How does one derive the Fourier transform of the Ramp function?
One approach could have been to see that the Ramp function ( http://mathworld.wolfram.com/RampFunction.html ) is the convolution of $2$ Heavisides (at $0$). Hence its Fourier transform should have been the product of the Fourier transforms of Heavisides. The Fourier transform of the Heaviside (http://mathworld.wolfram.com/HeavisideStepFunction.html) is, $frac{1}{2} [delta(t) - frac{1}{pi t} ] $. But its not clear to me as to how its square is the Fourier transform of the Ramp at $0$ which is $frac{i}{4pi} delta'(t) - frac{1}{4pi^2 t^2} $
- I would otherwise like to see a reference (or if someone can type in!) which derives the Fourier transform of the ramp function from scratch!
fourier-analysis fourier-series fourier-transform
asked Sep 9 '16 at 16:27
gradstudent
18517
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
One approach could have been to see that the Ramp function ( http://mathworld.wolfram.com/RampFunction.html ) is the convolution of $2$ Heavisides (at $0$). Hence its Fourier transform should have been the product of the Fourier transforms of Heavisides. The Fourier transform of the Heaviside (http://mathworld.wolfram.com/HeavisideStepFunction.html) is, $frac{1}{2} [delta(t) - frac{1}{pi t} ] $. But its not clear to me as to how its square is the Fourier transform of the Ramp at $0$ which is $frac{i}{4pi} delta'(t) - frac{1}{4pi^2 t^2} $
- I would otherwise like to see a reference (or if someone can type in!) which derives the Fourier transform of the ramp function from scratch!
fourier-analysis fourier-series fourier-transform
asked Sep 9 '16 at 16:27
gradstudent
18517
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
hint: The derivative of the ramp (vs. $t$) is a step function ( multiplied by the steepness of the ramp).
– G Cab
Sep 9 '16 at 16:34
added: the ramp is the convolution of a forward step (from $0$) with a backward step (from $t$).
– G Cab
Sep 9 '16 at 16:38
One big problem with the way that you want to use (square of the Fourier transform of the Heaviside function) is that the square of the Dirac delta is ill defined. Compare mathoverflow.net/q/48067/39447. This does not mean that this way is wrong, it is just very difficult to prove the equivalence of the result with that of a different approach.
– bers
Sep 12 '16 at 0:39
add a comment |
One approach could have been to see that the Ramp function ( http://mathworld.wolfram.com/RampFunction.html ) is the convolution of $2$ Heavisides (at $0$). Hence its Fourier transform should have been the product of the Fourier transforms of Heavisides. The Fourier transform of the Heaviside (http://mathworld.wolfram.com/HeavisideStepFunction.html) is, $frac{1}{2} [delta(t) - frac{1}{pi t} ] $. But its not clear to me as to how its square is the Fourier transform of the Ramp at $0$ which is $frac{i}{4pi} delta'(t) - frac{1}{4pi^2 t^2} $
- I would otherwise like to see a reference (or if someone can type in!) which derives the Fourier transform of the ramp function from scratch!
fourier-analysis fourier-series fourier-transform
asked Sep 9 '16 at 16:27
gradstudent
18517
One approach could have been to see that the Ramp function ( http://mathworld.wolfram.com/RampFunction.html ) is the convolution of $2$ Heavisides (at $0$). Hence its Fourier transform should have been the product of the Fourier transforms of Heavisides. The Fourier transform of the Heaviside (http://mathworld.wolfram.com/HeavisideStepFunction.html) is, $frac{1}{2} [delta(t) - frac{1}{pi t} ] $. But its not clear to me as to how its square is the Fourier transform of the Ramp at $0$ which is $frac{i}{4pi} delta'(t) - frac{1}{4pi^2 t^2} $
- I would otherwise like to see a reference (or if someone can type in!) which derives the Fourier transform of the ramp function from scratch!
fourier-analysis fourier-series fourier-transform
fourier-analysis fourier-series fourier-transform
asked Sep 9 '16 at 16:27
gradstudent
18517
asked Sep 9 '16 at 16:27
gradstudent
18517
asked Sep 9 '16 at 16:27
gradstudent
18517
asked Sep 9 '16 at 16:27
gradstudent
18517
asked Sep 9 '16 at 16:27
gradstudent
18517
18517
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
hint: The derivative of the ramp (vs. $t$) is a step function ( multiplied by the steepness of the ramp).
– G Cab
Sep 9 '16 at 16:34
added: the ramp is the convolution of a forward step (from $0$) with a backward step (from $t$).
– G Cab
Sep 9 '16 at 16:38
One big problem with the way that you want to use (square of the Fourier transform of the Heaviside function) is that the square of the Dirac delta is ill defined. Compare mathoverflow.net/q/48067/39447. This does not mean that this way is wrong, it is just very difficult to prove the equivalence of the result with that of a different approach.
– bers
Sep 12 '16 at 0:39
add a comment |
hint: The derivative of the ramp (vs. $t$) is a step function ( multiplied by the steepness of the ramp).
– G Cab
Sep 9 '16 at 16:34
added: the ramp is the convolution of a forward step (from $0$) with a backward step (from $t$).
– G Cab
Sep 9 '16 at 16:38
One big problem with the way that you want to use (square of the Fourier transform of the Heaviside function) is that the square of the Dirac delta is ill defined. Compare mathoverflow.net/q/48067/39447. This does not mean that this way is wrong, it is just very difficult to prove the equivalence of the result with that of a different approach.
– bers
Sep 12 '16 at 0:39
hint: The derivative of the ramp (vs. $t$) is a step function ( multiplied by the steepness of the ramp).
– G Cab
Sep 9 '16 at 16:34
hint: The derivative of the ramp (vs. $t$) is a step function ( multiplied by the steepness of the ramp).
– G Cab
Sep 9 '16 at 16:34
added: the ramp is the convolution of a forward step (from $0$) with a backward step (from $t$).
– G Cab
Sep 9 '16 at 16:38
added: the ramp is the convolution of a forward step (from $0$) with a backward step (from $t$).
– G Cab
Sep 9 '16 at 16:38
One big problem with the way that you want to use (square of the Fourier transform of the Heaviside function) is that the square of the Dirac delta is ill defined. Compare mathoverflow.net/q/48067/39447. This does not mean that this way is wrong, it is just very difficult to prove the equivalence of the result with that of a different approach.
– bers
Sep 12 '16 at 0:39
One big problem with the way that you want to use (square of the Fourier transform of the Heaviside function) is that the square of the Dirac delta is ill defined. Compare mathoverflow.net/q/48067/39447. This does not mean that this way is wrong, it is just very difficult to prove the equivalence of the result with that of a different approach.
– bers
Sep 12 '16 at 0:39
add a comment |
2 Answers
2
active
oldest
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"Frequency derivative" is a property of Fourier transform which is:
$$mathcal{F}{x(f(x)}=jfrac{d}{domega}F(omega)$$
Plug $f(x)=u(x)$ (i.e. heaviside function) whose FT is $F(omega)=pidelta(omega)-frac{j}{omega}$.
Since $text{ramp}(x)=xu(x)$ we get
$$mathcal{F}{text{ramp}(x)}=jfrac{d}{domega}left(pidelta(omega)-frac{j}{omega}right)=jpidelta'(omega)-frac{1}{omega^2}$$
If you want to represent it versus $f$, since $omega=2pi f$ it becomes
$$mathcal{F}{text{ramp}(x)}=(jpi)frac{1}{(2pi)^2}delta'(f)-frac{1}{4pi^2f^2}=frac{j}{4pi}delta'(f)-frac{1}{4pi^2f^2}$$
answered Sep 10 '16 at 2:23
msm
6,2092828
Thanks but I wanted to use the convolution picture. The ramp is the convolution of two Heavisides and hence FT of the ramp should be the product of the FTs of the two Heavisides. How do I use this picture to derive the FT of the ramp?
– gradstudent
Sep 11 '16 at 17:32
A ramp function can also be written as, $ (x-b)theta (x - a)$ and one could compute the integral, $FT [ (x-b)theta (x - a) ](t) = int_a ^infty e^{ixt} (x-b) dx$ by using an $epsilon$ regulator as, $lim_{epsilon rightarrow 0} [int_a ^infty e^{i(x+iepsilon)t} (x-b) dx ] $. This method does not seem to produce the $delta'$ term. Can you kindly explain why this is being missed?
– gradstudent
Sep 11 '16 at 18:07
@gradstudent "I would otherwise like to see ... the Fourier transform of the ramp function from scratch"
– msm
Sep 11 '16 at 22:07
Could you kindly explain (1) the error in my epsilon argument and (2) the FT of the Heaviside function and (3) How the convolution argument is to be used?
– gradstudent
Sep 12 '16 at 13:38
I would mention it is the principal value of $1/f$ and the finite part of $1/f^2$, since we are considering distributions.
– reuns
Jul 14 '17 at 9:51
|
show 1 more comment
In fact the convolution theorem is a bit more complicated than you think for distributions. To apply it you need to approximate your distributions with something like functions $in L^1$. So
$$1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0}$$
$$x 1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}$$
$$mathcal{F}[x 1_{ x > 0}] = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}]$$
$$ = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ]^2$$
$$ = lim_{a to 0^+} frac{1}{(a+2i pi xi)^2}$$
$$ = lim_{a to 0^+} frac{-1}{4pi^2} frac{1}{(xi-ia)^2}$$
$$ = lim_{a to 0^+} frac{1}{4pi^2} frac{d^2}{dxi^2} log(xi-ia)$$
$$ = lim_{a to 0^+} frac{1}{4pi^2}frac{d^2}{dxi^2}( -ipi 1_{xi
< 0} +log|xi|)$$
$$ = frac{i}{4pi} delta'(xi)- frac{1}{4 pi^2 } fp(frac{1}{xi^2})$$
where $fp(frac{1}{xi^2})$ is the finite part the second derivative of $-log |xi|$
answered Nov 20 at 2:32
reuns
19.6k21046
add a comment |
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2 Answers
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2 Answers
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active
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"Frequency derivative" is a property of Fourier transform which is:
$$mathcal{F}{x(f(x)}=jfrac{d}{domega}F(omega)$$
Plug $f(x)=u(x)$ (i.e. heaviside function) whose FT is $F(omega)=pidelta(omega)-frac{j}{omega}$.
Since $text{ramp}(x)=xu(x)$ we get
$$mathcal{F}{text{ramp}(x)}=jfrac{d}{domega}left(pidelta(omega)-frac{j}{omega}right)=jpidelta'(omega)-frac{1}{omega^2}$$
If you want to represent it versus $f$, since $omega=2pi f$ it becomes
$$mathcal{F}{text{ramp}(x)}=(jpi)frac{1}{(2pi)^2}delta'(f)-frac{1}{4pi^2f^2}=frac{j}{4pi}delta'(f)-frac{1}{4pi^2f^2}$$
answered Sep 10 '16 at 2:23
msm
6,2092828
Thanks but I wanted to use the convolution picture. The ramp is the convolution of two Heavisides and hence FT of the ramp should be the product of the FTs of the two Heavisides. How do I use this picture to derive the FT of the ramp?
– gradstudent
Sep 11 '16 at 17:32
A ramp function can also be written as, $ (x-b)theta (x - a)$ and one could compute the integral, $FT [ (x-b)theta (x - a) ](t) = int_a ^infty e^{ixt} (x-b) dx$ by using an $epsilon$ regulator as, $lim_{epsilon rightarrow 0} [int_a ^infty e^{i(x+iepsilon)t} (x-b) dx ] $. This method does not seem to produce the $delta'$ term. Can you kindly explain why this is being missed?
– gradstudent
Sep 11 '16 at 18:07
@gradstudent "I would otherwise like to see ... the Fourier transform of the ramp function from scratch"
– msm
Sep 11 '16 at 22:07
Could you kindly explain (1) the error in my epsilon argument and (2) the FT of the Heaviside function and (3) How the convolution argument is to be used?
– gradstudent
Sep 12 '16 at 13:38
I would mention it is the principal value of $1/f$ and the finite part of $1/f^2$, since we are considering distributions.
– reuns
Jul 14 '17 at 9:51
|
show 1 more comment
"Frequency derivative" is a property of Fourier transform which is:
$$mathcal{F}{x(f(x)}=jfrac{d}{domega}F(omega)$$
Plug $f(x)=u(x)$ (i.e. heaviside function) whose FT is $F(omega)=pidelta(omega)-frac{j}{omega}$.
Since $text{ramp}(x)=xu(x)$ we get
$$mathcal{F}{text{ramp}(x)}=jfrac{d}{domega}left(pidelta(omega)-frac{j}{omega}right)=jpidelta'(omega)-frac{1}{omega^2}$$
If you want to represent it versus $f$, since $omega=2pi f$ it becomes
$$mathcal{F}{text{ramp}(x)}=(jpi)frac{1}{(2pi)^2}delta'(f)-frac{1}{4pi^2f^2}=frac{j}{4pi}delta'(f)-frac{1}{4pi^2f^2}$$
answered Sep 10 '16 at 2:23
msm
6,2092828
Thanks but I wanted to use the convolution picture. The ramp is the convolution of two Heavisides and hence FT of the ramp should be the product of the FTs of the two Heavisides. How do I use this picture to derive the FT of the ramp?
– gradstudent
Sep 11 '16 at 17:32
A ramp function can also be written as, $ (x-b)theta (x - a)$ and one could compute the integral, $FT [ (x-b)theta (x - a) ](t) = int_a ^infty e^{ixt} (x-b) dx$ by using an $epsilon$ regulator as, $lim_{epsilon rightarrow 0} [int_a ^infty e^{i(x+iepsilon)t} (x-b) dx ] $. This method does not seem to produce the $delta'$ term. Can you kindly explain why this is being missed?
– gradstudent
Sep 11 '16 at 18:07
@gradstudent "I would otherwise like to see ... the Fourier transform of the ramp function from scratch"
– msm
Sep 11 '16 at 22:07
Could you kindly explain (1) the error in my epsilon argument and (2) the FT of the Heaviside function and (3) How the convolution argument is to be used?
– gradstudent
Sep 12 '16 at 13:38
I would mention it is the principal value of $1/f$ and the finite part of $1/f^2$, since we are considering distributions.
– reuns
Jul 14 '17 at 9:51
|
show 1 more comment
"Frequency derivative" is a property of Fourier transform which is:
$$mathcal{F}{x(f(x)}=jfrac{d}{domega}F(omega)$$
Plug $f(x)=u(x)$ (i.e. heaviside function) whose FT is $F(omega)=pidelta(omega)-frac{j}{omega}$.
Since $text{ramp}(x)=xu(x)$ we get
$$mathcal{F}{text{ramp}(x)}=jfrac{d}{domega}left(pidelta(omega)-frac{j}{omega}right)=jpidelta'(omega)-frac{1}{omega^2}$$
If you want to represent it versus $f$, since $omega=2pi f$ it becomes
$$mathcal{F}{text{ramp}(x)}=(jpi)frac{1}{(2pi)^2}delta'(f)-frac{1}{4pi^2f^2}=frac{j}{4pi}delta'(f)-frac{1}{4pi^2f^2}$$
answered Sep 10 '16 at 2:23
msm
6,2092828
"Frequency derivative" is a property of Fourier transform which is:
$$mathcal{F}{x(f(x)}=jfrac{d}{domega}F(omega)$$
Plug $f(x)=u(x)$ (i.e. heaviside function) whose FT is $F(omega)=pidelta(omega)-frac{j}{omega}$.
Since $text{ramp}(x)=xu(x)$ we get
$$mathcal{F}{text{ramp}(x)}=jfrac{d}{domega}left(pidelta(omega)-frac{j}{omega}right)=jpidelta'(omega)-frac{1}{omega^2}$$
If you want to represent it versus $f$, since $omega=2pi f$ it becomes
$$mathcal{F}{text{ramp}(x)}=(jpi)frac{1}{(2pi)^2}delta'(f)-frac{1}{4pi^2f^2}=frac{j}{4pi}delta'(f)-frac{1}{4pi^2f^2}$$
answered Sep 10 '16 at 2:23
msm
6,2092828
answered Sep 10 '16 at 2:23
msm
6,2092828
answered Sep 10 '16 at 2:23
msm
6,2092828
answered Sep 10 '16 at 2:23
msm
6,2092828
6,2092828
Thanks but I wanted to use the convolution picture. The ramp is the convolution of two Heavisides and hence FT of the ramp should be the product of the FTs of the two Heavisides. How do I use this picture to derive the FT of the ramp?
– gradstudent
Sep 11 '16 at 17:32
A ramp function can also be written as, $ (x-b)theta (x - a)$ and one could compute the integral, $FT [ (x-b)theta (x - a) ](t) = int_a ^infty e^{ixt} (x-b) dx$ by using an $epsilon$ regulator as, $lim_{epsilon rightarrow 0} [int_a ^infty e^{i(x+iepsilon)t} (x-b) dx ] $. This method does not seem to produce the $delta'$ term. Can you kindly explain why this is being missed?
– gradstudent
Sep 11 '16 at 18:07
@gradstudent "I would otherwise like to see ... the Fourier transform of the ramp function from scratch"
– msm
Sep 11 '16 at 22:07
Could you kindly explain (1) the error in my epsilon argument and (2) the FT of the Heaviside function and (3) How the convolution argument is to be used?
– gradstudent
Sep 12 '16 at 13:38
I would mention it is the principal value of $1/f$ and the finite part of $1/f^2$, since we are considering distributions.
– reuns
Jul 14 '17 at 9:51
|
show 1 more comment
Thanks but I wanted to use the convolution picture. The ramp is the convolution of two Heavisides and hence FT of the ramp should be the product of the FTs of the two Heavisides. How do I use this picture to derive the FT of the ramp?
– gradstudent
Sep 11 '16 at 17:32
A ramp function can also be written as, $ (x-b)theta (x - a)$ and one could compute the integral, $FT [ (x-b)theta (x - a) ](t) = int_a ^infty e^{ixt} (x-b) dx$ by using an $epsilon$ regulator as, $lim_{epsilon rightarrow 0} [int_a ^infty e^{i(x+iepsilon)t} (x-b) dx ] $. This method does not seem to produce the $delta'$ term. Can you kindly explain why this is being missed?
– gradstudent
Sep 11 '16 at 18:07
@gradstudent "I would otherwise like to see ... the Fourier transform of the ramp function from scratch"
– msm
Sep 11 '16 at 22:07
Could you kindly explain (1) the error in my epsilon argument and (2) the FT of the Heaviside function and (3) How the convolution argument is to be used?
– gradstudent
Sep 12 '16 at 13:38
I would mention it is the principal value of $1/f$ and the finite part of $1/f^2$, since we are considering distributions.
– reuns
Jul 14 '17 at 9:51
Thanks but I wanted to use the convolution picture. The ramp is the convolution of two Heavisides and hence FT of the ramp should be the product of the FTs of the two Heavisides. How do I use this picture to derive the FT of the ramp?
– gradstudent
Sep 11 '16 at 17:32
Thanks but I wanted to use the convolution picture. The ramp is the convolution of two Heavisides and hence FT of the ramp should be the product of the FTs of the two Heavisides. How do I use this picture to derive the FT of the ramp?
– gradstudent
Sep 11 '16 at 17:32
A ramp function can also be written as, $ (x-b)theta (x - a)$ and one could compute the integral, $FT [ (x-b)theta (x - a) ](t) = int_a ^infty e^{ixt} (x-b) dx$ by using an $epsilon$ regulator as, $lim_{epsilon rightarrow 0} [int_a ^infty e^{i(x+iepsilon)t} (x-b) dx ] $. This method does not seem to produce the $delta'$ term. Can you kindly explain why this is being missed?
– gradstudent
Sep 11 '16 at 18:07
A ramp function can also be written as, $ (x-b)theta (x - a)$ and one could compute the integral, $FT [ (x-b)theta (x - a) ](t) = int_a ^infty e^{ixt} (x-b) dx$ by using an $epsilon$ regulator as, $lim_{epsilon rightarrow 0} [int_a ^infty e^{i(x+iepsilon)t} (x-b) dx ] $. This method does not seem to produce the $delta'$ term. Can you kindly explain why this is being missed?
– gradstudent
Sep 11 '16 at 18:07
@gradstudent "I would otherwise like to see ... the Fourier transform of the ramp function from scratch"
– msm
Sep 11 '16 at 22:07
@gradstudent "I would otherwise like to see ... the Fourier transform of the ramp function from scratch"
– msm
Sep 11 '16 at 22:07
Could you kindly explain (1) the error in my epsilon argument and (2) the FT of the Heaviside function and (3) How the convolution argument is to be used?
– gradstudent
Sep 12 '16 at 13:38
Could you kindly explain (1) the error in my epsilon argument and (2) the FT of the Heaviside function and (3) How the convolution argument is to be used?
– gradstudent
Sep 12 '16 at 13:38
I would mention it is the principal value of $1/f$ and the finite part of $1/f^2$, since we are considering distributions.
– reuns
Jul 14 '17 at 9:51
I would mention it is the principal value of $1/f$ and the finite part of $1/f^2$, since we are considering distributions.
– reuns
Jul 14 '17 at 9:51
|
show 1 more comment
In fact the convolution theorem is a bit more complicated than you think for distributions. To apply it you need to approximate your distributions with something like functions $in L^1$. So
$$1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0}$$
$$x 1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}$$
$$mathcal{F}[x 1_{ x > 0}] = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}]$$
$$ = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ]^2$$
$$ = lim_{a to 0^+} frac{1}{(a+2i pi xi)^2}$$
$$ = lim_{a to 0^+} frac{-1}{4pi^2} frac{1}{(xi-ia)^2}$$
$$ = lim_{a to 0^+} frac{1}{4pi^2} frac{d^2}{dxi^2} log(xi-ia)$$
$$ = lim_{a to 0^+} frac{1}{4pi^2}frac{d^2}{dxi^2}( -ipi 1_{xi
< 0} +log|xi|)$$
$$ = frac{i}{4pi} delta'(xi)- frac{1}{4 pi^2 } fp(frac{1}{xi^2})$$
where $fp(frac{1}{xi^2})$ is the finite part the second derivative of $-log |xi|$
answered Nov 20 at 2:32
reuns
19.6k21046
add a comment |
In fact the convolution theorem is a bit more complicated than you think for distributions. To apply it you need to approximate your distributions with something like functions $in L^1$. So
$$1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0}$$
$$x 1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}$$
$$mathcal{F}[x 1_{ x > 0}] = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}]$$
$$ = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ]^2$$
$$ = lim_{a to 0^+} frac{1}{(a+2i pi xi)^2}$$
$$ = lim_{a to 0^+} frac{-1}{4pi^2} frac{1}{(xi-ia)^2}$$
$$ = lim_{a to 0^+} frac{1}{4pi^2} frac{d^2}{dxi^2} log(xi-ia)$$
$$ = lim_{a to 0^+} frac{1}{4pi^2}frac{d^2}{dxi^2}( -ipi 1_{xi
< 0} +log|xi|)$$
$$ = frac{i}{4pi} delta'(xi)- frac{1}{4 pi^2 } fp(frac{1}{xi^2})$$
where $fp(frac{1}{xi^2})$ is the finite part the second derivative of $-log |xi|$
answered Nov 20 at 2:32
reuns
19.6k21046
add a comment |
In fact the convolution theorem is a bit more complicated than you think for distributions. To apply it you need to approximate your distributions with something like functions $in L^1$. So
$$1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0}$$
$$x 1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}$$
$$mathcal{F}[x 1_{ x > 0}] = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}]$$
$$ = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ]^2$$
$$ = lim_{a to 0^+} frac{1}{(a+2i pi xi)^2}$$
$$ = lim_{a to 0^+} frac{-1}{4pi^2} frac{1}{(xi-ia)^2}$$
$$ = lim_{a to 0^+} frac{1}{4pi^2} frac{d^2}{dxi^2} log(xi-ia)$$
$$ = lim_{a to 0^+} frac{1}{4pi^2}frac{d^2}{dxi^2}( -ipi 1_{xi
< 0} +log|xi|)$$
$$ = frac{i}{4pi} delta'(xi)- frac{1}{4 pi^2 } fp(frac{1}{xi^2})$$
where $fp(frac{1}{xi^2})$ is the finite part the second derivative of $-log |xi|$
answered Nov 20 at 2:32
reuns
19.6k21046
In fact the convolution theorem is a bit more complicated than you think for distributions. To apply it you need to approximate your distributions with something like functions $in L^1$. So
$$1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0}$$
$$x 1_{ x > 0} = lim_{a to 0^+} e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}$$
$$mathcal{F}[x 1_{ x > 0}] = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ast e^{-ax} 1_{x > 0}]$$
$$ = lim_{a to 0^+} mathcal{F}[e^{-ax} 1_{x > 0} ]^2$$
$$ = lim_{a to 0^+} frac{1}{(a+2i pi xi)^2}$$
$$ = lim_{a to 0^+} frac{-1}{4pi^2} frac{1}{(xi-ia)^2}$$
$$ = lim_{a to 0^+} frac{1}{4pi^2} frac{d^2}{dxi^2} log(xi-ia)$$
$$ = lim_{a to 0^+} frac{1}{4pi^2}frac{d^2}{dxi^2}( -ipi 1_{xi
< 0} +log|xi|)$$
$$ = frac{i}{4pi} delta'(xi)- frac{1}{4 pi^2 } fp(frac{1}{xi^2})$$
where $fp(frac{1}{xi^2})$ is the finite part the second derivative of $-log |xi|$
answered Nov 20 at 2:32
reuns
19.6k21046
edited Nov 20 at 2:45
answered Nov 20 at 2:32
reuns
19.6k21046
answered Nov 20 at 2:32
reuns
19.6k21046
answered Nov 20 at 2:32
reuns
19.6k21046
19.6k21046
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hint: The derivative of the ramp (vs. $t$) is a step function ( multiplied by the steepness of the ramp).
– G Cab
Sep 9 '16 at 16:34
added: the ramp is the convolution of a forward step (from $0$) with a backward step (from $t$).
– G Cab
Sep 9 '16 at 16:38
One big problem with the way that you want to use (square of the Fourier transform of the Heaviside function) is that the square of the Dirac delta is ill defined. Compare mathoverflow.net/q/48067/39447. This does not mean that this way is wrong, it is just very difficult to prove the equivalence of the result with that of a different approach.
– bers
Sep 12 '16 at 0:39