Mixing
Fourier Transform Syntax and Conventions Clarification
$begingroup$
One common definition of a fourier transform for function f(x) is
$$F(v)=int_{-infty}^infty f(tau)e^{2pi itau v}dtau$$
I know some definitions have an extra sqrt(2*pi). I shall ignore that.
However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.
From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).
- Is that right?
- If so, is this the correct fourier transform for f(g(x)) with respect to x?
$$F(v)=int_{-infty}^infty f(g(tau))e^{2pi i tau v}dtau$$
fourier-transform convention
asked Jan 3 at 19:03
roobeeroobee
53
$endgroup$
add a comment |
$begingroup$
One common definition of a fourier transform for function f(x) is
$$F(v)=int_{-infty}^infty f(tau)e^{2pi itau v}dtau$$
I know some definitions have an extra sqrt(2*pi). I shall ignore that.
However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.
From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).
- Is that right?
- If so, is this the correct fourier transform for f(g(x)) with respect to x?
$$F(v)=int_{-infty}^infty f(g(tau))e^{2pi i tau v}dtau$$
fourier-transform convention
asked Jan 3 at 19:03
roobeeroobee
53
$endgroup$
$begingroup$
take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
$endgroup$
– alexjo
Jan 3 at 19:07
$begingroup$
@alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
$endgroup$
– roobee
Jan 3 at 19:11
add a comment |
$begingroup$
One common definition of a fourier transform for function f(x) is
$$F(v)=int_{-infty}^infty f(tau)e^{2pi itau v}dtau$$
I know some definitions have an extra sqrt(2*pi). I shall ignore that.
However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.
From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).
- Is that right?
- If so, is this the correct fourier transform for f(g(x)) with respect to x?
$$F(v)=int_{-infty}^infty f(g(tau))e^{2pi i tau v}dtau$$
fourier-transform convention
asked Jan 3 at 19:03
roobeeroobee
53
$endgroup$
One common definition of a fourier transform for function f(x) is
$$F(v)=int_{-infty}^infty f(tau)e^{2pi itau v}dtau$$
I know some definitions have an extra sqrt(2*pi). I shall ignore that.
However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.
From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).
- Is that right?
- If so, is this the correct fourier transform for f(g(x)) with respect to x?
$$F(v)=int_{-infty}^infty f(g(tau))e^{2pi i tau v}dtau$$
fourier-transform convention
fourier-transform convention
asked Jan 3 at 19:03
roobeeroobee
53
asked Jan 3 at 19:03
roobeeroobee
53
asked Jan 3 at 19:03
roobeeroobee
53
asked Jan 3 at 19:03
roobeeroobee
53
asked Jan 3 at 19:03
roobeeroobee
53
53
$begingroup$
take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
$endgroup$
– alexjo
Jan 3 at 19:07
$begingroup$
@alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
$endgroup$
– roobee
Jan 3 at 19:11
add a comment |
$begingroup$
take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
$endgroup$
– alexjo
Jan 3 at 19:07
$begingroup$
@alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
$endgroup$
– roobee
Jan 3 at 19:11
$begingroup$
take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
$endgroup$
– alexjo
Jan 3 at 19:07
$begingroup$
take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
$endgroup$
– alexjo
Jan 3 at 19:07
$begingroup$
@alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
$endgroup$
– roobee
Jan 3 at 19:11
$begingroup$
@alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
$endgroup$
– roobee
Jan 3 at 19:11
add a comment |
1 Answer
1
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oldest
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$begingroup$
The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$
With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$
For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$
For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$
There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$
answered Jan 3 at 19:35
alexjoalexjo
12.3k1430
$endgroup$
$begingroup$
You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
$endgroup$
– roobee
Jan 3 at 23:01
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$
With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$
For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$
For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$
There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$
answered Jan 3 at 19:35
alexjoalexjo
12.3k1430
$endgroup$
$begingroup$
You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
$endgroup$
– roobee
Jan 3 at 23:01
add a comment |
$begingroup$
The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$
With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$
For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$
For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$
There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$
answered Jan 3 at 19:35
alexjoalexjo
12.3k1430
$endgroup$
$begingroup$
You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
$endgroup$
– roobee
Jan 3 at 23:01
add a comment |
$begingroup$
The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$
With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$
For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$
For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$
There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$
answered Jan 3 at 19:35
alexjoalexjo
12.3k1430
$endgroup$
The Fourier transform of $f(x)$ is
$$
mathcal F{f(x)}=F(nu)=int_{-infty}^infty f(x),mathrm e^{-2pi i x nu},mathrm d x
$$
With this convention, the inverse Fourier transform is
$$
mathcal F^{-1}{F(nu)}=f(f)=int_{-infty}^infty F(nu),mathrm e^{+2pi i x nu},mathrm d nu
$$
For $f(x+1)$ we have (time shift property)
$$
mathcal F{f(x+1)}=int_{-infty}^infty f(x+1),mathrm e^{-2pi i (x+1) nu},mathrm d x=F(nu),mathrm e^{+2pi i nu}
$$
For $f(g(x))=h(x)$ we have to evaluate
$$
mathcal F{f(g(x))})=mathcal F{h(x)}=H(nu)=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x
$$
There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists
$$
mathcal F{f(g(x))}=int_{-infty}^infty f(g(x)),mathrm e^{-2pi i xnu},mathrm d x=int_{-infty}^infty f(y),mathrm e^{-2pi i g^{-1}(y)nu}frac{1}{|g'(y)|},mathrm d y
$$
answered Jan 3 at 19:35
alexjoalexjo
12.3k1430
answered Jan 3 at 19:35
alexjoalexjo
12.3k1430
answered Jan 3 at 19:35
alexjoalexjo
12.3k1430
answered Jan 3 at 19:35
alexjoalexjo
12.3k1430
12.3k1430
$begingroup$
You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
$endgroup$
– roobee
Jan 3 at 23:01
add a comment |
$begingroup$
You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
$endgroup$
– roobee
Jan 3 at 23:01
$begingroup$
You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
$endgroup$
– roobee
Jan 3 at 23:01
$begingroup$
You answered question 2 but not did not directly answer question 1. However, judging by your answer I shall assume you answered yes to question 1.
$endgroup$
– roobee
Jan 3 at 23:01
add a comment |
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$begingroup$
take a look here en.wikipedia.org/wiki/Fourier_transform#Other_conventions
$endgroup$
– alexjo
Jan 3 at 19:07
$begingroup$
@alexjo Yes, I am aware of those conventions. However, all those conventions are defined for f(x). Not for f(x+1) or f(g(x))
$endgroup$
– roobee
Jan 3 at 19:11