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Filtering of noise

Module by: a.i. trivedi

Summary: this module discusses the effects of filtering of noise

We will see the effects on the power spectrum when noise is passed through a filter.

Filtered output:

If h(t)h(t) size 12{h \( t \) } {} is the impulse response, filtered output no(t)no(t) size 12{n rSub { size 8{o} } \( t \) } {} obtained by convolution.

n o t = ∫ − ∞ ∞ n i τ h t − τ dτ = ∫ − ∞ t n i τ h t − τ dτ n o t = ∫ − ∞ ∞ n i τ h t − τ dτ = ∫ − ∞ t n i τ h t − τ dτ size 12{n rSub { size 8{o} } left (t right )= Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {n rSub { size 8{i} } left (τ right )h left (t - τ right )dτ} = Int cSub { size 8{ - infinity } } cSup { size 8{t} } {n rSub { size 8{i} } left (τ right )h left (t - τ right )dτ} } {} (1)
  • Approximate continuous time by discrete interval.
n o t = lim Δτ → 0 ∑ k = − ∞ k = n i kΔτ h t − kΔτ Δτ n o t = lim Δτ → 0 ∑ k = − ∞ k = n i kΔτ h t − kΔτ Δτ size 12{n rSub { size 8{o} } left (t right )= {"lim"} cSub { size 8{Δτ rightarrow 0} } Sum cSub { size 8{k= - infinity } } cSup { size 8{k= { size 6{t} } wideslash { size 6{Δτ} } } } {n rSub { size 8{i} } left (kΔτ right )h left (t - kΔτ right )Δτ} } {} (2)
  • In the interval impulse response is deterministic and known, but nini size 12{n rSub { size 8{i} } } {} depends on sample chosen,
  • For white Gaussian noise nini size 12{n rSub { size 8{i} } } {} in interval k is different from and independent of that in interval l.
  • Thus output at any time t0t0 size 12{t rSub { size 8{0} } } {} is a linear superposition of independent Gaussian variables: hence it is Gaussian.
  • If we split filter H(f)H(f) size 12{H \( f \) } {} into H1(f)H1(f) size 12{H rSub { size 8{1} } \( f \) } {} and H2(f)H2(f) size 12{H rSub { size 8{2} } \( f \) } {}, output of H1(f)H1(f) size 12{H rSub { size 8{1} } \( f \) } {} is Gaussian but NOT white (due to filtering)
Figure 1
Figure 1 (graphics1.png)
  • Hence output nono size 12{n rSub { size 8{o} } } {} of H2(f)H2(f) size 12{H rSub { size 8{2} } \( f \) } {} is also non-white.
  • output no'no' size 12{n rSub { size 8{o} } rSup { size 8{'} } } {} of H1(f)H1(f) size 12{H rSub { size 8{1} } \( f \) } {} is non-white, thus no'no' size 12{n rSub { size 8{o} } rSup { size 8{'} } } {} in intervals k and l are dependent, but despite of this nono size 12{n rSub { size 8{o} } } {} is Gaussian.
  • Hence superposition of non-white dependent Gaussian variables also gives Gaussian output.

 Spectral components:

  • Let spectral component k be
n k t = a k cos 2πkΔ ft + b k sin 2πkΔ ft = c k cos 2πkΔ ft + θ k n k t = a k cos 2πkΔ ft + b k sin 2πkΔ ft = c k cos 2πkΔ ft + θ k size 12{n rSub { size 8{k} } left (t right )=a rSub { size 8{k} } "cos"2πkΔ ital "ft"+b rSub { size 8{k} } "sin"2πkΔ ital "ft"=c rSub { size 8{k} } "cos" left (2πkΔ ital "ft"+θ rSub { size 8{k} } right )} {} (3)
  • Then though coefficients are random over ensemble, they are deterministic for a specific case and depend on the particular sample chosen.
  • Hence spectral component is stationary but not ergodic.
  • Its normalized power is
P k = n k t 2 ¯ = a k 2 ¯ cos 2 2πkΔ ft + b k 2 ¯ sin 2 2πkΔ ft + 2a k b k ¯ cos 2πkΔ ft sin 2πkΔ ft P k = n k t 2 ¯ = a k 2 ¯ cos 2 2πkΔ ft + b k 2 ¯ sin 2 2πkΔ ft + 2a k b k ¯ cos 2πkΔ ft sin 2πkΔ ft size 12{P rSub { size 8{k} } = {overline { left [n rSub { size 8{k} } left (t right ) right ] rSup { size 8{2} } }} = {overline {a rSub { size 8{k} } rSup { size 8{2} } }} "cos" rSup { size 8{2} } 2πkΔ ital "ft"+ {overline {b rSub { size 8{k} } rSup { size 8{2} } }} "sin" rSup { size 8{2} } 2πkΔ ital "ft"+ {overline {2a rSub { size 8{k} } b rSub { size 8{k} } }} "cos"2πkΔ ital "ft""sin"2πkΔ ital "ft"} {} (4)
  • As nk(t)nk(t) size 12{n rSub { size 8{k} } \( t \) } {} is stationary, t can be any time. Choose t as t1t1 size 12{t rSub { size 8{1} } } {}when cosine is 1, sine is 0. Then
P k = a k 2 ¯ P k = a k 2 ¯ size 12{P rSub { size 8{k} } = {overline {a rSub { size 8{k} } rSup { size 8{2} } }} } {} (5)

and similarly

P k = b k 2 ¯ P k = b k 2 ¯ size 12{P rSub { size 8{k} } = {overline {b rSub { size 8{k} } rSup { size 8{2} } }} } {} (6)
P k = 2G n ( kΔf ) Δf = − 2G n ( kΔf ) Δf = a k 2 ¯ = b k 2 ¯ = a k 2 ¯ + b k 2 ¯ 2 = c k 2 ¯ 2 P k = 2G n ( kΔf ) Δf = − 2G n ( kΔf ) Δf = a k 2 ¯ = b k 2 ¯ = a k 2 ¯ + b k 2 ¯ 2 = c k 2 ¯ 2 size 12{P rSub { size 8{k} } =2G rSub { size 8{n} } \( kΔf \) Δf= - 2G rSub { size 8{n} } \( kΔf \) Δf= {overline {a rSub { size 8{k} } rSup { size 8{2} } }} = {overline {b rSub { size 8{k} } 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} } = { { {overline {c rSub { size 8{k} } rSup { size 8{2} } }} } over {2} } } {} (7)

Since

a k 2 ¯ = b k 2 ¯ a k 2 ¯ = b k 2 ¯ size 12{ {overline {a rSub { size 8{k} } rSup { size 8{2} } }} = {overline {b rSub { size 8{k} } rSup { size 8{2} } }} } {} (8)
  • Then pkpk size 12{p rSub { size 8{k} } } {} can be written as showing a constant term and a sin-cos time dependent term.
P k = a k 2 ¯ cos 2 2πkΔ ft + sin 2 2πkΔ ft + 2a k b k ¯ cos 2πkΔ ft sin 2πkΔ ft P k = a k 2 ¯ cos 2 2πkΔ ft + sin 2 2πkΔ ft + 2a k b k ¯ cos 2πkΔ ft sin 2πkΔ ft size 12{P rSub { size 8{k} } = {overline {a rSub { size 8{k} } rSup { size 8{2} } }} left ("cos" rSup { size 8{2} } 2πkΔ ital "ft"+"sin" rSup { size 8{2} } 2πkΔ ital "ft" right )+ {overline {2a rSub { size 8{k} } b rSub { size 8{k} } }} "cos"2πkΔ ital "ft""sin"2πkΔ ital "ft"} {} (9)

thus

P k = a k 2 ¯ + 2a k b k ¯ cos 2πkΔ ft sin 2πkΔf P k = a k 2 ¯ + 2a k b k ¯ cos 2πkΔ ft sin 2πkΔf size 12{P rSub { size 8{k} } = {overline {a rSub { size 8{k} } rSup { size 8{2} } }} + {overline {2a rSub { size 8{k} } b rSub { size 8{k} } }} "cos"2πkΔ ital "ft""sin"2πkΔf} {} (10)
  • But pkpk size 12{p rSub { size 8{k} } } {} is stationary and independent of time, so
a k b k ¯ = 0 a k b k ¯ = 0 size 12{ {overline {a rSub { size 8{k} } b rSub { size 8{k} } }} =0} {} (11)

Hence the coefficients are seen to be uncorrelated.

  • Also at t1t1 size 12{t rSub { size 8{1} } } {}, we have nk(t1)=aknk(t1)=ak size 12{n rSub { size 8{k} } \( t rSub { size 8{1} } \) =a rSub { size 8{k} } } {}. But nknk size 12{n rSub { size 8{k} } } {} is Gaussian as it can be considered output of a narrowband filter whose input is Gaussian.

Thus akak size 12{a rSub { size 8{k} } } {} and bkbk size 12{b rSub { size 8{k} } } {} are Gaussian.

  • Also akak size 12{a rSub { size 8{k} } } {} is output of filter in a non zero frequency interval k, it is not DC, thus mean value of ak=0ak=0 size 12{a rSub { size 8{k} } =0} {}. Hence coefficients are Zero-mean Gaussian.
  • Taking product of two samples nk(t)nk(t) size 12{n rSub { size 8{k} } \( t \) } {} and nl(t)nl(t) size 12{n rSub { size 8{l} } \( t \) } {} where
n k t = a k cos 2πkΔ ft + b k sin 2πkΔ ft n k t = a k cos 2πkΔ ft + b k sin 2πkΔ ft size 12{n rSub { size 8{k} } left (t right )=a rSub { size 8{k} } "cos"2πkΔ ital "ft"+b rSub { size 8{k} } "sin"2πkΔ ital "ft"} {} (12)
n l t = a l cos 2πlΔ ft + b l sin 2πlΔ ft n l t = a l cos 2πlΔ ft + b l sin 2πlΔ ft size 12{n rSub { size 8{l} } left (t right )=a rSub { size 8{l} } "cos"2πlΔ ital "ft"+b rSub { size 8{l} } "sin"2πlΔ ital "ft"} {} (13)

The ensemble averages of the result must be independent of time as each component is stationary, hence

a k a l ¯ = a k b l ¯ = b k a l ¯ = b k b l ¯ = 0