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When passed through a narrowband filter noise can also be represented in terms of quadrature components
Where ${f}_{o}$ corresponds to $k=K$ and lies in the centre of the band.
Letting ${f}_{o}=\mathrm{K\Delta f}$ and using ${\mathrm{2\pi f}}_{o}t-\mathrm{2\pi K\Delta}\text{ft}=0$
We add the above to the arguments in the equation 1.
Using identities for the sine and cosine of sum of two angles, we get the equation in terms of ${n}_{c}$ and ${n}_{s}$ , where
${n}_{c}$ and ${n}_{s}$ are stationary random processes represented as superposition of components.
It can be shown that ${n}_{c}$ and ${n}_{s}$ are Gaussian, zero mean equal variance uncorrelated variables.
$n(t)$ of frequency $\mathrm{k\Delta f}$ gives rise to ${n}_{c}$ and ${n}_{s}$ of frequency $f-{f}_{o}$ within band -B/2 to B/2 and thus change insignificantly during one ${f}_{o}$ cycle. ${n}_{c}$ and ${n}_{s}$ represent amplitude variations of two slowly changing quadrature phasor components, the complete phasor is
$r\left(t\right)={\left[{n}_{c}^{2}\left(t\right)+{n}_{s}^{2}\left(t\right)\right]}^{\frac{1}{2}}$
The endpoint of r wanders randomly with passage of time.
Select spectral components corresponding to $k=K+\lambda $ and $k=K-\lambda $ where $\lambda $ is an integer. $k=K$ corresponds to frequency ${f}_{o}$ , hence selected components correspond to ${f}_{o}+\mathrm{\lambda \Delta f}$ and ${f}_{o}-\mathrm{\lambda \Delta f}$ , and these frequencies generate 4 power spectral lines.
Select from ${n}_{c}(t)$ that part ${\mathrm{\Delta n}}_{c}(t)$ corresponding to the frequencies we have selected above
${\mathrm{\Delta n}}_{c}\left(t\right)={a}_{K-\lambda}\text{cos}2\text{\pi \lambda}\Delta \text{ft}-{b}_{K-\lambda}\text{sin}2\text{\pi \lambda}\Delta \text{ft}+{a}_{K+\lambda}\text{cos}2\text{\pi \lambda}\Delta \text{ft}+{b}_{K+\lambda}\text{sin}2\text{\pi \lambda}\Delta \text{ft}$
All 4 terms are at same frequency and represent uncorrelated processes. Then the power ${P}_{\lambda}$ of ${\mathrm{\Delta n}}_{c}(t)$ is the ensemble average of ${\left[{\mathrm{\Delta n}}_{c}(t)\right]}^{2}$ and this may be calculated at any time ${t}_{1}$ . Choosing ${t}_{1}$ such that $\mathrm{\lambda \Delta}{\text{ft}}_{1}$ is an integer,
We then find
So that
For continuous frequency variable, this becomes
Similar equation also results for ${G}_{\text{ns}}$ The psd's of the orthogonal components are shown below, and are obtained by shifting the +ve and -ve parts of plot of ${G}_{n}$ from $+{f}_{o}$ and $-{f}_{o}$ to $x=0$ and adding the displaced plots.
Thus the power of $n(t)$ and the orthogonal components are equal
Example: For white noise filtered by a rectangular BPF centered at ${f}_{o}$ with $\text{BW}=B$ ,
Hence ${G}_{n}({f}_{o}+f)={G}_{n}({f}_{o}-f)$ and
and are twice the magnitude of ${G}_{n}({f}_{o}+f)$ .
The power of the orthogonal components is:
And that of $n(t)$ is
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