Ten-rays model

The model of ten rays is designed as from the geometry of five rays, which models the losses between a transmitting antenna and a receiving antenna. Generally the receiving antenna varies the height in relation to the transmitting antenna, and each ray that reaches the receiver has its respective reflection. The five direct rays are seen from a superior perspective and the reflected see themselves from a lateral focus of the antennas, so these are just below the direct rays. The first two direct rays have two reflections, Where each one collides with each edification that separates the street or SW (Single wall reflected). Other ones two where each ray collides two times against the wall DW (Double wall reflected. And finally a ray that goes directly to the receiver known as LOS (Line of sight).

Mathematical deduction

Top view of 10 rays

Analysis for antennas of heights different located in any point of the street

Side view of two rays transmitted to the receiver with antennas of different heights.

For the mathematical modeling of the propagation of ten rays, it is taken into consideration a side view and starts modeling the first two rays (line of view and your respect reflection). It is taken into consideration that the antennas have different heights then $h_{Tx}\neq {h_{Rx}}$, and have a direct distance $d$ that separates the two antennas; the first ray is formed applying Pythagoras’s theorem:

$R_{0}={\sqrt {d^{2}+(h_{t}-h_{r})^{2}}}$

The second ray or the reflected ray is done in a similar way to the first ray, but in this case it adds the heights of the antennas for form the right triangle by the reflection of the height of the transmitter.

$R_{0}'={\sqrt {d^{2}+(h_{t}+h_{r})^{2}}}$

In the deduction of the third ray it is necessary find the angle between the direct distance $d$ and the distance of line of view $R_{0}$_.

Lightning modeling with a wall collision

$\cos \theta ={\frac {h_{t}-h_{r}}{R_{0}}}$

Seeing the model with a side view, it is necessary find a flat distance between the transmitter and the receiver called $d'$.

$d'={\sqrt {d^{2}-(w_{r2}-w_{t2})^{2}}}$

Now the height is deducted from the subtraction of the wall with respect to the height of the receiver called $z$ for likeness of triangles:

${\frac {z}{a}}={\frac {h_{t}}{d'}}$

$z={\frac {{h_{t}}a}{d'}}$

For likeness of triangles it can be deduced the distance since where the ray clashes the wall until the perpendicular of receiver called $a$, getting:

${\frac {a}{d}}'={\frac {w_{r2}}{w_{t2}+w_{r2}}}$

$a={\frac {d'w_{r2}}{w_{t2}+w_{r2}}}$

The third ray is defined as a model of two rays, by which is:

$R_{1}={\frac {{\sqrt {(h_{t}-h_{r}-z)^{2}+(d'-a)^{2}}}+{\sqrt {z^{2}+a^{2}}}}{\cos \theta }}$

Since a side view is evidenced the reflected ray that there in $R_{1}$ and it is find of the next way:

$R_{1}'={\frac {{\sqrt {(h_{t}+h_{r}+z)^{2}+(d'-a)^{2}}}+{\sqrt {(2h_{r}+z)^{2}+a^{2}}}}{\cos \theta }}$

As there are two rays that shock a time on the wall, then the fifth ray is found which is equal to the third ray.

$R_{2}=R_{1}\,$

It's equalized the sixth ray with the fourth, because have the same features.

$R_{2}'=R_{1}'$

Side view of two transmitted beams reflected from one wall to the other side and reflected to the receiver on antennas of different heights at any point on the street.

To model the rays that hit on the wall two times is used the Pythagoras’s theorem, because to the direct distance $d$ and the sums of the distances between the receiver to each wall with double of the distance of the transmitter to the wall $w_{t2}$, this divides on the angle formed between the direct distance and the reflected ray.

$R_{3}={\frac {\sqrt {d^{2}+(w_{r2}+2w_{t2}+w_{r1})^{2}}}{\cos \theta }}$

For the eighth ray is calculated a set of variables that allow deduct the full equation, which is composed for distances and heights that were found by likeness of triangles.

In first instance it taken the flat distance between the wall of the second hit and the receiver:

$x={\frac {d'w_{r1}}{w_{r2}+w_{r1}}}$

It is found the distance between the transmitter and the wall in the first hit.

$h_{p2}=h_{r}+z_{2}$

$y={\frac {d'w_{t2}}{w_{r2}+w_{r1}}}$

It is found the distance between the height of the wall of the second hit with respect the first hit, getting:

$z_{1}={\frac {h_{t}(d'-(x+y))}{d'}}$

Deducting also the distance between the height of the wall of the second hit with respect the receiver:

$z_{2}={\frac {h_{t}x}{d'}}$

Calculating the height of the wall where occurs the first hit:

$h_{p1}=h_{r}+z_{1}+z_{2}$

Calculating the height of the wall where occurs the second hit:

$h_{p2}=h_{r}+z_{2}$

With these settings is calculated the equation for the eighth ray:

$R_{3}'={\frac {{\sqrt {y^{2}+(h_{t}+h_{p1})^{2}}}+{\sqrt {(d'-(x+y))^{2}+(h_{p1}+h_{p2}^{2}}}+{\sqrt {x^{2}+(h_{r}+h_{p2})^{2}}}}{\cos \theta }}$

For the ninth ray, the equation is the same of the seventh ray due to its features:

$R_{4}=R_{3}$

For the tenth ray, the equation is the same of the eighth ray because to its form of the reflected ray:

$R_{4}'=R_{3}'$

Losses for trajectory on free space

Modeling of the losses by free space trajectory in the 6-ray model when the distances of the wall and the heights are different.

It is considered a transmitted signal through of the space free to a receiver located to a distance d of the transmitter.

Assuming that there are no obstacles between the transmitter and the receiver, the signal it spreads along of a straight line between the two. The model of associated ray with this transmission is called line of sight (LOS), and the signal receiver corresponding is called signal of LOS or ray.

The losses of trajectory of the model of ten rays in the free space, it is defined as:

$p_{0}(t)={\sqrt {(G_{i}G_{R})}}{\frac {\lambda }{4\pi }}\left({\frac {\exp(j2\pi R_{0}/{\lambda })}{R_{0}}}+\Gamma {\frac {\exp(j2\pi R_{0}'/{\lambda })}{R_{0}'}}\right)$

$p_{1}(t)={\sqrt {(G_{i}G_{R})}}{\frac {\lambda }{4\pi }}~\Gamma _{1}\left({\frac {\exp(j2\pi R_{1}/\lambda )}{R_{1}}}+\Gamma {\frac {\exp(j2\pi R_{1}'/\lambda )}{R_{1}'}}\right)$

$p_{2}(t)={\sqrt {(G_{i}G_{R})}}{\frac {\lambda }{4\pi }}~\Gamma _{1}\left({\frac {\exp(j2\pi R_{2}/\lambda )}{R_{2}}}+\Gamma {\frac {\exp(j2\pi R_{2}'/\lambda )}{R_{2}'}}\right)$

$p_{3}(t)={\sqrt {(G_{i}G_{R})}}{\frac {\lambda }{4\pi }}~\Gamma _{1}\left({\frac {\exp(j2\pi R_{3}/\lambda )}{R_{3}}}+\Gamma {\frac {\exp(j2\pi R_{3}'/\lambda )}{R_{3}'}}\right)$

$p_{4}(t)={\sqrt {(G_{i}G_{R})}}{\frac {\lambda }{4\pi }}~\Gamma _{1}\left({\frac {\exp(j2\pi R_{4}/\lambda )}{R_{4}}}+\Gamma {\frac {\exp(j2\pi R_{4}'/\lambda )}{R_{4}'}}\right)$

$P_{L}~(dB)=20\log \left|\sum _{i=0}^{N}P_{i}\right|$

See also

• Six rays model
• Ray tracing (physics)
• Two-ray ground-reflection model