[SciPy-User] finding frequency of wav

[Linda]

Thanks for your reply. The explanation on fftfreq already made a few puzzle
pieces fall into place.

The signal I am trying to decode is a DSC transmission that is recorded in a
wav file. (Digital Selective Calling, used in marine radio)
It is a phase modulated digital signal: '1' is 2100Hz, '0' is 1300 Hz and
there's a carrier at 1700Hz. That should be all frequencies involved (apart
from noise). Currently I am used generated, clean signals. But probably I
should get a clean '10101010'-signal first to try my work on.

Since the bitrate is set at 1200bits/sec, the bit length would be
samplerate/1200 = 18.4 samples at 22050.
I can double the samplerate to 44100, but that still leaves me at only 36.8
samples per chunk.
If I understand what you say correctly, I would need at least 55 (64)
samples in each chunk?

I'm not sure what chunk[3] would have been, I should have used a
dotting-signal instead of an unknown message to try this on. I will try this
again with more useful data this afternoon.

cheers,
Linda


On Thu, May 27, 2010 at 15:28, Fabrice Silva <silva at lma.cnrs-mrs.fr> wrote:

> Le jeudi 27 mai 2010 à 10:09 +0200, Linda a écrit :
> > Hello all,
>
> > I have a digital signal where the bits in it are encoded with
> > frequencies 1300 and 2100 Hz. The message is sent as a wav-file with a
> > samplerate of 22050.
> > My goal is to find the bits again so I can decode the message in it,
> > for that I have chopped the wav up in pieces of 18 samples, which
> > would be the bitlength (at 1200 Bit/s > 22050/1200=18.375). So I have
> > a list of chunks of length 18. I thought I could just fft each chunks
> > and find the max of the chunk-spectrum, to find out the bitfrequency
> > in the chunk (and thus the bitvalue)
>
> Correct me if I am wrong. You are cutting your signal into chunks that
> you expect to contain at least one period of the lower coding frequency.
> You then perform a fft on a very small signal (18 samples) which gives
> you (without zero padding) an estimation of the Fourier transform of
> your chunk computed on only 18 frequencies, i.e. with a really bad
> frequential resolution. It is possible if your coding frequencies are
> not too close. A raw Rayleight criteria leads to cut your signal into at
> least N=2*Fe/df_min where df_min is the minimal spacing between two
> coding frequencies df_min=2100-1300 here thus N=55 (so 64 to have a
> power of 2).
> >
> > But somehow I am stuck in the numbers, I was hopeing you could give me
> > a hint. here is what I have:
>
> > chunks[3] #this is one of the wavchunks, there should be a bit hidden in
> here
> > Out[98]:
> >   array([ 2, -1,  1, -2,  2, -2,  2, -1,  0,  0,  0,  1, -2,  2, -1,  0,
>  0,  0], dtype=int16)
> > test = fft(chunks[3]) # spectrum of the chunk, the peak should give me
> the value of the bitfrequency 1300 of 2100 Hz?
> > test
> > Out[100]:
> > array([ 1.00000000 +0.00000000e+00j,  1.00000000 +2.37564698e-01j,
> >         1.46791111 +4.90375770e-01j,  2.50000000 +8.66025404e-01j,
> >         2.65270364 -7.37891832e-01j,  1.00000000 +3.01762603e+00j,
> >        -0.50000000 -2.59807621e+00j,  1.00000000 -2.41609109e+00j,
> >         4.87938524 +1.43601897e+01j,  7.00000000 -6.88706904e-15j,
> >         4.87938524 -1.43601897e+01j,  1.00000000 +2.41609109e+00j,
> >        -0.50000000 +2.59807621e+00j,  1.00000000 -3.01762603e+00j,
> >         2.65270364 +7.37891832e-01j,  2.50000000 -8.66025404e-01j,
> >         1.46791111 -4.90375770e-01j,  1.00000000 -2.37564698e-01j])
> >
> >
> > I am unsure how to proceed from here, so I would really appreciate any
> > tips.. I found fftfreq, but I am not sure how to use it? I read
> > fftfreq? but I don't see how the example even uses the 'fourier'
> > variable in the fftfreq there?
> >
> Fftfreq is a function that constructs the frequency vector associated to
> the data computed by the fft algorithm. It is aware of how fft orders
> the frequency bins, and transform it in a more convenient way (it
> 'anti-aliases', centering the results on zero frequency).
>
> import numpy as np
> import matplotlib.pyplot as plt
> chunks[3]=....
> test = np.fft.fft(chunks[3])
> frequencies = np.fft.fftfreq(len(test), d=1./22050.) # d is the sampling
> period
> plt.plot(frequencies, np.abs(test), 'o')
> plt.show()
>
> but you won't see any things on this fft. I am suspicious due to the
> fact that the signal to noise ratio seems rather low leading to strong
> peak at Fe/2
> In chunk[3], what do you expect to be the bit?
>
> Fabricio
>
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