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Are Laplace Transforms a Special Case of Fourier Transforms?
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A Laplace Transform is based on the integral:
$F(xi) = int_0^infty f(x) e^ -xi x,dx.$
In a roundabout way, a Fourier transform can get to $hatf(xi) = int_-infty^infty f(x) e^- 2pi i x xi,dx,$
Also, they both seem to use convolutions and transposes in "indirect" forms of "multiplication."
laplace-transform integral-transforms
asked Aug 8 '11 at 16:31
Tom Au
1,2401332
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up vote
12
down vote
favorite
A Laplace Transform is based on the integral:
$F(xi) = int_0^infty f(x) e^ -xi x,dx.$
In a roundabout way, a Fourier transform can get to $hatf(xi) = int_-infty^infty f(x) e^- 2pi i x xi,dx,$
Also, they both seem to use convolutions and transposes in "indirect" forms of "multiplication."
laplace-transform integral-transforms
asked Aug 8 '11 at 16:31
Tom Au
1,2401332
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
A Laplace Transform is based on the integral:
$F(xi) = int_0^infty f(x) e^ -xi x,dx.$
In a roundabout way, a Fourier transform can get to $hatf(xi) = int_-infty^infty f(x) e^- 2pi i x xi,dx,$
Also, they both seem to use convolutions and transposes in "indirect" forms of "multiplication."
laplace-transform integral-transforms
asked Aug 8 '11 at 16:31
Tom Au
1,2401332
A Laplace Transform is based on the integral:
$F(xi) = int_0^infty f(x) e^ -xi x,dx.$
In a roundabout way, a Fourier transform can get to $hatf(xi) = int_-infty^infty f(x) e^- 2pi i x xi,dx,$
Also, they both seem to use convolutions and transposes in "indirect" forms of "multiplication."
laplace-transform integral-transforms
asked Aug 8 '11 at 16:31
Tom Au
1,2401332
edited Aug 6 at 23:16
asked Aug 8 '11 at 16:31
Tom Au
1,2401332
asked Aug 8 '11 at 16:31
Tom Au
1,2401332
asked Aug 8 '11 at 16:31
Tom Au
1,2401332
1,2401332
add a comment |Â
add a comment |Â
1 Answer
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Your formula for the Laplace transform is wrong. It should be
$F(xi) = int_0^infty f(x) e^-xi x, dx$. But yes, when $xi$ is imaginary you have (up to normalization) the Fourier transform of $f$ (considered as a function on $(-infty, infty)$ which is 0 on $(-infty, 0)$).
answered Aug 8 '11 at 16:41
Robert Israel
304k22201443
Fixed the Laplace transform per your text.
– Tom Au
Aug 8 '11 at 16:53
1
@Robert: Could you elaborate? I've often wondered why these two transforms seem to be treated as quite separate things when there's an obvious connection between them.
– joriki
Aug 8 '11 at 17:29
1
As Robert said. But the properties of the two transforms are so different that (to me) the answer to the question in your title is a resounding: No.
– Did
Aug 8 '11 at 17:31
2
For example, the Bromwich integral for inverting the Laplace transform is really an inverse Fourier transform. See chapter 9 of Dettman, "Applied Complex Variables" books.google.ca/books?id=vmZ6PVtaexwC
– Robert Israel
Aug 8 '11 at 18:27
See also the Paley-Wiener theorems (e.g. Rudin, "Real and Complex Analysis", sec. 19.1 and 19.2). Remarkably, Rudin does not mention the name Laplace.
– Robert Israel
Aug 10 '11 at 7:52
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Your formula for the Laplace transform is wrong. It should be
$F(xi) = int_0^infty f(x) e^-xi x, dx$. But yes, when $xi$ is imaginary you have (up to normalization) the Fourier transform of $f$ (considered as a function on $(-infty, infty)$ which is 0 on $(-infty, 0)$).
answered Aug 8 '11 at 16:41
Robert Israel
304k22201443
Fixed the Laplace transform per your text.
– Tom Au
Aug 8 '11 at 16:53
1
@Robert: Could you elaborate? I've often wondered why these two transforms seem to be treated as quite separate things when there's an obvious connection between them.
– joriki
Aug 8 '11 at 17:29
1
As Robert said. But the properties of the two transforms are so different that (to me) the answer to the question in your title is a resounding: No.
– Did
Aug 8 '11 at 17:31
2
For example, the Bromwich integral for inverting the Laplace transform is really an inverse Fourier transform. See chapter 9 of Dettman, "Applied Complex Variables" books.google.ca/books?id=vmZ6PVtaexwC
– Robert Israel
Aug 8 '11 at 18:27
See also the Paley-Wiener theorems (e.g. Rudin, "Real and Complex Analysis", sec. 19.1 and 19.2). Remarkably, Rudin does not mention the name Laplace.
– Robert Israel
Aug 10 '11 at 7:52
 |Â
show 1 more comment
up vote
5
down vote
accepted
Your formula for the Laplace transform is wrong. It should be
$F(xi) = int_0^infty f(x) e^-xi x, dx$. But yes, when $xi$ is imaginary you have (up to normalization) the Fourier transform of $f$ (considered as a function on $(-infty, infty)$ which is 0 on $(-infty, 0)$).
answered Aug 8 '11 at 16:41
Robert Israel
304k22201443
Fixed the Laplace transform per your text.
– Tom Au
Aug 8 '11 at 16:53
1
@Robert: Could you elaborate? I've often wondered why these two transforms seem to be treated as quite separate things when there's an obvious connection between them.
– joriki
Aug 8 '11 at 17:29
1
As Robert said. But the properties of the two transforms are so different that (to me) the answer to the question in your title is a resounding: No.
– Did
Aug 8 '11 at 17:31
2
For example, the Bromwich integral for inverting the Laplace transform is really an inverse Fourier transform. See chapter 9 of Dettman, "Applied Complex Variables" books.google.ca/books?id=vmZ6PVtaexwC
– Robert Israel
Aug 8 '11 at 18:27
See also the Paley-Wiener theorems (e.g. Rudin, "Real and Complex Analysis", sec. 19.1 and 19.2). Remarkably, Rudin does not mention the name Laplace.
– Robert Israel
Aug 10 '11 at 7:52
 |Â
show 1 more comment
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Your formula for the Laplace transform is wrong. It should be
$F(xi) = int_0^infty f(x) e^-xi x, dx$. But yes, when $xi$ is imaginary you have (up to normalization) the Fourier transform of $f$ (considered as a function on $(-infty, infty)$ which is 0 on $(-infty, 0)$).
answered Aug 8 '11 at 16:41
Robert Israel
304k22201443
Your formula for the Laplace transform is wrong. It should be
$F(xi) = int_0^infty f(x) e^-xi x, dx$. But yes, when $xi$ is imaginary you have (up to normalization) the Fourier transform of $f$ (considered as a function on $(-infty, infty)$ which is 0 on $(-infty, 0)$).
answered Aug 8 '11 at 16:41
Robert Israel
304k22201443
answered Aug 8 '11 at 16:41
Robert Israel
304k22201443
answered Aug 8 '11 at 16:41
Robert Israel
304k22201443
answered Aug 8 '11 at 16:41
Robert Israel
304k22201443
304k22201443
Fixed the Laplace transform per your text.
– Tom Au
Aug 8 '11 at 16:53
1
@Robert: Could you elaborate? I've often wondered why these two transforms seem to be treated as quite separate things when there's an obvious connection between them.
– joriki
Aug 8 '11 at 17:29
1
As Robert said. But the properties of the two transforms are so different that (to me) the answer to the question in your title is a resounding: No.
– Did
Aug 8 '11 at 17:31
2
For example, the Bromwich integral for inverting the Laplace transform is really an inverse Fourier transform. See chapter 9 of Dettman, "Applied Complex Variables" books.google.ca/books?id=vmZ6PVtaexwC
– Robert Israel
Aug 8 '11 at 18:27
See also the Paley-Wiener theorems (e.g. Rudin, "Real and Complex Analysis", sec. 19.1 and 19.2). Remarkably, Rudin does not mention the name Laplace.
– Robert Israel
Aug 10 '11 at 7:52
 |Â
show 1 more comment
Fixed the Laplace transform per your text.
– Tom Au
Aug 8 '11 at 16:53
1
@Robert: Could you elaborate? I've often wondered why these two transforms seem to be treated as quite separate things when there's an obvious connection between them.
– joriki
Aug 8 '11 at 17:29
1
As Robert said. But the properties of the two transforms are so different that (to me) the answer to the question in your title is a resounding: No.
– Did
Aug 8 '11 at 17:31
2
For example, the Bromwich integral for inverting the Laplace transform is really an inverse Fourier transform. See chapter 9 of Dettman, "Applied Complex Variables" books.google.ca/books?id=vmZ6PVtaexwC
– Robert Israel
Aug 8 '11 at 18:27
See also the Paley-Wiener theorems (e.g. Rudin, "Real and Complex Analysis", sec. 19.1 and 19.2). Remarkably, Rudin does not mention the name Laplace.
– Robert Israel
Aug 10 '11 at 7:52
Fixed the Laplace transform per your text.
– Tom Au
Aug 8 '11 at 16:53
Fixed the Laplace transform per your text.
– Tom Au
Aug 8 '11 at 16:53
1
1
@Robert: Could you elaborate? I've often wondered why these two transforms seem to be treated as quite separate things when there's an obvious connection between them.
– joriki
Aug 8 '11 at 17:29
@Robert: Could you elaborate? I've often wondered why these two transforms seem to be treated as quite separate things when there's an obvious connection between them.
– joriki
Aug 8 '11 at 17:29
1
1
As Robert said. But the properties of the two transforms are so different that (to me) the answer to the question in your title is a resounding: No.
– Did
Aug 8 '11 at 17:31
As Robert said. But the properties of the two transforms are so different that (to me) the answer to the question in your title is a resounding: No.
– Did
Aug 8 '11 at 17:31
2
2
For example, the Bromwich integral for inverting the Laplace transform is really an inverse Fourier transform. See chapter 9 of Dettman, "Applied Complex Variables" books.google.ca/books?id=vmZ6PVtaexwC
– Robert Israel
Aug 8 '11 at 18:27
For example, the Bromwich integral for inverting the Laplace transform is really an inverse Fourier transform. See chapter 9 of Dettman, "Applied Complex Variables" books.google.ca/books?id=vmZ6PVtaexwC
– Robert Israel
Aug 8 '11 at 18:27
See also the Paley-Wiener theorems (e.g. Rudin, "Real and Complex Analysis", sec. 19.1 and 19.2). Remarkably, Rudin does not mention the name Laplace.
– Robert Israel
Aug 10 '11 at 7:52
See also the Paley-Wiener theorems (e.g. Rudin, "Real and Complex Analysis", sec. 19.1 and 19.2). Remarkably, Rudin does not mention the name Laplace.
– Robert Israel
Aug 10 '11 at 7:52
 |Â
show 1 more comment
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