How do I find the Fourier transform of $mathcal{F}[log(a^2+s^2)](s)$
For $a>0$ i have managed to show that this is the Fourier transform of the function.
$$
mathcal{F}[e^{-a|x|}](s) = frac {2a}{sqrt{2{pi}}(a^2+s^2)}.
$$
How do I now use this to find the Fourier transform of:
$$
mathcal{F}[log(a^2+s^2)](s)?
$$
I have tried to apply the inversion formula for Fourier transforms but I haven't had any success. Any help would be greatly appreciated.
fourier-analysis
edited May 13 '15 at 21:54
Davide Giraudo
125k16150261
asked May 13 '15 at 18:29
seansean
534
|
show 2 more comments
For $a>0$ i have managed to show that this is the Fourier transform of the function.
$$
mathcal{F}[e^{-a|x|}](s) = frac {2a}{sqrt{2{pi}}(a^2+s^2)}.
$$
How do I now use this to find the Fourier transform of:
$$
mathcal{F}[log(a^2+s^2)](s)?
$$
I have tried to apply the inversion formula for Fourier transforms but I haven't had any success. Any help would be greatly appreciated.
fourier-analysis
edited May 13 '15 at 21:54
Davide Giraudo
125k16150261
asked May 13 '15 at 18:29
seansean
534
What is the derivative of $log(a^2+s^2)$ with respect to $s$? What do you know about the Fourier transform of a primitive?
– Jack D'Aurizio
May 13 '15 at 18:33
so we have $frac{d}{ds}(log(a^2+s^2)=frac{2s}{a^2+s^2}$ but im not sure what you mean for the next part
– sean
May 13 '15 at 18:42
any chance you could just show me how to do it please? kinda desperate my exam is tomorrow
– sean
May 13 '15 at 18:55
Do you know the formula for the Fourier transform of the derivatives of a function?
– Lost
May 13 '15 at 18:57
unfortunately not.
– sean
May 13 '15 at 18:58
|
show 2 more comments
For $a>0$ i have managed to show that this is the Fourier transform of the function.
$$
mathcal{F}[e^{-a|x|}](s) = frac {2a}{sqrt{2{pi}}(a^2+s^2)}.
$$
How do I now use this to find the Fourier transform of:
$$
mathcal{F}[log(a^2+s^2)](s)?
$$
I have tried to apply the inversion formula for Fourier transforms but I haven't had any success. Any help would be greatly appreciated.
fourier-analysis
edited May 13 '15 at 21:54
Davide Giraudo
125k16150261
asked May 13 '15 at 18:29
seansean
534
For $a>0$ i have managed to show that this is the Fourier transform of the function.
$$
mathcal{F}[e^{-a|x|}](s) = frac {2a}{sqrt{2{pi}}(a^2+s^2)}.
$$
How do I now use this to find the Fourier transform of:
$$
mathcal{F}[log(a^2+s^2)](s)?
$$
I have tried to apply the inversion formula for Fourier transforms but I haven't had any success. Any help would be greatly appreciated.
fourier-analysis
fourier-analysis
edited May 13 '15 at 21:54
Davide Giraudo
125k16150261
asked May 13 '15 at 18:29
seansean
534
edited May 13 '15 at 21:54
Davide Giraudo
125k16150261
asked May 13 '15 at 18:29
seansean
534
edited May 13 '15 at 21:54
Davide Giraudo
125k16150261
edited May 13 '15 at 21:54
Davide Giraudo
125k16150261
edited May 13 '15 at 21:54
Davide Giraudo
125k16150261
125k16150261
asked May 13 '15 at 18:29
seansean
534
asked May 13 '15 at 18:29
seansean
534
asked May 13 '15 at 18:29
seansean
534
534
What is the derivative of $log(a^2+s^2)$ with respect to $s$? What do you know about the Fourier transform of a primitive?
– Jack D'Aurizio
May 13 '15 at 18:33
so we have $frac{d}{ds}(log(a^2+s^2)=frac{2s}{a^2+s^2}$ but im not sure what you mean for the next part
– sean
May 13 '15 at 18:42
any chance you could just show me how to do it please? kinda desperate my exam is tomorrow
– sean
May 13 '15 at 18:55
Do you know the formula for the Fourier transform of the derivatives of a function?
– Lost
May 13 '15 at 18:57
unfortunately not.
– sean
May 13 '15 at 18:58
|
show 2 more comments
What is the derivative of $log(a^2+s^2)$ with respect to $s$? What do you know about the Fourier transform of a primitive?
– Jack D'Aurizio
May 13 '15 at 18:33
so we have $frac{d}{ds}(log(a^2+s^2)=frac{2s}{a^2+s^2}$ but im not sure what you mean for the next part
– sean
May 13 '15 at 18:42
any chance you could just show me how to do it please? kinda desperate my exam is tomorrow
– sean
May 13 '15 at 18:55
Do you know the formula for the Fourier transform of the derivatives of a function?
– Lost
May 13 '15 at 18:57
unfortunately not.
– sean
May 13 '15 at 18:58
What is the derivative of $log(a^2+s^2)$ with respect to $s$? What do you know about the Fourier transform of a primitive?
– Jack D'Aurizio
May 13 '15 at 18:33
What is the derivative of $log(a^2+s^2)$ with respect to $s$? What do you know about the Fourier transform of a primitive?
– Jack D'Aurizio
May 13 '15 at 18:33
so we have $frac{d}{ds}(log(a^2+s^2)=frac{2s}{a^2+s^2}$ but im not sure what you mean for the next part
– sean
May 13 '15 at 18:42
so we have $frac{d}{ds}(log(a^2+s^2)=frac{2s}{a^2+s^2}$ but im not sure what you mean for the next part
– sean
May 13 '15 at 18:42
any chance you could just show me how to do it please? kinda desperate my exam is tomorrow
– sean
May 13 '15 at 18:55
any chance you could just show me how to do it please? kinda desperate my exam is tomorrow
– sean
May 13 '15 at 18:55
Do you know the formula for the Fourier transform of the derivatives of a function?
– Lost
May 13 '15 at 18:57
Do you know the formula for the Fourier transform of the derivatives of a function?
– Lost
May 13 '15 at 18:57
unfortunately not.
– sean
May 13 '15 at 18:58
unfortunately not.
– sean
May 13 '15 at 18:58
|
show 2 more comments
1 Answer
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The derivative of $log(a^2+s^2)$ is $frac{2s}{a^2+s^2}$. Now, we know that the fourier transform of $f'(x)$ is $iomega hat{f}(omega)$. So we can find the fourier transform of $frac{2s}{a^2+s^2}$ and then divide it by $iomega$ to find our answer. We can use the Residue Theorem to evaluate $int_{-infty}^{infty} frac{2s}{a^2+s^2}e^{-iomega s}ds$.
There are poles at $pm ia$. Depending on the sign of $omega$, we either do our contour integral in the upper half or lower half of the plane. Let's do it in the lower half plane. The only residue in our domain is at $-ia$. Our residue is :$$frac{2s}{2s}e^{-i omega s}=e^{-i omega s}$$ For $s=-ia$, we get $e^{-omega a}$
With the Residue Theorem, we get $-2 pi i e^{-omega a}$ (minus because we are in the lower half plane), which is the Fourier Transform of the derivative. Now we divide by $i omega$ to find the FT of the original function, and we get : $$frac{-2 pi}{omega}e^{- omega a}$$ which we could rewrite as ($omega= 2 pi xi $ , $xi$ is the frequency) $$frac{-1}{xi}e^{- omega a}$$
Hope this helps !
Note 1) : that this question was never answered for 3 years is not a reason to never answer it.
Note 2) : this is my first answer. If I did any mistake, please say it in the comments and I will correct my answer. Thank you !
answered Dec 3 '18 at 22:31
PoujhPoujh
552416
add a comment |
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The derivative of $log(a^2+s^2)$ is $frac{2s}{a^2+s^2}$. Now, we know that the fourier transform of $f'(x)$ is $iomega hat{f}(omega)$. So we can find the fourier transform of $frac{2s}{a^2+s^2}$ and then divide it by $iomega$ to find our answer. We can use the Residue Theorem to evaluate $int_{-infty}^{infty} frac{2s}{a^2+s^2}e^{-iomega s}ds$.
There are poles at $pm ia$. Depending on the sign of $omega$, we either do our contour integral in the upper half or lower half of the plane. Let's do it in the lower half plane. The only residue in our domain is at $-ia$. Our residue is :$$frac{2s}{2s}e^{-i omega s}=e^{-i omega s}$$ For $s=-ia$, we get $e^{-omega a}$
With the Residue Theorem, we get $-2 pi i e^{-omega a}$ (minus because we are in the lower half plane), which is the Fourier Transform of the derivative. Now we divide by $i omega$ to find the FT of the original function, and we get : $$frac{-2 pi}{omega}e^{- omega a}$$ which we could rewrite as ($omega= 2 pi xi $ , $xi$ is the frequency) $$frac{-1}{xi}e^{- omega a}$$
Hope this helps !
Note 1) : that this question was never answered for 3 years is not a reason to never answer it.
Note 2) : this is my first answer. If I did any mistake, please say it in the comments and I will correct my answer. Thank you !
answered Dec 3 '18 at 22:31
PoujhPoujh
552416
add a comment |
The derivative of $log(a^2+s^2)$ is $frac{2s}{a^2+s^2}$. Now, we know that the fourier transform of $f'(x)$ is $iomega hat{f}(omega)$. So we can find the fourier transform of $frac{2s}{a^2+s^2}$ and then divide it by $iomega$ to find our answer. We can use the Residue Theorem to evaluate $int_{-infty}^{infty} frac{2s}{a^2+s^2}e^{-iomega s}ds$.
There are poles at $pm ia$. Depending on the sign of $omega$, we either do our contour integral in the upper half or lower half of the plane. Let's do it in the lower half plane. The only residue in our domain is at $-ia$. Our residue is :$$frac{2s}{2s}e^{-i omega s}=e^{-i omega s}$$ For $s=-ia$, we get $e^{-omega a}$
With the Residue Theorem, we get $-2 pi i e^{-omega a}$ (minus because we are in the lower half plane), which is the Fourier Transform of the derivative. Now we divide by $i omega$ to find the FT of the original function, and we get : $$frac{-2 pi}{omega}e^{- omega a}$$ which we could rewrite as ($omega= 2 pi xi $ , $xi$ is the frequency) $$frac{-1}{xi}e^{- omega a}$$
Hope this helps !
Note 1) : that this question was never answered for 3 years is not a reason to never answer it.
Note 2) : this is my first answer. If I did any mistake, please say it in the comments and I will correct my answer. Thank you !
answered Dec 3 '18 at 22:31
PoujhPoujh
552416
add a comment |
The derivative of $log(a^2+s^2)$ is $frac{2s}{a^2+s^2}$. Now, we know that the fourier transform of $f'(x)$ is $iomega hat{f}(omega)$. So we can find the fourier transform of $frac{2s}{a^2+s^2}$ and then divide it by $iomega$ to find our answer. We can use the Residue Theorem to evaluate $int_{-infty}^{infty} frac{2s}{a^2+s^2}e^{-iomega s}ds$.
There are poles at $pm ia$. Depending on the sign of $omega$, we either do our contour integral in the upper half or lower half of the plane. Let's do it in the lower half plane. The only residue in our domain is at $-ia$. Our residue is :$$frac{2s}{2s}e^{-i omega s}=e^{-i omega s}$$ For $s=-ia$, we get $e^{-omega a}$
With the Residue Theorem, we get $-2 pi i e^{-omega a}$ (minus because we are in the lower half plane), which is the Fourier Transform of the derivative. Now we divide by $i omega$ to find the FT of the original function, and we get : $$frac{-2 pi}{omega}e^{- omega a}$$ which we could rewrite as ($omega= 2 pi xi $ , $xi$ is the frequency) $$frac{-1}{xi}e^{- omega a}$$
Hope this helps !
Note 1) : that this question was never answered for 3 years is not a reason to never answer it.
Note 2) : this is my first answer. If I did any mistake, please say it in the comments and I will correct my answer. Thank you !
answered Dec 3 '18 at 22:31
PoujhPoujh
552416
The derivative of $log(a^2+s^2)$ is $frac{2s}{a^2+s^2}$. Now, we know that the fourier transform of $f'(x)$ is $iomega hat{f}(omega)$. So we can find the fourier transform of $frac{2s}{a^2+s^2}$ and then divide it by $iomega$ to find our answer. We can use the Residue Theorem to evaluate $int_{-infty}^{infty} frac{2s}{a^2+s^2}e^{-iomega s}ds$.
There are poles at $pm ia$. Depending on the sign of $omega$, we either do our contour integral in the upper half or lower half of the plane. Let's do it in the lower half plane. The only residue in our domain is at $-ia$. Our residue is :$$frac{2s}{2s}e^{-i omega s}=e^{-i omega s}$$ For $s=-ia$, we get $e^{-omega a}$
With the Residue Theorem, we get $-2 pi i e^{-omega a}$ (minus because we are in the lower half plane), which is the Fourier Transform of the derivative. Now we divide by $i omega$ to find the FT of the original function, and we get : $$frac{-2 pi}{omega}e^{- omega a}$$ which we could rewrite as ($omega= 2 pi xi $ , $xi$ is the frequency) $$frac{-1}{xi}e^{- omega a}$$
Hope this helps !
Note 1) : that this question was never answered for 3 years is not a reason to never answer it.
Note 2) : this is my first answer. If I did any mistake, please say it in the comments and I will correct my answer. Thank you !
answered Dec 3 '18 at 22:31
PoujhPoujh
552416
edited Dec 3 '18 at 22:57
answered Dec 3 '18 at 22:31
PoujhPoujh
552416
answered Dec 3 '18 at 22:31
PoujhPoujh
552416
answered Dec 3 '18 at 22:31
PoujhPoujh
552416
552416
add a comment |
add a comment |
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What is the derivative of $log(a^2+s^2)$ with respect to $s$? What do you know about the Fourier transform of a primitive?
– Jack D'Aurizio
May 13 '15 at 18:33
so we have $frac{d}{ds}(log(a^2+s^2)=frac{2s}{a^2+s^2}$ but im not sure what you mean for the next part
– sean
May 13 '15 at 18:42
any chance you could just show me how to do it please? kinda desperate my exam is tomorrow
– sean
May 13 '15 at 18:55
Do you know the formula for the Fourier transform of the derivatives of a function?
– Lost
May 13 '15 at 18:57
unfortunately not.
– sean
May 13 '15 at 18:58