HIGH IREQUENCY AMP/JEIER 39 



than if there were held in the x direction only and equation (11) were vah'd. 

 Combining (16) with (13) we obtain 



w tie re 



1 1 



.o^t)-^)" (^(^ + 'y:^"'''" '''' 



I- = ;r"2 ^2 77 (^8) 



Za eoPn Mo I o 



In solving (17) it is most convenient to represent /3 in terms of (So and a 

 new variable 5 



IS = ^o(l + 5) (19) 



Thus, (14) becomes 



1 1 1 



+ 



,„.v'(,,.v-.= 



Solving for d, we obtain 



The positive sign inside of the brackets always gives a real value of d 

 and hence unattenuated waves. The negative sign inside the brackets 

 gives unattenuated waves for small values of U/b. However, when 



j) > I (^^) 



there are two waves with a phase constant 0o and with equal and opposite 

 attenuation constants. 



Suppose we let Um be the minimum value of U for which there is gain. 

 From (22), 



t/A/2 = b'/8 (23) 



From (21) we have for the increasing wave 



The gain in db/wavelength is 



db/wavelength - 20(2x) logio e\8\ (25) 



= 54.6 I 5 I 



i = jj 



lUj[A/i + s[r^) - 1 ) - 1 



