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

Psd after filtering:

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

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

f c k = c k P k e c k 2 / 2P k c k 0 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}} {} , f θ k = 1 π θ k π size 12{f left (θrSub { size 8{k} } right )= { {1} over {2π} } ~ -π<=θrSub { size 8{k} }<=π} {}

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

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∠ϕrSub { size 8{k} } } {}

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

The power associated with the input component is

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} } } {}

As H kΔf size 12{ lline H left (kΔf right ) rline } {} is a deterministic function, H kΔf a k 2 ¯ = H kΔf 2 a k 2 ¯ 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 b k 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 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} } } {}

And the power spectral densities are related by

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 )} {}

Where the kΔf size 12{kΔf} {} has been replaced by f size 12{f} {} as a continuous variable as Δf size 12{Δf} {} tends to 0.

Superposition of noises:

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

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

Power of the sum of n 1 size 12{n rSub { size 8{1} } } {} and n 2 size 12{n rSub { size 8{2} } } {} will be p 1 + p 2 + 2 E n 1 n 2 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 n 1 size 12{n rSub { size 8{1} } } {} and n 2 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 k th 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 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"2πleft (kΔf+f rSub { size 8{o} } right )t+ { {b rSub { size 8{k} } } over {2} } "sin"2πleft (kΔf+f rSub { size 8{o} } right )t} {} # + { {a rSub { size 8{k} } } over {2} } "cos"2πleft (kΔf - f rSub { size 8{o} } right )t+ { {b rSub { size 8{k} } } over {2} } "sin"2πleft (kΔf+f rSub { size 8{o} } right )t {} } } {}

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 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} } } {}

Considering power spectral components at kΔf size 12{kΔf} {} and lΔf size 12{lΔf} {} , let the mixing frequency be f 0 = k + l Δf 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 = f 0 kΔf = lΔf f 0 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 size 12{n rSub { size 8{p1} } left (t right )= { {a rSub { size 8{k} } } over {2} } "cos"2πpΔital "ft" - { {b rSub { size 8{k} } } over {2} } "sin"2πpΔital "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"2πpΔital "ft"+ { {b rSub { size 8{l} } } over {2} } "sin"2πpΔital "ft"} {}

But as a k a l ¯ = a k b l ¯ = b k a l ¯ = b k b l ¯ = 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 E n p1 t n p2 t = 0 size 12{E left [n rSub { size 8{p1} } left (t right )n rSub { size 8{p2} } left (t right ) right ]=0} {}

and E n p1 t + n p2 t 2 = E n p1 t 2 + E n p2 t 2 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.

Questions & Answers

what is the stm
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How we are making nano material?
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write examples of Nano molecule?
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The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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what king of growth are you checking .?
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
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why?
Adin
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biomolecules are e building blocks of every organics and inorganic materials.
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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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what does nano mean?
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absolutely yes
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there is no specific books for beginners but there is book called principle of nanotechnology
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how can I make nanorobot?
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what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
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fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Noise in communications. OpenStax CNX. Jul 07, 2008 Download for free at http://cnx.org/content/col10549/1.1
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