• Compute length of the line-of-sight.
  d ²+ (h_t-h_r)² = 1,000,784.
  Taking the square root gives the length: 1000.3919 meter

• Compute length of the ground reflected wave.
  d ²+ (h_t+h_r)² = 1,001,024.
  Taking the square root gives the length: 1000.5119 meter

• Compute the phase difference due to path length differences.

  
  1000.5119 meter - 1000.3919 meter = 0.12 meter
  At a wavelength l = c / f = 0.3 meter, this phase difference correspondes to 0.4 l, or 144 degrees.

• Compare this with the approximation.
  

  For large propagation ranges, i.e., for large d, one finds the approximate expression

  ,

  This is 2p/l times 0.12 meter.

  How significant is the difference?

  The approximation is quite accurate.

• At which distance would the phase difference be 2p?

  For d = 2 h_th_r/l = 400 meter. As the reflection coefficient R is typically -1 (180 degree phase rotation), at this range the line-of-sight and the ground reflection cancel each other. Poor reception has to be anticipated at this spot.

• Find the path loss at 500 and 1,000 meters from

  

  • For d = 1000 m: P_R/P_T = (0.3)²/(4000p)² *[2 sin (2p* 30 * 2 /(0.3 *1000) ) ]²
    P_R/P_T = (3/40,000p)² *[2 sin (0.4 * p ) ]²
    P_R/P_T = 2.38 10^-5 *[2 sin (1.256) ) ]²
    = 2.38 10^-5 *[2 * 0.95) ]²= [2.38 10^-5]² *[2 sin (1.256) ) ]²= [5.7 10^-10* (1.9)²=2.06 10^-9
    The attenuation is 4.86 10⁸ (87 dB)

  • For d = 500 m: P_R/P_T = (0.3)²/(2000p)² *[2 sin (2p* 30 * 2 /(0.3 *500) ) ]²
    P_R/P_T = (3/10,000p)² *[2 sin (0.2 * p ) ]²
    P_R/P_T = 9.12 10^-9 *[2 sin (0.8) ) ]²
    = 9.12 10^-9 *[2 * 0.59) ]²= 9.12 10^-9 * 1.38 = 1.26 10^-8 The attenuation is 7.95 10⁷ (79 dB)

JPL's Wireless Communication Reference Website © Jean-Paul M.G. Linnartz, 1998.