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# Filtering of noise - 2

Module by: a.i. trivedi. E-mail the author

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

## PSD after filtering:

The relation between a ,b and c and φφ size 12{φ} {} which describe the noise components can be seen to be identical with that between X,Y and R and θθ size 12{θ} {}.

Hence pdf of c is Rayleigh and that of θθ size 12{θ} {} is uniform.

fck=ckPkeck2/2Pkck0fck=ckPkeck2/2Pkck0 size 12{f left (c rSub { size 8{k} } right )= { {c rSub { size 8{k} } } over {P rSub { size 8{k} } } } e rSup { size 8{ - c rSub { size 6{k} } rSup { size 6{2} } /2P rSub { size 6{k} } } } ~c rSub {k} size 12{ > = 0}} {}, fθk=1πθkπfθk=1πθkπ size 12{f left (θ rSub { size 8{k} } right )= { {1} over {2π} } ~ - π <= θ rSub { size 8{k} } <= π} {}

Let a spectral component of noise be the input to a filter whose transfer function at frequency kΔfkΔf size 12{kΔf} {} is

H kΔf = H kΔf e jϕk = H kΔf ϕ k H kΔf = H kΔf e jϕk = H kΔf ϕ k size 12{H left (kΔf right )= lline H left (kΔf right ) rline e rSup { size 8{jϕk} } = lline H left (kΔf right ) rline ∠ϕ rSub { size 8{k} } } {}
(1)

The output spectral component of noise is

n ko t = H kΔf a k cos 2πkΔ ft + ϕ k + H kΔf b k sin 2πkΔ ft + ϕ k n ko t = H kΔf a k cos 2πkΔ ft + ϕ k + H kΔf b k sin 2πkΔ ft + ϕ k size 12{n rSub { size 8{ ital "ko"} } left (t right )= lline H left (kΔf right ) rline a rSub { size 8{k} } "cos" left (2πkΔ ital "ft"+ϕ rSub { size 8{k} } right )+ lline H left (kΔf right ) rline b rSub { size 8{k} } "sin" left (2πkΔ ital "ft"+ϕ rSub { size 8{k} } right )} {}
(2)

The power associated with the input component is

P ki = a k 2 ¯ + b k 2 ¯ 2 P ki = a k 2 ¯ + b k 2 ¯ 2 size 12{P rSub { size 8{ ital "ki"} } = { { {overline {a rSub { size 8{k} } rSup { size 8{2} } }} + {overline {b rSub { size 8{k} } rSup { size 8{2} } }} } over {2} } } {}
(3)

As HkΔfHkΔf size 12{ lline H left (kΔf right ) rline } {}is a deterministic function, HkΔfak2¯=HkΔf2ak2¯HkΔfak2¯=HkΔf2ak2¯ size 12{ {overline { left [ lline H left (kΔf right ) rline a rSub { size 8{k} } right ] rSup { size 8{2} } }} = lline H left (kΔf right ) rline rSup { size 8{2} } {overline {a rSub { size 8{k} } rSup { size 8{2} } }} } {}

Similarly for bkbk size 12{b rSub { size 8{k} } } {}, and thus the power associated with noise output is

P ko = H kΔf 2 a k 2 ¯ + b k 2 ¯ 2 P ko = H kΔf 2 a k 2 ¯ + b k 2 ¯ 2 size 12{P rSub { size 8{ ital "ko"} } = lline H left (kΔf right ) rline rSup { size 8{2} } { { {overline {a rSub { size 8{k} } rSup { size 8{2} } }} + {overline {b rSub { size 8{k} } rSup { size 8{2} } }} } over {2} } } {}
(4)

And the power spectral densities are related by

G no f = H f 2 G ni f G no f = H f 2 G ni f size 12{G rSub { size 8{ ital "no"} } left (f right )= lline H left (f right ) rline rSup { size 8{2} } G rSub { size 8{ ital "ni"} } left (f right )} {}
(5)

Where the kΔfkΔf size 12{kΔf} {} has been replaced by ff size 12{f} {} as a continuous variable as ΔfΔf size 12{Δf} {}tends to 0.

## Superposition of noises:

Noise can be represented as superposition of (orthogonal) harmonics of ΔfΔf size 12{Δf} {} therefore total power is the result of superposition of component powers.

Consider Two processes n1n1 size 12{n rSub { size 8{1} } } {} and n2n2 size 12{n rSub { size 8{2} } } {} with overlapping spectral components.

Power of the sum of n1n1 size 12{n rSub { size 8{1} } } {} and n2n2 size 12{n rSub { size 8{2} } } {} will be p1+p2+2En1n2p1+p2+2En1n2 size 12{p rSub { size 8{1} } +p rSub { size 8{2} } +2E left [n rSub { size 8{1} } n rSub { size 8{2} } right ]} {} and since n1n1 size 12{n rSub { size 8{1} } } {} and n2n2 size 12{n rSub { size 8{2} } } {}are uncorrelated, the last term = 0.

Then these noises also obey the superposition of powers rule.

## Mixing of noise with a sinusoid

If kthkth size 12{k rSup { size 8{ ital "th"} } } {} component of noise is mixed with a sinusoid

n k t cos 2πf o t = a k 2 cos kΔf + f o t + b k 2 sin kΔf + f o t + a k 2 cos kΔf f o t + b k 2 sin kΔf + f o t n k t cos 2πf o t = a k 2 cos kΔf + f o t + b k 2 sin kΔf + f o t + a k 2 cos kΔf f o t + b k 2 sin kΔf + f o t alignl { stack { size 12{n rSub { size 8{k} } left (t right )"cos"2πf rSub { size 8{o} } t= { {a rSub { size 8{k} } } over {2} } "cos"2π left (kΔf+f rSub { size 8{o} } right )t+ { {b rSub { size 8{k} } } over {2} } "sin"2π left (kΔf+f rSub { size 8{o} } right )t} {} # + { {a rSub { size 8{k} } } over {2} } "cos"2π left (kΔf - f rSub { size 8{o} } right )t+ { {b rSub { size 8{k} } } over {2} } "sin"2π left (kΔf+f rSub { size 8{o} } right )t {} } } {}
(6)

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

G n f + f o = G n f f o = G n f 4 G n f + f o = G n f f o = G n f 4 size 12{G rSub { size 8{n} } left (f+f rSub { size 8{o} } right )=G rSub { size 8{n} } left (f - f rSub { size 8{o} } right )= { {G rSub { size 8{n} } left (f right )} over {4} } } {}
(7)

Considering power spectral components at kΔfkΔf size 12{kΔf} {} and lΔflΔf size 12{lΔf} {}, let the mixing frequency be f0=k+lΔff0=k+lΔf size 12{f rSub { size 8{0} } = { size 8{1} } wideslash { size 8{2} } left (k+l right )Δf} {}. This will generate 2 difference frequency components at the same frequency: pΔf=f0kΔf=lΔff0pΔf=f0kΔf=lΔff0 size 12{pΔf=f rSub { size 8{0} } - kΔf=lΔf - f rSub { size 8{0} } } {}

Then difference frequency components are

n p1 t = a k 2 cos 2πpΔ ft b k 2 sin 2πpΔ ft n p1 t = a k 2 cos 2πpΔ ft b k 2 sin 2πpΔ ft size 12{n rSub { size 8{p1} } left (t right )= { {a rSub { size 8{k} } } over {2} } "cos"2πpΔ ital "ft" - { {b rSub { size 8{k} } } over {2} } "sin"2πpΔ ital "ft"} {}
(8)
n p2 t = a l 2 cos 2πpΔ ft + b l 2 sin 2πpΔ ft n p2 t = a l 2 cos 2πpΔ ft + b l 2 sin 2πpΔ ft size 12{n rSub { size 8{p2} } left (t right )= { {a rSub { size 8{l} } } over {2} } "cos"2πpΔ ital "ft"+ { {b rSub { size 8{l} } } over {2} } "sin"2πpΔ ital "ft"} {}
(9)

But as akal¯=akbl¯=bkal¯=bkbl¯=0akal¯=akbl¯=bkal¯=bkbl¯=0 size 12{ {overline {a rSub { size 8{k} } a rSub { size 8{l} } }} = {overline {a rSub { size 8{k} } b rSub { size 8{l} } }} = {overline {b rSub { size 8{k} } a rSub { size 8{l} } }} = {overline {b rSub { size 8{k} } b rSub { size 8{l} } }} =0} {}, We find Enp1tnp2t=0Enp1tnp2t=0 size 12{E left [n rSub { size 8{p1} } left (t right )n rSub { size 8{p2} } left (t right ) right ]=0} {}

and Enp1t+np2t2=Enp1t2+Enp2t2Enp1t+np2t2=Enp1t2+Enp2t2 size 12{E left lbrace left [n rSub { size 8{p1} } left (t right )+n rSub { size 8{p2} } left (t right ) right ] rSup { size 8{2} } right rbrace =E left lbrace left [n rSub { size 8{p1} } left (t right ) right ] rSup { size 8{2} } right rbrace +E left lbrace left [n rSub { size 8{p2} } left (t right ) right ] rSup { size 8{2} } right rbrace } {}

Thus superposition of power applies even after shifting due to mixing.

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