REAL GENERATORS 



Since the problem has left-right symmetry, it is a reasonable guess that 

 the resistances in the black-box will have left-right symmetry too. A series 

 resistance alone, or a parallel resistance alone, have this symmetry, but 

 clearly they will not do. What about a combination of the two ? A sym- 

 metrical arrangement worth trying is the star or T network, and another is 

 the delta or pi network. Since we have seen that any delta can be replaced 

 by an appropriate star, clearly if we can get one to do what is required, the 

 other will do it too. 



Let us then investigate the T network used as an attenuator. In Figure 2.28, 

 if we write down the resistance looking in at the input terminals of the 



r-W\'< 



RM 1 \R2 s/?r--r; 



Figure 2.28 

 attenuator with the load connected, and equate it to the generator resistance 



the matching conditions are met. Further, if the transmission factor of the 

 attenuator is d, then 



Kin ~ VpQ ' Fin 



and it is not too hard to see that this equals 



R2(Ri + r) 

 r R^ + R^ + r 



Ri' + r- R,iRr' + r) 



^1 + /?2 + i?/ + r 



In general we know r and 6 and have to find R^ and i?2- Solving the equations 



we find 



l-d 

 R,=r 



and R2 = r 



1 + 6 

 20 



1-02 



A variant of the 7 attenuator which appears at first sight more complicated 

 than it really is, is the 'balanced' or H form (Figure 2.29). All that has been 

 done to derive this is : 



(1) to take half of R^ and connect it in series with the other side of the 

 generator; 



(2) to take half of R^ and connect it in series with the other side of the 

 load. 



17 



