Is Fourier transform a generalisation of Fourier series?
$begingroup$
Is the Fourier transform a generalisation of a Fourier series or an a different concept?
I.e. Can Fourier transforms be used with periodic functions and will it reduce down to the Fourier series when this is done. Or can Fourier transforms not deal with Fourier series at all?
fourier-analysis fourier-series
asked Jun 2 '15 at 8:17
Quantum spaghettificationQuantum spaghettification
55511023
$endgroup$
add a comment |
$begingroup$
Is the Fourier transform a generalisation of a Fourier series or an a different concept?
I.e. Can Fourier transforms be used with periodic functions and will it reduce down to the Fourier series when this is done. Or can Fourier transforms not deal with Fourier series at all?
fourier-analysis fourier-series
asked Jun 2 '15 at 8:17
Quantum spaghettificationQuantum spaghettification
55511023
$endgroup$
add a comment |
$begingroup$
Is the Fourier transform a generalisation of a Fourier series or an a different concept?
I.e. Can Fourier transforms be used with periodic functions and will it reduce down to the Fourier series when this is done. Or can Fourier transforms not deal with Fourier series at all?
fourier-analysis fourier-series
asked Jun 2 '15 at 8:17
Quantum spaghettificationQuantum spaghettification
55511023
$endgroup$
Is the Fourier transform a generalisation of a Fourier series or an a different concept?
I.e. Can Fourier transforms be used with periodic functions and will it reduce down to the Fourier series when this is done. Or can Fourier transforms not deal with Fourier series at all?
fourier-analysis fourier-series
fourier-analysis fourier-series
asked Jun 2 '15 at 8:17
Quantum spaghettificationQuantum spaghettification
55511023
asked Jun 2 '15 at 8:17
Quantum spaghettificationQuantum spaghettification
55511023
asked Jun 2 '15 at 8:17
Quantum spaghettificationQuantum spaghettification
55511023
asked Jun 2 '15 at 8:17
Quantum spaghettificationQuantum spaghettification
55511023
asked Jun 2 '15 at 8:17
Quantum spaghettificationQuantum spaghettification
55511023
55511023
add a comment |
add a comment |
4 Answers
4
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votes
$begingroup$
Fourier transform can represent periodic functions. If the Fourier series coefficient of $f(t)$ is $a_k$
$$F(omega)=2pi sum _{k=-infty}^infty a_k delta(omega-komega _ 0)$$
Moreover, Fourier transform can be thought of as the limit of the fourier series as the period approaches infinity (a function that repeats once every infinity, so it happens only once)
answered Jun 2 '15 at 8:45
grdgfgrgrdgfgr
2,866722
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add a comment |
$begingroup$
From a moderately abstract point of view, the Fourier transform of a periodic function exists in the sense of distribution theory. It consists of a finite combination of Dirac deltas at the frequencies represented by the function. Similarly, for a square integrable function satisfying a "symmetric" boundary condition on a compact interval (e.g. homogeneous Dirichlet or homogeneous Neumann boundary conditions), the Fourier transform consists of a possibly infinite combination of Dirac deltas at the frequencies represented by the function. The weights on the Dirac deltas are exactly the Fourier coefficients (up to a normalization constant depending only on your choice of convention for the Fourier transform).
answered Jun 2 '15 at 15:58
IanIan
67.6k25387
$endgroup$
add a comment |
$begingroup$
From a more abstract point of view, Fourier transform and Fourier series are instances of harmonic analysis on locally compact groups. For the Fourier transform, the concerned group is $mathbb{R}$ (where $f$ is defined) and the Fourier transform $hat f$ is defined on its dual group, which happens to be also $mathbb{R}$. In the case of Fourier series the groups are $mathbb{T}=mathbb{R}/mathbb{Z}$ and the dual group is $mathbb{Z}$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 2 '15 at 15:02
Julián AguirreJulián Aguirre
68.1k24094
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add a comment |
$begingroup$
Fourier series is just the representation of your periodic signal in sine and cosine waves. It just decomposes your signal into these sine-cosine waves. The reason you can't do it for an aperiodic signal is that you would lose out some information while decomposing it.
That is where the Fourier transform comes into the picture. It TRANSFORMS your aperiodic signal into the CONTINUOUS frequency domain where it can be easily represented.
In fact, Fourier series coefficients of a periodic function are sampled values of the Fourier transform of one period of the function (make rest of the function non-zero, thus making it aperiodic, thereby calculating Fourier transform).
answered Jan 9 at 19:44
Madhav ThakkerMadhav Thakker
11
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Fourier transform can represent periodic functions. If the Fourier series coefficient of $f(t)$ is $a_k$
$$F(omega)=2pi sum _{k=-infty}^infty a_k delta(omega-komega _ 0)$$
Moreover, Fourier transform can be thought of as the limit of the fourier series as the period approaches infinity (a function that repeats once every infinity, so it happens only once)
answered Jun 2 '15 at 8:45
grdgfgrgrdgfgr
2,866722
$endgroup$
add a comment |
$begingroup$
Fourier transform can represent periodic functions. If the Fourier series coefficient of $f(t)$ is $a_k$
$$F(omega)=2pi sum _{k=-infty}^infty a_k delta(omega-komega _ 0)$$
Moreover, Fourier transform can be thought of as the limit of the fourier series as the period approaches infinity (a function that repeats once every infinity, so it happens only once)
answered Jun 2 '15 at 8:45
grdgfgrgrdgfgr
2,866722
$endgroup$
add a comment |
$begingroup$
Fourier transform can represent periodic functions. If the Fourier series coefficient of $f(t)$ is $a_k$
$$F(omega)=2pi sum _{k=-infty}^infty a_k delta(omega-komega _ 0)$$
Moreover, Fourier transform can be thought of as the limit of the fourier series as the period approaches infinity (a function that repeats once every infinity, so it happens only once)
answered Jun 2 '15 at 8:45
grdgfgrgrdgfgr
2,866722
$endgroup$
Fourier transform can represent periodic functions. If the Fourier series coefficient of $f(t)$ is $a_k$
$$F(omega)=2pi sum _{k=-infty}^infty a_k delta(omega-komega _ 0)$$
Moreover, Fourier transform can be thought of as the limit of the fourier series as the period approaches infinity (a function that repeats once every infinity, so it happens only once)
answered Jun 2 '15 at 8:45
grdgfgrgrdgfgr
2,866722
answered Jun 2 '15 at 8:45
grdgfgrgrdgfgr
2,866722
answered Jun 2 '15 at 8:45
grdgfgrgrdgfgr
2,866722
answered Jun 2 '15 at 8:45
grdgfgrgrdgfgr
2,866722
2,866722
add a comment |
add a comment |
$begingroup$
From a moderately abstract point of view, the Fourier transform of a periodic function exists in the sense of distribution theory. It consists of a finite combination of Dirac deltas at the frequencies represented by the function. Similarly, for a square integrable function satisfying a "symmetric" boundary condition on a compact interval (e.g. homogeneous Dirichlet or homogeneous Neumann boundary conditions), the Fourier transform consists of a possibly infinite combination of Dirac deltas at the frequencies represented by the function. The weights on the Dirac deltas are exactly the Fourier coefficients (up to a normalization constant depending only on your choice of convention for the Fourier transform).
answered Jun 2 '15 at 15:58
IanIan
67.6k25387
$endgroup$
add a comment |
$begingroup$
From a moderately abstract point of view, the Fourier transform of a periodic function exists in the sense of distribution theory. It consists of a finite combination of Dirac deltas at the frequencies represented by the function. Similarly, for a square integrable function satisfying a "symmetric" boundary condition on a compact interval (e.g. homogeneous Dirichlet or homogeneous Neumann boundary conditions), the Fourier transform consists of a possibly infinite combination of Dirac deltas at the frequencies represented by the function. The weights on the Dirac deltas are exactly the Fourier coefficients (up to a normalization constant depending only on your choice of convention for the Fourier transform).
answered Jun 2 '15 at 15:58
IanIan
67.6k25387
$endgroup$
add a comment |
$begingroup$
From a moderately abstract point of view, the Fourier transform of a periodic function exists in the sense of distribution theory. It consists of a finite combination of Dirac deltas at the frequencies represented by the function. Similarly, for a square integrable function satisfying a "symmetric" boundary condition on a compact interval (e.g. homogeneous Dirichlet or homogeneous Neumann boundary conditions), the Fourier transform consists of a possibly infinite combination of Dirac deltas at the frequencies represented by the function. The weights on the Dirac deltas are exactly the Fourier coefficients (up to a normalization constant depending only on your choice of convention for the Fourier transform).
answered Jun 2 '15 at 15:58
IanIan
67.6k25387
$endgroup$
From a moderately abstract point of view, the Fourier transform of a periodic function exists in the sense of distribution theory. It consists of a finite combination of Dirac deltas at the frequencies represented by the function. Similarly, for a square integrable function satisfying a "symmetric" boundary condition on a compact interval (e.g. homogeneous Dirichlet or homogeneous Neumann boundary conditions), the Fourier transform consists of a possibly infinite combination of Dirac deltas at the frequencies represented by the function. The weights on the Dirac deltas are exactly the Fourier coefficients (up to a normalization constant depending only on your choice of convention for the Fourier transform).
answered Jun 2 '15 at 15:58
IanIan
67.6k25387
edited Oct 1 '17 at 21:50
answered Jun 2 '15 at 15:58
IanIan
67.6k25387
answered Jun 2 '15 at 15:58
IanIan
67.6k25387
answered Jun 2 '15 at 15:58
IanIan
67.6k25387
67.6k25387
add a comment |
add a comment |
$begingroup$
From a more abstract point of view, Fourier transform and Fourier series are instances of harmonic analysis on locally compact groups. For the Fourier transform, the concerned group is $mathbb{R}$ (where $f$ is defined) and the Fourier transform $hat f$ is defined on its dual group, which happens to be also $mathbb{R}$. In the case of Fourier series the groups are $mathbb{T}=mathbb{R}/mathbb{Z}$ and the dual group is $mathbb{Z}$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 2 '15 at 15:02
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
$begingroup$
From a more abstract point of view, Fourier transform and Fourier series are instances of harmonic analysis on locally compact groups. For the Fourier transform, the concerned group is $mathbb{R}$ (where $f$ is defined) and the Fourier transform $hat f$ is defined on its dual group, which happens to be also $mathbb{R}$. In the case of Fourier series the groups are $mathbb{T}=mathbb{R}/mathbb{Z}$ and the dual group is $mathbb{Z}$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 2 '15 at 15:02
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
$begingroup$
From a more abstract point of view, Fourier transform and Fourier series are instances of harmonic analysis on locally compact groups. For the Fourier transform, the concerned group is $mathbb{R}$ (where $f$ is defined) and the Fourier transform $hat f$ is defined on its dual group, which happens to be also $mathbb{R}$. In the case of Fourier series the groups are $mathbb{T}=mathbb{R}/mathbb{Z}$ and the dual group is $mathbb{Z}$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 2 '15 at 15:02
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
From a more abstract point of view, Fourier transform and Fourier series are instances of harmonic analysis on locally compact groups. For the Fourier transform, the concerned group is $mathbb{R}$ (where $f$ is defined) and the Fourier transform $hat f$ is defined on its dual group, which happens to be also $mathbb{R}$. In the case of Fourier series the groups are $mathbb{T}=mathbb{R}/mathbb{Z}$ and the dual group is $mathbb{Z}$.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 2 '15 at 15:02
Julián AguirreJulián Aguirre
68.1k24094
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
answered Jun 2 '15 at 15:02
Julián AguirreJulián Aguirre
68.1k24094
answered Jun 2 '15 at 15:02
Julián AguirreJulián Aguirre
68.1k24094
answered Jun 2 '15 at 15:02
Julián AguirreJulián Aguirre
68.1k24094
68.1k24094
add a comment |
add a comment |
$begingroup$
Fourier series is just the representation of your periodic signal in sine and cosine waves. It just decomposes your signal into these sine-cosine waves. The reason you can't do it for an aperiodic signal is that you would lose out some information while decomposing it.
That is where the Fourier transform comes into the picture. It TRANSFORMS your aperiodic signal into the CONTINUOUS frequency domain where it can be easily represented.
In fact, Fourier series coefficients of a periodic function are sampled values of the Fourier transform of one period of the function (make rest of the function non-zero, thus making it aperiodic, thereby calculating Fourier transform).
answered Jan 9 at 19:44
Madhav ThakkerMadhav Thakker
11
$endgroup$
add a comment |
$begingroup$
Fourier series is just the representation of your periodic signal in sine and cosine waves. It just decomposes your signal into these sine-cosine waves. The reason you can't do it for an aperiodic signal is that you would lose out some information while decomposing it.
That is where the Fourier transform comes into the picture. It TRANSFORMS your aperiodic signal into the CONTINUOUS frequency domain where it can be easily represented.
In fact, Fourier series coefficients of a periodic function are sampled values of the Fourier transform of one period of the function (make rest of the function non-zero, thus making it aperiodic, thereby calculating Fourier transform).
answered Jan 9 at 19:44
Madhav ThakkerMadhav Thakker
11
$endgroup$
add a comment |
$begingroup$
Fourier series is just the representation of your periodic signal in sine and cosine waves. It just decomposes your signal into these sine-cosine waves. The reason you can't do it for an aperiodic signal is that you would lose out some information while decomposing it.
That is where the Fourier transform comes into the picture. It TRANSFORMS your aperiodic signal into the CONTINUOUS frequency domain where it can be easily represented.
In fact, Fourier series coefficients of a periodic function are sampled values of the Fourier transform of one period of the function (make rest of the function non-zero, thus making it aperiodic, thereby calculating Fourier transform).
answered Jan 9 at 19:44
Madhav ThakkerMadhav Thakker
11
$endgroup$
Fourier series is just the representation of your periodic signal in sine and cosine waves. It just decomposes your signal into these sine-cosine waves. The reason you can't do it for an aperiodic signal is that you would lose out some information while decomposing it.
That is where the Fourier transform comes into the picture. It TRANSFORMS your aperiodic signal into the CONTINUOUS frequency domain where it can be easily represented.
In fact, Fourier series coefficients of a periodic function are sampled values of the Fourier transform of one period of the function (make rest of the function non-zero, thus making it aperiodic, thereby calculating Fourier transform).
answered Jan 9 at 19:44
Madhav ThakkerMadhav Thakker
11
answered Jan 9 at 19:44
Madhav ThakkerMadhav Thakker
11
answered Jan 9 at 19:44
Madhav ThakkerMadhav Thakker
11
answered Jan 9 at 19:44
Madhav ThakkerMadhav Thakker
11
11
add a comment |
add a comment |
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