642 THE BELL SYSTEM TECHNICAL JOURNAL, JULY 1951 



Also, at the beginning of the helix, a-c. velocity and the a-c. convection 

 current must be zero. This means that 



• ^ + ^ + ^ = (3.5) 



Ol 02 03 



Eio E20 Ezo ^ . ,. 



-^ + -TT + -72- = (3.6) 



Ol O2 O3 



For the case we have considered, /Si = jSo , jSi real, 



63 = 5ie+^'^^'^/^^ 



and our equations become 



£10 + JE20 + -E30 = -Eo 



iiio + -c,2o6 -r -C,30 = 



iiio -h ii20 ^ + -c,30 = U 



We easily see that the solution is 



jEio = £20 = -E30 = 3 -Eo (3.7) 



If E is the field at a distance 2 along the helix 



E = \ Eoe-'^''ie''' + e''' + e'^') (3.8) 



In Fig. 3.1, 



20 logi 



_E 



Eo 



is plotted vs CN, a factor proportional to distance. 



We see that initially the ampUtude does not change. This is necessarily 

 so. The strength of the field can grow only through the electron stream 

 giving energy to the circuit. The electron stream can give energy to the 

 circuit only if it has an a-c. convection current. Initially the electron stream 

 is unmodulated and hence it can give energy to the circuit only after it has 

 traveled far enough to become modulated. 



In the case we have considered, the amplitudes of the three wave com- 

 ponents of the field are initially equal. Now, £1 increases with distance, 

 while £2 decreases with distance and £3 is unattenuated. Hence, if the tube 

 is long enough, £2 and £3 will be negligible near the output of the tube; 

 and the field at the output, a distance t from the input, will be very nearly 



E = lEoe-^^'^e'^^ 



