Ten-rays model

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 ${\displaystyle h_{Tx}\neq {h_{Rx}}}$, and they have a direct distance d that separates the two antennas; The first ray is formed applying Pitágoras theorem:

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

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.

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

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

${\displaystyle \cos \theta ={\frac {d}{R_{0}}}}$

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

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

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

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

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

By likeness of triangles we can deduce the distance from where collides the ray to wall until the perpendicular of the receiver called ${\displaystyle a}$, getting:

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

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

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

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

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

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

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

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

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

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

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 ${\displaystyle d}$ and the sum of the distances between the receiver to each wall with double of distance of the transmitter to the wall ${\displaystyle w_{t2}}$, this divides on the angle formed between the direct distance and the reflected ray.

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

With these parameters is calculate the equation for the eighth ray:

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

For the ninth ray, equation is the same as the seventh ray due to its characteristics:

${\displaystyle R_{4}=R_{3}}$

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

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

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.[2]

The trajectory losses of the ten-ray model in free space is defined as:

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

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

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

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

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

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