How to avoid symbolic function when working with results files

The data_temp structure stores all the results for multiple probes for multiple point.

Say we have N probes and M point the structure would be data_temp(N).results(2:1:M)

My goal is to evaluate the integral over one periode of the PF_interface function which is built using multiple results values for all given point.

I could'nt find a way to do it without the symbolic expression but it takes ages to computes which does'nt work when M>1000, thus I'm quite sure it is not the optimal solution. Is there another way around ?

If not I already tried the following:

• going from exact fraction numbers to decimal value (hence de vpa) didn't do anything
• without the m for loop using matrix computation and replacing m by : in my calls, it did not change anything.

The code runs "fast" for x iteration and then slows down by a factor of 100 to 1000x without explanation

Any other way to optimise it ?

I'm using 2013b version for compatibility issues, so no vpaintegral function available for example.

Thanks for your help

syms func_t(t,amp,phase)
func_t(t,amp,phase)=amp*cos(t+phase);

PF_RMS=zeros(1,length(data_temp(1).resultat));
for m=1:length(data_temp(1).resultat)
    tic
    PF_interface = 0;
    for ii=1:3
        PF_interface = PF_interface + vpa(func_t(t,data_temp(ii).resultat(1,m),data_temp(ii).resultat(2,m)).*func_t(t,data_temp(ii+3).resultat(1,m),data_temp(ii+3).resultat(2,m)),8);
    end
    PF_interface=vpa(PF_interface.^2,8);
    PF_RMS(m) = sqrt(1/(2*pi)*double(int(PF_interface,t,0,2*pi)));
    toc
end

[Dyuman Joshi]

Are you using vpa() on symbolic expressions? Why?

[Maxime Eckstein]

To validate my code I wanted to check that the coefficients in the function where consistent with the entry data, but I runed it with and without the vpa parts nothing changes regarding computation time

[Maxime Eckstein]

I tried with arrays, would this be the same as the upper code and would there be a possibility to get rid of M for loop ?

I've tried it on small sample it seems to yield the same results as the original code

tt=linspace(0,2*pi,10000);
PF_RMS_2=zeros(1,length(data_temp(1).resultat));
for m=1:length(data_temp(1).resultat)
    tic
    PF_interface = 0;
    for ii=1:3
        PF_interface = PF_interface + data_temp(ii).resultat(1,m).*cos(tt+data_temp(ii).resultat(2,m)).*data_temp(ii+3).resultat(1,m).*cos(tt+data_temp(ii+3).resultat(2,m));
    end
    PF_interface=PF_interface.^2;
    PF_RMS_2(m) = sqrt(1/(2*pi)*trapz(tt,PF_interface));
    toc
end

[Dyuman Joshi]

Could you attach the data you are working with? Use the paperclip button to attach.

[Maxime Eckstein]

it would be something like that

[Star Strider]

If yfou are doing a Fourier transform, there are easier ways. Specifically, see fft and fourier depending on what approach you want to take.

[Star Strider]

The ‘resultat’ field has one non-zero entry, so it a sort of impulse function —

LD = load('data_temp.mat')

LD = struct with fields:
              data: {1×174 cell}
          metadata: [1×1 struct]
    evrn_interface: {9×2 cell}
                tt: [0 6.2838e-04 0.0013 0.0019 0.0025 0.0031 0.0038 0.0044 0.0050 0.0057 0.0063 0.0069 0.0075 0.0082 0.0088 0.0094 0.0101 0.0107 0.0113 0.0119 0.0126 0.0132 … ] (1×10000 double)
      interfaceIdx: 9
         data_temp: [1×6 struct]
           dataIdx: 174
          PF_RMS_2: [1.6433e-18 2.5817e-10 1.2083e-08 9.5448e-07 3.9415e-07 3.7248e-07 2.6259e-06 2.2613e-05 1.1703e-05 1.5983e-05 4.9306e-05 6.8085e-05 3.4939e-05 3.3271e-05 … ] (1×301 double)
      PF_interface: [5.3306 5.3311 5.3316 5.3321 5.3327 5.3332 5.3337 5.3342 5.3347 5.3352 5.3357 5.3362 5.3367 5.3372 5.3377 5.3382 5.3387 5.3392 5.3396 5.3401 5.3406 5.3411 … ] (1×10000 double)
                 m: 301
                ii: 3

data_temp = LD.data_temp

data_temp = 1×6 struct array with fields:
    valeur_label
    base
    resultat

valeur_label = data_temp.valeur_label

valeur_label = 2×5 char array
    'Ampl '
    'Phase'

base = data_temp.base;
resultat = data_temp.resultat;
resultat = squeeze(resultat)

resultat = 2×301

0.0000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1.6952 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

figure
plot(base, resultat)
grid
xlim([0 100])

.

[Maxime Eckstein]

I'm sorry, I did save the full workspace by mistake so multiple file not related to the question, I updated the file

In that case yes the first sensor has 0 amplitude because of the setup used, all the other are non zero.

[Star Strider]

I figured that.

The Fourier transforms of those data are not going to be very informative, regardless.