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The Fourier Series is the representation of continuous-time, periodic signals in terms of complex exponentials. The Dirichletconditions suggest that discontinuous signals may have a Fourier Series representation so long as there are a finite number ofdiscontinuities. This seems counter-intuitive, however, as complex exponentials are continuous functions. It does not seempossible to exactly reconstruct a discontinuous function from a set of continuous ones. In fact, it is not. However, it can beif we relax the condition of exactly and replace it with the idea of almost everywhere. This is to say that the reconstruction isexactly the same as the original signal except at a finite number of points. These points, not necessarily suprisingly, occur atthe points of discontinuities.

Introduction

The Fourier Series is the representation of continuous-time, periodic signals in terms of complex exponentials. The Dirichlet conditions suggest that discontinuous signals may have a Fourier Series representationso long as there are a finite number of discontinuities. This seems counter-intuitive, however, as complex exponentials are continuous functions. It does not seem possible to exactly reconstruct adiscontinuous function from a set of continuous ones. In fact, it is not. However, it can be if we relax the conditionof 'exactly' and replace it with the idea of 'almost everywhere'. This is to say that the reconstruction isexactly the same as the original signal except at a finite number of points. These points, not necessarily surprisingly,occur at the points of discontinuities.

History

In the late 1800s, many machines were built to calculate Fourier coefficients and re-synthesize:

f N t n N N c n ω 0 n t
Albert Michelson (an extraordinary experimental physicist) built a machine in 1898 that could compute c n up to n ± 79 , and he re-synthesized
f 79 t n 79 -79 c n ω 0 n t
The machine performed very well on all tests except thoseinvolving discontinuous functions . When a square wave, like that shown in [link] , was inputed into the machine, "wiggles" around the discontinuities appeared, and even as the numberof Fourier coefficients approached infinity, the wiggles never disappeared - these can be seen in the last plot in [link] . J. Willard Gibbs first explained this phenomenon in 1899, and therefore thesediscontinuous points are referred to as Gibbs Phenomenon .

Explanation

We begin this discussion by taking a signal with a finite number of discontinuities (like a square pulse ) and finding its Fourier Series representation. We thenattempt to reconstruct it from these Fourier coefficients. What we find is that the more coefficients we use, the morethe signal begins to resemble the original. However, around the discontinuities, we observe rippling that does not seem tosubside. As we consider even more coefficients, we notice that the ripples narrow, but do not shorten. As we approachan infinite number of coefficients, this rippling still does not go away. This is when we apply the idea of almosteverywhere. While these ripples remain (never dropping below 9% of the pulse height), the area inside them tends to zero,meaning that the energy of this ripple goes to zero. This means that their width is approaching zero and we can assertthat the reconstruction is exactly the original except at the points of discontinuity. Since the Dirichlet conditionsassert that there may only be a finite number of discontinuities, we can conclude that the principle of almosteverywhere is met. This phenomenon is a specific case of nonuniform convergence .

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
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
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
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
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
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.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Signals and systems. OpenStax CNX. Aug 14, 2014 Download for free at http://legacy.cnx.org/content/col10064/1.15
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