RADAR ANTENNAS 229 



the maximum power flow per unit area occurs in a direction normal to it and 

 is given by 



_ 3W w^atts ,,_. 



SttH meter^ 



where T'F is the total radiated power. If W had been radiated equally 

 in all directions the power flow per unit area would be 



p ^ W_ watts .gv 



47rr2 meters^ 



It follows that the gain of the small current element is 



p 



Gdiople = -^— = 1-5 (9) 



The effective area of the dipole will now be calculated. When it is used 

 to receive a plane linearly polarized electromagnetic wave, the available 

 output power is equal to the induced voltage squared divided by four times 

 the radiation resistance. Thus 



Pn = ^ Watts (10) 



4i?rad 



where E is the effective value of the electric field of the wave, i is the length 

 of the current element and i?rad is the radiation resistance of the current 



element. From Fig. 5 we see that i?rad — , ohms. Since the power 



A" 



flow per unit area is equal to the electric field squared divided by the im- 



pedance of free space, in other words Po — tt— we have 

 u. 1 ZOir 



P ^X^ 



^dipoie = ^ = ^- meter" (11) 



We combine formulas (9) and (11) to find that 



6^dipole _ 4t 

 ■^dipole A 



Since, as proved in 2.2 this ratio is the same for all antennas, it follows that 

 for any antenna 



^=^ (12) 



