even the expression is 



5, = E tan-i , ^"^ _ ■ (7a) 



' These being the factors that correspond to physically realizable network elements, 

 they belong with the physically realizable factor of the exponent. 



