Monthly Archives: September 2016

Reflection vs Scattering in Ray-Tracing

In ray-tracing simulations one is faced with a complexity at points of intersection between rays and objects. How to handle the reflection, should it be specular or scattered. In this post we consider the two extreme cases; pure reflection vs scattering.

For this we consider that in the case of specular reflection the power of the ray is reduced by 3dB (that is R=0.5) and the ray continues as it would if there was no interaction with the object (off course direction would be changed with angle of reflection being equal to angle of incidence).

In the case of scattered ray we assume that the point of interaction between the ray and the object is a point source from where the rays are regenerated. Therefore there is a rapid drop in the E-field strength as the ray propagates away from the point of interaction.

clear all
close all

Eo=1;
r1=1:100;
r2=101:200;
r=[r1 r2];
R=0.5;

Er1=Eo./r1;
Er2=R*Eo./r2;
Er=[Er1 Er2];

Es1=Eo./r1;
Es2=Es1(end)./(r2-r1(end));
Es=[Es1 Es2];

plot(r,10*log10(Er),'b')
hold on
plot(r,10*log10(Es),'r:')
hold off

legend('Reflection','Scattering')
xlabel('Distance (m)')
ylabel('E-field (V/m)')

Reflected vs Scattered Ray

The simulation code above considers that a ray travels unobstructed for 100 m and at this point comes in contact with an object and is reflected (reflection coefficient of 0.5 is assumed) or scattered. The ray then again travels for another 100 m without coming in contact with an object.

It is observed that there is a rapid decrease in E-field strength of the scattered ray beyond 100 m. This is because the E-field strength at the point of interaction is much lower than that at the source and when this is considered to be a point source, re-generating the rays, the resulting E-field decays quite rapidly.

We assume that the reflection is specular in most of our simulations as this is closer to reality than assuming a fully scattered ray.

How to Find Point of Intersection of Two Lines

Finding the point of intersection of two lines has many important application such as in Ray-Tracing Simulation.  Two lines always intersect at some point unless they are absolutely parallel, like the rails of a railway track. We start with writing the equations of the two lines in slope-intercept form.

y1=b1+m1*x1

y2=b2+m2*x2

straight-lines

Here m1 and m2 are the slopes of the two lines and b1 and b2 are their y-intercepts. At the point of intesection y1=y2, so we have.

b1+m1*x1=b2+m2*x2

But at  the point of intersection x1=x2 as well, so replacing x1 and x2 with x we have.

b1+m1*x=b2+m2*x

or

b1-b2=-x*(m1-m2)

or

x=-(b1-b2)/(m1-m2)

Once the x-component of the point of intersection is found we can easily find the y-component by substituting x in any of the two line equations above.

y=b1+m1*x

In future posts we would like to discuss the cases of intersection of two surfaces and the intersection of two volumes.