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Introduction to noise and noise filtering.

We have mentioned that communications are, to varying degrees, subject to interference and noise. It's time tobe more precise about what these quantities are and how they differ.

Interference represents man-made signals. Telephone lines are subject to power-line interference (in the UnitedStates a distorted 60 Hz sinusoid). Cellular telephone channels are subject to adjacent-cell phone conversations usingthe same signal frequency. The problem with such interference is that it occupies the same frequency band as the desiredcommunication signal, and has a similar structure.

Suppose interference occupied a different frequency band; how would the receiver remove it?

If the interferer's spectrum does not overlap that of our communications channel—the interferer isout-of-band—we need only use a bandpass filter that selects our transmission band and removes other portions ofthe spectrum.

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We use the notation i t to represent interference. Because interference has man-madestructure, we can write an explicit expression for it that may contain some unknown aspects (how large it is, for example).

Noise signals have little structure and arise from both human and natural sources. Satellite channels are subjectto deep space noise arising from electromagnetic radiation pervasive in the galaxy. Thermal noise plagues all electronic circuits that contain resistors. Thus, in receiving small amplitude signals, receiveramplifiers will most certainly add noise as they boost the signal's amplitude. All channels are subject to noise, and weneed a way of describing such signals despite the fact we can't write a formula for the noise signal like we can forinterference. The most widely used noise model is white noise . It is defined entirely by its frequency-domain characteristics.

  • White noise has constant power at all frequencies.
  • At each frequency, the phase of the noise spectrum is totally uncertain: It can be any value in between 0 and 2 π , and its value at any frequency is unrelated to the phase atany other frequency.
  • When noise signals arising from two different sources add, the resultant noise signal has a power equal to the sum of thecomponent powers.

Because of the emphasis here on frequency-domain power, we are led to define the power spectrum . Because of Parseval's Theorem , we define the power spectrum P s f of a non-noise signal s t to be the magnitude-squared of its Fourier transform.

P s f S f 2
Integrating the power spectrum over any range of frequencies equals the power the signal contains in that band. Because signals must have negative frequency components that mirror positive frequency ones, we routinely calculate the power in aspectral band as the integral over positive frequencies multiplied by two.
Power in f 1 f 2 2 f f 1 f 2 P s f
Using the notation n t to represent a noise signal's waveform, we define noise in terms of its powerspectrum. For white noise, the power spectrum equals the constant N 0 2 . With this definition, the power in a frequency band equals N 0 f 2 f 1 .

When we pass a signal through a linear, time-invariant system, theoutput's spectrum equals the product of the system's frequency response and the input's spectrum. Thus, the powerspectrum of the system's output is given by

P y f H f 2 P x f
This result applies to noise signals as well. When we pass white noise through a filter, the output is also a noise signal but withpower spectrum H f 2 N 0 2 .

Questions & Answers

can someone help me with some logarithmic and exponential equations.
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Fundamentals of electrical engineering i. OpenStax CNX. Aug 06, 2008 Download for free at http://legacy.cnx.org/content/col10040/1.9
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