Ten-rays model

Top view of 10 rays

The ten rays model is a model applied to the transmissions in the urban area, to generate a model of ten rays typically four rays more are added to the model of six rays, these are (R3 and R4 bouncing on both sides of the wall); This incorporate all paths with one, two, or three reflections: specifically, there is the LOS, the groundreflected (GR), the single-wall (SW) reflected, the double-wall (DW) reflected, the triple-wall (TW) reflected, the wall-ground (WG) reflected and the ground-wall (GW) reflected paths. Where each of the mentioned paths bounces on both sides of the wall (two reflections). As from experimental data, it shown that a model of ten reflection rays closely approximates signal propagation through the dielectric canyon.

As example for this model it is assume: rectilinear streets with buildings along both sides of the street and transmitter and receiver antenna heights that are well below the tops of the buildings. The building-lined streets act as a dielectric canyon to the propagating signal. Theoretically, an infinite number of rays can be reflected off the building fronts to arrive at the receiver; in addition, rays may also be back-reflected from buildings behind the transmitter or receiver. However, since some of the signal energy is dissipated with each reflection, signal paths corresponding to more than three reflections can generally be ignored. When the street layout is relatively straight, back reflections are usually negligible .

Mathematical deduction

Top view of 10 rays

Analysis for antennas of heights different heights lo-cated in street's any point

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

For the mathematical modeling of the propagation of ten rays, One has in account a side view and this starts off modeling the two first rays (line by sight and his respective reflection), Considering that antennas have different heights, Then $h_{Tx}\neq {h_{Rx}}$, and they have a direct distance d that separates the two antennas; The first ray is formed applying Pitágoras theorem:

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

The second ray or the reflected ray is made in a similar way to the first, but in this case the heights of the antennas to form the right angled triangle for the reflection of the height of the transmitter are added up.

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

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 $R_{0}$_.

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

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

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

Now we deduce the remaining height of the wall from the height of the receiver called $z$ by the similarity of triangles:

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

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

By likeness of triangles we can deduce the distance from where collides the ray to wall until the perpendicular of the 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:

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

Taking a side view it is achieves to evidence the reflected ray that there in $R_{1}$ and is find as following manner:

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

As exist two rays that collide once on the wall, then is find the fifth ray, equating it to the third.

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

Similarly, is equalized the sixth ray with the fourth ray, since they have the same characteristics.

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

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 collide with the wall twice, is used the Pythagoras theorem because of the direct distance $d$ and the sum of the distances between the receiver to each wall with double of 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 calculate a series of variables that allow to deduce the complete equation, which is composed by distances and heights that were found by likeness of triangles.

In first instance is take the flat distance between the wall of the second shock and the receiver:

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

Is found the flat distance between the transmitter and the wall in the first shock.

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

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

Finding the distance between the height of the wall of the second shock with respect to the first shock, is obtain:

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

Deducing also the distance between the height of the wall of the second shock with respect to 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 shock:

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

With these parameters is calculate 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, equation is the same as the seventh ray due to its characteristics:

$R_{4}=R_{3}$

For the tenth ray, the equation is the same as the eighth ray due to its reflected ray shape:

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

Losses for trajectory of 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.

Is considered a signal transmitted through free space to a receiver located at a distance d from the transmitter.

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

The trajectory losses of the ten-ray model in free space 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