TRANSMISSION LINE EQUATIONS 139 



Second Method: Compute the propagation constants 7i , 72 , ••• jm 

 and the square matrices Nijr) given by (1.23) as in the second method for 

 the infinite line. Then 



cosh -iT = 2 iV(7r) cosh XJr 



sinh .vr Zo = sinh xT TV~^ 



= S iV(7^) sinh :V7r7rF~^ (1.30) 



sinh .vr' Z-' = sinh .iT' T'~'Y 



It 



where iV'(7r) is the transposed of i\''(7r), A^(7r) being defined by (1.23), 

 and the summations extend from r = 1 to r = m. 



Wlien the transmission Hne consists of perfectly conducting wires strung 

 on perfect insulators over a perfectly conducting earth the magnetic and 

 electrostatic fields are related so as to make Z equal to 70 F where 



7o = /co/c, 



CO being 27r times the frequency and c the speed of light. 



It is interesting to apply the first method of solution to this line. Even 

 though the proof of the first method, which is given in §1.10, does not cover 

 this case there seems to be little doubt that the correct answer is obtained. 



We have 



ZY = 7o/ 

 Choosing 7 = 70 gives 7? = and therefore S = 0. It follows that 



r = 7J, z<, = r~'z=: y-^z 



cosh xY = cosh (.V7o/) = cosh X'^o I 

 sinh xV Zo — sinh x-yo 'U ^ 

 sinh xV Z~ = sinh xyo loZ 



When these are put into equations (1.25) the expressions for v{x) and iix) 

 in a perfect transmission line are obtained: 



v{x) = cosh xyov(o) — " Zi{o) 



To (1.31) 



i{x) = —Jo sinh xjoZ" v{o) + cosh xyoi{.o) 



