REFLEX OSCILLATORS 515 



bounds a region of zero power commonly called the "sink," since all the 

 frequency contours converge into it. The other zero power boundary is 

 the outer boundary of the chart, Gi = 0, which, of course, is an open circuit 

 load. The power contours on this chart occur in pairs, except the maximum 

 power contour which is single. These correspond to coupling greater than 

 and less than the optimum. 



The value of Gi for any given power contour for A0 = may be deter- 

 mined by referring to Fig. 9. We are assuming no resonator loss so we use 

 the curve for which Gulje = 0. From (9.5), ii p = 1 we have Gt/N^ye = 

 2.5/X- which, substituted in (9.3), gives XJi{X) = 1.25. This is just the 

 condition for maximum power output with no resonator loss. From this 

 it can be seen that we have chosen a set of normalized coordinates. Hence, 

 in using Fig. 9, we have p = H/Hm, where Hm = .394 is the maximum gen- 

 eralized efficiency. Thus, for any given value of p we let H in Fig. 9 have 

 the value .394/> and determine the two values of Gi corresponding to that 

 contour. 



From Fig. 32 we can construct several other charts describing the per- 

 formance of reflex oscillators under other conditions. For instance, sup- 

 pose we make M other than zero. Such a condition commonly occurs in 

 use either through erroneous adjustment of. the repeller or through inten- 

 tional use of the electronic tuning of the oscillator. We can construct a 

 new chart for this condition using Fig. 32. Consider first the constant 

 power contours. Suppose we consider the old contour of value pn lying 

 along a conductance line Gin . To get a new contour, we can change the 

 label from pn to pm = pn cos A0, and we move the contour to a conductance 

 line Gn = Gm cos A0. That this is correct can be seen by substituting these 

 values in (9.9). Consider a given frequency contour lying along Bi . 

 We shift each point of this contour along a constant conductance line Gi„ 

 an amount B^ = Gin tan M. It will be observed that this satisfied (9.11). 

 In Fig. 33 this has been done for tan M — \, cos A0 = ■s/ll'l. 



Now let us consider the effect of resonator loss. Suppose we have a 

 shunt resonator conductance Gr . Let 



G. = Gnhe. (9.12) 



Then, if the total conductance is G„ , the fraction of the power produced 

 which goes to the load is 



/ = {Gn - G,)/Gn = Gi/(Gi + G,) (9.13) 



accordingly, we multiply each power contour label by the fraction/. Then 

 we move all contour points along constant susceptance lines to new values 



G„. = Gn- G2 (9.14) 



In Fig. 34, this has been done to the contours of Fig. 32, for G-z = .3. 



