Matrix factorisation of the Fourier matrix
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I am currently reading a paper Low Communication FMM-Accelerated FFT on GPUs
In that I am not able to understand the definition of the twiddle factor matrix $T_{P, M}$.
The Fourier matrix $F_N$ is defined as:
$[F_N]_{i, j} = omega_N^{ij}$
According to the paper, the Fourier matrix is factorised as :
$ F_N = Pi_{M, P}(I_M otimes F_P) Pi_{P, M} T_{P, M} (I_P otimes F_M) Pi_{M, P}
$
Where $Pi$ is the block to cyclic operator.
The paper defines $T_{P,M}$ as the diagonal matrix of the twiddle factors.
Mathematically:
$[T_{P, M}]_{i, j} = delta_{i, j} omega_{N}^{(imod M) cdot lfloor i/M rfloor}
$
I feel that the above equation is wrong. Because if the matrix is diagonal, then $i < min(P, M)$ meaning that $lfloor i/M rfloor$ will always be 0. This would mean that the $T$ is a diagonal matrix of 1s only.
Can someone help me figure out the correct equation for $T$?
matrix-equations fourier-transform fast-fourier-transform
$endgroup$
add a comment |
$begingroup$
I am currently reading a paper Low Communication FMM-Accelerated FFT on GPUs
In that I am not able to understand the definition of the twiddle factor matrix $T_{P, M}$.
The Fourier matrix $F_N$ is defined as:
$[F_N]_{i, j} = omega_N^{ij}$
According to the paper, the Fourier matrix is factorised as :
$ F_N = Pi_{M, P}(I_M otimes F_P) Pi_{P, M} T_{P, M} (I_P otimes F_M) Pi_{M, P}
$
Where $Pi$ is the block to cyclic operator.
The paper defines $T_{P,M}$ as the diagonal matrix of the twiddle factors.
Mathematically:
$[T_{P, M}]_{i, j} = delta_{i, j} omega_{N}^{(imod M) cdot lfloor i/M rfloor}
$
I feel that the above equation is wrong. Because if the matrix is diagonal, then $i < min(P, M)$ meaning that $lfloor i/M rfloor$ will always be 0. This would mean that the $T$ is a diagonal matrix of 1s only.
Can someone help me figure out the correct equation for $T$?
matrix-equations fourier-transform fast-fourier-transform
$endgroup$
add a comment |
$begingroup$
I am currently reading a paper Low Communication FMM-Accelerated FFT on GPUs
In that I am not able to understand the definition of the twiddle factor matrix $T_{P, M}$.
The Fourier matrix $F_N$ is defined as:
$[F_N]_{i, j} = omega_N^{ij}$
According to the paper, the Fourier matrix is factorised as :
$ F_N = Pi_{M, P}(I_M otimes F_P) Pi_{P, M} T_{P, M} (I_P otimes F_M) Pi_{M, P}
$
Where $Pi$ is the block to cyclic operator.
The paper defines $T_{P,M}$ as the diagonal matrix of the twiddle factors.
Mathematically:
$[T_{P, M}]_{i, j} = delta_{i, j} omega_{N}^{(imod M) cdot lfloor i/M rfloor}
$
I feel that the above equation is wrong. Because if the matrix is diagonal, then $i < min(P, M)$ meaning that $lfloor i/M rfloor$ will always be 0. This would mean that the $T$ is a diagonal matrix of 1s only.
Can someone help me figure out the correct equation for $T$?
matrix-equations fourier-transform fast-fourier-transform
$endgroup$
I am currently reading a paper Low Communication FMM-Accelerated FFT on GPUs
In that I am not able to understand the definition of the twiddle factor matrix $T_{P, M}$.
The Fourier matrix $F_N$ is defined as:
$[F_N]_{i, j} = omega_N^{ij}$
According to the paper, the Fourier matrix is factorised as :
$ F_N = Pi_{M, P}(I_M otimes F_P) Pi_{P, M} T_{P, M} (I_P otimes F_M) Pi_{M, P}
$
Where $Pi$ is the block to cyclic operator.
The paper defines $T_{P,M}$ as the diagonal matrix of the twiddle factors.
Mathematically:
$[T_{P, M}]_{i, j} = delta_{i, j} omega_{N}^{(imod M) cdot lfloor i/M rfloor}
$
I feel that the above equation is wrong. Because if the matrix is diagonal, then $i < min(P, M)$ meaning that $lfloor i/M rfloor$ will always be 0. This would mean that the $T$ is a diagonal matrix of 1s only.
Can someone help me figure out the correct equation for $T$?
matrix-equations fourier-transform fast-fourier-transform
matrix-equations fourier-transform fast-fourier-transform
edited Feb 3 at 0:13
Keith McClary
8481412
8481412
asked Feb 2 at 16:00
97amarnathk97amarnathk
1014
1014
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