Extracting multiple envelopes from a signal

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First Last 2017-8-11

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编辑： Cedric 2017-10-20

I have a signal which when observed carefully shows a three individual sinusoidal signals (figures attached). I want these three individual signals to be extracted in terms of amplitude, frequency and phase. Please help!

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John BG 2017-8-11

would you please be so kind to supply the signal?

in a .mat file

thanks in advance, awaiting answer

First Last 2017-8-11

MultipleSignalEnvelope.mat

Sure! Here is the .mat file. Thanks!

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Cedric 2017-10-17

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Quick cheap trick gets 2/3. Was going to bed, no time for more but that could be a starting point.

 function main
    load( 'MultipleSignalEnvelope', 'ITDc0' ) ;
    plot( ITDc0, 'b.' ) ;
    grid on ;
    hold on ;
    t = 1 : numel( ITDc0 ) ;
    A = 1.2 ;
    P = fminsearch( @objective, zeros(3,1) ) ;
    plot( A*sin((t+P)/360*2*pi).' ) ;
    function mindif = objective( P )
        mindif = sum( min( abs( ITDc0-A*sin((t+P)./360*2*pi )))) ;
    end
 end

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Cedric 2017-10-20

编辑：Cedric 2017-10-20

在 MATLAB Online 中打开

You are probably using a version of MATLAB older than 2016b. From 2016b on, MATLAB supports automatic expansion. We can check this quickly by evaluating:

 >> [1;2;3] .* [10,20,30,40]
 ans =
    10    20    30    40
    20    40    60    80
    30    60    90   120

On your computer it should generate an error(?) What happens here is that MATLAB expands automatically the operands "along the dimension of each other" in a fashion that allows the operation. I'm not wording that too well, but you can understand looking at the output, which is what you would get with:

1 1 1               10 20 30 40
2 2 2      .*       10 20 30 40 
3 3 3               10 20 30 40

On former versions of MATLAB, we had/have to use BSXFUN for this ( an explicit call to a function that does an implicit expansion ;) ). I cannot test, but if you replace t+p by bsxfun( @plus, t, P ), that should solve a first expansion issue, and then same thing for ITDc0-A.... So overall you should replace:

ITDc0-A*sin((t+P)./360*2*pi)

bsxfun( @minus, ITDc0, A*sin( bsxfun( @plus, t, P)./360*2*pi )

Given my first explanation and what you understand of the approach, you probably understand that P is a vertical vector of three "phases" (term that I am loosing loosely here because there are multiplicative factors), t+P (now built using BSXFUN) is an array of three horizontal time vectors stacked vectically that will "de-phase" the sine, ITDc0-A.. a vector of differences between your signal and the three sines, and the FMIN call is aiming at finding the three phases that minimizes the min absolute difference between the signal and the three sines.

Now this is not a good approach; it has nothing to do with digital signal processing, it is just minimizing something that is not even that smart because it fails 1/3. But it was a fun attempt for the 5 minutes that I had.

First Last 2017-10-20

Do you think we can use the same logic for Amplitude too?

Cedric 2017-10-20

编辑：Cedric 2017-10-20

You will have to try. Define P as a vector with 6 components instead of 3, the first three being amplitudes and the next three being "phases". In the objective function, replace A with P(1:3) and P with P(4:6). Also, update the initial conditions. Instead of zeros(3,1), try to start e.g.. with three unit amplitudes and three null phases: [ones(3,1);zeros(3,1)].

Note that you could use FMINCON and define constraints, e.g. upper and lower bounds for the amplitude and phases.

Also note that you'd better try to see if you could get help from someone with experience in digital signal processing, because this approach is just a quick trick. If you reword your question in a new thread and nobody answers, you may be able to hire someone for not that much on sites that market freelancers.

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