106 



THE BELL SYSTEM TECHNICAL JOURNAL, JANUARY 1951 



Za4 = 



25 



+ 



2R, 



(1 - Ml) 



+ 



2X2(1 + M2) 



(1 - Ml) 

 4klX, (1 + M2) 



+ 



[1 + ^2 - 2kl] 



X, (1 - Ml) (Zl + i?6 + X3) 



Eq. (17) 



From Equation (17) the circuit of Fig. 10(a) can be developed in a manner 

 similar to the way Fig. 9(a) was developed from Equation (12). Furthermore, 

 Fig. 10(a) can be rearranged in the forxH shown in Fig. 10(b) . 



4)^3^X2 Re 



Xa 2R2 



A kjXzZL g 4*32X2X3 

 X3 



(a) 2X2(1 + ^2-2^3^) 



2R 



T 



EQUIVALENT 



CIRCUIT OF 



TRANSFORMER 



MULTIPLY-*— 



BY 1 



/ X3^ 

 /4X2/fe3' 



(b) 



Fig. 10 — Equivalent circuit of equation (17). 



A comparison of Fig. 9(b) with Fig. 10(b) shows these two circuits to be 

 essentially the same except for the point of view. If it is remembered that 

 Zn in Figs. 8, 9 and 10 has been chosen to include the elements in the RC 

 network which couples the plate of one vacuum tube to the grid of the 

 other, then from consideration of Figs. 9(b) and 10(b) it is obvious that 

 Fig. 6(b) is the equivalent circuit of Fig. 6(a). 



If Equation (6) or Equation (7) were studied it would be found that for 

 all conditions of circuit stability the current h would flow in the same 

 general direction in which it would be expected to flow in any circuit mesh 

 where all impedances were positive. But Zn can equal a negative impedance. 

 Hence, Zn must be a reversed voltage type of negative impedance equal to 

 — VII. For an impedance to be negative either the voltage V at some 

 frequency must be the reverse in polarity of that measured across a like 

 positive impedance or the current must flow in the reverse direction. If the 



