CALCULATIONS FOR OPTICAL-INTERFERENCE REGION 



85 



The value of P2/P1 [equation (3)] is represented 

 graphically in Figure 25. 



Figure 25. Radio gain in decibels versus height h^. 

 Horizontal polarization. 



Radar Gain: Two- Way Transmission 



Knowing the values of 20 log A, the corresponding 

 values of Pi./ Pi are given by equation (5). Taking 

 Gj = G2 = 13.5 db, 167r/9 = 5.58, X = 1.5, 



10 log— = 27-1- 7.5 -h 17 -f- 40 log ^ - 20 log 1.5, 



Pi 



= -|-401ogA-|-48. 



5.6.4 Type III. Radio Gain Versus 



Distance for Given Antenna Heights 



A radar used over the sea has a wavelength of 

 1.5 meters. The transmitter is 30 meters above 

 sea level and the target is at an altitude of 1,000 

 meters above the sea. The power gain of the trans- 

 mitting antenna is 13.5 db, the polarization hori- 

 zontal. The one-way radio gain is to be found as a 

 function of distance. Also, the radar gain at the 

 radar set by echo from the target of cross section 

 o- = 10 square meters is to be calculated. 



Radio Gain: One- Way Transmission (see Figure 26). 



1. The number r from Figure 15 in Chapter 6 or 

 equation (115) is found to be 9.403. 



2. u = hi/h = 33.3. 



3. For r > 2 and n = 1, p is approximately equal 

 to 1/r. Hence we start with -p = 0.1. 



4. Equation (121), with p = 0.1 and u = 33.3, 

 gives y = 2.76. 



5. From equation (112), fi = 9.445. 



6. ?i = R/r = 1.005. Hence the target is on the 

 first (i.e., lowest) lobe. Moving in the direction of 

 increasing distance, the target soon approaches the 

 first maximum. To get points beyond, i.e., n < 1, 

 we need greater values of p but it is not necessary or 

 desirable to go below about n = 0.8, since the curve 

 beyond n = 0.8 is generally in the transition region 

 in which the curve is more easily and more accurately 

 obtained by joining the optical and diffractive curves. 

 The diffractive part of the calcidation is given in 

 Section 5.7. Accordingly, in Table 2, the values of p 

 are taken only slightly above p = 0.1 (correspond- 

 ing to n = 1) and are diminished to find points at 

 the nearer distances. 



7. To find A, we need first the free-space value Ao, 

 which is given in Figure 3 in Chapter 2 as a function 

 of d. Since dy = 4.12 V30 = 22.6 km, the value of 

 d corresponding to y = 2.76 is d = vd^ = 62.3 km, 

 and 20 log ^0 = - HI. 



8. To find the value of 



lOloi 



^a- 







we need D. Since n is practically unity, correspond- 

 ing to the first maximum, sin- (0/2) may be taken as 

 unity. Hence the radical reduces to 1 + D. Calcula- 

 tion using equation (103) and Figure 17 gives 

 D = 0.98, and hence 



1 + D ^ 1.98, 

 20 log {1 + D) = 6. 



Since the transmitting antenna gain Gi is 13.5 db, 

 and assuming the receiving antenna gam to be db, 

 10 log (Po/Pi) has the value 20 log A + 13.5 or 



10 log p' 



= 4o^^(l- 



D)- + 4D sin^ - CnG. + 0.0 db 



llldb -f6db +13.5db 



= - 105-f 13.5 = - 91.5 db. 



