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Fourier series coefficients from the dft

If the signal to be analyzed is periodic, the Fourier integral in [link] does not converge to a function (it may to a distribution). This function is usually expanded in a Fourier series to define itsspectrum or a frequency description. We will sample this function and show how to approximately calculate the Fourier series coefficients usingthe DFT of the samples.

Consider a periodic signal f ˜ ( t ) = f ˜ ( t + P ) with N samples taken every T seconds to give T n ˜ ( t ) for integer n such that N T = P . The Fourier series expansion of f ˜ ( t ) is

f ˜ ( t ) = k = - C ( k ) e 2 π k t / P

with the coefficients given in [link] . Samples of this are

f ˜ ( T n ) = k = - C ( k ) e 2 π k T n / P = k = - C ( k ) e 2 π k n / N

which is broken into a sum of sums as

f ˜ ( T n ) = - k = 0 N - 1 C ( k + N ) e 2 π ( k + N ) n / N = k = 0 N - 1 - C ( k + N ) e 2 π k n / N .

But the inverse DFT is of the form

f ˜ ( T n ) = 1 N k = 0 N - 1 F ( k ) e j 2 π n k / N

therefore,

DFT { f ˜ ( T n ) } = N C ( k + N ) = N C p ( k ) .

and we have our result of the relation of the Fourier coefficients to the DFT of a sampled periodic signal. Once again aliasing is a result ofsampling.

Shannon's sampling theorem

Given a signal modeled as a real (sometimes complex) valued function of a real variable (usually time here), we define a bandlimited function as anyfunction whose Fourier transform or spectrum is zero outside of some finite domain

| F ( ω ) | = 0 for | ω | > W

for some W < . The sampling theorem states that if f ( t ) is sampled

f s ( n ) = f ( T n )

such that T < 2 π / W , then f ( t ) can be exactly reconstructed (interpolated) from its samples f s ( n ) using

f ( t ) = n = - f s ( n ) sin ( π t / T - π n ) π t / T - π n .

This is more compactly written by defining the sinc function as

sinc ( x ) = sin ( x ) x

which gives the sampling formula Equation 53 from Least Squared Error Design of FIR Filters the form

f ( t ) = n f s ( n ) sinc ( π t / T - π n ) .

The derivation of Equation 53 from Least Squared Error Design of FIR Filters or Equation 56 from Least Squared Error Design of FIR Filters can be done a number of ways. One of the quickest uses infinite sequences of delta functions and will bedeveloped later in these notes. We will use a more direct method now to better see the assumptions and restrictions.

We first note that if f ( t ) is bandlimited and if T < 2 π / W then there is no overlap or aliasing in F p ( ω ) . In other words, we can write [link] as

f ( t ) = 1 2 π - F ( ω ) e j ω t d ω = 1 2 π - π / T π / T F p ( ω ) e j ω t d ω

but

F p ( ω ) = F ( ω + 2 π / T ) = T n f ( T n ) e - j ω T n

therefore,

f ( t ) = 1 2 π - π / T π / T T n f ( T n ) e - j ω T n e j ω t d ω
= T 2 π n f ( T n ) - π / T π / T e j ( t - T n ) ω d ω
= n f ( T n ) sin ( π T t - π n ) π T t - π n

which is the sampling theorem. An alternate derivation uses a rectangle function and its Fourier transform, the sinc function, together withconvolution and multiplication. A still shorter derivation uses strings of delta function with convolutions and multiplications. This isdiscussed later in these notes.

There are several things to notice about this very important result. First, note that although f ( t ) is defined for all t from only its samples, it does require an infinite number of them to exactly calculate f ( t ) . Also note that this sum can be thought of as an expansion of f ( t ) in terms of an orthogonal set of basis function which are the sinc functions. One can show that the coefficients in this expansion of f ( t ) calculated by an inner product are simply samples of f ( t ) . In other words, the sinc functions span the space of bandlimited functions with avery simple calculation of the expansion coefficients. One can ask the question of what happens if a signal is “under sampled". What happens ifthe reconstruction formula in Equation 12 from Continuous Time Signals is used when there is aliasing and Equation 57 from Least Squarred Error Design of FIR Filters is not true. We will not pursue that just now. In any case, there are many variations and generalizations of this result thatare quite interesting and useful.

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
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8m/s²
Aishat
What is Thermodynamics
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Someone
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Someone
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Source:  OpenStax, Brief notes on signals and systems. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10565/1.7
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