584 
14 
sy Z,— Zo 
T= Zaza and 
my Z,— Zo 
2 Zo +2 
It is the quantity E(l) which is equal to FE, that is important. 
This represents the output voltage of the cable and therefore the input volt- 
age for the amplifiers. 
mais E,Z,)(1 + r,) 
eT AEE ale” = ri tee [2] 
The assumption will now be made that the line has no dissipation; 
i.e., R =G=0.* Since C is approximately 40 x 107? farad per foot, G is 
unimportant above 20 or 30 cycles per second. The approximation is not as 
good for R since this quantity is approximately 0.01 ohm per foot for Number 
20 copper wire, while L is approximately 1.2 x 107% henry per foot. However, 
good agreement with experiment is obtained even when RF is neglected. This 
approximation simplifies the equation considerably. 
a= jwVLC; Z,= V4 [3] 
Z, is now a pure resistance. . 
E,Z)(1 + 72) 
eas anne 
(eee 7 teen a 
Tay antiaes 
It is to be noted that even for a dissipationless line there will 
be frequency distortion; i.e., EH, is still a function of the frequency w/2n. 
However, if the line were terminated at both ends by impedances equal to the 
surge impedance, j.e., if Z, = Z, = Z,), then r, =r, = 0. 
Bj Big ets [5] 
and 
og = B, ai" = 2a gee) [6] 
The amplitude is now independent of the frequency, while the phase 
is altered only by having a constant subtracted from the time. Thus if the 
dissipationless line is terminated by its surge impedance, it becomes a 
=1 
aG, £10 4 mho per foot for the cables used. 
