Mixing
Building an Integral Transform from an Orthonormal Basis on the L2 Circle
Clash Royale CLAN TAG#URR8PPP
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Background
It is well known that the orthonormal basis for $L_2[-pi, pi]$ is $Omega = e^-jmt _m in mathbbZ$. We extend this to $L_2(mathbbR)$ via the Fourier Transform, which is an integral transform of the form:
$$ f(omega) = displaystyleintlimits_-infty^infty f(t) K(omega, t) dt = displaystyleintlimits_-infty^infty f(t) e^-j omega t dt $$
I.e. where $K(omega, t) = e^-jomega t$ We know this kernel is unitary, so the extension to $L_2(mathbbR)$ is simply:
$$ f(t) = displaystyleintlimits_-infty^infty f(omega) K^ast(omega, t) domega = displaystyleintlimits_-infty^infty f(omega) e^j omega t domega $$
Extension
Now suppose I apply some unitary transform $U$ to $Omega$, giving us the new orthonormal basis for $L_2[-pi, pi]$ denoted by $Omega_U = Ue^-jmt _m in mathbbZ$. I want to extend this to $L_2(mathbbR)$ in the same way. So this would mean:
$$ f(omega) = displaystyleintlimits_-infty^infty f(t) K_U(omega, t) dt = displaystyleintlimits_-infty^infty f(t) big(Ue^-j omega tbig) dt $$
Which means the extension to $L_2(mathbbR)$ takes the form:
$$ f(t) = displaystyleintlimits_-infty^infty f(omega) K_U^ast(omega, t) domega = displaystyleintlimits_-infty^infty f(omega) big(Ue^-j omega tbig)^ast domega = displaystyleintlimits_-infty^infty f(omega) big(U^-1e^j omega tbig) domega$$
In other words:
$K_U(omega, t) = Ue^-jomega t$ and $K_U^ast(omega, t) = U^-1e^jomega t$
Questions
1.) Is this the correct way to build an integral transform from an orthonormal basis? If not, where did I go wrong?
2.) If $Omega$ was some arbitrary orthonormal basis $e_m_m in mathbbZ$ for some space $L_2[a,b]$, would this still be the way to go to extend the basis into an integral transform for all of $mathbbR$?
3.) If $Omega$ is no longer orthogonal, can I still do these two extensions (one with $U$ and one without)?
This is of course assuming we can actually compute the integrals.
Thanks!
fourier-analysis orthonormal integral-transforms change-of-basis
asked Aug 2 at 16:50
The Dude
15310
add a comment |Â
up vote
0
down vote
favorite
Background
It is well known that the orthonormal basis for $L_2[-pi, pi]$ is $Omega = e^-jmt _m in mathbbZ$. We extend this to $L_2(mathbbR)$ via the Fourier Transform, which is an integral transform of the form:
$$ f(omega) = displaystyleintlimits_-infty^infty f(t) K(omega, t) dt = displaystyleintlimits_-infty^infty f(t) e^-j omega t dt $$
I.e. where $K(omega, t) = e^-jomega t$ We know this kernel is unitary, so the extension to $L_2(mathbbR)$ is simply:
$$ f(t) = displaystyleintlimits_-infty^infty f(omega) K^ast(omega, t) domega = displaystyleintlimits_-infty^infty f(omega) e^j omega t domega $$
Extension
Now suppose I apply some unitary transform $U$ to $Omega$, giving us the new orthonormal basis for $L_2[-pi, pi]$ denoted by $Omega_U = Ue^-jmt _m in mathbbZ$. I want to extend this to $L_2(mathbbR)$ in the same way. So this would mean:
$$ f(omega) = displaystyleintlimits_-infty^infty f(t) K_U(omega, t) dt = displaystyleintlimits_-infty^infty f(t) big(Ue^-j omega tbig) dt $$
Which means the extension to $L_2(mathbbR)$ takes the form:
$$ f(t) = displaystyleintlimits_-infty^infty f(omega) K_U^ast(omega, t) domega = displaystyleintlimits_-infty^infty f(omega) big(Ue^-j omega tbig)^ast domega = displaystyleintlimits_-infty^infty f(omega) big(U^-1e^j omega tbig) domega$$
In other words:
$K_U(omega, t) = Ue^-jomega t$ and $K_U^ast(omega, t) = U^-1e^jomega t$
Questions
1.) Is this the correct way to build an integral transform from an orthonormal basis? If not, where did I go wrong?
2.) If $Omega$ was some arbitrary orthonormal basis $e_m_m in mathbbZ$ for some space $L_2[a,b]$, would this still be the way to go to extend the basis into an integral transform for all of $mathbbR$?
3.) If $Omega$ is no longer orthogonal, can I still do these two extensions (one with $U$ and one without)?
This is of course assuming we can actually compute the integrals.
Thanks!
fourier-analysis orthonormal integral-transforms change-of-basis
asked Aug 2 at 16:50
The Dude
15310
The individual exponentials are not in $L^2$. What meaning does unitary transformation have when you are applying it to objects that are not in the space?
– DisintegratingByParts
Aug 3 at 13:20
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Background
It is well known that the orthonormal basis for $L_2[-pi, pi]$ is $Omega = e^-jmt _m in mathbbZ$. We extend this to $L_2(mathbbR)$ via the Fourier Transform, which is an integral transform of the form:
$$ f(omega) = displaystyleintlimits_-infty^infty f(t) K(omega, t) dt = displaystyleintlimits_-infty^infty f(t) e^-j omega t dt $$
I.e. where $K(omega, t) = e^-jomega t$ We know this kernel is unitary, so the extension to $L_2(mathbbR)$ is simply:
$$ f(t) = displaystyleintlimits_-infty^infty f(omega) K^ast(omega, t) domega = displaystyleintlimits_-infty^infty f(omega) e^j omega t domega $$
Extension
Now suppose I apply some unitary transform $U$ to $Omega$, giving us the new orthonormal basis for $L_2[-pi, pi]$ denoted by $Omega_U = Ue^-jmt _m in mathbbZ$. I want to extend this to $L_2(mathbbR)$ in the same way. So this would mean:
$$ f(omega) = displaystyleintlimits_-infty^infty f(t) K_U(omega, t) dt = displaystyleintlimits_-infty^infty f(t) big(Ue^-j omega tbig) dt $$
Which means the extension to $L_2(mathbbR)$ takes the form:
$$ f(t) = displaystyleintlimits_-infty^infty f(omega) K_U^ast(omega, t) domega = displaystyleintlimits_-infty^infty f(omega) big(Ue^-j omega tbig)^ast domega = displaystyleintlimits_-infty^infty f(omega) big(U^-1e^j omega tbig) domega$$
In other words:
$K_U(omega, t) = Ue^-jomega t$ and $K_U^ast(omega, t) = U^-1e^jomega t$
Questions
1.) Is this the correct way to build an integral transform from an orthonormal basis? If not, where did I go wrong?
2.) If $Omega$ was some arbitrary orthonormal basis $e_m_m in mathbbZ$ for some space $L_2[a,b]$, would this still be the way to go to extend the basis into an integral transform for all of $mathbbR$?
3.) If $Omega$ is no longer orthogonal, can I still do these two extensions (one with $U$ and one without)?
This is of course assuming we can actually compute the integrals.
Thanks!
fourier-analysis orthonormal integral-transforms change-of-basis
asked Aug 2 at 16:50
The Dude
15310
Background
It is well known that the orthonormal basis for $L_2[-pi, pi]$ is $Omega = e^-jmt _m in mathbbZ$. We extend this to $L_2(mathbbR)$ via the Fourier Transform, which is an integral transform of the form:
$$ f(omega) = displaystyleintlimits_-infty^infty f(t) K(omega, t) dt = displaystyleintlimits_-infty^infty f(t) e^-j omega t dt $$
I.e. where $K(omega, t) = e^-jomega t$ We know this kernel is unitary, so the extension to $L_2(mathbbR)$ is simply:
$$ f(t) = displaystyleintlimits_-infty^infty f(omega) K^ast(omega, t) domega = displaystyleintlimits_-infty^infty f(omega) e^j omega t domega $$
Extension
Now suppose I apply some unitary transform $U$ to $Omega$, giving us the new orthonormal basis for $L_2[-pi, pi]$ denoted by $Omega_U = Ue^-jmt _m in mathbbZ$. I want to extend this to $L_2(mathbbR)$ in the same way. So this would mean:
$$ f(omega) = displaystyleintlimits_-infty^infty f(t) K_U(omega, t) dt = displaystyleintlimits_-infty^infty f(t) big(Ue^-j omega tbig) dt $$
Which means the extension to $L_2(mathbbR)$ takes the form:
$$ f(t) = displaystyleintlimits_-infty^infty f(omega) K_U^ast(omega, t) domega = displaystyleintlimits_-infty^infty f(omega) big(Ue^-j omega tbig)^ast domega = displaystyleintlimits_-infty^infty f(omega) big(U^-1e^j omega tbig) domega$$
In other words:
$K_U(omega, t) = Ue^-jomega t$ and $K_U^ast(omega, t) = U^-1e^jomega t$
Questions
1.) Is this the correct way to build an integral transform from an orthonormal basis? If not, where did I go wrong?
2.) If $Omega$ was some arbitrary orthonormal basis $e_m_m in mathbbZ$ for some space $L_2[a,b]$, would this still be the way to go to extend the basis into an integral transform for all of $mathbbR$?
3.) If $Omega$ is no longer orthogonal, can I still do these two extensions (one with $U$ and one without)?
This is of course assuming we can actually compute the integrals.
Thanks!
fourier-analysis orthonormal integral-transforms change-of-basis
asked Aug 2 at 16:50
The Dude
15310
asked Aug 2 at 16:50
The Dude
15310
asked Aug 2 at 16:50
The Dude
15310
asked Aug 2 at 16:50
The Dude
15310
15310
The individual exponentials are not in $L^2$. What meaning does unitary transformation have when you are applying it to objects that are not in the space?
– DisintegratingByParts
Aug 3 at 13:20
add a comment |Â
The individual exponentials are not in $L^2$. What meaning does unitary transformation have when you are applying it to objects that are not in the space?
– DisintegratingByParts
Aug 3 at 13:20
The individual exponentials are not in $L^2$. What meaning does unitary transformation have when you are applying it to objects that are not in the space?
– DisintegratingByParts
Aug 3 at 13:20
The individual exponentials are not in $L^2$. What meaning does unitary transformation have when you are applying it to objects that are not in the space?
– DisintegratingByParts
Aug 3 at 13:20
add a comment |Â
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The individual exponentials are not in $L^2$. What meaning does unitary transformation have when you are applying it to objects that are not in the space?
– DisintegratingByParts
Aug 3 at 13:20