The two ray ground reflection model is shown in figure above. It considers both the direct path and a ground reflected propagation path between transmitter and receiver.

In most mobile communication systems, the maximum T-R separation distance is at most a few tens of kilometres, and the earth may be assumed flat. The total received E-Field, ETOT is then a result of the direct line of sight component, ELOS, the free space propagating E- field is given by,

$E\left(d,t\right)=\ \frac{E_0d_0}{d}cos\{w_c\left\{t-\frac{d}{c}\right\}\} $

Where,

E0 is the free space E field at a reference distance d0 from the transmitter

|E(d,t)| = E0 d0/d represents the envelope of the E field at d meters from the transmitter.

Two propagating waves arrive at the receiver, namely

1. The direct wave that travels a distance d1 and
2. The ground reflected wave that travels a distance d2.

The E field due to the line of sight component at the receiver can be expressed as:

$E_{LOS}\left(d',t\right)=\ \frac{E_0d_0}{d'}cos\{w_c\left\{t-\frac{d'}{c}\right\}\} $

....(1)

And the E field for ground reflected wave, which has a propagation distance d’’, can be expressed as:

$E_g=\Gamma{}\ \frac{E_0d_0}{d''}cos\{w_c\left\{t-\frac{d''}{c}\right\}\}$

....(2)

According to laws of reflection in dielectrics,

θi = θ0

Eg = Γ Ei

Et = (1+Γ) Ei

Γ  = reflection coefficient for ground

For small values of θi (i.e. grazing incidence) the reflected wave is equal in amplitude and 180° out of phase with the incident wave. The resultant E field assuming perfect horizontal E field polarization and ground reflection (i.e.Γ⊥ =-1 and Et=0), in the vector sum of ELOS and Eg, and the resultant total E field envelope is given by

 |ETOT| = |ELOS + Eg|

Calculation of |ETOT|

The electric field can be expressed as a sum of equations (1) and (2)

$E_{TOT}\left(d,t\right)=\frac{E_0d_0}{d'}\cos{\left(w_c\left(t-\frac{d^{'}}{c}\right)\right)}+\left(-1\right)\frac{E_0d_0}{d^{''}}\cos{\left(w_c\left(t-\frac{d"}{c}\right)\right)} $

....(3)

It should be noted that as the distance d becomes large, the difference between the distances d’ and d” becomes very small, and the amplitudes of ELOS and Eg are nearly equal and differ only in phase.

|E0d0/d|=| E0d0/d’|=| E0d0/d”|

If the received E field is evaluated at some time, say at d”/c, equation (3) becomes

$E_{TOT}\left(d,t=\frac{d"}{c}\right)=\frac{E_0d_0}{d^{'}}\cos{\left({\omega{}}_c\left(\frac{d"-d'}{c}\right)\right)}-\frac{E_0d_0}{d"}cos0^\circ{} $

$E_{TOT}\left(d,t=\frac{d"}{c}\right)=\frac{E_0d_0}{d^{'}}\cos{\left(\frac{\Delta{}{\omega{}}_c}{c}\right)}-\frac{E_0d_0}{d"}cos0^\circ{} $

Where Δ= d”-d’= path difference

$E_{TOT}\left(d,t=\frac{d"}{c}\right)=\frac{E_0d_0}{d^{'}}\cos{\left(\frac{2\pi{}}{\lambda{}}\Delta{}\right)}-\frac{E_0d_0}{d"}cos0^\circ{} $

$E_{TOT}\left(d,t=\frac{d"}{c}\right)=\frac{E_0d_0}{d^{'}}\angle{}{\theta{}}_{\Delta{}}-\frac{E_0d_0}{d"} $

Where ${\theta{}}_{\Delta{}}\ =\ \frac{\Delta{}{\omega{}}_c}{c}\ =\ \frac{2\pi{}}{\lambda{}}\Delta{} $= phase difference

$E_{TOT}\left(d,t=\frac{d"}{c}\right)=\frac{E_0d_0}{d^{'}}(\angle{}{\theta{}}_{\Delta{}}-1) $

Phasor diagram for ETOT

From phasor diagram,

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}{\left\vert{}E_{TOT}\left(d\right)\right\vert{}}^2={\left(\frac{E_0d_0}{d}\right)}^2+{\left(\frac{E_0d_0}{d}\right)}^2+2{\left(\frac{E_0d_0}{d}\right)}^2cos⁡(180-{\theta{}}_{\Delta{}}) $

${\left\vert{}E_{TOT}\left(d\right)\right\vert{}}^2={2\left(\frac{E_0d_0}{d}\right)}^2-2{\left(\frac{E_0d_0}{d}\right)}^2cos⁡({\theta{}}_{\Delta{}})$

$\left\vert{}E_{TOT}\left(d\right)\right\vert{}=\frac{E_0d_0}{d}\sqrt{2(1-cos⁡({\theta{}}_{\Delta{}}))} $

$\left\vert{}E_{TOT}\left(d\right)\right\vert{}=\frac{E_0d_0}{d}\sqrt{2\times{}2sin{^2}(\frac{{\theta{}}_{\Delta{}}}{2})}$

$\left\vert{}E_{TOT}\left(d\right)\right\vert{}=\frac{{2E}_0d_0}{d}sin⁡(\frac{{\theta{}}_{\Delta{}}}{2})$

Calculation of θΔ

Using geometry of the figure above,

$\Delta{}=d"-d'=\frac{(d"{^2}-d'{^2})}{d"+d'} $

$\Delta{}=\frac{h_t{^2}+h_r{^2}+2h_th_r+d{^2}-h_t{^2}-h_r{^2}+2h_th_r-d{^2}}{d"+d'}$

$\Delta{}=\frac{4h_th_r}{d"+d'} $

Since Δd is much smaller than either d’ or d”, we may consider d’≈d”≈d

$\Delta{}=\frac{4h_th_r}{2d}$

$\Delta{}=\frac{2h_th_r}{d}$

${\theta{}}_{\Delta{}}=\left(\frac{2\pi{}}{\lambda{}}\Delta{}\right)=\ \frac{2\pi{}}{\lambda{}}\frac{2h_th_r}{d}$

If$\frac{{\theta{}}_{\Delta{}}}{2}$is small, typically less than 0.3 radians then

Sin $(\frac{{\theta{}}_{\Delta{}}}{2})\approx{}\ \frac{{\theta{}}_{\Delta{}}}{2}$

$\left\vert{}E_{TOT}\left(d\right)\right\vert{}=\frac{{2E}_0d_0}{d}\sin{\left(\frac{{\theta{}}_{\Delta{}}}{2}\right)}=\ \frac{{2E}_0d_0}{d}\frac{{\theta{}}_{\Delta{}}}{2} $

$\left\vert{}E_{TOT}\left(d\right)\right\vert{}=\ \frac{{2E}_0d_0}{d}\frac{2\pi{}}{\lambda{}}\frac{2h_th_r}{d}=\frac{k}{d{^2}}\ V/m $

Condition on d for above equation to be satisfied

$ ({\theta{}}_{\Delta{}}=\ \frac{2\pi{}}{\lambda{}}\frac{2h_th_r}{d}\lt0.3{\ radians{}})\ $

$ (d\gt\ \frac{20\pi{}h_th_r}{3\lambda{}}\approx{}\frac{20h_th_r}{\lambda{}})\ $