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Quadrature components of noise

Module by: a.i. trivedi

Summary: 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 ( t ) = n c t cos 2πf o t − n s t sin 2πf o t n ( t ) = n c t cos 2πf o t − n s t sin 2πf o t size 12{n \( t \) =n rSub { size 8{c} } left (t right )"cos"2πf rSub { size 8{o} } t - n rSub { size 8{s} } left (t right )"sin"2πf rSub { size 8{o} } t} {} (1)

Where fofo size 12{f rSub { size 8{o} } } {} corresponds to k=Kk=K size 12{k=K} {} and lies in the centre of the band.

Letting fo=KΔffo=KΔf size 12{f rSub { size 8{o} } =KΔf} {}and using 2πfot−2πKΔft=02πfot−2πKΔft=0 size 12{2πf rSub { size 8{o} } t - 2πKΔ ital "ft"=0} {}

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

n t = lim Δf → 0 ∑ k = 1 ∞ a k cos 2π f 0 + k − K Δf t + b k sin 2π f 0 + k − K Δf t n t = lim Δf → 0 ∑ k = 1 ∞ a k cos 2π f 0 + k − K Δf t + b k sin 2π f 0 + k − K Δf t size 12{n left (t right )= {"lim"} cSub { size 8{Δf rightarrow 0} } Sum cSub { size 8{k=1} } cSup { size 8{ infinity } } { left lbrace a rSub { size 8{k} } "cos"2π left [f rSub { size 8{0} } + left (k - K right )Δf right ]t+b rSub { size 8{k} } "sin"2π left [f rSub { size 8{0} } + left (k - K right )Δf right ]t right rbrace } } {} (2)

Using identities for the sine and cosine of sum of two angles, we get the equation in terms of ncnc size 12{n rSub { size 8{c} } } {} and nsns size 12{n rSub { size 8{s} } } {}, where

n c t = lim Δf → 0 ∑ k = 1 ∞ a k cos 2π k − K Δ ft + b k sin 2π k − K Δ ft n c t = lim Δf → 0 ∑ k = 1 ∞ a k cos 2π k − K Δ ft + b k sin 2π k − K Δ ft size 12{n rSub { size 8{c} } left (t right )= {"lim"} cSub { size 8{Δf rightarrow 0} } Sum cSub { size 8{k=1} } cSup { size 8{ infinity } } { left [a rSub { size 8{k} } "cos"2π left (k - K right )Δ ital "ft"+b rSub { size 8{k} } "sin"2π left (k - K right )Δ ital "ft" right ]} } {} (3)
n s t = lim Δf → 0 ∑ k = 1 ∞ a k sin 2π k − K Δ ft − b k cos 2π k − K Δ ft n s t = lim Δf → 0 ∑ k = 1 ∞ a k sin 2π k − K Δ ft − b k cos 2π k − K Δ ft size 12{n rSub { size 8{s} } left (t right )= {"lim"} cSub { size 8{Δf rightarrow 0} } Sum cSub { size 8{k=1} } cSup { size 8{ infinity } } { left [a rSub { size 8{k} } "sin"2π left (k - K right )Δ ital "ft" - b rSub { size 8{k} } "cos"2π left (k - K right )Δ ital "ft" right ]} } {} (4)

ncnc size 12{n rSub { size 8{c} } } {} and nsns size 12{n rSub { size 8{s} } } {} are stationary random processes represented as superposition of components.

It can be shown that ncnc size 12{n rSub { size 8{c} } } {} and nsns size 12{n rSub { size 8{s} } } {}are Gaussian, zero mean equal variance uncorrelated variables.

 Significance:

n(t)n(t) size 12{n \( t \) } {}of frequency kΔfkΔf size 12{kΔf} {}gives rise to ncnc size 12{n rSub { size 8{c} } } {} and nsns size 12{n rSub { size 8{s} } } {} of frequency f−fof−fo size 12{f - f rSub { size 8{o} } } {}within band -B/2 to B/2 and thus change insignificantly during one fofo size 12{f rSub { size 8{o} } } {}cycle. ncnc size 12{n rSub { size 8{c} } } {} and nsns size 12{n rSub { size 8{s} } } {} represent amplitude variations of two slowly changing quadrature phasor components, the complete phasor is

Figure 1
Figure 1 (clip_image011.png)

  rt=nc2t+ns2t12rt=nc2t+ns2t12 size 12{r left (t right )= left [n rSub { size 8{c} } rSup { size 8{2} } left (t right )+n rSub { size 8{s} } rSup { size 8{2} } left (t right ) right ] rSup { size 8{ { {1} over {2} } } } } {}

θ t = tan − 1 n s t / n c t θ t = tan − 1 n s t / n c t size 12{θ left (t right )="tan" rSup { size 8{ - 1} } left [ {n rSub { size 8{s} } left (t right )} slash {n rSub { size 8{c} } left (t right )} right ]} {} (5)

The endpoint of r wanders randomly with passage of time.

PSD of orthogonal noise

 Select spectral components corresponding to k=K+λk=K+λ size 12{k=K+λ} {} and k=K−λk=K−λ size 12{k=K - λ} {} where λλ size 12{λ} {} is an integer. k=Kk=K size 12{k=K} {}corresponds to frequency fofo size 12{f rSub { size 8{o} } } {}, hence selected components correspond to fo+λΔffo+λΔf size 12{f rSub { size 8{o} } +λΔf} {} and fo−λΔffo−λΔf size 12{f rSub { size 8{o} } - λΔf} {} , and these frequencies generate 4 power spectral lines.

Figure 2
Figure 2 (clip_image001.png)

Select from nc(t)nc(t) size 12{n rSub { size 8{c} } \( t \) } {} that part Δnc(t)Δnc(t) size 12{Δn rSub { size 8{c} } \( t \) } {}corresponding to the frequencies we have selected above

  Δnct=aK−λcos2πλΔft−bK−λsin2πλΔft+aK+λcos2πλΔft+bK+λsin2πλΔftΔnct=aK−λcos2πλΔft−bK−λsin2πλΔft+aK+λcos2πλΔft+bK+λsin2πλΔft size 12{Δn rSub { size 8{c} } left (t right )=a rSub { size 8{K - λ} } "cos"2 ital "πλ"Δ ital "ft" - b rSub { size 8{K - λ} } "sin"2 ital "πλ"Δ ital "ft"+a rSub { size 8{K+λ} } "cos"2 ital "πλ"Δ ital "ft"+b rSub { size 8{K+λ} } "sin"2 ital "πλ"Δ ital "ft"} {}

All 4 terms are at same frequency and represent uncorrelated processes. Then the power PλPλ size 12{P rSub { size 8{λ} } } {} of Δnc(t)Δnc(t) size 12{Δn rSub { size 8{c} } \( t \) } {} is the ensemble average of Δnc(t)2Δnc(t)2 size 12{ left [Δn rSub { size 8{c} } \( t \) right ] rSup { size 8{2} } } {} and this may be calculated at any time t1t1 size 12{t rSub { size 8{1} } } {}. Choosing t1t1 size 12{t rSub { size 8{1} } } {} such that λΔft1λΔft1 size 12{λΔ ital "ft" rSub { size 8{1} } } {} is an integer,

Δn c t = a K − λ + a K + λ Δn c t = a K − λ + a K + λ size 12{Δn rSub { size 8{c} } left (t right )=a rSub { size 8{K - λ} } +a rSub { size 8{K+λ} } } {} (6)
P λ = E Δn c t 1 2 = E a K − λ + a K + λ 2 = a K − λ 2 ¯ + a K + λ 2 ¯ P λ = E Δn c t 1 2 = E a K − λ + a K + λ 2 = a K − λ 2 ¯ + a K + λ 2 ¯ size 12{P rSub { size 8{λ} } =E left lbrace left [Δn rSub { size 8{c} } left (t rSub { size 8{1} } right ) right ] rSup { size 8{2} } right rbrace =E left [ left (a rSub { size 8{K - λ} } +a rSub { size 8{K+λ} } right ) rSup { size 8{2} } right ]= {overline {a rSub { size 8{ {} rSub { size 6{K - λ} } } } rSup {2} }} size 12{+ {overline {a rSub { {} rSub { size 6{K+λ} } } rSup {2} }} }} {} (7)

We then find

P λ = 2G nc λΔf Δf = 2G n K − λ Δf Δf = 2G n K + λ Δf Δf P λ = 2G nc λΔf Δf = 2G n K − λ Δf Δf = 2G n K + λ Δf Δf size 12{P rSub { size 8{λ} } =2G rSub { size 8{ ital "nc"} } left (λΔf right )Δf=2G rSub { size 8{n} } left [ left (K - λ right )Δf right ]Δf=2G rSub { size 8{n} } left [ left (K+λ right )Δf right ]Δf} {} (8)

So that

G nc λΔf = G n K − λ Δf = G n K + λ Δf G nc λΔf = G n K − λ Δf = G n K + λ Δf size 12{G rSub { size 8{ ital "nc"} } left (λΔf right )=G rSub { size 8{n} } left [ left (K - λ right )Δf right ]=G rSub { size 8{n} } left [ left (K+λ right )Δf right ]} {} (9)

For continuous frequency variable, this becomes

G nc f = G n f 0 − f + G n f 0 + f G nc f = G n f 0 − f + G n f 0 + f size 12{G rSub { size 8{ ital "nc"} } left (f right )=G rSub { size 8{n} } left (f rSub { size 8{0} } - f right )+G rSub { size 8{n} } left (f rSub { size 8{0} } +f right )} {} (10)

Similar equation also results for GnsGns size 12{G rSub { size 8{ ital "ns"} } } {}The psd's of the orthogonal components are shown below, and are obtained by shifting the +ve and -ve parts of plot of GnGn size 12{G rSub { size 8{n} } } {}from +fo+fo size 12{+f rSub { size 8{o} } } {}and −fo−fo size 12{ - f rSub { size 8{o} } } {}to x=0x=0 size 12{x=0} {} and adding the displaced plots.

Thus the power of n(t)n(t) size 12{n \( t \) } {}and the orthogonal components are equal

 

Figure 3