Re: N step fft in D language
On Sunday, 15 September 2013 at 20:58:54 UTC, Kadir Erdem Demir
wrote:
On Sunday, 15 September 2013 at 15:39:14 UTC, John Colvin wrote:
On Sunday, 15 September 2013 at 15:15:28 UTC, Kadir Erdem
Demir wrote:
I am using fft function from std.numeric
Complex!double[] resultfft = fft(timeDomainAmplitudeVal);
The parameter timeDomainAmplitudeVal is audio amplitude data.
Sample rate 44100 hz and there is 131072(2^16) samples
I am seeing that resultfft has the same size as
timeDomainAmplitudeVal(131072) which does not fits my
project(also makes no sense).
That's what the FFT does. See here:
http://stackoverflow.com/questions/4364823/how-to-get-frequency-from-fft-result
I believe I am well aware of the things which are explained in
the link. There is a sentence in link which says : The first
bin in the FFT is DC (0 Hz), the second bin is Fs / N, where Fs
is the sample rate and N is the size of the FFT.
My question how can I determine the N which is the size of
FFT ?
In fftw library one can define N like :
fftw_create_plan(N, FFTW_FORWARD, FFTW_ESTIMATE);
In D do we have a way to do that ?
Hi!
Haven't tried this, but looking at the docs it looks like you
have to:
auto mySample = getSampleData();
auto fftData0 = fft(mySample[0..512]);
auto fftData1 = fft(mySample[512..1024]);
...
This would give you the frequency correlation data for two
windows in time, and by reading that stackoverflow I assume for
256 frequencies each.
Docs also says there's a class you can reuse if you know the max
size, which should be faster.
auto ffter = new Fft(512);
foreach(chunk; getSampleByChunk(512))
{
auto fftData = ffter.fft(chunk);
...
}
The FFT size, N, is the same as the number of samples you
provide it. So the more data you provide, the finer correlation
detail you get. I think it makes sense?
Hmm, what it seems you have done is pass the entire sample
though, which gives you an FFT of size 131072. So you get
phenomenally good detail in frequency correlation, but just one
huge time window—the entire sample! I think what you're simply
looking for is chopping it up into smaller pieces and run fft on
each—that's at least what most audio tools do I believe.
kind regards
k
Re: N step fft in D language
I think you are not aswering his question (but maybe I am wrong). If you want a Fourier transform with less frequencies than temporal samples you can perform a fft to get a result of same length like this : 9 2 7 6 1 8 (amplitude) 0 2 4 6 8 10 (frequency) Then transform it like this : 11 13 9 (amplitude) (integrate the amplitudes) 1 5 9 (frequency) (according to your frequencies) This is a trivial example...I hope it's clear enougth...
N step fft in D language
I am using fft function from std.numeric Complex!double[] resultfft = fft(timeDomainAmplitudeVal); The parameter timeDomainAmplitudeVal is audio amplitude data. Sample rate 44100 hz and there is 131072(2^16) samples I am seeing that resultfft has the same size as timeDomainAmplitudeVal(131072) which does not fits my project(also makes no sense) . I need to be able to divide FFT to N equally spaced frequencies. And I need this N to be defined by me . Is there anyway to implement this with std.numeric.fft or can you have any advices for fft library? Ps: I will be glad to hear if some DSP libraries exist also
Re: N step fft in D language
On Sunday, 15 September 2013 at 15:15:28 UTC, Kadir Erdem Demir wrote: I am using fft function from std.numeric Complex!double[] resultfft = fft(timeDomainAmplitudeVal); The parameter timeDomainAmplitudeVal is audio amplitude data. Sample rate 44100 hz and there is 131072(2^16) samples I am seeing that resultfft has the same size as timeDomainAmplitudeVal(131072) which does not fits my project(also makes no sense). That's what the FFT does. See here: http://stackoverflow.com/questions/4364823/how-to-get-frequency-from-fft-result
Re: N step fft in D language
On Sunday, 15 September 2013 at 15:15:28 UTC, Kadir Erdem Demir wrote: I am using fft function from std.numeric Complex!double[] resultfft = fft(timeDomainAmplitudeVal); The parameter timeDomainAmplitudeVal is audio amplitude data. Sample rate 44100 hz and there is 131072(2^16) samples I am seeing that resultfft has the same size as timeDomainAmplitudeVal(131072) which does not fits my project(also makes no sense) . I need to be able to divide FFT to N equally spaced frequencies. And I need this N to be defined by me . Is there anyway to implement this with std.numeric.fft or can you have any advices for fft library? Ps: I will be glad to hear if some DSP libraries exist also When you perform a classic DFT the size of the resulting vector is the same since it is reversible (square matrix see http://en.wikipedia.org/wiki/DFT_matrix). If you want to get a larger vector, it is useless see Nyquist–Shannon sampling theorem. Otherwise I don't know any library which provide this kind of fft (with the possibility to made the matrix rectangular). You got 2 option : - taking the good samples in the produced vector. (complexity O(nlog n)) - shrinking your vector first (O(n/k log (n/k))
Re: N step fft in D language
On Sunday, 15 September 2013 at 15:39:14 UTC, John Colvin wrote: On Sunday, 15 September 2013 at 15:15:28 UTC, Kadir Erdem Demir wrote: I am using fft function from std.numeric Complex!double[] resultfft = fft(timeDomainAmplitudeVal); The parameter timeDomainAmplitudeVal is audio amplitude data. Sample rate 44100 hz and there is 131072(2^16) samples I am seeing that resultfft has the same size as timeDomainAmplitudeVal(131072) which does not fits my project(also makes no sense). That's what the FFT does. See here: http://stackoverflow.com/questions/4364823/how-to-get-frequency-from-fft-result I believe I am well aware of the things which are explained in the link. There is a sentence in link which says : The first bin in the FFT is DC (0 Hz), the second bin is Fs / N, where Fs is the sample rate and N is the size of the FFT. My question how can I determine the N which is the size of FFT ? In fftw library one can define N like : fftw_create_plan(N, FFTW_FORWARD, FFTW_ESTIMATE); In D do we have a way to do that ?
Re: N step fft in D language
On Sunday, 15 September 2013 at 20:58:54 UTC, Kadir Erdem Demir wrote: On Sunday, 15 September 2013 at 15:39:14 UTC, John Colvin wrote: On Sunday, 15 September 2013 at 15:15:28 UTC, Kadir Erdem Demir wrote: I am using fft function from std.numeric Complex!double[] resultfft = fft(timeDomainAmplitudeVal); The parameter timeDomainAmplitudeVal is audio amplitude data. Sample rate 44100 hz and there is 131072(2^16) samples I am seeing that resultfft has the same size as timeDomainAmplitudeVal(131072) which does not fits my project(also makes no sense). That's what the FFT does. See here: http://stackoverflow.com/questions/4364823/how-to-get-frequency-from-fft-result I believe I am well aware of the things which are explained in the link. There is a sentence in link which says : The first bin in the FFT is DC (0 Hz), the second bin is Fs / N, where Fs is the sample rate and N is the size of the FFT. My question how can I determine the N which is the size of FFT ? In fftw library one can define N like : fftw_create_plan(N, FFTW_FORWARD, FFTW_ESTIMATE); In D do we have a way to do that ? The only time i've ever come across an fft size is as the window width for a spectrogram/rolling FFT (such as one might use for a real-time fft filter in audio processing). Is that what you're looking for?
Re: N step fft in D language
On Sunday, 15 September 2013 at 20:58:54 UTC, Kadir Erdem Demir
wrote:
On Sunday, 15 September 2013 at 15:39:14 UTC, John Colvin wrote:
On Sunday, 15 September 2013 at 15:15:28 UTC, Kadir Erdem
Demir wrote:
I am using fft function from std.numeric
Complex!double[] resultfft = fft(timeDomainAmplitudeVal);
The parameter timeDomainAmplitudeVal is audio amplitude data.
Sample rate 44100 hz and there is 131072(2^16) samples
I am seeing that resultfft has the same size as
timeDomainAmplitudeVal(131072) which does not fits my
project(also makes no sense).
That's what the FFT does. See here:
http://stackoverflow.com/questions/4364823/how-to-get-frequency-from-fft-result
I believe I am well aware of the things which are explained in
the link. There is a sentence in link which says : The first
bin in the FFT is DC (0 Hz), the second bin is Fs / N, where Fs
is the sample rate and N is the size of the FFT.
My question how can I determine the N which is the size of
FFT ?
In fftw library one can define N like :
fftw_create_plan(N, FFTW_FORWARD, FFTW_ESTIMATE);
In D do we have a way to do that ?
Others are correct, the FFT result is always the same length as
the input and the 'N' is the number of samples you have to
transform. See this page:
http://www.fftw.org/fftw2_doc/fftw_2.html
and the example from it:
#include fftw.h
...
{
fftw_complex in[N], out[N]; // == N = number of samples
fftw_plan p;
...
p = fftw_create_plan(N, FFTW_FORWARD, FFTW_ESTIMATE);
...
fftw_one(p, in, out);
...
fftw_destroy_plan(p);
}
As you're probably aware, the FFT result will be symmetrical
around frequency N/2. However, the result array is not centred
around elelment N/2 but rather the symmetry is at each end,
wrapping from N back to 0. Like this:
||
|| ||
||||||
You need to shift it to get it centred around N/2
|
|||
..|..
Cheers,
G.
