BELOW THE INTERFERENCE REGION 



107 



To convert 20 log A to 20 log A by ecjuation (175), 

 we need 20 log !h imd 20 log gi, wbioh are given 

 bj' Figure 41 : 



20 log /?! = 41.3, 



201ogffi= 1.5. 

 Hence 



20 log .4 = - 1 70, 

 and by equation (182) 



10 log— = 

 Pi 



170 + 90+ 10 = -04. 



= io-iSv. 



Since Pi is 10^ watts, 



p, = 10= X lo-''-^ 



Radar: Substituting in equation (183), the radar 

 gain is given by 



10 log (P2/P1) = -340 + 2(96) + 7.5 + 16 - 9.5, 

 = -134db 

 or 



P, = Pi X 10-^^1 



The power output Pi is 10^ watts, so that the max- 

 imum received power 



P2 



10" 



The minimum detectable power of the set is given as 

 1.6 X 10"^ = 10"'-Svatt, so that under the given 

 conditions the power returned b}^ the target would 

 be slightly below the threshold of detection. 



Type II. Gain versus receiver (or target) height 

 is to be found for given distance, given wavelength. 

 and given transmitter height : A radar has an antenna 

 height of hi = 30 meters, a wavelength of X = 1.5 

 meters and a distance from a receiver (or target) of 

 d = 100 km. Assiuning a receiver antenna gain 

 ^12 = 1, the variation of P2/P1 at the receiver with 

 receiver height is to be found. Also, assuming a 

 target of cross section c = 50 sq meters, the varia- 

 tion of P2/P1 at the radar receiver with target height 

 is to be found. The radar antenna has a gain of 

 13.5 db. 



One-ivay: 



sd = 3.9 (from Figure 40). 



From Figure 37, for the fixed ^-aiue of sd = 3.9, we 

 find a correspondence between values of 20 log A and 

 e/i2, listed in Table 5 below. By means of Figure 41 

 or equation (159), eh^ is changed to fh and by means 

 of eciuation (176), A to A. From Figure 41, it is seen 

 that 20 log hi = 29.5 and 20 log gi = 0. To change A 

 to P2/P1, the transmitter gain of 13.5 db and the 

 receiver gain of db must be taken into account, 

 according to ec[uation (182). The result is given in 



Table 5. The values of 10 log (P2/P1) are plotted in 

 Figure 25, together with the results found with the 

 same data in Section 5.6.3 for the optical-inter- 

 ference region. 



Table 5* 



See also Table 1 and Figure 25. 



Radar: 



P-> 



10 log — = 40 log A + 27 + 7.5 + 10 log a 

 ^' -20 log 1.5, 



= 40 log A + 27 + 7.5 + 17 - 3.5, 



= 40 log A +48. 

 Type III. Gain ver.sus distance is to be found, 

 with antenna heights and wavelength given: Using 

 the same data given in Section 5.6.4, the gain as a 

 function of distance in the diffraction region is to be 

 found. The result has been plotted in Figure 26. 

 The polarization is horizontal. 

 hi = 30 meters 

 h = 1000 meters 

 X = 1.5 meters 



From Figure 41, 



Gi = 22.4 (13.5 db) 

 (?2 (one-way) = 1 (0 db) 

 (?2 (radar) = 22.4 (13.5 db) 

 (7 = 10 square meters 



eh-i = 12.5. 



Referring to Figure 37, we find a correspondence 

 between A and sd. Values of A are to be assumed. 

 To change sd to d, use Figure 40. To change A to A , 

 use equation (176). From Figure 41, 



20 log 30 = 29.5, 



201ogffi =0, 



20 log A = 20 log A + 29.5 + 0. 



The radio gain is then given by : 



One-way: 



10 log ^ 

 Pi 



20 log A + 13.5 + 0; 



The radar gain is then given by: 

 Radar: 



10 log— = 40 log A + 7.5 + 27 + 10 - 3.5 

 Pi 



= 40 log A + 41. 



