Mixing
Is the inverse Fourier transform always defined?
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The Fourier transform of a continuous-time function is defined if the function is absolutely integrable, otherwise it does not exist. What about the inverse Fourier transform? If I make up any function of $omega$, can I always invert it to some time-domain counterpart?
Would that mean that every function in the frequency domain will have a corresponding function in the time domain but not vice versa?
fourier-analysis fourier-transform signal-processing
asked Feb 1 at 9:59
UndertherainbowUndertherainbow
321417
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add a comment |
$begingroup$
The Fourier transform of a continuous-time function is defined if the function is absolutely integrable, otherwise it does not exist. What about the inverse Fourier transform? If I make up any function of $omega$, can I always invert it to some time-domain counterpart?
Would that mean that every function in the frequency domain will have a corresponding function in the time domain but not vice versa?
fourier-analysis fourier-transform signal-processing
asked Feb 1 at 9:59
UndertherainbowUndertherainbow
321417
$endgroup$
add a comment |
$begingroup$
The Fourier transform of a continuous-time function is defined if the function is absolutely integrable, otherwise it does not exist. What about the inverse Fourier transform? If I make up any function of $omega$, can I always invert it to some time-domain counterpart?
Would that mean that every function in the frequency domain will have a corresponding function in the time domain but not vice versa?
fourier-analysis fourier-transform signal-processing
asked Feb 1 at 9:59
UndertherainbowUndertherainbow
321417
$endgroup$
The Fourier transform of a continuous-time function is defined if the function is absolutely integrable, otherwise it does not exist. What about the inverse Fourier transform? If I make up any function of $omega$, can I always invert it to some time-domain counterpart?
Would that mean that every function in the frequency domain will have a corresponding function in the time domain but not vice versa?
fourier-analysis fourier-transform signal-processing
fourier-analysis fourier-transform signal-processing
asked Feb 1 at 9:59
UndertherainbowUndertherainbow
321417
asked Feb 1 at 9:59
UndertherainbowUndertherainbow
321417
edited Feb 1 at 15:29
Undertherainbow
asked Feb 1 at 9:59
UndertherainbowUndertherainbow
321417
asked Feb 1 at 9:59
UndertherainbowUndertherainbow
321417
asked Feb 1 at 9:59
UndertherainbowUndertherainbow
321417
321417
add a comment |
add a comment |
2 Answers
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Well, you can't just make up any function of $omega.$ Just as for the Fourier transform itself, the function has to be nice enough, e.g. $L^1$ or $L^2.$ Since the Fourier transform can be extended to tempered distributions the functions can actually grow polynomially at the infinities.
Perhaps you haven't seen it yet, but the inverse Fourier transform is almost the same as the "forward" Fourier transform. They only differ in a sign and perhaps a factor (depending on which definition is used; there are a few variants). Therefore the same rules of what can be transformed apply the inverse transform as to the "forward" transform.
answered Feb 1 at 15:56
md2perpemd2perpe
8,43911028
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add a comment |
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Yes the inverse Fourier Transform can always generate its time domain counterpart. The frequency domain representation is simply a set of bin frequencies each with its coefficient and phase which can always be parsed to synthesize a time domain signal
answered Feb 1 at 14:18
Scott StenslandScott Stensland
1367
$endgroup$
1
$begingroup$
The OP is rather asking if we "make up any function" (not knowing in advance it's the Fourier transform of some other function) , if the inverse Fourier transform exists.
$endgroup$
– leonbloy
Feb 1 at 16:00
add a comment |
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2 Answers
2
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2 Answers
2
active
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$begingroup$
Well, you can't just make up any function of $omega.$ Just as for the Fourier transform itself, the function has to be nice enough, e.g. $L^1$ or $L^2.$ Since the Fourier transform can be extended to tempered distributions the functions can actually grow polynomially at the infinities.
Perhaps you haven't seen it yet, but the inverse Fourier transform is almost the same as the "forward" Fourier transform. They only differ in a sign and perhaps a factor (depending on which definition is used; there are a few variants). Therefore the same rules of what can be transformed apply the inverse transform as to the "forward" transform.
answered Feb 1 at 15:56
md2perpemd2perpe
8,43911028
$endgroup$
add a comment |
$begingroup$
Well, you can't just make up any function of $omega.$ Just as for the Fourier transform itself, the function has to be nice enough, e.g. $L^1$ or $L^2.$ Since the Fourier transform can be extended to tempered distributions the functions can actually grow polynomially at the infinities.
Perhaps you haven't seen it yet, but the inverse Fourier transform is almost the same as the "forward" Fourier transform. They only differ in a sign and perhaps a factor (depending on which definition is used; there are a few variants). Therefore the same rules of what can be transformed apply the inverse transform as to the "forward" transform.
answered Feb 1 at 15:56
md2perpemd2perpe
8,43911028
$endgroup$
add a comment |
$begingroup$
Well, you can't just make up any function of $omega.$ Just as for the Fourier transform itself, the function has to be nice enough, e.g. $L^1$ or $L^2.$ Since the Fourier transform can be extended to tempered distributions the functions can actually grow polynomially at the infinities.
Perhaps you haven't seen it yet, but the inverse Fourier transform is almost the same as the "forward" Fourier transform. They only differ in a sign and perhaps a factor (depending on which definition is used; there are a few variants). Therefore the same rules of what can be transformed apply the inverse transform as to the "forward" transform.
answered Feb 1 at 15:56
md2perpemd2perpe
8,43911028
$endgroup$
Well, you can't just make up any function of $omega.$ Just as for the Fourier transform itself, the function has to be nice enough, e.g. $L^1$ or $L^2.$ Since the Fourier transform can be extended to tempered distributions the functions can actually grow polynomially at the infinities.
Perhaps you haven't seen it yet, but the inverse Fourier transform is almost the same as the "forward" Fourier transform. They only differ in a sign and perhaps a factor (depending on which definition is used; there are a few variants). Therefore the same rules of what can be transformed apply the inverse transform as to the "forward" transform.
answered Feb 1 at 15:56
md2perpemd2perpe
8,43911028
answered Feb 1 at 15:56
md2perpemd2perpe
8,43911028
answered Feb 1 at 15:56
md2perpemd2perpe
8,43911028
answered Feb 1 at 15:56
md2perpemd2perpe
8,43911028
8,43911028
add a comment |
add a comment |
$begingroup$
Yes the inverse Fourier Transform can always generate its time domain counterpart. The frequency domain representation is simply a set of bin frequencies each with its coefficient and phase which can always be parsed to synthesize a time domain signal
answered Feb 1 at 14:18
Scott StenslandScott Stensland
1367
$endgroup$
1
$begingroup$
The OP is rather asking if we "make up any function" (not knowing in advance it's the Fourier transform of some other function) , if the inverse Fourier transform exists.
$endgroup$
– leonbloy
Feb 1 at 16:00
add a comment |
$begingroup$
Yes the inverse Fourier Transform can always generate its time domain counterpart. The frequency domain representation is simply a set of bin frequencies each with its coefficient and phase which can always be parsed to synthesize a time domain signal
answered Feb 1 at 14:18
Scott StenslandScott Stensland
1367
$endgroup$
1
$begingroup$
The OP is rather asking if we "make up any function" (not knowing in advance it's the Fourier transform of some other function) , if the inverse Fourier transform exists.
$endgroup$
– leonbloy
Feb 1 at 16:00
add a comment |
$begingroup$
Yes the inverse Fourier Transform can always generate its time domain counterpart. The frequency domain representation is simply a set of bin frequencies each with its coefficient and phase which can always be parsed to synthesize a time domain signal
answered Feb 1 at 14:18
Scott StenslandScott Stensland
1367
$endgroup$
Yes the inverse Fourier Transform can always generate its time domain counterpart. The frequency domain representation is simply a set of bin frequencies each with its coefficient and phase which can always be parsed to synthesize a time domain signal
answered Feb 1 at 14:18
Scott StenslandScott Stensland
1367
answered Feb 1 at 14:18
Scott StenslandScott Stensland
1367
answered Feb 1 at 14:18
Scott StenslandScott Stensland
1367
answered Feb 1 at 14:18
Scott StenslandScott Stensland
1367
1367
1
$begingroup$
The OP is rather asking if we "make up any function" (not knowing in advance it's the Fourier transform of some other function) , if the inverse Fourier transform exists.
$endgroup$
– leonbloy
Feb 1 at 16:00
add a comment |
1
$begingroup$
The OP is rather asking if we "make up any function" (not knowing in advance it's the Fourier transform of some other function) , if the inverse Fourier transform exists.
$endgroup$
– leonbloy
Feb 1 at 16:00
1
1
$begingroup$
The OP is rather asking if we "make up any function" (not knowing in advance it's the Fourier transform of some other function) , if the inverse Fourier transform exists.
$endgroup$
– leonbloy
Feb 1 at 16:00
$begingroup$
The OP is rather asking if we "make up any function" (not knowing in advance it's the Fourier transform of some other function) , if the inverse Fourier transform exists.
$endgroup$
– leonbloy
Feb 1 at 16:00
add a comment |
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