Signal power with 1/3 octave filter

Hi, I'm trying to calculate the power in dB of a signal and express it in 1/3 octave. I mean, I apply the 1/3 octave filter to the signal and then calculate the power for each band. The problem is that when I compare my program with a professional one I get very different results. Below my code:

[x, Fs] = audioread('audio/pink/signal.wav');            
C = 25;

%%octave
    lx = length(x);
    % filter design
    Fc = [20 25 31.5 40 50 63 80 100 125 160 200 250 315 400 ...
        500 630 800 1000 1250 1600 2000 2500 3150 4000 5000 ... 
        6300 8000 10000 12500 16000 20000];% frequenze centrali ANSI
    Fc = Fc(Fc<Fs/2);
    fl = Fc*2^(-1/6); % lower freq
    fu = Fc*2^(1/6); % upper freq
    fu(fu>Fs/2) = Fs/2 -1; 
    numBands = length(Fc);
    ord = 12;
    z = zeros(2*ord, numBands);
    p = zeros(2*ord, numBands);
    k = zeros(1, numBands);
    for n = 1:numBands
        [z(:,n),p(:,n),k(n)] = cheby1(ord,10,[fl(n) fu(n)]/(Fs/2));
    end

      % apply filter
      x_oct = zeros(lx, numBands);
      s = zeros(ord, 6, numBands);

      for i = 1:numBands
          s(:, :, i) = zp2sos(z(:,i),p(:,i),k(i));
          x_oct(:, i) = sosfilt(s(:,:,i),x); 
      end

      x_pow_oct = zeros(1, numBands);
     for k = 1:numBands
          fftx = fft(x_oct(:,k));        
          fftxsquared = fftx'*fftx;          
          s1 = sum(fftxsquared/lx);
          x_pow_oct(k) = 10*log10(s1) + C;
     end

%%end function

maxV = max(x_pow_oct);
f = [0 Fc(5:5:30)];
figure(150)
subplot(1,2,1)
bar(x_pow_oct);
title('x_oct_pow');
set(gca,'XTickLabel',{round(f)});
xlabel('Central frequencies');
ylabel('Power');
ylim([10 maxV+10]);
xlim([0 numBands+1]);

To show the differences with the professional software, below my result (first image) and the professional one (seond image).