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I have a cellular signal calculation function, which calculates the signal given the distance from the antenna. Without the constants, the function is basically: $f(d)=1/(d^α)$ where $α$ is a parameter.

I need to find the mean of a part of a circle containing points which are the function values. I will show on an example:

circles cut

In this picture, assume the center of function $f$ is the center of the yellow circle. Each point inside the green circle has distance $d$ from the center, and a value $f(d)$.

I want to find the mean of the values for all the points in the green circle(excluding the blue area, which is determained by the yellow circle!)

I also want to find the mean of the values for all the points in the blue circle area, which I assume is done by substracting.

I saw here that it can be done for a full circle. But how to do it for a part of a circle?

BestR
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1 Answers1

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Let $O$ be the center of the yellow disc and $\rho_0$ its radius. Let the blue disc have its center at distance $R$ from $O$, and let $\rho$ be its radius. Consider now concentric circles $\gamma_r$ of variable radius $r$ centered at $O$. For given $r\in[R-\rho,R+\rho]$ find by trigonometric reasoning the angle $2\alpha(r)$ of the arc that the blue disc cuts out from $\gamma_r$. Then the integrals $$\int_{R-\rho}^{\rho_0} f(r)\>2\alpha(r)\>r\>dr,\qquad \int_{\rho_0}^{R+\rho} f(r)\>2\alpha(r)\>r\>dr$$ give the "total signal energy" received in the two parts of the blue disc. Divide by the corresponding areas, and you obtain the average signal strengths.

Christian Blatter
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  • Thank you!! How do you explain those results? Are they proven somewhere? – BestR Nov 02 '18 at 06:44
  • @BestArmy: The formulas in my answer are "intuitively obvious". For a formal proof you need the formula for variable transformations in multiple integrals. – Christian Blatter Nov 02 '18 at 19:00