# 0.3 Filtering of noise - 2

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this module considers the effect on power spectrum of noise after ffiltering

## Psd after filtering:

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.

$f\left({c}_{k}\right)=\frac{{c}_{k}}{{P}_{k}}{e}^{-{c}_{k}^{2}/{2P}_{k}}{c}_{k}0$ , $f\left({\theta }_{k}\right)=\frac{1}{2\pi }-\pi \le {\theta }_{k}\le \pi$

Let a spectral component of noise be the input to a filter whose transfer function at frequency $\mathrm{k\Delta f}$ is

$H\left(\mathrm{k\Delta f}\right)=\mid H\left(\mathrm{k\Delta f}\right)\mid {e}^{\mathrm{j\varphi k}}=\mid H\left(\mathrm{k\Delta f}\right)\mid \angle {\varphi }_{k}$

The output spectral component of noise is

${n}_{\text{ko}}\left(t\right)=\mid H\left(\mathrm{k\Delta f}\right)\mid {a}_{k}\text{cos}\left(2\pi k\Delta \text{ft}+{\varphi }_{k}\right)+\mid H\left(\mathrm{k\Delta f}\right)\mid {b}_{k}\text{sin}\left(2\pi k\Delta \text{ft}+{\varphi }_{k}\right)$

The power associated with the input component is

${P}_{\text{ki}}=\frac{\overline{{a}_{k}^{2}}+\overline{{b}_{k}^{2}}}{2}$

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

${P}_{\text{ko}}={\mid H\left(\mathrm{k\Delta f}\right)\mid }^{2}\frac{\overline{{a}_{k}^{2}}+\overline{{b}_{k}^{2}}}{2}$

And the power spectral densities are related by

${G}_{\text{no}}\left(f\right)={\mid H\left(f\right)\mid }^{2}{G}_{\text{ni}}\left(f\right)$

Where the $\mathrm{k\Delta f}$ has been replaced by $f$ as a continuous variable as $\mathrm{\Delta f}$ tends to 0.

## Superposition of noises:

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.

## Mixing of noise with a sinusoid

If ${k}^{\text{th}}$ component of noise is mixed with a sinusoid

$\begin{array}{}{n}_{k}\left(t\right)\text{cos}{2\pi f}_{o}t=\frac{{a}_{k}}{2}\text{cos}2\pi \left(\mathrm{k\Delta f}+{f}_{o}\right)t+\frac{{b}_{k}}{2}\text{sin}2\pi \left(\mathrm{k\Delta f}+{f}_{o}\right)t\\ +\frac{{a}_{k}}{2}\text{cos}2\pi \left(\mathrm{k\Delta f}-{f}_{o}\right)t+\frac{{b}_{k}}{2}\text{sin}2\pi \left(\mathrm{k\Delta f}+{f}_{o}\right)t\end{array}$

Sum and difference frequency noise spectral components with 1/2 amplitude are generated and

${G}_{n}\left(f+{f}_{o}\right)={G}_{n}\left(f-{f}_{o}\right)=\frac{{G}_{n}\left(f\right)}{4}$

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

${n}_{\mathrm{p1}}\left(t\right)=\frac{{a}_{k}}{2}\text{cos}2\pi p\Delta \text{ft}-\frac{{b}_{k}}{2}\text{sin}2\pi p\Delta \text{ft}$
${n}_{\mathrm{p2}}\left(t\right)=\frac{{a}_{l}}{2}\text{cos}2\pi p\Delta \text{ft}+\frac{{b}_{l}}{2}\text{sin}2\pi p\Delta \text{ft}$

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.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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I'm not sure why it wrote it the other way
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I got X =-6
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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