Discrete Fourier Transforms, showing the transform of a product of a series
$begingroup$
If $a=(a_{1}, ldots,a_{n})$. Define $F_{a}( lambda)= n^{-1/2} sumlimits_{t=1}^{n} a_{t}e^{-it lambda}$.
Let $lbrace x_{1}, ldots,x_{n} rbrace$ and $lbrace y_{1},ldots,y_{n}rbrace$ be real numbers. Let $z_{t}=x_{t}y_{t}$. I want to show the following equality,
$F_{z}(2 pi j /n)= n^{-1/2} sumlimits_{ k in D_{n}} F_{X}( 2 pi j/n)F_{Y}(2 pi (j-k)/n)$, where $D_{n} = lbrace j in mathbb{Z} : 2 pi j/ n in (- pi, pi] rbrace$.
analysis fourier-analysis fourier-transform
asked Dec 9 '18 at 0:00
mich95mich95
6,88011126
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add a comment |
$begingroup$
If $a=(a_{1}, ldots,a_{n})$. Define $F_{a}( lambda)= n^{-1/2} sumlimits_{t=1}^{n} a_{t}e^{-it lambda}$.
Let $lbrace x_{1}, ldots,x_{n} rbrace$ and $lbrace y_{1},ldots,y_{n}rbrace$ be real numbers. Let $z_{t}=x_{t}y_{t}$. I want to show the following equality,
$F_{z}(2 pi j /n)= n^{-1/2} sumlimits_{ k in D_{n}} F_{X}( 2 pi j/n)F_{Y}(2 pi (j-k)/n)$, where $D_{n} = lbrace j in mathbb{Z} : 2 pi j/ n in (- pi, pi] rbrace$.
analysis fourier-analysis fourier-transform
asked Dec 9 '18 at 0:00
mich95mich95
6,88011126
$endgroup$
$begingroup$
See how the terms $e^{-i(t_1-t_2) lambda}$ appear in $F_x(lambda)F_y(u-lambda)$ then for $0 ne t_1-t_2 $ dividing $ n!$ : $sum_{k= 1}^{n!} e^{-i(t_1-t_2) 2pi k/n!} =0$. You can replace $n!$ by $lcm(1,ldots,n)$ but not by $n$.
$endgroup$
– reuns
Dec 9 '18 at 1:17
$begingroup$
I tried to go from either ways to the other, but miserabely failed. Would you be able to help more ?
$endgroup$
– mich95
Dec 9 '18 at 1:20
$begingroup$
Where are you stuck in what I said ?
$endgroup$
– reuns
Dec 9 '18 at 1:21
$begingroup$
I do not get any $n!$...
$endgroup$
– mich95
Dec 9 '18 at 2:19
add a comment |
$begingroup$
If $a=(a_{1}, ldots,a_{n})$. Define $F_{a}( lambda)= n^{-1/2} sumlimits_{t=1}^{n} a_{t}e^{-it lambda}$.
Let $lbrace x_{1}, ldots,x_{n} rbrace$ and $lbrace y_{1},ldots,y_{n}rbrace$ be real numbers. Let $z_{t}=x_{t}y_{t}$. I want to show the following equality,
$F_{z}(2 pi j /n)= n^{-1/2} sumlimits_{ k in D_{n}} F_{X}( 2 pi j/n)F_{Y}(2 pi (j-k)/n)$, where $D_{n} = lbrace j in mathbb{Z} : 2 pi j/ n in (- pi, pi] rbrace$.
analysis fourier-analysis fourier-transform
asked Dec 9 '18 at 0:00
mich95mich95
6,88011126
$endgroup$
If $a=(a_{1}, ldots,a_{n})$. Define $F_{a}( lambda)= n^{-1/2} sumlimits_{t=1}^{n} a_{t}e^{-it lambda}$.
Let $lbrace x_{1}, ldots,x_{n} rbrace$ and $lbrace y_{1},ldots,y_{n}rbrace$ be real numbers. Let $z_{t}=x_{t}y_{t}$. I want to show the following equality,
$F_{z}(2 pi j /n)= n^{-1/2} sumlimits_{ k in D_{n}} F_{X}( 2 pi j/n)F_{Y}(2 pi (j-k)/n)$, where $D_{n} = lbrace j in mathbb{Z} : 2 pi j/ n in (- pi, pi] rbrace$.
analysis fourier-analysis fourier-transform
analysis fourier-analysis fourier-transform
asked Dec 9 '18 at 0:00
mich95mich95
6,88011126
asked Dec 9 '18 at 0:00
mich95mich95
6,88011126
asked Dec 9 '18 at 0:00
mich95mich95
6,88011126
asked Dec 9 '18 at 0:00
mich95mich95
6,88011126
asked Dec 9 '18 at 0:00
mich95mich95
6,88011126
6,88011126
$begingroup$
See how the terms $e^{-i(t_1-t_2) lambda}$ appear in $F_x(lambda)F_y(u-lambda)$ then for $0 ne t_1-t_2 $ dividing $ n!$ : $sum_{k= 1}^{n!} e^{-i(t_1-t_2) 2pi k/n!} =0$. You can replace $n!$ by $lcm(1,ldots,n)$ but not by $n$.
$endgroup$
– reuns
Dec 9 '18 at 1:17
$begingroup$
I tried to go from either ways to the other, but miserabely failed. Would you be able to help more ?
$endgroup$
– mich95
Dec 9 '18 at 1:20
$begingroup$
Where are you stuck in what I said ?
$endgroup$
– reuns
Dec 9 '18 at 1:21
$begingroup$
I do not get any $n!$...
$endgroup$
– mich95
Dec 9 '18 at 2:19
add a comment |
$begingroup$
See how the terms $e^{-i(t_1-t_2) lambda}$ appear in $F_x(lambda)F_y(u-lambda)$ then for $0 ne t_1-t_2 $ dividing $ n!$ : $sum_{k= 1}^{n!} e^{-i(t_1-t_2) 2pi k/n!} =0$. You can replace $n!$ by $lcm(1,ldots,n)$ but not by $n$.
$endgroup$
– reuns
Dec 9 '18 at 1:17
$begingroup$
I tried to go from either ways to the other, but miserabely failed. Would you be able to help more ?
$endgroup$
– mich95
Dec 9 '18 at 1:20
$begingroup$
Where are you stuck in what I said ?
$endgroup$
– reuns
Dec 9 '18 at 1:21
$begingroup$
I do not get any $n!$...
$endgroup$
– mich95
Dec 9 '18 at 2:19
$begingroup$
See how the terms $e^{-i(t_1-t_2) lambda}$ appear in $F_x(lambda)F_y(u-lambda)$ then for $0 ne t_1-t_2 $ dividing $ n!$ : $sum_{k= 1}^{n!} e^{-i(t_1-t_2) 2pi k/n!} =0$. You can replace $n!$ by $lcm(1,ldots,n)$ but not by $n$.
$endgroup$
– reuns
Dec 9 '18 at 1:17
$begingroup$
See how the terms $e^{-i(t_1-t_2) lambda}$ appear in $F_x(lambda)F_y(u-lambda)$ then for $0 ne t_1-t_2 $ dividing $ n!$ : $sum_{k= 1}^{n!} e^{-i(t_1-t_2) 2pi k/n!} =0$. You can replace $n!$ by $lcm(1,ldots,n)$ but not by $n$.
$endgroup$
– reuns
Dec 9 '18 at 1:17
$begingroup$
I tried to go from either ways to the other, but miserabely failed. Would you be able to help more ?
$endgroup$
– mich95
Dec 9 '18 at 1:20
$begingroup$
I tried to go from either ways to the other, but miserabely failed. Would you be able to help more ?
$endgroup$
– mich95
Dec 9 '18 at 1:20
$begingroup$
Where are you stuck in what I said ?
$endgroup$
– reuns
Dec 9 '18 at 1:21
$begingroup$
Where are you stuck in what I said ?
$endgroup$
– reuns
Dec 9 '18 at 1:21
$begingroup$
I do not get any $n!$...
$endgroup$
– mich95
Dec 9 '18 at 2:19
$begingroup$
I do not get any $n!$...
$endgroup$
– mich95
Dec 9 '18 at 2:19
add a comment |
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$begingroup$
See how the terms $e^{-i(t_1-t_2) lambda}$ appear in $F_x(lambda)F_y(u-lambda)$ then for $0 ne t_1-t_2 $ dividing $ n!$ : $sum_{k= 1}^{n!} e^{-i(t_1-t_2) 2pi k/n!} =0$. You can replace $n!$ by $lcm(1,ldots,n)$ but not by $n$.
$endgroup$
– reuns
Dec 9 '18 at 1:17
$begingroup$
I tried to go from either ways to the other, but miserabely failed. Would you be able to help more ?
$endgroup$
– mich95
Dec 9 '18 at 1:20
$begingroup$
Where are you stuck in what I said ?
$endgroup$
– reuns
Dec 9 '18 at 1:21
$begingroup$
I do not get any $n!$...
$endgroup$
– mich95
Dec 9 '18 at 2:19