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Inverse Fourier transform of Lorentzians and sign function
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I'm trying to caculate the inverse Fourier transform of
$$ G(omega) = dfrac(omega+a)^2+b^2((omega-c)^2+b^2)((omega+c)^2+b^2) mathrmsgn(omega-d)$$
It is the product of two Lorentzians and sign function $sgn$. The constants $a,b,c,d$ are all real values.
I'm using complex integration to solve the FT integral by brute force:
$$ g(t) = int_-infty^infty dfracdomega2pi e^-iomega t G(omega)
= int_-infty^infty dfracdomega2pi e^-iomega tdfrac(omega+a)^2+b^2((omega-c)^2+b^2)((omega+c)^2+b^2) mathrmsgn(omega-d)$$
There are poles at $omega=c-ib,c+ib,-c-ib,-c+ib$.
One semicircle contour goes anticlockwise on the upper half of the complex plane. The other semicircle contour goes clockwise on the lower half of the complex plane. Then calculating the residues inside each closed contour:
$2pi i sum$ (residues in upper half of plane) + $2pi i sum$ (residues in lower half of plane)
There are two cases:
for $t>0$, $e^−iomega t$ converges to zero at infinity on the lower half plane, so residues are
$$ R_1 = 2pi i left(
e^-ict-bt
dfrac((c+a-ib)^2+b^2)sgn(c-d-ib)-2ib((2c-ib)^2+b^2) \
+ e^ict-bt
dfrac((-c+a-ib)^2+b^2)sgn(-c-d-ib)-2ib((-2c-ib)^2+b^2)
right)$$for $t<0$, $e^−iomega t$ converges to zero at infinity on the upper half plane, so residues are
$$ R_2 = 2pi i left(e^-ict+bt
dfrac((c+a+ib)^2+b^2)sgn(c-d+ib)2ib((2c+ib)^2+b^2) \
+ e^ict+bt
dfrac((-c+a+ib)^2+b^2)sgn(-c-d+ib)2ib((-2c+ib)^2+b^2)
right)$$
This means $g(t)=frac12pi(H(t)R_1 + H(-t)R_2)$ where $H$ is the Heaviside function.
Does my working look correct? Have I missed anything? Below I plot the analytical $g(t)$ alongside the Python IFFT result of $mathcalF^-1(G(omega))$ but they don't agree. (I checked my Python code by performing IFFTs on known FT pairs, so there's nothing wrong with the algorithm.)
complex-analysis fourier-analysis fourier-transform residue-calculus complex-integration
asked Jul 18 at 10:12
Medulla Oblongata
271110
add a comment |Â
up vote
2
down vote
favorite
I'm trying to caculate the inverse Fourier transform of
$$ G(omega) = dfrac(omega+a)^2+b^2((omega-c)^2+b^2)((omega+c)^2+b^2) mathrmsgn(omega-d)$$
It is the product of two Lorentzians and sign function $sgn$. The constants $a,b,c,d$ are all real values.
I'm using complex integration to solve the FT integral by brute force:
$$ g(t) = int_-infty^infty dfracdomega2pi e^-iomega t G(omega)
= int_-infty^infty dfracdomega2pi e^-iomega tdfrac(omega+a)^2+b^2((omega-c)^2+b^2)((omega+c)^2+b^2) mathrmsgn(omega-d)$$
There are poles at $omega=c-ib,c+ib,-c-ib,-c+ib$.
One semicircle contour goes anticlockwise on the upper half of the complex plane. The other semicircle contour goes clockwise on the lower half of the complex plane. Then calculating the residues inside each closed contour:
$2pi i sum$ (residues in upper half of plane) + $2pi i sum$ (residues in lower half of plane)
There are two cases:
for $t>0$, $e^−iomega t$ converges to zero at infinity on the lower half plane, so residues are
$$ R_1 = 2pi i left(
e^-ict-bt
dfrac((c+a-ib)^2+b^2)sgn(c-d-ib)-2ib((2c-ib)^2+b^2) \
+ e^ict-bt
dfrac((-c+a-ib)^2+b^2)sgn(-c-d-ib)-2ib((-2c-ib)^2+b^2)
right)$$for $t<0$, $e^−iomega t$ converges to zero at infinity on the upper half plane, so residues are
$$ R_2 = 2pi i left(e^-ict+bt
dfrac((c+a+ib)^2+b^2)sgn(c-d+ib)2ib((2c+ib)^2+b^2) \
+ e^ict+bt
dfrac((-c+a+ib)^2+b^2)sgn(-c-d+ib)2ib((-2c+ib)^2+b^2)
right)$$
This means $g(t)=frac12pi(H(t)R_1 + H(-t)R_2)$ where $H$ is the Heaviside function.
Does my working look correct? Have I missed anything? Below I plot the analytical $g(t)$ alongside the Python IFFT result of $mathcalF^-1(G(omega))$ but they don't agree. (I checked my Python code by performing IFFTs on known FT pairs, so there's nothing wrong with the algorithm.)
complex-analysis fourier-analysis fourier-transform residue-calculus complex-integration
asked Jul 18 at 10:12
Medulla Oblongata
271110
Residue theory works on analytic functions, and I don't think the Lorentzian times a sign function is analytic. What you could do is compute the Lorentzian's IFT first, then use that the (inverse) fourier transform of a product is a convolution of their (inverse) fourier transforms
– Calvin Khor
Jul 18 at 10:19
Yes, I initially tried to use the FT tables in Wikipedia (en.wikipedia.org/wiki/…) to separately transform the Lorentzian functions and sign function into the time domain, then do convolution on those results (i.e. using property 108: $F(omega)G(omega) to f(t)*g(t)$). But it doesn't seem possible to analytically solve the convolution. I want to calculate a purely analytical result.
– Medulla Oblongata
Jul 18 at 10:30
I'm not aware of the available tables,literature etc, but you might be able to find something if you note that convolutions against $mathcal F^-1 (-ioperatornamesgn)$ are known as Hilbert transforms.
– Calvin Khor
Jul 18 at 10:35
@CalvinKhor BTW could you expand on your comment and post it as an answer below? Thanks
– Medulla Oblongata
Jul 18 at 11:04
Okay, I'll try to do so when I have some time, but if you (or someone else) wants to do it, feel free to so.
– Calvin Khor
Jul 18 at 11:08
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to caculate the inverse Fourier transform of
$$ G(omega) = dfrac(omega+a)^2+b^2((omega-c)^2+b^2)((omega+c)^2+b^2) mathrmsgn(omega-d)$$
It is the product of two Lorentzians and sign function $sgn$. The constants $a,b,c,d$ are all real values.
I'm using complex integration to solve the FT integral by brute force:
$$ g(t) = int_-infty^infty dfracdomega2pi e^-iomega t G(omega)
= int_-infty^infty dfracdomega2pi e^-iomega tdfrac(omega+a)^2+b^2((omega-c)^2+b^2)((omega+c)^2+b^2) mathrmsgn(omega-d)$$
There are poles at $omega=c-ib,c+ib,-c-ib,-c+ib$.
One semicircle contour goes anticlockwise on the upper half of the complex plane. The other semicircle contour goes clockwise on the lower half of the complex plane. Then calculating the residues inside each closed contour:
$2pi i sum$ (residues in upper half of plane) + $2pi i sum$ (residues in lower half of plane)
There are two cases:
for $t>0$, $e^−iomega t$ converges to zero at infinity on the lower half plane, so residues are
$$ R_1 = 2pi i left(
e^-ict-bt
dfrac((c+a-ib)^2+b^2)sgn(c-d-ib)-2ib((2c-ib)^2+b^2) \
+ e^ict-bt
dfrac((-c+a-ib)^2+b^2)sgn(-c-d-ib)-2ib((-2c-ib)^2+b^2)
right)$$for $t<0$, $e^−iomega t$ converges to zero at infinity on the upper half plane, so residues are
$$ R_2 = 2pi i left(e^-ict+bt
dfrac((c+a+ib)^2+b^2)sgn(c-d+ib)2ib((2c+ib)^2+b^2) \
+ e^ict+bt
dfrac((-c+a+ib)^2+b^2)sgn(-c-d+ib)2ib((-2c+ib)^2+b^2)
right)$$
This means $g(t)=frac12pi(H(t)R_1 + H(-t)R_2)$ where $H$ is the Heaviside function.
Does my working look correct? Have I missed anything? Below I plot the analytical $g(t)$ alongside the Python IFFT result of $mathcalF^-1(G(omega))$ but they don't agree. (I checked my Python code by performing IFFTs on known FT pairs, so there's nothing wrong with the algorithm.)
complex-analysis fourier-analysis fourier-transform residue-calculus complex-integration
asked Jul 18 at 10:12
Medulla Oblongata
271110
I'm trying to caculate the inverse Fourier transform of
$$ G(omega) = dfrac(omega+a)^2+b^2((omega-c)^2+b^2)((omega+c)^2+b^2) mathrmsgn(omega-d)$$
It is the product of two Lorentzians and sign function $sgn$. The constants $a,b,c,d$ are all real values.
I'm using complex integration to solve the FT integral by brute force:
$$ g(t) = int_-infty^infty dfracdomega2pi e^-iomega t G(omega)
= int_-infty^infty dfracdomega2pi e^-iomega tdfrac(omega+a)^2+b^2((omega-c)^2+b^2)((omega+c)^2+b^2) mathrmsgn(omega-d)$$
There are poles at $omega=c-ib,c+ib,-c-ib,-c+ib$.
One semicircle contour goes anticlockwise on the upper half of the complex plane. The other semicircle contour goes clockwise on the lower half of the complex plane. Then calculating the residues inside each closed contour:
$2pi i sum$ (residues in upper half of plane) + $2pi i sum$ (residues in lower half of plane)
There are two cases:
for $t>0$, $e^−iomega t$ converges to zero at infinity on the lower half plane, so residues are
$$ R_1 = 2pi i left(
e^-ict-bt
dfrac((c+a-ib)^2+b^2)sgn(c-d-ib)-2ib((2c-ib)^2+b^2) \
+ e^ict-bt
dfrac((-c+a-ib)^2+b^2)sgn(-c-d-ib)-2ib((-2c-ib)^2+b^2)
right)$$for $t<0$, $e^−iomega t$ converges to zero at infinity on the upper half plane, so residues are
$$ R_2 = 2pi i left(e^-ict+bt
dfrac((c+a+ib)^2+b^2)sgn(c-d+ib)2ib((2c+ib)^2+b^2) \
+ e^ict+bt
dfrac((-c+a+ib)^2+b^2)sgn(-c-d+ib)2ib((-2c+ib)^2+b^2)
right)$$
This means $g(t)=frac12pi(H(t)R_1 + H(-t)R_2)$ where $H$ is the Heaviside function.
Does my working look correct? Have I missed anything? Below I plot the analytical $g(t)$ alongside the Python IFFT result of $mathcalF^-1(G(omega))$ but they don't agree. (I checked my Python code by performing IFFTs on known FT pairs, so there's nothing wrong with the algorithm.)
complex-analysis fourier-analysis fourier-transform residue-calculus complex-integration
asked Jul 18 at 10:12
Medulla Oblongata
271110
asked Jul 18 at 10:12
Medulla Oblongata
271110
asked Jul 18 at 10:12
Medulla Oblongata
271110
asked Jul 18 at 10:12
Medulla Oblongata
271110
271110
Residue theory works on analytic functions, and I don't think the Lorentzian times a sign function is analytic. What you could do is compute the Lorentzian's IFT first, then use that the (inverse) fourier transform of a product is a convolution of their (inverse) fourier transforms
– Calvin Khor
Jul 18 at 10:19
Yes, I initially tried to use the FT tables in Wikipedia (en.wikipedia.org/wiki/…) to separately transform the Lorentzian functions and sign function into the time domain, then do convolution on those results (i.e. using property 108: $F(omega)G(omega) to f(t)*g(t)$). But it doesn't seem possible to analytically solve the convolution. I want to calculate a purely analytical result.
– Medulla Oblongata
Jul 18 at 10:30
I'm not aware of the available tables,literature etc, but you might be able to find something if you note that convolutions against $mathcal F^-1 (-ioperatornamesgn)$ are known as Hilbert transforms.
– Calvin Khor
Jul 18 at 10:35
@CalvinKhor BTW could you expand on your comment and post it as an answer below? Thanks
– Medulla Oblongata
Jul 18 at 11:04
Okay, I'll try to do so when I have some time, but if you (or someone else) wants to do it, feel free to so.
– Calvin Khor
Jul 18 at 11:08
add a comment |Â
Residue theory works on analytic functions, and I don't think the Lorentzian times a sign function is analytic. What you could do is compute the Lorentzian's IFT first, then use that the (inverse) fourier transform of a product is a convolution of their (inverse) fourier transforms
– Calvin Khor
Jul 18 at 10:19
Yes, I initially tried to use the FT tables in Wikipedia (en.wikipedia.org/wiki/…) to separately transform the Lorentzian functions and sign function into the time domain, then do convolution on those results (i.e. using property 108: $F(omega)G(omega) to f(t)*g(t)$). But it doesn't seem possible to analytically solve the convolution. I want to calculate a purely analytical result.
– Medulla Oblongata
Jul 18 at 10:30
I'm not aware of the available tables,literature etc, but you might be able to find something if you note that convolutions against $mathcal F^-1 (-ioperatornamesgn)$ are known as Hilbert transforms.
– Calvin Khor
Jul 18 at 10:35
@CalvinKhor BTW could you expand on your comment and post it as an answer below? Thanks
– Medulla Oblongata
Jul 18 at 11:04
Okay, I'll try to do so when I have some time, but if you (or someone else) wants to do it, feel free to so.
– Calvin Khor
Jul 18 at 11:08
Residue theory works on analytic functions, and I don't think the Lorentzian times a sign function is analytic. What you could do is compute the Lorentzian's IFT first, then use that the (inverse) fourier transform of a product is a convolution of their (inverse) fourier transforms
– Calvin Khor
Jul 18 at 10:19
Residue theory works on analytic functions, and I don't think the Lorentzian times a sign function is analytic. What you could do is compute the Lorentzian's IFT first, then use that the (inverse) fourier transform of a product is a convolution of their (inverse) fourier transforms
– Calvin Khor
Jul 18 at 10:19
Yes, I initially tried to use the FT tables in Wikipedia (en.wikipedia.org/wiki/…) to separately transform the Lorentzian functions and sign function into the time domain, then do convolution on those results (i.e. using property 108: $F(omega)G(omega) to f(t)*g(t)$). But it doesn't seem possible to analytically solve the convolution. I want to calculate a purely analytical result.
– Medulla Oblongata
Jul 18 at 10:30
Yes, I initially tried to use the FT tables in Wikipedia (en.wikipedia.org/wiki/…) to separately transform the Lorentzian functions and sign function into the time domain, then do convolution on those results (i.e. using property 108: $F(omega)G(omega) to f(t)*g(t)$). But it doesn't seem possible to analytically solve the convolution. I want to calculate a purely analytical result.
– Medulla Oblongata
Jul 18 at 10:30
I'm not aware of the available tables,literature etc, but you might be able to find something if you note that convolutions against $mathcal F^-1 (-ioperatornamesgn)$ are known as Hilbert transforms.
– Calvin Khor
Jul 18 at 10:35
I'm not aware of the available tables,literature etc, but you might be able to find something if you note that convolutions against $mathcal F^-1 (-ioperatornamesgn)$ are known as Hilbert transforms.
– Calvin Khor
Jul 18 at 10:35
@CalvinKhor BTW could you expand on your comment and post it as an answer below? Thanks
– Medulla Oblongata
Jul 18 at 11:04
@CalvinKhor BTW could you expand on your comment and post it as an answer below? Thanks
– Medulla Oblongata
Jul 18 at 11:04
Okay, I'll try to do so when I have some time, but if you (or someone else) wants to do it, feel free to so.
– Calvin Khor
Jul 18 at 11:08
Okay, I'll try to do so when I have some time, but if you (or someone else) wants to do it, feel free to so.
– Calvin Khor
Jul 18 at 11:08
add a comment |Â
1 Answer
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Assume that $b neq 0 land c neq 0$. Then the denominator doesn't have multiple roots, and partial fraction decomposition gives a sum of four terms of the form $C/(w pm c pm ib)$. Thus it is sufficient to find the Fourier transform of $operatornamesgn(w+d)/(w pm c pm i b)$, which reduces to finding the transform of $H(w)/(w+w_0)$, with $w_0$ not on the real axis.
Assume further that $w_0$ is not on the imaginary axis, which is equivalent to $c neq pm d$. We have
$$I = int_-infty^infty frac H(w) w+w_0 e^-itw dw =
e^tau_0 int_tau_0^itinfty frac e^-tau tau dtau, \
tau_0 = i w_0 t, ; w_0 notin mathbb R, ; i w_0 notin mathbb R.$$
When the origin is inside the sector bounded by the rays from $tau_0$ to $itinfty$ and from $tau_0$ to $infty$, we'll have
$$e^-tau_0 I - Gamma(0, tau_0) =
-2 pi i operatornamesgn t
operatornameRes_tau=0 frac e^-tau tau.$$
Otherwise the difference $e^-tau_0 I - Gamma(0, tau_0)$ will be zero. The condition for the origin to be inside the sector is
$$operatornameRe tau_0 < 0 land \
((t < 0 land operatornameIm tau_0 > 0) lor
(t > 0 land operatornameIm tau_0 < 0)),$$
which simplifies to
$$operatornameRe w_0 < 0 land t operatornameIm w_0 > 0,$$
and we obtain
$$I = e^i w_0 t(Gamma(0, i w_0 t) -
2 pi i operatornamesgn t ,
H(-operatornameRe w_0) H(t operatornameIm w_0)).$$
answered Jul 19 at 16:44
Maxim
2,090113
Thanks @Maxim. Just to clarify, does the $Gamma$ refer to the gamma function en.wikipedia.org/wiki/Gamma_function ?
– Medulla Oblongata
Jul 22 at 7:04
More precisely, the incomplete gamma function.
– Maxim
Jul 22 at 12:59
and $w_0=pm cpm ib$ I assume
– Medulla Oblongata
Jul 23 at 11:13
Plus/minus $d$. $H(w)$ is the unit step function; $H(w)/(w - d pm c pm i b)$ accounts for the part where $operatornamesgn (w + d) = 1$, and $-H(w)/(-w - d pm c pm i b)$ accounts for $operatornamesgn (w + d) = -1$.
– Maxim
Jul 23 at 12:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Assume that $b neq 0 land c neq 0$. Then the denominator doesn't have multiple roots, and partial fraction decomposition gives a sum of four terms of the form $C/(w pm c pm ib)$. Thus it is sufficient to find the Fourier transform of $operatornamesgn(w+d)/(w pm c pm i b)$, which reduces to finding the transform of $H(w)/(w+w_0)$, with $w_0$ not on the real axis.
Assume further that $w_0$ is not on the imaginary axis, which is equivalent to $c neq pm d$. We have
$$I = int_-infty^infty frac H(w) w+w_0 e^-itw dw =
e^tau_0 int_tau_0^itinfty frac e^-tau tau dtau, \
tau_0 = i w_0 t, ; w_0 notin mathbb R, ; i w_0 notin mathbb R.$$
When the origin is inside the sector bounded by the rays from $tau_0$ to $itinfty$ and from $tau_0$ to $infty$, we'll have
$$e^-tau_0 I - Gamma(0, tau_0) =
-2 pi i operatornamesgn t
operatornameRes_tau=0 frac e^-tau tau.$$
Otherwise the difference $e^-tau_0 I - Gamma(0, tau_0)$ will be zero. The condition for the origin to be inside the sector is
$$operatornameRe tau_0 < 0 land \
((t < 0 land operatornameIm tau_0 > 0) lor
(t > 0 land operatornameIm tau_0 < 0)),$$
which simplifies to
$$operatornameRe w_0 < 0 land t operatornameIm w_0 > 0,$$
and we obtain
$$I = e^i w_0 t(Gamma(0, i w_0 t) -
2 pi i operatornamesgn t ,
H(-operatornameRe w_0) H(t operatornameIm w_0)).$$
answered Jul 19 at 16:44
Maxim
2,090113
Thanks @Maxim. Just to clarify, does the $Gamma$ refer to the gamma function en.wikipedia.org/wiki/Gamma_function ?
– Medulla Oblongata
Jul 22 at 7:04
More precisely, the incomplete gamma function.
– Maxim
Jul 22 at 12:59
and $w_0=pm cpm ib$ I assume
– Medulla Oblongata
Jul 23 at 11:13
Plus/minus $d$. $H(w)$ is the unit step function; $H(w)/(w - d pm c pm i b)$ accounts for the part where $operatornamesgn (w + d) = 1$, and $-H(w)/(-w - d pm c pm i b)$ accounts for $operatornamesgn (w + d) = -1$.
– Maxim
Jul 23 at 12:39
add a comment |Â
up vote
1
down vote
Assume that $b neq 0 land c neq 0$. Then the denominator doesn't have multiple roots, and partial fraction decomposition gives a sum of four terms of the form $C/(w pm c pm ib)$. Thus it is sufficient to find the Fourier transform of $operatornamesgn(w+d)/(w pm c pm i b)$, which reduces to finding the transform of $H(w)/(w+w_0)$, with $w_0$ not on the real axis.
Assume further that $w_0$ is not on the imaginary axis, which is equivalent to $c neq pm d$. We have
$$I = int_-infty^infty frac H(w) w+w_0 e^-itw dw =
e^tau_0 int_tau_0^itinfty frac e^-tau tau dtau, \
tau_0 = i w_0 t, ; w_0 notin mathbb R, ; i w_0 notin mathbb R.$$
When the origin is inside the sector bounded by the rays from $tau_0$ to $itinfty$ and from $tau_0$ to $infty$, we'll have
$$e^-tau_0 I - Gamma(0, tau_0) =
-2 pi i operatornamesgn t
operatornameRes_tau=0 frac e^-tau tau.$$
Otherwise the difference $e^-tau_0 I - Gamma(0, tau_0)$ will be zero. The condition for the origin to be inside the sector is
$$operatornameRe tau_0 < 0 land \
((t < 0 land operatornameIm tau_0 > 0) lor
(t > 0 land operatornameIm tau_0 < 0)),$$
which simplifies to
$$operatornameRe w_0 < 0 land t operatornameIm w_0 > 0,$$
and we obtain
$$I = e^i w_0 t(Gamma(0, i w_0 t) -
2 pi i operatornamesgn t ,
H(-operatornameRe w_0) H(t operatornameIm w_0)).$$
answered Jul 19 at 16:44
Maxim
2,090113
Thanks @Maxim. Just to clarify, does the $Gamma$ refer to the gamma function en.wikipedia.org/wiki/Gamma_function ?
– Medulla Oblongata
Jul 22 at 7:04
More precisely, the incomplete gamma function.
– Maxim
Jul 22 at 12:59
and $w_0=pm cpm ib$ I assume
– Medulla Oblongata
Jul 23 at 11:13
Plus/minus $d$. $H(w)$ is the unit step function; $H(w)/(w - d pm c pm i b)$ accounts for the part where $operatornamesgn (w + d) = 1$, and $-H(w)/(-w - d pm c pm i b)$ accounts for $operatornamesgn (w + d) = -1$.
– Maxim
Jul 23 at 12:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Assume that $b neq 0 land c neq 0$. Then the denominator doesn't have multiple roots, and partial fraction decomposition gives a sum of four terms of the form $C/(w pm c pm ib)$. Thus it is sufficient to find the Fourier transform of $operatornamesgn(w+d)/(w pm c pm i b)$, which reduces to finding the transform of $H(w)/(w+w_0)$, with $w_0$ not on the real axis.
Assume further that $w_0$ is not on the imaginary axis, which is equivalent to $c neq pm d$. We have
$$I = int_-infty^infty frac H(w) w+w_0 e^-itw dw =
e^tau_0 int_tau_0^itinfty frac e^-tau tau dtau, \
tau_0 = i w_0 t, ; w_0 notin mathbb R, ; i w_0 notin mathbb R.$$
When the origin is inside the sector bounded by the rays from $tau_0$ to $itinfty$ and from $tau_0$ to $infty$, we'll have
$$e^-tau_0 I - Gamma(0, tau_0) =
-2 pi i operatornamesgn t
operatornameRes_tau=0 frac e^-tau tau.$$
Otherwise the difference $e^-tau_0 I - Gamma(0, tau_0)$ will be zero. The condition for the origin to be inside the sector is
$$operatornameRe tau_0 < 0 land \
((t < 0 land operatornameIm tau_0 > 0) lor
(t > 0 land operatornameIm tau_0 < 0)),$$
which simplifies to
$$operatornameRe w_0 < 0 land t operatornameIm w_0 > 0,$$
and we obtain
$$I = e^i w_0 t(Gamma(0, i w_0 t) -
2 pi i operatornamesgn t ,
H(-operatornameRe w_0) H(t operatornameIm w_0)).$$
answered Jul 19 at 16:44
Maxim
2,090113
Assume that $b neq 0 land c neq 0$. Then the denominator doesn't have multiple roots, and partial fraction decomposition gives a sum of four terms of the form $C/(w pm c pm ib)$. Thus it is sufficient to find the Fourier transform of $operatornamesgn(w+d)/(w pm c pm i b)$, which reduces to finding the transform of $H(w)/(w+w_0)$, with $w_0$ not on the real axis.
Assume further that $w_0$ is not on the imaginary axis, which is equivalent to $c neq pm d$. We have
$$I = int_-infty^infty frac H(w) w+w_0 e^-itw dw =
e^tau_0 int_tau_0^itinfty frac e^-tau tau dtau, \
tau_0 = i w_0 t, ; w_0 notin mathbb R, ; i w_0 notin mathbb R.$$
When the origin is inside the sector bounded by the rays from $tau_0$ to $itinfty$ and from $tau_0$ to $infty$, we'll have
$$e^-tau_0 I - Gamma(0, tau_0) =
-2 pi i operatornamesgn t
operatornameRes_tau=0 frac e^-tau tau.$$
Otherwise the difference $e^-tau_0 I - Gamma(0, tau_0)$ will be zero. The condition for the origin to be inside the sector is
$$operatornameRe tau_0 < 0 land \
((t < 0 land operatornameIm tau_0 > 0) lor
(t > 0 land operatornameIm tau_0 < 0)),$$
which simplifies to
$$operatornameRe w_0 < 0 land t operatornameIm w_0 > 0,$$
and we obtain
$$I = e^i w_0 t(Gamma(0, i w_0 t) -
2 pi i operatornamesgn t ,
H(-operatornameRe w_0) H(t operatornameIm w_0)).$$
answered Jul 19 at 16:44
Maxim
2,090113
edited Jul 20 at 20:21
answered Jul 19 at 16:44
Maxim
2,090113
answered Jul 19 at 16:44
Maxim
2,090113
answered Jul 19 at 16:44
Maxim
2,090113
2,090113
Thanks @Maxim. Just to clarify, does the $Gamma$ refer to the gamma function en.wikipedia.org/wiki/Gamma_function ?
– Medulla Oblongata
Jul 22 at 7:04
More precisely, the incomplete gamma function.
– Maxim
Jul 22 at 12:59
and $w_0=pm cpm ib$ I assume
– Medulla Oblongata
Jul 23 at 11:13
Plus/minus $d$. $H(w)$ is the unit step function; $H(w)/(w - d pm c pm i b)$ accounts for the part where $operatornamesgn (w + d) = 1$, and $-H(w)/(-w - d pm c pm i b)$ accounts for $operatornamesgn (w + d) = -1$.
– Maxim
Jul 23 at 12:39
add a comment |Â
Thanks @Maxim. Just to clarify, does the $Gamma$ refer to the gamma function en.wikipedia.org/wiki/Gamma_function ?
– Medulla Oblongata
Jul 22 at 7:04
More precisely, the incomplete gamma function.
– Maxim
Jul 22 at 12:59
and $w_0=pm cpm ib$ I assume
– Medulla Oblongata
Jul 23 at 11:13
Plus/minus $d$. $H(w)$ is the unit step function; $H(w)/(w - d pm c pm i b)$ accounts for the part where $operatornamesgn (w + d) = 1$, and $-H(w)/(-w - d pm c pm i b)$ accounts for $operatornamesgn (w + d) = -1$.
– Maxim
Jul 23 at 12:39
Thanks @Maxim. Just to clarify, does the $Gamma$ refer to the gamma function en.wikipedia.org/wiki/Gamma_function ?
– Medulla Oblongata
Jul 22 at 7:04
Thanks @Maxim. Just to clarify, does the $Gamma$ refer to the gamma function en.wikipedia.org/wiki/Gamma_function ?
– Medulla Oblongata
Jul 22 at 7:04
More precisely, the incomplete gamma function.
– Maxim
Jul 22 at 12:59
More precisely, the incomplete gamma function.
– Maxim
Jul 22 at 12:59
and $w_0=pm cpm ib$ I assume
– Medulla Oblongata
Jul 23 at 11:13
and $w_0=pm cpm ib$ I assume
– Medulla Oblongata
Jul 23 at 11:13
Plus/minus $d$. $H(w)$ is the unit step function; $H(w)/(w - d pm c pm i b)$ accounts for the part where $operatornamesgn (w + d) = 1$, and $-H(w)/(-w - d pm c pm i b)$ accounts for $operatornamesgn (w + d) = -1$.
– Maxim
Jul 23 at 12:39
Plus/minus $d$. $H(w)$ is the unit step function; $H(w)/(w - d pm c pm i b)$ accounts for the part where $operatornamesgn (w + d) = 1$, and $-H(w)/(-w - d pm c pm i b)$ accounts for $operatornamesgn (w + d) = -1$.
– Maxim
Jul 23 at 12:39
add a comment |Â
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Residue theory works on analytic functions, and I don't think the Lorentzian times a sign function is analytic. What you could do is compute the Lorentzian's IFT first, then use that the (inverse) fourier transform of a product is a convolution of their (inverse) fourier transforms
– Calvin Khor
Jul 18 at 10:19
Yes, I initially tried to use the FT tables in Wikipedia (en.wikipedia.org/wiki/…) to separately transform the Lorentzian functions and sign function into the time domain, then do convolution on those results (i.e. using property 108: $F(omega)G(omega) to f(t)*g(t)$). But it doesn't seem possible to analytically solve the convolution. I want to calculate a purely analytical result.
– Medulla Oblongata
Jul 18 at 10:30
I'm not aware of the available tables,literature etc, but you might be able to find something if you note that convolutions against $mathcal F^-1 (-ioperatornamesgn)$ are known as Hilbert transforms.
– Calvin Khor
Jul 18 at 10:35
@CalvinKhor BTW could you expand on your comment and post it as an answer below? Thanks
– Medulla Oblongata
Jul 18 at 11:04
Okay, I'll try to do so when I have some time, but if you (or someone else) wants to do it, feel free to so.
– Calvin Khor
Jul 18 at 11:08