1
$\begingroup$


I try to make a third-octave spectrum from a time signal and I have some troubles with the algorithm. I'm not an expert in signal processing so I don't know what is wrong.

Here what my algorithm looks like :

Time signal -> Z transform of the signal -> First octave band filtering -> Quadratic sum of the result's values => First octave band value.
Under factor 2 sampling ->Z transform of the signal -> Second octave band filtering -> Quadratic sum of the result's values =>Second octave band.

Now I have this, I try to test it on Matlab with a time signal about 2 millions points.

I have this code to calculate all the octave's bands :

    fc = fMin*2.^((0:bmax)*bw);             %Center frequencies.
    fl = fc*2^(-bw/2);                      %Lower cutoffs.
    fu = fc*2^(+bw/2);                      %Upper cutoffs.

And this one to calculate the Z transform of the time signal :

    x = [1 2 3];
    syms z;
    d = z.^(-1*(0:numel(x)-1));
    zTransform = sum(x.*d)

My main problem is, I'm not completely sure that my algorithm is right and if it is, how can I filter the z transform equation that the code above gives me with each third-octave band filters?

Thanks in advance, it's been a while I'm on this problem.

|improve this question
$\endgroup$
  • $\begingroup$ How is that different from the question you asked before? $\endgroup$ – jojek♦ Jun 1 '18 at 9:08
1
$\begingroup$

maybe either the bandpass filters or the peaking EQ filters in the audio EQ cookbook can help. bandwidth is defined in octaves in that. the output, $y$, of each 1/3 octave BPF filter should be squared and low-pass filtered to get the mean square, $\overline{y^2}$. and then

$$dB = 10 \log_{10}\Big(\,\overline{y^2}\,\Big)$$

is applied to that to get dB.

are you planning on doing this to the whole audible spectrum? 31 bands?

|improve this answer
$\endgroup$
  • $\begingroup$ Yes, I would like to filter the signal from 10Hz to 10kHz (maybe 20kHz later). In fact, the z transform calcul gives me an equation in output. I don't know what can I do with it. Is there not a better way, have a table in output ? It will be easy after to filter this $\endgroup$ – Pandaglas Jun 1 '18 at 7:52
  • $\begingroup$ if you have something that works, go with it. if it doesn't work or you don't understand what is necessary to make it work, you might need to try something else. i wouldn't bother with the bottom octave from 10 to 20 Hz. you should consider what the conventional 31 band frequencies are $\endgroup$ – robert bristow-johnson Jun 1 '18 at 8:02
  • $\begingroup$ Actually, this calcul works with few points. But when I try to insert a table with 2 millions points, my computer is out of memory... To filter a z transform spectrum, I need to have a table of datas in output, no ? $\endgroup$ – Pandaglas Jun 1 '18 at 8:11
  • $\begingroup$ What do you mean by "You should consider what the conventional 31 band frequencies are" ? $\endgroup$ – Pandaglas Jun 1 '18 at 8:11
  • $\begingroup$ you're doing a 1/3 Octave spectrum analyzer over 10 octaves, right? including both fenceposts, that is 30+1 bands. they have standard frequencies: ...100, 125, 160, 200, 250, 320, 400, 500, 630, 800, 1000, 1250, ... $\endgroup$ – robert bristow-johnson Jun 1 '18 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.