complementary distribution function

The scope is to evaluate the CDF (complementary distribution function) of a sum of two signals (one Gaussian, one CW). 𝑐𝑑𝑓𝑖 stands for cdf of interferer and should be a function with 𝜎 and 𝑎 parameters. Beyond are the considerations leading to the need of cdf evaluation. There are two signals adding in the propagation channel. First one is AWGN, having its pdf: 𝑝𝑛 (𝑥) =

1 √2𝜋𝜎

𝑒

−

𝑥2 2𝜎 2

Second is a CW (sinus) interferer, having some arcsin pdf:

1

𝑝𝑠 (𝑥) = 𝜋 ∙

1 √𝑎 2 −𝑥 2

, with 𝑥 2 < 𝑎2

This is not standard arcsin pdf (which can be seen in the link below), but a similar one http://www.math.uah.edu/stat/special/Arcsine.html . Arcsiuns distribution is a particular case of beta distribution. Both signal are zero average, so there are only two parameters: σ for normal distribution and a for arcsinus – which is in fact the amplitude of the CW interferer. Also, one can observe that 𝑝𝑠 (𝑥) has an elementary primitive, namely

1 𝑥 𝑎𝑟𝑐𝑠𝑖𝑛 . 𝜋 𝑎

Pdf of the interferer can be obtained from the convolution of the two distributions: 𝑝𝑖 = 𝑝𝑛 ∗ 𝑝𝑠 So interferer pdf is then: 𝑝𝑖 (𝑥) =

1 √2𝜋 3 𝜎

𝑎

∫

(𝑥−𝑡)2 − 𝑒 2𝜎2

−𝑎 √𝑡

2

− 𝑎2

𝑑𝑡

probably transcendental. Integration limits are finite since out of −𝑎. . . +𝑎 sinus distribution is zero (the pdf is bounded). Maybe it will be possible to use some of convolution properties in relation with an integral transform (the transform of convolution is the product of convolution terms transform). For determining influence of interferer over BER there is necessary to calculate either tail function, either cdf: 𝑥

𝑐𝑑𝑓𝑖 (𝑥) = ∫ 𝑝𝑖 (𝑡)𝑑𝑡 −∞

There is not mandatory an exact calculation, a decent approximation should be OK. Here comes a suggestion of approximation of the convolution product (not necessarily the best one ):

1) Normal distribution 2

Instead of 𝑝𝑛 (𝑥) =

𝑥 1 − 2 2𝜎 𝑒 √2𝜋𝜎

we can use approximation 𝑎𝑛 (𝑥) =

1 1 2 4 √2𝜋𝜎 1+( 𝑥 ) +1( 𝑥 ) √2𝜎

4 √2𝜎

2) Sinus pdf 1

Instead of 𝑝𝑠 (𝑥) = 𝜋 ∙

1 √𝑎 2 −𝑥 2

we can use approximation 𝑎𝑠 (𝑥) =

1 𝑎 ∙ 𝜋 𝑎 2 −𝑥 2

3) Now convolution integral is 4) 𝑝𝑑𝑓(𝑥) = 𝑎𝑛 (𝑥) ∗ 𝑎𝑠 (𝑥) ≅

𝑎 1 1 𝑎 𝑑𝑡 ∫ 2 4 ∙ √2𝜋3 𝜎𝑎 −𝑎 1+1 (𝑥−𝑡) +1 (𝑥−𝑡) 𝑎 2 −𝑡 2 2

𝜎

8

𝜎

We used first function as sliding part, since second function is bounded to (−𝑎, 𝑎). If this is not solvable, further simplification can be done: 1. Quit the 4th grade term 2. Adjust the 4th grade term in order to lead denominator to a perfect square 5) CDF, our final result: 𝑥

𝑐𝑑𝑓(𝑥) = ∫ 𝑝𝑑𝑓(𝑡)𝑑𝑡 −∞

Attached pictures are graphs of approximate functions vs original functions, in order to create a view of approximation functions.

Figure 1- approximation of sin distribution (red) vs original (blue)

Figure 2 – Normal bell curve (green) vs various approximations