# Propagation over a Plane Earth

If we consider the effect of the earth surface, the expressions for the received signal become more complicated than in case of free space propagation. The main effect is that signals reflected off the earth surface may (partially) cancel the line of sight wave.

## Model

For (theoretical) isotropic antennas above a plane earth, the received electric field strength E is

with Rc the reflection coefficient and E0 the field strength for propagation in free space. For AM broadcasting propagtion models these full expressions are relavant and have been used to create frequency plans for mediumwave and long wave broadcasting.

For VHF and higher frequency, this expression can be interpreted as the complex sum of a direct line-of-sight wave, a ground-reflected wave and a surface wave. The phasor sum of the first and second term is known as the `space wave'.

For a horizontally-polarized wave incident on the surface of a perfectly smooth earth,

,

where er is the relative dielectric constant of the earth, Y is the angle of incidence (between the radio ray and the earth surface) and x = s/(2 p fc e0) with s the conductivity of the ground and e0 the dielectric constant of vacuum. So, x = s/(we0)=18 109s/f.

For vertical polarization

 Surface Conductivity s Relative Dielectric Constant er Dry Poor Ground 10-3 4-7 Average Ground 5 10-3 15 Wet Good Ground 2 10-2 25-30 Fresh Water 10-2 81 Sea Water 5 81

### Exercise

Show that the reflection coefficient tends to -1 for angles close to 0. Verify that for horizontal polarization, abs(Rc) > 0.9 for Y < 10 degrees. For vertical polarization, abs( Rc) > 0.5 for Y < 5 degrees and abs( Rc) > 0.9 for Y < 1 degree. Use the special purpose embedded calculator.

## UHF Mobile Communication

The relative amplitude F(.) of the surface wave is very small for most cases of mobile UHF communication (F(.) << 1). Its contribution is relevant only a few wavelengths above the ground. So, mostly a two-ray model is considered.

 Two-ray model for UHF propagation over a plane reflecting earth. Transmit antenna height : ht Receive antenna height : hr Path length : d Reflection coefficient : R Line-of-sight (red) and a ground reflected wave (orange) arrive at the receive antenna. These two rays are out of phase due to differences in path length and due to phase shifts at the reflection area.

The phase difference between the direct and the ground-reflected wave can be found from the two-ray approximation considering only a line-of-sight and a ground reflection. Denoting the transmit and receive antenna heights as hT and hR, respectively, the phase difference can be expressed as

### Exercise

Consider a base station antenna with hT= 30 meters and a mobile antenna with hR= 2 meters. The mobile is d = 1,000 meters (1 km) separated from the base station. The carrier frequency is 1 GHz.
• Compute length of the line-of-sight.
• Compute length of the ground reflected wave.
• Compute the phase difference due to path length differences.
• Compare this with the approximation proposed in the next paragraph. How significant is the difference?
• At which distance would the phase difference be 2p?
Do the exercise and use a JavaTM calculator

The exact expression for the phase difference is not very convenient for further analysis. For large propagation ranges, i.e., for large d, one can use an approximation based on

This results in the approximate, but mostly quite accurate phase difference

,

This, and an approximation for the reflection coefficient give an often used two-ray model for plane earth propagation. For large d, (d >> 5hT hR ), the reflection coefficient tends to -1, so the received signal power becomes

Here, the distance

acts as a turnover point. For this distance, the argument of the sine becomes equal to p/4. For carrier frequencies around 900 MHz or 1 GHz, a base station height of 30 meter and a mobile antenna height of 2 meter, the turnover distance is about 100 meters. Hence for macro-cellular systems, distances beyond the turnover distance are more relevant.

The full path loss expression shows an interference pattern of the line-of-sight and the ground-reflected wave for relatively short ranges, and a rapid decay of the signal power beyond the turnover distance. See also: a plot of loss versus distance for various antenna heights.

For propagation distances substantially beyond the turnover point path loss tends to the fourth power distance law:

### Exercise

Discuss the effect of path loss on the performance of a cellular radio network.
Is it good to have signals that attenuate rapidly with increasing distance?

## Egli's Model (1957)

Experiments confirm that in macro-cellular links over smooth, plane terrain, the received signal power (expressed in dB) decreases with "40 log d". Also a "6 dB/octave" height gain is experienced: doubling the antenna height increases the received power by a factor 4. These two effects are correctly predicted by the two-ray model.

However, in contrast to the theoretical plane earth loss, Egli measured a significant increase of the path loss with the carrier frequency fc for ranges 1< d < 50 km. He proposed the semi-empirical model

i.e., he introduced a frequency dependent empirical correction (40 MHz/fc)2 for carrier frequencies 30 MHz < fc < 1 GHz.

For communication at short range, this formula looses its accuracy because the reflection coefficient is not necessarily close to -1. For d << hT hR / 4l, free space propagation is more appropriate, but a number of significant reflections against the earth surface must be anticipated. However, these are often hard to distinguish from reflections against obstacles in the vicinity of the antenna site. Moreover, in streets with high buildings, guided propagation may occur.

### Input parameters:

Carrier Frequency:  MHz
Distance: meter
Tx ant. height:  meter
Rx ant. height:  meter

Attenuation:  dB