In a mobile radio channel, a single direct path between the Base Station and a mobile is rarely the only physical means for propagation, and hence the Free space propagation model is in most cases is inaccurate when used alone. The 2-ray ground reflection model shown in Figure 2 is a useful propagation model that is based on geometric optics, and considers both the direct path and a ground reflected propagation path between transmitter and receiver. This model has been found to be reasonably accurate for predicting the large-scale signal strength over distances of several kilometers for mobile radio systems that use tall towers (heights which exceed 50 m), as well as for line-of-sight microcell channels in urban environments.

The model consists of a Tx and Rx antenna each of height $h_t$ and $h_r$ respectively. The signal reaches the receiver by two multipath components: The LOS component and the reflected component. For simplicity, it is assumed that the entire impinging signal on the surface is totally reflected back , i.e. there is no $E_T$ ( transmitted E-field component passing through the surface). We assume perpendicular polarized wave reasons of which will be discussed later.

Part 1:

To derive an expression $E_{TOT}$, i.e. total electric field at the receiver antenna, If $E_i$ is the incident E field and $E_g$ is the reflected E field component, $\Gamma$ is the reflection coefficient , then we proceed as ,

$E_g = \Gamma \ E_i$

$E_T = (1 + \Gamma) E_i$

And $\theta_i = \theta_o$

For simplicity we assume the following that $E_T = 0$ and the E component is perpendicularly polarized.

$E (d, t) = \frac{E_o \ d_o }{d} \ cos [ w_c ( t - \frac{d}{c})]$

$E_{Los} = \frac{E_o \ d_o }{d^{'}} \ cos [ w_c ( t - \frac{d^{'}}{c})]$

$E_g = \Gamma \ E_i$

Assuming reflection in near receiver,

Hence,

$E_g = \Gamma \ \frac{E_o \ d_o}{d^{”}} \ cos \ [ w_c (t - \frac{d^{”}}{c})]$

$E_{TOT} = \ E_{Los} + E_g$

$= \frac{E_o \ d_o}{d^{'}} \ cos \ [ \ w_c (t - \frac{ d^{"}}{c})]$

$= -\frac{E_o \ d^o}{d^{”}} \ cos \ [ w_c ( t - \frac{ d^{”}}{c})] \cdots(1)$

As $\Gamma$ is perpendicularly polarized $\Gamma = -1$

Part 2:

To find out the path difference and phase difference, time delay between the two components.

We use taylor series for the expansion in the latter part . Taylor series goes as follows:

$(1 + r)^n = 1 + n r + \frac{n(n-1) r^2}{2!} p - - - -$

Path difference.

$\triangle = d” – d’$

$= \sqrt{ d^2 + (h_t + h_r)^2} - \sqrt{ d^2 + (h_t – h_r)^2}$

$= [d^2 + (h_t + h_r)^2]^{1/2} – [d^2 + (h_t – h_r)^2]^{1/2}$

$= d[ 1 + \frac{(h_t + h_r)^2}{d^2}]^{1/2} – d [ 1 + \frac{(h_t – h_r)^2}{d^2}]^{1/2}$

Using Taylor series expansion.

$\triangle = d [1 + \frac{(h_t + h_r)^2}{2d^2}] – d [1 + \frac{(h_t – h_r)^2}{2d^2}]$

$= d [ \frac{(h_t + h_r)^2}{2d^2} - \frac{ (h_t + h_r)^2}{2d^2}]$

$= \frac{1}{2d} [ 4h_t \ h_r]$

$\triangle = \frac{2(h_t \ h_r)}{d}$

The path difference $\theta_{\triangle}$ between the 2E field component.

$\theta_{\triangle} = \frac{2 \pi \triangle}{ \lambda} = \frac{2 \pi \triangle}{ c/f} = \frac{ 2 \pi f \triangle}{c} = \frac{w_c \triangle}{c}$

$\theta_{\triangle} = \frac{w_c}{c} \times \frac{2 h_t \ h_r}{d}$

The time delay between arrival of waves

$\tau_d = \frac{ \triangle}{c} = \frac{\theta_{\triangle}}{w_c}$

Part 3:

To find T total at a particular time interval t for which $t = \frac{d^{”}}{c}$

Substituting in [1]

$E_{TOT} = \frac{E_o d_o}{d^{’}} \ cos \ [w_c(\frac{d^{”} – d’}{c})] - \frac{E_o \ d_o}{d^{”}} \ cos \ 0^°$

$= \ \frac{E_o \ d_o}{d_o^{’}} \ cos \ \theta_{\triangle} - \frac{E_o \ d_o}{d^{”}} $

$= \ \frac{E_o \ d_o}{d} (cos \ \theta_{\triangle} – 1)$

$ [ \because \ \frac{E_o \ d_o}{d^{”}} \approx \ \frac{E_o \ d^o}{d^{'}} \approx \ \frac{E_o \ d_o}{d}] $

Part 4:

To find expression of $E_{Tot}$ using vector algebra.

$E_{Los} = A \ cos \theta_{ \triangle } \ \ \hat x + A \ sin \theta_{ \triangle} \ \hat y$

$E_g = - A \hat x$

$E_{TOT} = E_{Los} + E_g$

$= A \ cos \theta_{\triangle} \ \hat x^ + A \ sin \ \theta_{\triangle} \ \hat y - A \ \hat x$

$E_{TOT} = A \ \hat x (cos \ \theta_{\triangle} – 1) + A sin \ \theta_{\triangle} \ \hat y$

$| E_{TOT} | = \sqrt{A^2 (cos \theta_{\triangle} – 1)^2 + A^2 sin^2 \theta_{\triangle}} $

$= A \sqrt{cos^2 \theta_{\triangle} – 2 cos \theta_{\triangle} + 1 + sin^2 \theta_{\triangle}} $

= $ A \sqrt{ 2 – 2 \ cos \theta_{\triangle}}$

= $A \sqrt{2(1 – cos \theta_{\triangle})}$

= $A \sqrt{2.2 \ sin^2 \ \theta_{\triangle}/2}$

= $2A sin \ \frac{\theta_{\triangle}}{2}$

= $2 \frac{E_o \ d_o}{d} \ sin \frac{\theta_{\triangle} }{2}$

Whenever d is very large $\theta_{\triangle}$ is very small hence,

$sin \ \frac{\theta_{\triangle} }{2} \approx \ \frac{\theta_{\triangle} }{2}$

Hence

$|E_{TOT}| = \frac{2 \ E_o d_o}{d} . \frac{\theta_{\triangle}}{2}$

= $ \frac{2 E_o \ d_o}{d} \ \frac{2 \pi \triangle}{2 \lambda}$

= $ \frac{2 E_o d_o}{d} . \frac{2 \pi \ h_t \ h_r}{ \lambda d}$

= $\frac{4 \pi \ E_o \ d_o \ h_t \ h_r}{ \lambda} \times \frac{1}{d^2}$

$|E_{TOT}| \approx \ \frac{k}{d^2} \ v/m$

From this expression we get,

$\lambda = \frac{4 \pi \ E_o \ d_o \ h_t \ h_r}{ |E_{tot}|} \times \frac{1}{d^2}$

$\because$ There is a line of sight component, substituting the expression of $\lambda$ into friss free space equation.

$P_r (d) = \frac{P_t \ G_t \ G_r \ \lambda^2}{(4 \pi)^2 \ d^2 \ (L)} $

= $\frac{P_t \ G_t \ G_r}{(4 \pi)^2 \ d^2} \frac{(4 \pi)^2 E_o^2 d_o^2 (h_t \ h_r)^2}{ |E_{tot}|^2}$ $\times \frac{1}{d^4}$

= $\frac{P_t \ G_t \ G_r}{d^2} \frac{E_o^2 \ d_o^2 (h_t \ h_r)^2}{ |E_{tot}|^2} \times \frac{1}{d^4}$

$|E_{tot}|^2 = \frac{E_o^2 \ d_o^2}{d^2}$

$P_r(d) = P_t \frac{G_t \ G_r (h_t \ h_r)^2}{d^4}$

Thus we conclude that, whenever there are reflective components present, the power received is inversely proportional to the fourth power of the distance between Tx and Rx.

There is a definite reason for assuming only perpendicular polarized waves: It is because Brewster angle does not exist for these type of waves and hence the model will work for all angles of incidence.

Brewster angle: The Brewster angle is the angle at which no reflection occurs in the medium of origin. It occurs when the incident angle $\theta_{B}$ is such that the reflection coefficient $\Gamma_{ \|}$ is equal to zero.The Brewster angle is given by the value of $\theta_{B}$ which satisfies

$$\sin \left(\theta_{B}\right)=\sqrt{\frac{\varepsilon_{1}}{\varepsilon_{1}+\varepsilon_{2}}}$$

For the case when the first medium is free space and the second medium has a relative permittivity $\varepsilon_{r},$The above equation can be expressed as

$$ \sin \left(\theta_{B}\right)=\frac{\sqrt{\varepsilon_{r}-1}}{\sqrt{\varepsilon_{r}^{2}-1}} $$

The following graphs show the reflection coefficients for different angles of incidence for two different materials of relative permittivity 4 and 12.

In the case of parallel polarization, it is seen that when the angle of incidence is approximately $12^o$, there is no reflection in the source medium. This angle is called Brewster angle. The model will fail at this Brewster angle. Note that the Brewster angle occurs only for vertical (i.e. parallel) polarization. For perpendicular polarization, reflection is occuring at every angle of incidence. Hence, perpendicular polarization is assumed.

Examples:

1) A mobile is located 5 km away from the base station and uses a vertical $\lambda/4$ monopole antenna with a gain of 2.55 dB, to receive cellular signal. The E field at 1 km away from the transmitter is measured to be $10^{-3}$ v/m, the carrier frequency is 900 MHz.

A] Find the length and the effective aperture of the receiving antenna.

B] Find the received power at the mobile using 2 ray ground reflection model assuming height of 50 m and receiving antenna is 1.5m above the ground.

$\rightarrow$ T-R separation distance $=5 \mathrm{km}$

E-field at a distance of $1 \mathrm{km}=10^{-3} \mathrm{V} / \mathrm{m}$

Frequency of operation, $f=900 \mathrm{MHz}$

$\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{900 \times 10^{6}}=0.333 \mathrm{m}$

Length of the antenna, $L=\lambda / 4=0.333 / 4=0.0833 \mathrm{m}=8.33 \mathrm{cm}$

Gain of antenna $=1.8=2.55 \mathrm{dB}$

b) Since $d= \sqrt{h_{t} h_{r}}$ . the electric field is given by,

$E_{R}(d)=\frac{2 E_{0} d_{0}}{d} \frac{2 \pi h_{t} h_{r}}{\lambda d}=\frac{k}{d^{2}} \mathrm{V} / \mathrm{m}$

$=\frac{2 \times 10^{-3} \times 1 \times 10^{3}}{5 \times 10^{3}}\left[\frac{2 \pi(50)(1.5)}{0.333\left(5 \times 10^{3}\right)}\right]$

$=11.3 .1 \times 10^{-6} \mathrm{V} / \mathrm{m}$

The received power at a distance d can be,

$P_{r}(d)=\frac{\left(113.1 \times 10^{-6}\right)^{2}}{377}\left[\frac{1.8(0.333)^{2}}{4 \pi}\right]$

$P_{r}(d=5 \mathrm{km})=5.4 \times 10^{-13} \mathrm{W}=-122.68 \mathrm{dBW}$ or $-92.68 \mathrm{dBm}$

2) Given that $P_t$ = 50w, $G_r$ = 2, $h_t$ = 50 m, $h_r$ = 1.5m, d = 10km.

Find the power received at a distance of 10 km to 2 ray ground reflection mode 1.

$\rightarrow P_r(d) = \frac{P_t \ G_t \ G_r (h_t \ h_r)^2}{d^4}$

= 56.25 PW.

$P_r(d)$ = -102.49 dBw

= - 72.49 dBw