108 



CALCULATION OF RADIO GAIN 



These equations are evaluated in Table G and the 

 one-way values are plotted in Figure 26. 



Therefore 



Table 6 



Type lY. The determination of contours along 

 which the radio gain (or A) i.s constant (the coverage 

 problem): A radar has a wavelength of 0.107 meter 

 and a power output of 750 kw. Assume a receiver in 

 space with a minimum detectable power of 10"'° 

 watt. The maximum possible distance between the 

 radar transmitter whose elevation is 100 meters and 

 the receiver for varying heights of the receiver is to 

 be found. The gain of the radar antenna is 10,000 

 (or 40 decibels), the gain of the receiver will be 

 assumed to be 30 decibels. 



For the radar problem, a target of radar cross 

 section o- = 50 sqiuire meters is assumed to take the 

 place of the receiver. The minimum detectable 

 power of the radar is taken as 10"'°w^att; the range 

 of the set for varying altitudes of the target will be 

 calculated. 



One-way: The radio gain sought is the ratio 

 of the minimum detectable power to the power 

 output, or 



P,_ 10-'° 



Pi 



= 4x 



750 X 10' 3 



10'- 



or 



101og(P2/Pi) = -160+ 1 = -159db. 



From equation (3), 



20 \ogA = - 159 - 40 



30, 



= - 229 db. 



From Figure 41, 



20 log g. 



18.5, 

 20 log /)i = 40. 



20 log A 



-229 - 40 

 -287.5 db. 



18.5, 



Referring to Figure 37, the pairs of values of sd and 

 e/io along the contour 20 log A = —287 are given 

 in Table 7. By means of Figures 40 and 41, sd and 

 eJh are changed to d and /)2- The points found are 

 to the right of the curve for ehi = 7.3, so that they 

 correspond to points in the diffraction region. 



Table 7 



Radar: The value of 10 log {P2/P1) is the same as 

 for the one-way calculation, —159 db. This must 

 be changed to 20 log .4 by equation (5), 



20 log A = -141 db, 

 and, as above, 



20 log i = -141 - 40 - 18.5, 

 = -199.5 db. 



Referring to Figure 37, we see that the contour 

 20 log i = - 199.5 is to the left of eh = 7.3 [see 

 caution in equation (181)]. Therefore it is not possible 

 to get the necessary power return Po from the given 

 target so long as it is below the line of sight. The 

 desired contour would lie above the line of sight. 

 The determination of the contour is discussed in 

 Section 5.6.5. 



5.7.4 



Sea Water, VHF, 

 Vertical Polarization 



1. Graphical Aids. Graphical aids are given in 

 this section, which, as in Section 5.7.3, are valid for 

 all practical distances when both antennas are low 

 [h < h, (see Figure 35)1 and 2hih < < Xrf. If one 

 or both antennas ai-e elevated, they will give the 

 A-alue of the radio gain for the first mode, which is a 

 good approximation for the result found by summing 

 all the modes when the receiver is well within the 

 diffraction region, i.e., when d > dj^. [If one antenna 

 is low- {h < h,.) and one U'cll elevated {h > 40/?,.), 

 the result found by using se\'eral modes is gi^•en in 

 Section 5.7.5.1 



