What is the Fourier transform of the identity function? The 2019 Stack Overflow Developer Survey Results Are InWhat's the Fourier transform of these functions?Fourier transform of the identity function $f(x)=x$Fourier Transform Dirac DeltaDirac delta distribution and fourier transformFourier transform of distribution without physicist's $delta$-functionfourier transform normalization constantComputing Fourier TransformFormal derivation of the Fourier transform of Dirac delta using a distributionFourier transform using Dirac'sFourier Transform of Heaviside Step Function
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What is the Fourier transform of the identity function?
The 2019 Stack Overflow Developer Survey Results Are InWhat's the Fourier transform of these functions?Fourier transform of the identity function $f(x)=x$Fourier Transform Dirac DeltaDirac delta distribution and fourier transformFourier transform of distribution without physicist's $delta$-functionfourier transform normalization constantComputing Fourier TransformFormal derivation of the Fourier transform of Dirac delta using a distributionFourier transform using Dirac'sFourier Transform of Heaviside Step Function
$begingroup$
How to find the Fourier transform of $x mapsto x$ using distribution $delta$?
Since $FT(1)=sqrt2pi delta(k)$ then $FT(x cdot 1)=sqrt2pi i delta'(k)$
But also since $1=d/dx (x)$ then $FT(x)=FT(1)/(ik)=delta(k)/(ik)$
Are both of these correct?
fourier-analysis fourier-transform distribution-theory dirac-delta
edited Mar 30 at 16:26
Rodrigo de Azevedo
13.2k41962
asked Mar 30 at 16:15
SørënSørën
1009
$endgroup$
add a comment |
$begingroup$
How to find the Fourier transform of $x mapsto x$ using distribution $delta$?
Since $FT(1)=sqrt2pi delta(k)$ then $FT(x cdot 1)=sqrt2pi i delta'(k)$
But also since $1=d/dx (x)$ then $FT(x)=FT(1)/(ik)=delta(k)/(ik)$
Are both of these correct?
fourier-analysis fourier-transform distribution-theory dirac-delta
edited Mar 30 at 16:26
Rodrigo de Azevedo
13.2k41962
asked Mar 30 at 16:15
SørënSørën
1009
$endgroup$
$begingroup$
You are not really allowed to write $delta(k)/k$ in distribution theory, but the best interpretation if this expressions is $-delta'(k)$.
$endgroup$
– md2perpe
Mar 30 at 16:22
$begingroup$
Your first derivation is correct.
$endgroup$
– md2perpe
Mar 30 at 16:26
$begingroup$
Your second derivation lacks a term $C , delta(k)$: $$sqrt2pi , delta(k) = FT(1) = FT(fracddxx) = ik , FT(x)$$ if and only if $$-sqrt2pi , delta'(k) + C , delta(k) = i , FT(x).$$
$endgroup$
– md2perpe
Mar 30 at 16:29
add a comment |
$begingroup$
How to find the Fourier transform of $x mapsto x$ using distribution $delta$?
Since $FT(1)=sqrt2pi delta(k)$ then $FT(x cdot 1)=sqrt2pi i delta'(k)$
But also since $1=d/dx (x)$ then $FT(x)=FT(1)/(ik)=delta(k)/(ik)$
Are both of these correct?
fourier-analysis fourier-transform distribution-theory dirac-delta
edited Mar 30 at 16:26
Rodrigo de Azevedo
13.2k41962
asked Mar 30 at 16:15
SørënSørën
1009
$endgroup$
How to find the Fourier transform of $x mapsto x$ using distribution $delta$?
Since $FT(1)=sqrt2pi delta(k)$ then $FT(x cdot 1)=sqrt2pi i delta'(k)$
But also since $1=d/dx (x)$ then $FT(x)=FT(1)/(ik)=delta(k)/(ik)$
Are both of these correct?
fourier-analysis fourier-transform distribution-theory dirac-delta
fourier-analysis fourier-transform distribution-theory dirac-delta
edited Mar 30 at 16:26
Rodrigo de Azevedo
13.2k41962
asked Mar 30 at 16:15
SørënSørën
1009
edited Mar 30 at 16:26
Rodrigo de Azevedo
13.2k41962
asked Mar 30 at 16:15
SørënSørën
1009
edited Mar 30 at 16:26
Rodrigo de Azevedo
13.2k41962
edited Mar 30 at 16:26
Rodrigo de Azevedo
13.2k41962
edited Mar 30 at 16:26
Rodrigo de Azevedo
13.2k41962
13.2k41962
asked Mar 30 at 16:15
SørënSørën
1009
asked Mar 30 at 16:15
SørënSørën
1009
asked Mar 30 at 16:15
SørënSørën
1009
1009
$begingroup$
You are not really allowed to write $delta(k)/k$ in distribution theory, but the best interpretation if this expressions is $-delta'(k)$.
$endgroup$
– md2perpe
Mar 30 at 16:22
$begingroup$
Your first derivation is correct.
$endgroup$
– md2perpe
Mar 30 at 16:26
$begingroup$
Your second derivation lacks a term $C , delta(k)$: $$sqrt2pi , delta(k) = FT(1) = FT(fracddxx) = ik , FT(x)$$ if and only if $$-sqrt2pi , delta'(k) + C , delta(k) = i , FT(x).$$
$endgroup$
– md2perpe
Mar 30 at 16:29
add a comment |
$begingroup$
You are not really allowed to write $delta(k)/k$ in distribution theory, but the best interpretation if this expressions is $-delta'(k)$.
$endgroup$
– md2perpe
Mar 30 at 16:22
$begingroup$
Your first derivation is correct.
$endgroup$
– md2perpe
Mar 30 at 16:26
$begingroup$
Your second derivation lacks a term $C , delta(k)$: $$sqrt2pi , delta(k) = FT(1) = FT(fracddxx) = ik , FT(x)$$ if and only if $$-sqrt2pi , delta'(k) + C , delta(k) = i , FT(x).$$
$endgroup$
– md2perpe
Mar 30 at 16:29
$begingroup$
You are not really allowed to write $delta(k)/k$ in distribution theory, but the best interpretation if this expressions is $-delta'(k)$.
$endgroup$
– md2perpe
Mar 30 at 16:22
$begingroup$
You are not really allowed to write $delta(k)/k$ in distribution theory, but the best interpretation if this expressions is $-delta'(k)$.
$endgroup$
– md2perpe
Mar 30 at 16:22
$begingroup$
Your first derivation is correct.
$endgroup$
– md2perpe
Mar 30 at 16:26
$begingroup$
Your first derivation is correct.
$endgroup$
– md2perpe
Mar 30 at 16:26
$begingroup$
Your second derivation lacks a term $C , delta(k)$: $$sqrt2pi , delta(k) = FT(1) = FT(fracddxx) = ik , FT(x)$$ if and only if $$-sqrt2pi , delta'(k) + C , delta(k) = i , FT(x).$$
$endgroup$
– md2perpe
Mar 30 at 16:29
$begingroup$
Your second derivation lacks a term $C , delta(k)$: $$sqrt2pi , delta(k) = FT(1) = FT(fracddxx) = ik , FT(x)$$ if and only if $$-sqrt2pi , delta'(k) + C , delta(k) = i , FT(x).$$
$endgroup$
– md2perpe
Mar 30 at 16:29
add a comment |
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$begingroup$
You are not really allowed to write $delta(k)/k$ in distribution theory, but the best interpretation if this expressions is $-delta'(k)$.
$endgroup$
– md2perpe
Mar 30 at 16:22
$begingroup$
Your first derivation is correct.
$endgroup$
– md2perpe
Mar 30 at 16:26
$begingroup$
Your second derivation lacks a term $C , delta(k)$: $$sqrt2pi , delta(k) = FT(1) = FT(fracddxx) = ik , FT(x)$$ if and only if $$-sqrt2pi , delta'(k) + C , delta(k) = i , FT(x).$$
$endgroup$
– md2perpe
Mar 30 at 16:29