Convert Real and Imaginary part from FFT to bandsreal and imaginary part in $sin z$ where z is complexCalculating the real and imaginary parts of a holomorphic functionReal and imaginary part from trigonometric formReal and imaginary part functionsFinding the complex function from its real and imaginary partWhat does the imaginary part of a set of polar coordinates represent before and after funning an FFT?Real part and imaginary part of an expression under sqrtReal and imaginary partFind Imaginary and real partProve the even part of a complex signal can has even real part and odd imaginary part
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Convert Real and Imaginary part from FFT to bands
real and imaginary part in $sin z$ where z is complexCalculating the real and imaginary parts of a holomorphic functionReal and imaginary part from trigonometric formReal and imaginary part functionsFinding the complex function from its real and imaginary partWhat does the imaginary part of a set of polar coordinates represent before and after funning an FFT?Real part and imaginary part of an expression under sqrtReal and imaginary partFind Imaginary and real partProve the even part of a complex signal can has even real part and odd imaginary part
$begingroup$
how can I convert real and imaginary part from FFT to octave bands? I know how to do it for Absolute value, but dont know for Re, Im parts.
Thanks
complex-analysis complex-numbers fast-fourier-transform octave
asked Mar 17 at 13:28
user148733user148733
1
$endgroup$
add a comment |
$begingroup$
how can I convert real and imaginary part from FFT to octave bands? I know how to do it for Absolute value, but dont know for Re, Im parts.
Thanks
complex-analysis complex-numbers fast-fourier-transform octave
asked Mar 17 at 13:28
user148733user148733
1
$endgroup$
$begingroup$
Pick any functions $h_m : 0ldots N-1 to [0,1]$ such that $sum_m=1^M H_m(k) = 1$ then $x= sum_m=1^M y_m$ where $y_m = FFT^-1[Y_m], Y_m(k) =H_m(k) X(k), X = FFT[x]$. The octaves bands are obtained when $H_m+1(k) approx 1_[2^m,2^m+1]+1_N-[2^m,2^m+1]$
$endgroup$
– reuns
Mar 17 at 18:40
$begingroup$
what exactly means 1[2m,2m+1]+1N−[2m,2m+1]?
$endgroup$
– user148733
Mar 22 at 13:04
$begingroup$
And do you mena hm=Hm?
$endgroup$
– user148733
Mar 22 at 13:10
add a comment |
$begingroup$
how can I convert real and imaginary part from FFT to octave bands? I know how to do it for Absolute value, but dont know for Re, Im parts.
Thanks
complex-analysis complex-numbers fast-fourier-transform octave
asked Mar 17 at 13:28
user148733user148733
1
$endgroup$
how can I convert real and imaginary part from FFT to octave bands? I know how to do it for Absolute value, but dont know for Re, Im parts.
Thanks
complex-analysis complex-numbers fast-fourier-transform octave
complex-analysis complex-numbers fast-fourier-transform octave
asked Mar 17 at 13:28
user148733user148733
1
asked Mar 17 at 13:28
user148733user148733
1
asked Mar 17 at 13:28
user148733user148733
1
asked Mar 17 at 13:28
user148733user148733
1
asked Mar 17 at 13:28
user148733user148733
1
1
$begingroup$
Pick any functions $h_m : 0ldots N-1 to [0,1]$ such that $sum_m=1^M H_m(k) = 1$ then $x= sum_m=1^M y_m$ where $y_m = FFT^-1[Y_m], Y_m(k) =H_m(k) X(k), X = FFT[x]$. The octaves bands are obtained when $H_m+1(k) approx 1_[2^m,2^m+1]+1_N-[2^m,2^m+1]$
$endgroup$
– reuns
Mar 17 at 18:40
$begingroup$
what exactly means 1[2m,2m+1]+1N−[2m,2m+1]?
$endgroup$
– user148733
Mar 22 at 13:04
$begingroup$
And do you mena hm=Hm?
$endgroup$
– user148733
Mar 22 at 13:10
add a comment |
$begingroup$
Pick any functions $h_m : 0ldots N-1 to [0,1]$ such that $sum_m=1^M H_m(k) = 1$ then $x= sum_m=1^M y_m$ where $y_m = FFT^-1[Y_m], Y_m(k) =H_m(k) X(k), X = FFT[x]$. The octaves bands are obtained when $H_m+1(k) approx 1_[2^m,2^m+1]+1_N-[2^m,2^m+1]$
$endgroup$
– reuns
Mar 17 at 18:40
$begingroup$
what exactly means 1[2m,2m+1]+1N−[2m,2m+1]?
$endgroup$
– user148733
Mar 22 at 13:04
$begingroup$
And do you mena hm=Hm?
$endgroup$
– user148733
Mar 22 at 13:10
$begingroup$
Pick any functions $h_m : 0ldots N-1 to [0,1]$ such that $sum_m=1^M H_m(k) = 1$ then $x= sum_m=1^M y_m$ where $y_m = FFT^-1[Y_m], Y_m(k) =H_m(k) X(k), X = FFT[x]$. The octaves bands are obtained when $H_m+1(k) approx 1_[2^m,2^m+1]+1_N-[2^m,2^m+1]$
$endgroup$
– reuns
Mar 17 at 18:40
$begingroup$
Pick any functions $h_m : 0ldots N-1 to [0,1]$ such that $sum_m=1^M H_m(k) = 1$ then $x= sum_m=1^M y_m$ where $y_m = FFT^-1[Y_m], Y_m(k) =H_m(k) X(k), X = FFT[x]$. The octaves bands are obtained when $H_m+1(k) approx 1_[2^m,2^m+1]+1_N-[2^m,2^m+1]$
$endgroup$
– reuns
Mar 17 at 18:40
$begingroup$
what exactly means 1[2m,2m+1]+1N−[2m,2m+1]?
$endgroup$
– user148733
Mar 22 at 13:04
$begingroup$
what exactly means 1[2m,2m+1]+1N−[2m,2m+1]?
$endgroup$
– user148733
Mar 22 at 13:04
$begingroup$
And do you mena hm=Hm?
$endgroup$
– user148733
Mar 22 at 13:10
$begingroup$
And do you mena hm=Hm?
$endgroup$
– user148733
Mar 22 at 13:10
add a comment |
0
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$begingroup$
Pick any functions $h_m : 0ldots N-1 to [0,1]$ such that $sum_m=1^M H_m(k) = 1$ then $x= sum_m=1^M y_m$ where $y_m = FFT^-1[Y_m], Y_m(k) =H_m(k) X(k), X = FFT[x]$. The octaves bands are obtained when $H_m+1(k) approx 1_[2^m,2^m+1]+1_N-[2^m,2^m+1]$
$endgroup$
– reuns
Mar 17 at 18:40
$begingroup$
what exactly means 1[2m,2m+1]+1N−[2m,2m+1]?
$endgroup$
– user148733
Mar 22 at 13:04
$begingroup$
And do you mena hm=Hm?
$endgroup$
– user148733
Mar 22 at 13:10