What RF Views mean ?
The map of how signal strength due to a signal strength source would vary on a floor plan. The signal strength source we talk of in this would essentially be for those sources which operate at Radio Frequency
What are signal strength source ?
For the discussion in this blog, it would be access point (AP) or a wireless signal strength producing device. Access point (AP) are devices which are used for allowing wireless access these days.
Floor plan :
For the discussion in this blog, we shall consider only Rectangular floor plans.
Formula for free space propogation signal strength at a point P :
The signal strength decreases with distance. Ideally, the signal strength should decrease by square of the distance, however, depending on the environment, the signal strength could decrease even more sharper than that.
Lets assume for simplicity,
S(d) = SS[1 - K*(d^2)], ... (1)
where S(d) is signal strength at distance 'd' from the point where AP is kept.
SS is the transmitting power of the AP.
Attenuation due to obstacles :
The above formula (1) is useful and meaningful in free space. If there are obstacles between the line joining a point P and point where AP is kept, then Signal strength at P would be less than (1). The reason is that the wave suffers loss of intensity due to attenuation when it encounters an obstacle. Each obstacle could be of various sizes and shapes. The attenuation depends on the material, the shape of the material and the dimensions of the material. However, for simplicity, we shall assume that all obstacles of a material will have same attenaution coefficient no matter what their shape or dimensions are. Lets represent this by AC(m) for an obstacle of material 'm'. Concrete will have higher AC than cardboard because of its hard inner structure.
Signal strength at a point P :
To find the signal strength at a point P would be simply
S(distance(P, Placement of AP)) - Sigma (AC(m))
Sigma(AC(m)) represents the sum of attenuation coefficients of all the obstacles that lie in line joining the point where AP is placed and the point P where we want to calculate the signal strength.
A real example :
A floor plan could be considered to be rectangle. To find the signal strength on this floor plan and to create the RF Map, we need to find the signal strength at various points on this plan. This could be done by that thefloor plan is divided evenly into small squares. Let the number of such squares be 'P'. If P is high then the RF views become granular.
Each such square is referred to as a pixel. The Signal is computed at center of each the pixel. Fot the pixel, the signal strength at every point in that pixel would be the same signal strength as at the center of the pixel.
Also, let there be around a 'O' obstacles on the floor plan.
Let that the number of APs be 'D'.
Attempt 1 : KISS Stratergy : The nearly brute force algorithm
Let AP be placed at 'L'
- For every AP
- For every pixel
- Find distance between AP and pixel. Calculate SS(pixel)
- For all the obstacles
- Find if the obstacle intersects line joining L and pixel's center.
- If obstacle does intersect, decrement AC(obstacle's material) from SS(pixel).
This indirectly means that this algorithm cannot scale if the number of obstacles increases. In one of the floor plan that I worked with, the time it took to compute RF views for a device which was placed on a floor plan with number of obstacles of around 4 K was 10 seconds, when number of pixels used were 28K. The number of APs that were placed on that floor plan were about 10, so the time it took to compute the RF Views were too large. As could be seen, this algo does not scale with obstacles.
Proposed algorithm to find attenuation
Step 1 : Associate each obstacle with some pixels
Each obstacle would be assumed to lie completely on some number of pixels.
Let there be an obstacle Oi which lies on pixels P1, P2, … , Pj.
For such case, we define a term ‘footprint’ :
Obstacle Oi has its footprint on Pixel P1, P2, … Pj.
Step 2 : Maintain for each pixel the set of Obstacles
We maintain for each pixel a set of obstacles which have footprint on that Pixel. Lets define it as SetObs(P) for a pixel P for simplicity.
Step 3 : Find attenuation at a pixel ‘P’
Let the AP be kept at point ‘p’ and let the center point of Pixel ‘P’ be ‘d’.
Join points ‘d’ and ‘p’. Find the pixels which have their boundaries intersected by this line. Let the set of such pixels be { Pm, Pm+1, … , Pm+n}
Find the set of obstacles which is union of SetObs(Pm), SetObs(Pm+1), … , SetObs(Pm+n). Let this be a set ‘S’. These would be the obstacles which lie on the line joining the AP and the Pixel ‘P’.
Set S = SetObs(Pm) U SetObs(Pm+1) U … U SetObs(Pm+n)
Find the attenuation because of these obstacles. Add the attenuations. This is the attenuation we wanted.
Analysis of proposed algorithm
Step 1 & 2
Finding the footprint per obstacle and subsequent creation of SetObs(Pixel P) for all the pixels in those footprint is a one time operation. It would be done just once, no matter how many devices are placed on the layout.
Step 3
This step is not linearly dependent on number of obstacles any more.
Results
The PoC is yet to be done.
Some one please help me :)
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