How to determine C/N0 and SNR

In this tutorial we will show a technique to determine CN0 and the behind story
-Suppose that I(m) and Q(m) is the results of tracking phase at mth millisecond
$$I(m)=\sum_{n=0}^{N-1}(s(m,n)+w(m,n))cos(2\pi f_{IF}n)PRN(n)$$
$$Q(m)=\sum_{n=0}^{N-1}(s(m,n)+w(m,n))sin(2\pi f_{IF}n)PRN(n)$$
We have
$$P_s=(\frac{1}{M}\sum_{m=0}^{M-1}I(m))^2$$
$$P_{total}=\frac{1}{M}\sum_{m=0}^{M-1}|I(m)+jQ(m)|^2$$
$$P_n=P_{total}-Ps$$
$$CN_0=10log_{10}(\frac{2P_sF_s}{P_nN})$$

Prove:
$$I(m)=\sum_{n=0}^{N-1}(s(m,n)+w(m,n))cos(2\pi f_{IF}n)PRN(n)$$
where $$s(m,n)=A_sD(m)PRN(n)cos(2\pi f_{IF}n)$$
Therefore $$I(m)=\sum_{n=0}^{N-1}(D(m)A_s\frac{1+cos(2\pi2f_{IF}n)}{2})+\sum_{n=0}^{N-1}w(m,n)cos(2\pi f_{IF}n)PRN(n)$$
$$I(m)=ND(m)\frac{A_s}2+\sum_{n=0}^{N-1}w(m,n)cos(2\pi f_{IF}n)PRN(n)$$
$$P_s=(\frac1M\sum_{m=0}^{M-1}I(m))^2=N\frac{As}{2}+\sum_{n=0}^{N-1}(\sum_{m=0}^{M-1}w(m,n))cos(2\pi f_{IF}n)PRN(n)=N\frac{A_s}{2}$$ Because of white Gaussian Noise property.

$$Q(m)=\sum_{n=0}^{N-1}(s(m,n)+w(m,n))sin(2\pi f_{IF}n)PRN(n)$$
$$=\sum_{n=0}^{N-1}D(m)A_s\frac{sin(2\pi 2f_{IF}n))}{2}+\sum_{n=0}^{N-1}w(m,n)sin(2\pi f_{IF}n)PRN(n)$$
$$=\sum_{n=0}^{N-1}w(m,n)sin(2\pi f_{IF}n)PRN(n)$$

And $$I^2(m)+Q^2(m)=(N\frac{A_s}{2})^2+2NA_s\sum_{n=0}^{N-1}w(m,n)cos(2\pi f_{IF}n)PRN(n)+(\sum_{n=0}^{N-1}w(m,n)cos(2\pi f_{IF}n)PRN(n))^2+(\sum_{n=0}^{N-1}w(m,n)sin(2\pi f_{IF}n)PRN(n))^2$$
$$P_{total}=\frac1M\sum_{m=0}^{M-1}(I^2(m)+Q^2(m))$$
$$=(\frac{NA_s}{2})^2+\sum_{n=0}^{N-1}(\sum_{m=0}^{M-1}w(m,n))cos(2\pi f_{IF}n)PRN(n)+\frac1M\sum_{m=0}^{M-1}(\sum_{n=0}^{N-1}w(m,n)cos(2\pi f_{IF}n)PRN(n))^2+\frac1M\sum_{m=0}^{M-1}(\sum_{n=0}^{N-1}w(m,n)sin(2\pi f_{IF}n)PRN(n))^2$$
$$\sum_{m=0}^{M-1}(\sum_{n=0}^{N-1}w(m,n)cos(2\pi f_{IF}n)PRN(n))^2$$
$$=\sum_{m=0}^{M-1}(\sum_{n=0}^{N-1}w^2(m,n)cos^2(2\pi f_{IF}n)+2(\sum_{i=0}^{N-2}\sum_{j=0,j>i}^{N-1}w(m,i)w(m,j)cos(2\pi f_{IF}i)cos(2\pi f_{IF}j)))$$
Note that
$$\sum_{i=0}^{N-2}\sum_{j=0,j>i}^{N-1}\sum_{m=0}^{M-1}w(m,i)w(m,j)(cos(2\pi f_{IF}i)cos(2\pi f_{IF}j))=0$$ because $$w(m,i)$$ and $$w(m,j)$$ is a different random variables
And
$$P_{total}=P_s+\frac1M\sum_{m=0}^{M-1}\sum_{n=0}^{N-1}w^2(m,n)=P_s+\sigma^2(w)$$
$$=P_s+BN_o=P_s+F_sN_0$$
From above equations, it's easy to show the desired result:
$$P_n=P_{total}-Ps=F_sN_0$$
And $$CN_0=10log_{10}(\frac{2P_sF_s}{P_nN})$$