What is the path attenuation between transmitter and receiver at a frequency of 1.2 GHz and a distance of 11,000. Given: f = 1.2GHz d = 11,000 mi. Calculate the path attenuation; dB = 37dB + 20 log f + 20 log d = 37 + 20 log 1.2^9 + 20 log 11,000 (0r would it be, 37 + 20 log 1.2 + 20 log 11,000) = 37 + 181.58 + 80.83 = 299.41 dB