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brandy
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Finite Fourier Transform on a 2d wave
How does the finite Fourier transform work exactly?
The transform of f(x) is
[itex]\widetilde{f}(\lambda_{n})[/itex] =[itex]\int[/itex][itex]^{L}_{0}[/itex] f(x) X[itex]_{n}[/itex] dx
If I had a 3d wave equation pde and I applied Finite Fourier transform on the pde for
z(x,y,t)=X(x)Y(y)T(t)
f(x)=z[itex]_{xx}[/itex]
[itex]\tilde{z_{xx}(\lambda_{n})}[/itex] =[itex]\int[/itex][itex]^{L}_{0}[/itex] z[itex]_{xx}[/itex] X[itex]_{n}[/itex](x) dx
f(x)=z[itex]_{yy}[/itex]
[itex]\tilde{z_{yy}(\lambda_{n})}[/itex] =[itex]\int[/itex][itex]^{L}_{0}[/itex] z[itex]_{yy}[/itex] Y[itex]_{n}[/itex](y) dy
etc for T
or
is the transform the same for everything, namely:
f(x)=z[itex]_{xx}[/itex]
[itex]\tilde{z_{xx}(\lambda_{n})}[/itex] =[itex]\int[/itex][itex]^{L}_{0}[/itex] z[itex]_{xx}[/itex] X[itex]_{n}[/itex](x) Y[itex]_{n}[/itex](y) dx
or is it a double integral?
or do you apply finite Fourier twice with respect to y and x.
How does the finite Fourier transform work exactly?
The transform of f(x) is
[itex]\widetilde{f}(\lambda_{n})[/itex] =[itex]\int[/itex][itex]^{L}_{0}[/itex] f(x) X[itex]_{n}[/itex] dx
If I had a 3d wave equation pde and I applied Finite Fourier transform on the pde for
z(x,y,t)=X(x)Y(y)T(t)
f(x)=z[itex]_{xx}[/itex]
[itex]\tilde{z_{xx}(\lambda_{n})}[/itex] =[itex]\int[/itex][itex]^{L}_{0}[/itex] z[itex]_{xx}[/itex] X[itex]_{n}[/itex](x) dx
f(x)=z[itex]_{yy}[/itex]
[itex]\tilde{z_{yy}(\lambda_{n})}[/itex] =[itex]\int[/itex][itex]^{L}_{0}[/itex] z[itex]_{yy}[/itex] Y[itex]_{n}[/itex](y) dy
etc for T
or
is the transform the same for everything, namely:
f(x)=z[itex]_{xx}[/itex]
[itex]\tilde{z_{xx}(\lambda_{n})}[/itex] =[itex]\int[/itex][itex]^{L}_{0}[/itex] z[itex]_{xx}[/itex] X[itex]_{n}[/itex](x) Y[itex]_{n}[/itex](y) dx
or is it a double integral?
or do you apply finite Fourier twice with respect to y and x.
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