How to reconstruct a sound with data in MATLAB?

Hi all! I have in matlab a vector with different pressures (2501 exactly) for one frequency (1856hz) that have been recorded with a microphone. Thus, I'm asking if I can go back and reconstruct the sound so I can hear it. Someone suggested to me to do it with Soudforge, but I don't know how to pass from matlab to soundforge. Can you help me, please? Thank you !

Answer 1

[Jan]

What format do your pressure values have? Assuming they have a range from -N to N (e.g. "max(abs(pressure))")

   sound(pressure / N, 1856)

[Daniel Shub]

I doubt your soundcard can handle a 1856 sample rate properly.

[Lucio]

I have to tray anyway !

[Daniel Shub]

If the pressure waveform has a large DC offset scaling by N will not work so well,

[Jan]

@Daniel: Correct. If there is any DC offset, it must be removed. I've associated a offset free signal according to the term 'pressure', which has no DC by definition.

@Lucio: If the sound card have problems, use RESAMPLE to onvert it to a standard frequency like 8000 or 8192Hz.

Answer 2

[Sean de Wolski]

like

doc sound %?

be more specific what you mean by "reconstruct".

[Lucio]

Hear the sound.

Answer 3

[Vieniava]

Try this:

fs=8000;
soundsc( pressure-mean(pressure), fs)

You should know sampling frequency used in data collecting - here I assumed typical 8kHz.

[Lucio]

And my pressure data where should I put it? And can I hear the sound on matlab?

[Jan]

Your pressure data are expected to be the values of the variable called "pressure". Perhaps you forgot to mention the format your pressure data??

You cannot hear the sound "on Matlab", but in the speakers of headphone of your computer. Just try it.

Answer 4

[Daniel Shub]

It is not totally clear what you are starting with: Assuming x is your pressure waveform and is an array 2501x1 samples and the samples were acquired at a rate of 1856 samples per second. If the sample rate is in fact 1856, then unless your soundcard is connected to a lowpass filter, you will get all sorts of audible aliasing if you present the sound with a sample rate of 1856. The solution is to upsample the pressure waveform and present it at a higher sample rate. To do this you can do:

 y = resample(x, round(44.1e3/1856), 1);

to get a signal with a sample rate of round(44.1e3/1856)*1856. To then present the sound with an arbitrary level you can do:

soundsc(y-mean(y), round(44.1e3/1856)*1856);

You need to remove the mean value for the scaling to work properly. Some microphones have large bias voltages and this will swamp the small deviations in pressure/voltage that you want to hear.

To get the actual level you need to know the sensitivity of your microphone, the maximum output voltage of your soundcard, the sensitivity of your transducer (headphones/speakers), and the gain of any amplifiers or attenuators.

[Lucio]

Thank you for your answer ! :)

No, the 1856hz is not the sampling frequency which is in fact 133hz, it is the frequency of the signal. The microphone recorded different frequencies (from 1 to 2051), and for each frequency, we got 2501 pressures. Here we are talking about the 1856hz frequency which we are studying in the lab.

Thus, we have for the 1856hz an array of 2501*1 like you said.

Hope it's more clear.

[Daniel Shub]

Unless you are studying the Nyquist theorem in your lab, then sampling a 1856 Hz signal at 133 Hz does not make any sense. You need to sample at at least twice the frequency of the highest frequency signal.

[Lucio]

Yes I know that, but I didn't the recording, another team did it, and I only have the data, and they said they recorded it with a sample frequency of 133hz. Or maybe I can choose another frequency under that so the rendrering could be better, but I still don't know how to do it, event theorically :(

Answer 5

[Daniel Shub]

It appears that the sampling rate is 133 Hz and the signal frequency is 1856 Hz. This means the signal will be aliased to a frequency less than 66.5 Hz. Unless you have a very good transducer (e.g., headphones or speaker), you will not be able to hear this signal. A better approach would be to use the signal to amplitude modulate a higher frequency carrier. Assuming your pressure signal is "x."

fs = round(44.1e3/133)*133;

x = x-mean(x);
x = x./sqrt(sum(x.^2));
x = resample(x, round(44.1e3/133), 1);

soundsc(x.*sin(2*pi*8e3*((0:(length(x)-1))/fs)), fs);

[Lucio]

Oh thank you, I'll tray it and let you know the results ;)