# 0.5 Quadrature components of noise

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This module describes the quadrature components of narrowband noise

## Narrowband representation

When passed through a narrowband filter noise can also be represented in terms of quadrature components

$n\left(t\right)={n}_{c}\left(t\right)\text{cos}{2\pi f}_{o}t-{n}_{s}\left(t\right)\text{sin}{2\pi f}_{o}t$

Where ${f}_{o}$ corresponds to $k=K$ and lies in the centre of the band.

Letting ${f}_{o}=\mathrm{K\Delta f}$ and using ${2\pi f}_{o}t-2\pi K\Delta \text{ft}=0$

We add the above to the arguments in the equation 1.

$n\left(t\right)=\underset{\mathrm{\Delta f}\to 0}{\text{lim}}\sum _{k=1}^{\infty }\left\{{a}_{k}\text{cos}2\pi \left[{f}_{0}+\left(k-K\right)\mathrm{\Delta f}\right]t+{b}_{k}\text{sin}2\pi \left[{f}_{0}+\left(k-K\right)\mathrm{\Delta f}\right]t\right\}$

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

${n}_{c}\left(t\right)=\underset{\mathrm{\Delta f}\to 0}{\text{lim}}\sum _{k=1}^{\infty }\left[{a}_{k}\text{cos}2\pi \left(k-K\right)\Delta \text{ft}+{b}_{k}\text{sin}2\pi \left(k-K\right)\Delta \text{ft}\right]$
${n}_{s}\left(t\right)=\underset{\mathrm{\Delta f}\to 0}{\text{lim}}\sum _{k=1}^{\infty }\left[{a}_{k}\text{sin}2\pi \left(k-K\right)\Delta \text{ft}-{b}_{k}\text{cos}2\pi \left(k-K\right)\Delta \text{ft}\right]$

${n}_{c}$ 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.

## significance:

$n\left(t\right)$ 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

$r\left(t\right)={\left[{n}_{c}^{2}\left(t\right)+{n}_{s}^{2}\left(t\right)\right]}^{\frac{1}{2}}$

$\theta \left(t\right)={\text{tan}}^{-1}\left[{n}_{s}\left(t\right)/{n}_{c}\left(t\right)\right]$

The endpoint of r wanders randomly with passage of time.

## Psd of orthogonal noise

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 ${n}_{c}\left(t\right)$ that part ${\mathrm{\Delta n}}_{c}\left(t\right)$ corresponding to the frequencies we have selected above

${\mathrm{\Delta n}}_{c}\left(t\right)={a}_{K-\lambda }\text{cos}2\text{πλ}\Delta \text{ft}-{b}_{K-\lambda }\text{sin}2\text{πλ}\Delta \text{ft}+{a}_{K+\lambda }\text{cos}2\text{πλ}\Delta \text{ft}+{b}_{K+\lambda }\text{sin}2\text{πλ}\Delta \text{ft}$

All 4 terms are at same frequency and represent uncorrelated processes. Then the power ${P}_{\lambda }$ of ${\mathrm{\Delta n}}_{c}\left(t\right)$ is the ensemble average of ${\left[{\mathrm{\Delta n}}_{c}\left(t\right)\right]}^{2}$ and this may be calculated at any time ${t}_{1}$ . Choosing ${t}_{1}$ such that $\mathrm{\lambda \Delta }{\text{ft}}_{1}$ is an integer,

${\mathrm{\Delta n}}_{c}\left(t\right)={a}_{K-\lambda }+{a}_{K+\lambda }$
${P}_{\lambda }=E\left\{{\left[{\mathrm{\Delta n}}_{c}\left({t}_{1}\right)\right]}^{2}\right\}=E\left[{\left({a}_{K-\lambda }+{a}_{K+\lambda }\right)}^{2}\right]=\overline{{a}_{{}_{K-\lambda }}^{2}}+\overline{{a}_{{}_{K+\lambda }}^{2}}$

We then find

${P}_{\lambda }={2G}_{\text{nc}}\left(\mathrm{\lambda \Delta f}\right)\mathrm{\Delta f}={2G}_{n}\left[\left(K-\lambda \right)\mathrm{\Delta f}\right]\mathrm{\Delta f}={2G}_{n}\left[\left(K+\lambda \right)\mathrm{\Delta f}\right]\mathrm{\Delta f}$

So that

${G}_{\text{nc}}\left(\mathrm{\lambda \Delta f}\right)={G}_{n}\left[\left(K-\lambda \right)\mathrm{\Delta f}\right]={G}_{n}\left[\left(K+\lambda \right)\mathrm{\Delta f}\right]$

For continuous frequency variable, this becomes

${G}_{\text{nc}}\left(f\right)={G}_{n}\left({f}_{0}-f\right)+{G}_{n}\left({f}_{0}+f\right)$

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 $n\left(t\right)$ and the orthogonal components are equal

Example: For white noise filtered by a rectangular BPF centered at ${f}_{o}$ with $\text{BW}=B$ ,

$\begin{array}{}{G}_{n}\left(f\right)=\frac{\eta }{2}{f}_{o}-\frac{B}{2}\le \mid f\mid \le {f}_{o}+\frac{B}{2},\\ {G}_{n}\left(f\right)=0\text{elsewhere}\end{array}$

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

${G}_{\text{nc}}\left(f\right)={G}_{\text{ns}}\left(f\right)={G}_{n}\left({f}_{o}-f\right)+{G}_{n}\left({f}_{o}+f\right)=\frac{\eta }{2}+\frac{\eta }{2}=\eta \mid f\mid \le \frac{B}{2}$

and are twice the magnitude of ${G}_{n}\left({f}_{o}+f\right)$ .

The power of the orthogonal components is:

${\sigma }_{\text{nc}}^{2}={\sigma }_{\text{ns}}^{2}={\int }_{-B/2}^{B/2}{G}_{\text{nc}}\left(f\right)\text{df}=\mathrm{\eta B}$

And that of $n\left(t\right)$ is

${\sigma }_{n}^{2}={\int }_{-\text{fo}-B/2}^{-\text{fo}+B/2}{G}_{n}\left(f\right)\text{df}+{\int }_{\text{fo}-B/2}^{\text{fo}+B/2}{G}_{n}\left(f\right)\text{df}=2\frac{\eta }{2}B=\mathrm{\eta B}$

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