STABILIZED FEEDBACK OSCILLATORS 



465 



which requires the product K1K2 to be negative and, hence, that both 

 image impedances be reactances of the same sign. Since the condition 

 expressed by equation (15) is dependent upon the values of the tube 

 resistances, it follows that frequency stabilization cannot be obtained 

 at frequencies in an attenuation band. 



The problem of devising stabilized reactance type oscillator circuits 

 therefore resolves itself into that of obtaining band-pass coupling net- 

 works which have phase constants of 180 degrees at frequencies within 

 the pass-bands. Evidently there is a multiplicity of known filter 

 structures that meet the requirements and also many other networks of 

 similar character including all-pass reactance networks. Since each 

 half -section of a filter gives a phase shift of 90 degrees in the pass-band, 

 it follows that the coupling network should be equivalent to at least 

 three half-sections. More complex networks may be used, but with 

 networks equivalent to more than six half-sections oscillations may 

 occur at more than one frequency. 



Illustrative Examples 

 The principles discussed in the foregoing section will be illustrated 

 by a consideration of the circuit shown in Fig. 3, in which the coupling 



Fig. 3— Oscillator with plate circuit stabilizing impedance. 



network is a simple ladder network having four reactive branches. 

 A screen-grid vacuum tube is assumed so that the grid-to-plate capaci- 

 tance is negligibly small and all feedback is confined to the coupling 

 network. For this circuit, the determinant D has the value 



D = XsiXoX, + X1X2 -f X2X0). (16) 



The several minors are 



^11.22 = Xi + X,-\- X3, (17) 



2^11 = Xs{Xr-i-X,), (18) 



£>22 = Xo(Xi + X2 + X3) + Xi{X2 + X3) (19) 



