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The relation between a ,b and c and $\phi $ which describe the noise components can be seen to be identical with that between X,Y and R and $\theta $ .
Hence pdf of c is Rayleigh and that of $\theta $ is uniform.
$f\left({c}_{k}\right)=\frac{{c}_{k}}{{P}_{k}}{e}^{-{c}_{k}^{2}/{\mathrm{2P}}_{k}}{c}_{k}0$ , $f\left({\theta}_{k}\right)=\frac{1}{\mathrm{2\pi}}-\pi \le {\theta}_{k}\le \pi $
Let a spectral component of noise be the input to a filter whose transfer function at frequency $\mathrm{k\Delta f}$ is
The output spectral component of noise is
The power associated with the input component is
As $\mid H\left(\mathrm{k\Delta f}\right)\mid $ is a deterministic function, $\overline{{\left[\mid H\left(\mathrm{k\Delta f}\right)\mid {a}_{k}\right]}^{2}}={\mid H\left(\mathrm{k\Delta f}\right)\mid}^{2}\overline{{a}_{k}^{2}}$
Similarly for ${b}_{k}$ , and thus the power associated with noise output is
And the power spectral densities are related by
Where the $\mathrm{k\Delta f}$ has been replaced by $f$ as a continuous variable as $\mathrm{\Delta f}$ tends to 0.
Noise can be represented as superposition of (orthogonal) harmonics of $\mathrm{\Delta f}$ therefore total power is the result of superposition of component powers.
Consider Two processes ${n}_{1}$ and ${n}_{2}$ with overlapping spectral components.
Power of the sum of ${n}_{1}$ and ${n}_{2}$ will be ${p}_{1}+{p}_{2}+2E\left[{n}_{1}{n}_{2}\right]$ and since ${n}_{1}$ and ${n}_{2}$ are uncorrelated, the last term = 0.
Then these noises also obey the superposition of powers rule.
If ${k}^{\text{th}}$ component of noise is mixed with a sinusoid
Sum and difference frequency noise spectral components with 1/2 amplitude are generated and
Considering power spectral components at $\mathrm{k\Delta f}$ and $\mathrm{l\Delta f}$ , let the mixing frequency be ${f}_{0}=\left(k+l\right)\mathrm{\Delta f}$ . This will generate 2 difference frequency components at the same frequency: $\mathrm{p\Delta f}={f}_{0}-\mathrm{k\Delta f}=\mathrm{l\Delta f}-{f}_{0}$
Then difference frequency components are
But as $\overline{{a}_{k}{a}_{l}}=\overline{{a}_{k}{b}_{l}}=\overline{{b}_{k}{a}_{l}}=\overline{{b}_{k}{b}_{l}}=0$ , We find $E\left[{n}_{\mathrm{p1}}\left(t\right){n}_{\mathrm{p2}}\left(t\right)\right]=0$
and $E\left\{{\left[{n}_{\mathrm{p1}}\left(t\right)+{n}_{\mathrm{p2}}\left(t\right)\right]}^{2}\right\}=E\left\{{\left[{n}_{\mathrm{p1}}\left(t\right)\right]}^{2}\right\}+E\left\{{\left[{n}_{\mathrm{p2}}\left(t\right)\right]}^{2}\right\}$
Thus superposition of power applies even after shifting due to mixing.
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