Non linear ODE system matlab fourier analysis problem

  1. Hello all.

    I am studying a system and want to investigate how the frequency of y(3) varies under different conditions. However, my the fft I perform on it tells me the frequency is zero, which must be incorrect. I have tried a stack of things but can't see what the problem is. I'm relatively new to the matlab environment.

    The main code:

    close all; clear all;
    %--------------------------------------%
    Time control and initial conditions
    timeinmin=600; % Time in minutes
    tfin=timeinmin*60; % Time in seconds
    %--------------------------------------%
    Initial Levels and system Saturation
    totIKK=0.08; % Total IKKn at start
    totNFkB=0.08; % Total NFkB cytoplasmic at start
    T=1; % Saturation of the system
    yint=[totNFkB,0,0,0,0,0,totIKK,0,0,0,0,totIKK,totNFkB,T]; % Initial conditions
    %--------------------------------------%
    ODEs
    [t,y]=ode45(@nfkb,[0 tfin],yint); % calls Model
    %--------------------------------------%
    Extract Data from the ODE
    fs = 1; % The sampling freqency in Hz
    time = 0:1/fs:tfin; % Define time of interesrt
    Datapoints = tfin*fs; % # of Data points being used
    yi = interp1(t,y(:,3),time);
    %plot(time/60,yi,'o',t/60,y(:,3)) % test to ensure it is working
    %dbstop
    %--------------------------------------%
    Fourier Analysis from
    NFFT = 2^nextpow2(tfin); % Next power of 2 from length of y
    Y = fft(yi,NFFT)/Datapoints;
    f = fs/2*linspace(0,1,NFFT/2+1);
    % Plot single-sided amplitude spectrum.
    plot(f,2*abs(Y(1:NFFT/2+1)))
    title('Single-Sided Amplitude Spectrum of y(t)')
    xlabel('Frequency (Hz)')
    ylabel('Amplitude - |Y(f)|')
    % Get value of frequency
    [B,IX] = sort(2*abs(Y(1:NFFT/2+1)));
    BFloor=0.1; %BFloor is the minimum amplitude value (ignore small values)
    Amplitudes=B(B>=BFloor) %find all amplitudes above the BFloor
    Frequencies=f(IX(1+end-numel(Amplitudes):end)) %frequency of the peaks

    The ODEs:

    function dydt=nfkb(t,y)
    parameters;
    dydt=[kd1*(y(13)-(y(3)+y(5))/kv-y(11)-y(1))-ka1*y(2)*y(1)-ki1*y(1)+...
    kdegc*(y(13)-(y(3)+y(5))/kv-y(11)-y(1))+ke1f*y(3)+kdegpin*y(11);
    kd1*(y(13)-(y(3)+y(5))/kv-y(11)-y(1))-ka1*y(2)*y(1)-ki2*y(2)+...
    kdegf*y(2)+ktria*y(6)-kc2*y(8)*y(2);
    kd1*y(5)-ka1*y(4)*y(3)+kv*ki1*y(1)-kv*ke1f*y(3);
    kd1*y(5)-ka1*y(4)*y(3)+kv*ki2*y(2)-kv*ke2*y(4)-kdegf*y(4);
    ka1*y(4)*y(3)-kd1*y(5)-kv*ke1c*y(5);
    kitria*(y(3).^h)/((y(3).^h)+(k.^h))-kdegtia*y(6);
    kp*(y(12)-y(8)-y(7))*kbA20/(kbA20+y(10)*y(14))-y(14)*ka*y(7);
    y(14)*ka*y(7)-ki*y(8);
    kitra*(y(3).^h)/((y(3).^h)+(k.^h))-kdegta*y(9);
    ktra*y(9)-kda*y(10);
    kc3*y(8)*(y(13)-(y(3)+y(5))/kv-y(11)-y(1))-kdegpin*y(11);
    0;
    0;
    0];

    The parameters:

    %rates
    kv=3.3; % C:N ratio
    kp=0.0006; % Conversion rate of IKKi
    ka=0.004; %activation rate of IKK
    ki=0.003; % Usage rate of IKK
    kda=0.0045; % Deg rate of A20 protein
    kbA20=0.0018; % A20 concentration
    kc2=0.074; % phosopho of IkB
    kc3=0.37; % phospho of IkB-NFkB
    kdegpin=0.1; % splitting of Ikb-NFkB due to phospho
    ka1=0.5; % nuclear IkB-NFkB formation
    kd1=0.0005; % dissisociation of nuclear IkB-NFkB
    kdegf=0.0005; % Deg rate of nuclear IkB
    kdegc=0.000022; % dissisociation of cyto IkB-NFkB
    ki1=0.0026; % nuclear entry of NF-kB
    ke1f=52*10^(-6); % nuclear exit of NF-kB
    ke1c=0.01; % Transport of IkB-NFkB out of nuc.
    ki2=0.00067; % nuclear entry of IkB
    ke2=3.35*10^(-4); % Nuclear exit of IkB
    h=2; % hill coefficient of IkB transcription
    k=0.065; % dissasociation constant for IkB tanscription with NFkB
    kitria=1.4*10^(-7); % trans rate of IkB
    ktria=0.5; % translation of IkB
    kdegtia=0.0003; % Deg rate of IkB RNA
    kitra=1.4*10^(-7); % Transcription rate of A20 RNA
    ktra=0.5; % transcription rate of A20
    kdegta=0.00048; % deg rate of A20 RNA

    Thank you in advance for any insights.
     
  2. jcsd
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