ELECTROMAGNETIC THEORY OF LINES AND SHIELDS 569 



This equation, being of the second order, possesses two independent 

 solutions : one for diverging cylindrical waves and the other for reflected 

 waves. The ratio oi E to H in the first case and its negative in the 

 second will be called the radial impedance offered by the medium to 

 cylindrical waves. 



In the next section we shall determine radial impedances in di- 

 electrics and metals and show that for all practical purposes the 

 attenuation of cylindrical waves in metals is exponential. The sig- 

 nificance of the radial impedance is the same as that of the charac- 

 teristic impedance of a transmission line. When a cylindrical wave 

 passes from one medium into another, a reflection takes place unless 

 the radial impedances are the same in the two media. Thus if Eq 

 and Hq are the impressed intensities (at the boundary between the two 

 media), Et and Hr the reflected and Et and Ht, the transmitted 

 intensities, we have 



E, + Er = Et and //o + Hr = Ht, (110) 



since both intensities must be continuous. On the other hand, if k is 

 the ratio of the impedance in the first medium to that in the second, 

 then equations (110) become 



kHo - kHr = Ht and Ho + Hr = Ht. (Ill) 



Solving we obtain 



2k 1 



Ht = ^^^^Ho and Et = ^^ ^o- (112) 



The reflection loss will be defined as 



R = 20 logiof^ decibels. (113) 



When a wave passes through a shield, it encounters two boundaries 

 and if the shield is electrically thick, that is, if the attenuation of the 

 wave in the shield is so great that secondary reflections can be dis- 

 regarded without introducing a serious error, the total reflection loss 

 is the sum of the losses at each boundary. The first loss can be com- 

 puted directly from (112) and the second from the same equation if 

 we replace k by its reciprocal. Thus, the total reflection loss for 

 electrically thick shields is 



R = 20 logio-^-^y^' decibels, (114) 



