The total distance between the transmitting and receiving antenna of a
microwave link at 10GHz, is 30 Km. the height of the Tx antenna is
above ground level is 20 m. the maximum acceptable total path loss is
169 dB. Furthermore there is hill located 10 km away from the
transmitter antenna with a height of 80m.

calculate the height of the receiver antenna for the path loss to be
just equal to the maximum acceptable value?

1. Antenna essentially operates in free space with no near field ground
losses because the wavelength is extremely small (3.3cm) compared to 20m
antenna height at the transmitter.

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
.008 degrees, thus the transmitter hardly "sees" it.

3. The path loss seen by an atenna at 0 feet is then 131.8 dB (Path Loss =
20log(4*pi*d/lambda)), which is much less than the desired 169dB maximum.

Answer: 0 Feet.

Definitely NOT 0 feet.

Even without the hill in the way the curvature of the earth means that the
radio horizon is at about 20km with the tx on a 20m mast. 0.6 Fresnel
clearance occurs at about 6km. and the path loss exceeds 180dB.!!
The Rx mast needs to be at about 25m to obtain a 0.6 fresnel zone clearance,
WITHOUT a hill.This would give about 150dB path loss.

Adding the 80m hill at 10km gives a single knife edge diffraction, that
increases the path loss enormously to about 200dB!!

This path loss does not change significantly until the Rx antenna height
gets so large that near line of sight is achieved. That is over 100m!! To
obtain the 169dB required figure the mast height would have to be about
250m!!!!!!!

Regards
Jeff