discrete time Fourier transform in relation with continuous time Fourier transform
Fourier Transforms of Sampled SignalsFernando Diego Sanz Gamiz
Computers are able to handle only finite number of data. Hence, if
we are to study and treat real world signals (i.e. functions ) in a computer, a way to characterize signals by a finite number
of data has to be found.
If we sample the values of the signal at periodic times we form the sequence . Does this sequence contain all the information relative to ?
We know from the sampling theorem that if the signal
is bandlimited, the sampled sequence allows us to recover the
original continuous
time signal provided the sampling frequency
is at least twice the maximum frequency of the signal. However,
real signals are not of finite bandwidth as this would imply the
signal to be of infinite
time duration. Therefore, a problem arise
of how well can we approximate the original signal by the
sampled sequence . In fact, we are interested in studying
the spectrum of the original signal based upon the samples .
While the relation between and the spectrum of
is widely used in communication and electronic engineering books, it
is difficult to find a rigorous proof. We cover here the gap between
engineering daily knowledge and rigorous mathematical proof of the
named relations establishing under what assumptions
those relations
are valid.
Relation between Discrete Time Fourier Transform and Continuous time Fourier transform
Theorem 1.
Let be a bounded variation (it can be, in particular,
piecewise smooth) function and let be its Fourier
transform 11in the present entry we will take the Fourier
transform of to be . Let . If converges
a.e as and is there
is such that a.e. then
(1) |
If additionally is continuous, the
right hand side of (1) converges uniformly to the left hand
side in any closed interval .
Proof.
By hypothesis we can form the function
This function is obviously periodic of period and bounded, hence it can be expanded in its Fourier series which converge in ; the Fourier theory shows that the convergence is uniformly in if is continuous.
where the coefficients are given by
As we can appeal the dominated convergence theorem to write
Now, as is of bounded variation, the Jordan theorem on Fourier transform inversion says that
and the result follows.
∎
There are, however, very which do not
satisfy the conditions required in the theorem above. Such is the
case of the pulse function
whose Fourier transform is -with the definition made in
footnote 1-. behaves as , so
will not converge in general 22for monotone
decreasing functions ”series behaves as integrals
”, that is, if
converges or diverges so does
; it will converge for those that make
the series an alternating (http://planetmath.org/AlternatingSeries) one, but not for the rest values of .
Therefore we need another result which somehow relates both sides of
eq 1.
As we have pointed out, the problem with the pulse function is that its Fourier transform does not decay rapid enough for the series to converge. So we will try smoothing the signal out so that its Fourier transform will decay faster and, hopefully, the series converges. We wish the smoothed version of to resemble the original signal, so uniform approximation seems reasonable. But, as we will see, for an infinite number of samples each of these might require a different degree of approximation and it could be impossible to find an uniform approximation for all the samples. So, we will focus on time limited signal, for which we have the following result.
Notation.
will denote the set of test functions on and
the set of rapidly decreasing functions on -the Schwartz
space-. The symbol above a function will denote its Fourier
transform. We know that and that the Fourier
transform is an isomorphism of onto itself. The product
of
two functions and at will be denoted by
. Convolution
will be denoted by .
Theorem 2.
Let be a test function and be an approximate identity. Let be a time limited bounded variation signal. If converges for all to a continuous function, then
(2) |
Proof.
Take the signal whose Fourier transform is . This signal satisfies, by hypothesis, the conditions of Theorem 1, so we can write
The right hand side is actually a sum of a finite number of terms
since the signal is time limited -being the
convolution of two compact supported functions-. This makes the
Fourier series a continuous function, which, together with the
hypothesis that is continuous, shows that
equality in the above equation is pointwise.
Now let and use the fact that is an approximate identity to obtain equation (2). ∎
The rationale behind choosing test functions in the last theorem is that, in most cases, will converge even though do not. This is because rapidly decreasing functions decay faster than for any . So, for example, the Fourier transform of the pulse function has been tamed enough to make the series converge.
Remark 1.
Title | discrete time Fourier transform in relation with continuous time Fourier transform |
Canonical name | DiscreteTimeFourierTransformInRelationWithContinuousTimeFourierTransform |
Date of creation | 2013-03-22 17:37:45 |
Last modified on | 2013-03-22 17:37:45 |
Owner | fernsanz (8869) |
Last modified by | fernsanz (8869) |
Numerical id | 10 |
Author | fernsanz (8869) |
Entry type | Theorem |
Classification | msc 94A20 |
Related topic | SamplingTheorem |
Related topic | ApproximatingFourierIntegralsWithDiscreteFourierTransforms |
Related topic | Distribution4 |
Related topic | SpaceOfRapidlyDecreasingFunctions |
Defines | DTFT |
Defines | Discrete Time Fourier Transform |