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Here is the definition of the Fourier Series and some properties and examples

The fourier series

The problem of expanding a finite length signal in a trigonometric series was posed and studied in the late 1700's by renown mathematicians such asBernoulli, d'Alembert, Euler, Lagrange, and Gauss. Indeed, what we now call the Fourier series and the formulas for the coefficients were used by Euler in1780. However, it was the presentation in 1807 and the paper in 1822 by Fourier stating that an arbitrary function could be represented by a series ofsines and cosines that brought the problem to everyone's attention and started serious theoretical investigations and practical applications that continue tothis day . The theoretical work has been at the center of analysis and the practicalapplications have been of major significance in virtually every field of quantitative science and technology. For these reasons and others, the Fourierseries is worth our serious attention in a study of signal processing.

Definition of the fourier series

We assume that the signal x ( t ) to be analyzed is well described by a real or complex valued function of a real variable t defined over a finite interval { 0 t T } . The trigonometric series expansion of x ( t ) is given by

x ( t ) = a ( 0 ) 2 + k = 1 a ( k ) cos ( 2 π T k t ) + b ( k ) sin ( 2 π T k t ) .

where the sines and cosines are the basis functions for the expansion.The energy or power in an electrical, mechanical, etc. system is a function of the square of voltage, current, velocity, pressure, etc. For this reason, thenatural setting for a representation of signals is the Hilbert space of L 2 [ 0 , T ] . This modern formulation of the problem is developed in . The sinusoidal basis functions in the trigonometric expansion form a completeorthogonal set in L 2 [ 0 , T ] . The orthogonality is easily seen from inner products

( cos ( 2 π T k t ) , cos ( 2 π T t ) ) = 0 T cos ( 2 π T k t ) cos ( 2 π T t ) ) t = δ ( k )
and
( cos ( 2 π T k t ) , sin ( 2 π T t ) ) = 0 T cos ( 2 π T k t ) sin ( 2 π T t ) ) t = 0
where the Kronecker delta function δ ( 0 ) = 1 and δ ( k 0 ) = 0 . Because of this, the k th coefficients in the series can be found by taking the inner product of x ( t ) with the k th basis functions. This gives for the coefficients
a ( k ) = 2 T 0 T x ( t ) cos ( 2 π T k t ) t
and
b ( k ) = 2 T 0 T x ( t ) sin ( 2 π T k t ) t
where T is the time interval of interest or the period of the periodic signal. Because of the orthogonality of the basis functions, a finite Fourier series formed by truncating the infiniteseries is an optimal least squared error approximation to x ( t ) . If the finite series is defined by
x ̂ ( t ) = a ( 0 ) 2 + k = 1 N a ( k ) cos ( 2 π T k t ) + b ( k ) sin ( 2 π T k t ) ,
the squared error is
ɛ = 1 T 0 T | x ( t ) x ̂ ( t ) | 2 t
which is minimized over all a ( k ) and b ( k ) by ( ) and ( ). This is an extraordinarily important property.

It follows that if x ( t ) L 2 [ 0 , T ] , then the series converges to x ( t ) in the sense that ɛ 0 as N . The question of point-wise convergence is more difficult. A sufficientcondition that is adequate for most application states: If f ( x ) is bounded, is piece-wise continuous, and has no more than a finite number of maxima over an interval, the Fourier series converges point-wise to f ( x ) at all points of continuity and to the arithmetic mean at points of discontinuities. If f ( x ) is continuous, the series converges uniformly at all points .

A useful condition states that if x ( t ) and its derivatives through the q th derivative are defined and have bounded variation, the Fourier coefficients a ( k ) and b ( k ) asymptotically drop off at least as fast as 1 k q + 1 as k . This ties global rates of convergence of the coefficients to local smoothnessconditions of the function.

The form of the Fourier series using both sines and cosines makes determination of the peak value or of the location of a particular frequencyterm difficult. A form that explicitly gives the peak value of the sinusoid of that frequency and the location or phase shift of that sinusoid is given by

x ( t ) = d ( 0 ) 2 + k = 1 d ( k ) cos ( 2 π T k t + θ ( k ) )
and, using Euler's relation and the usual electrical engineering notation of j = 1 ,
e j x = cos ( x ) + j sin ( x ) ,
the complex exponential form is obtained as
x ( t ) = k = c ( k ) e j 2 π T k t
where
c ( k ) = a ( k ) + j b ( k ) .
The coefficient equation is
c ( k ) = 1 T 0 T x ( t ) e j 2 π T k t t
The coefficients in these three forms are related by
| d | 2 = | c | 2 = a 2 + b 2
and
θ = a r g { c } = tan 1 ( b a )
It is easier to evaluate a signal in terms of c ( k ) or d ( k ) and θ ( k ) than in terms of a ( k ) and b ( k ) . The first two are polar representation of a complex value and the last isrectangular. The exponential form is easier to work with mathematically.

Although the function to be expanded is defined only over a specific finite region, the series converges to a function that is defined over the real lineand is periodic. It is equal to the original function over the region of definition and is a periodic extension outside of the region. Indeed, onecould artificially extend the given function at the outset and then the expansion would converge everywhere.

A geometric view

It can be very helpful to develop a geometric view of the Fourier series where x ( t ) is considered to be a vector and the basis functions are the coordinate or basis vectors. The coefficients become the projections of x ( t ) on the coordinates. The ideas of a measure of distance, size, and orthogonality are important and the definition of error is easy to picture.This is done in using Hilbert space methods.

Examples

  • An example of the Fourier series is the expansion of a square wave signal withperiod 2 π . The expansion is
    x ( t ) = 4 π [ sin ( t ) + 1 3 sin ( 3 t ) + 1 5 sin ( 5 t ) ] .
    Because x ( t ) is odd, there are no cosine terms (all a ( k ) = 0 ) and, because of its symmetries, there are no even harmonics (even k terms are zero). The function is well defined and bounded; its derivative is not, therefore, the coefficients drop off as 1 k .
  • A second example is a triangle wave of period 2 π . This is a continuous function where the square wave was not. The expansion ofthe triangle wave is
    x ( t ) = 4 π [ sin ( t ) + 1 3 2 sin ( 3 t ) + 1 5 2 sin ( 5 t ) + ] .
    Here the coefficients drop off as 1 k 2 since the function and its first derivative exist and are bounded.

Note the derivative of a triangle wave is a square wave. Examine the series coefficients to see this. There are many books and web sites on the Fourier series that give insight through examples and demos.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
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
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Abhi
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Source:  OpenStax, Principles of digital communications. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10805/1.1
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