Signum function and Fourier transform
I'm extracting a portion of my notes which I believe I might have copied wrongly.
Given this equation:
$$frac{G(omega)}{2icomega} [e^{icomega t}-e^{-icwt}]$$
I want to find the Fouerir transform of the above. I believe it involves the Fourier transform of the signum function.
The Signum function is:
$$sgn(x) = left{begin{matrix}
1 ,&x>0 \
0,&x=0 \
-1, &x<0
end{matrix}right.$$
On my notes it says the Fourier transform of sgn(x) to be
$$frac{1}{pi iomega}$$ (is this the correct?)
Secondly, how do I apply the fourier transform to the above equation?
I'm pretty lost so any help would be good.
pde fourier-analysis fourier-series
asked Jun 15 '15 at 10:54
Mathematicing
2,44121851
add a comment |
I'm extracting a portion of my notes which I believe I might have copied wrongly.
Given this equation:
$$frac{G(omega)}{2icomega} [e^{icomega t}-e^{-icwt}]$$
I want to find the Fouerir transform of the above. I believe it involves the Fourier transform of the signum function.
The Signum function is:
$$sgn(x) = left{begin{matrix}
1 ,&x>0 \
0,&x=0 \
-1, &x<0
end{matrix}right.$$
On my notes it says the Fourier transform of sgn(x) to be
$$frac{1}{pi iomega}$$ (is this the correct?)
Secondly, how do I apply the fourier transform to the above equation?
I'm pretty lost so any help would be good.
pde fourier-analysis fourier-series
asked Jun 15 '15 at 10:54
Mathematicing
2,44121851
What is $G$? The formula works out to $G(omega)frac{sin(comega t)}{comega}$, with no signum directly in sight.
– Henning Makholm
Jun 15 '15 at 10:57
G is the fourier transform of some g(x), both of which has not yet been defined.
– Mathematicing
Jun 15 '15 at 10:58
add a comment |
I'm extracting a portion of my notes which I believe I might have copied wrongly.
Given this equation:
$$frac{G(omega)}{2icomega} [e^{icomega t}-e^{-icwt}]$$
I want to find the Fouerir transform of the above. I believe it involves the Fourier transform of the signum function.
The Signum function is:
$$sgn(x) = left{begin{matrix}
1 ,&x>0 \
0,&x=0 \
-1, &x<0
end{matrix}right.$$
On my notes it says the Fourier transform of sgn(x) to be
$$frac{1}{pi iomega}$$ (is this the correct?)
Secondly, how do I apply the fourier transform to the above equation?
I'm pretty lost so any help would be good.
pde fourier-analysis fourier-series
asked Jun 15 '15 at 10:54
Mathematicing
2,44121851
I'm extracting a portion of my notes which I believe I might have copied wrongly.
Given this equation:
$$frac{G(omega)}{2icomega} [e^{icomega t}-e^{-icwt}]$$
I want to find the Fouerir transform of the above. I believe it involves the Fourier transform of the signum function.
The Signum function is:
$$sgn(x) = left{begin{matrix}
1 ,&x>0 \
0,&x=0 \
-1, &x<0
end{matrix}right.$$
On my notes it says the Fourier transform of sgn(x) to be
$$frac{1}{pi iomega}$$ (is this the correct?)
Secondly, how do I apply the fourier transform to the above equation?
I'm pretty lost so any help would be good.
pde fourier-analysis fourier-series
pde fourier-analysis fourier-series
asked Jun 15 '15 at 10:54
Mathematicing
2,44121851
asked Jun 15 '15 at 10:54
Mathematicing
2,44121851
asked Jun 15 '15 at 10:54
Mathematicing
2,44121851
asked Jun 15 '15 at 10:54
Mathematicing
2,44121851
asked Jun 15 '15 at 10:54
Mathematicing
2,44121851
2,44121851
What is $G$? The formula works out to $G(omega)frac{sin(comega t)}{comega}$, with no signum directly in sight.
– Henning Makholm
Jun 15 '15 at 10:57
G is the fourier transform of some g(x), both of which has not yet been defined.
– Mathematicing
Jun 15 '15 at 10:58
add a comment |
What is $G$? The formula works out to $G(omega)frac{sin(comega t)}{comega}$, with no signum directly in sight.
– Henning Makholm
Jun 15 '15 at 10:57
G is the fourier transform of some g(x), both of which has not yet been defined.
– Mathematicing
Jun 15 '15 at 10:58
What is $G$? The formula works out to $G(omega)frac{sin(comega t)}{comega}$, with no signum directly in sight.
– Henning Makholm
Jun 15 '15 at 10:57
What is $G$? The formula works out to $G(omega)frac{sin(comega t)}{comega}$, with no signum directly in sight.
– Henning Makholm
Jun 15 '15 at 10:57
G is the fourier transform of some g(x), both of which has not yet been defined.
– Mathematicing
Jun 15 '15 at 10:58
G is the fourier transform of some g(x), both of which has not yet been defined.
– Mathematicing
Jun 15 '15 at 10:58
add a comment |
1 Answer
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oldest
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$$x(t) = sgn(t)$$
$$frac{dx(t)}{dt} = 2delta(t)$$
Now take fourier transform in both side
$$(jomega)X(omega) = 2$$
$$ X(omega) = frac{2}{jomega}$$
$$sgn(t) rightleftharpoons frac{2}{jomega}$$
answered Mar 27 at 8:23
Ashok Saini
1215
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
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votes
$$x(t) = sgn(t)$$
$$frac{dx(t)}{dt} = 2delta(t)$$
Now take fourier transform in both side
$$(jomega)X(omega) = 2$$
$$ X(omega) = frac{2}{jomega}$$
$$sgn(t) rightleftharpoons frac{2}{jomega}$$
answered Mar 27 at 8:23
Ashok Saini
1215
add a comment |
$$x(t) = sgn(t)$$
$$frac{dx(t)}{dt} = 2delta(t)$$
Now take fourier transform in both side
$$(jomega)X(omega) = 2$$
$$ X(omega) = frac{2}{jomega}$$
$$sgn(t) rightleftharpoons frac{2}{jomega}$$
answered Mar 27 at 8:23
Ashok Saini
1215
add a comment |
$$x(t) = sgn(t)$$
$$frac{dx(t)}{dt} = 2delta(t)$$
Now take fourier transform in both side
$$(jomega)X(omega) = 2$$
$$ X(omega) = frac{2}{jomega}$$
$$sgn(t) rightleftharpoons frac{2}{jomega}$$
answered Mar 27 at 8:23
Ashok Saini
1215
$$x(t) = sgn(t)$$
$$frac{dx(t)}{dt} = 2delta(t)$$
Now take fourier transform in both side
$$(jomega)X(omega) = 2$$
$$ X(omega) = frac{2}{jomega}$$
$$sgn(t) rightleftharpoons frac{2}{jomega}$$
answered Mar 27 at 8:23
Ashok Saini
1215
answered Mar 27 at 8:23
Ashok Saini
1215
answered Mar 27 at 8:23
Ashok Saini
1215
answered Mar 27 at 8:23
Ashok Saini
1215
1215
add a comment |
add a comment |
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What is $G$? The formula works out to $G(omega)frac{sin(comega t)}{comega}$, with no signum directly in sight.
– Henning Makholm
Jun 15 '15 at 10:57
G is the fourier transform of some g(x), both of which has not yet been defined.
– Mathematicing
Jun 15 '15 at 10:58