Area under the curve

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Bhuvan Khoshoo 2019-10-24

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评论： Bhuvan Khoshoo 2019-10-29

MATLAB_Code.m

I have follwing curve resulting from plotting current in capacitor versus time. (Matlab code is attached).

I want to calculate the charge stored in the capacitor. For that I need to calculate the positive area of the curve. As you can see in the zoomed in image, the current plotted has pulses (switching frequency = 20kHz), and I would like to calculate the charge stored in one fundamental cycle (fundamental frequency = 60Hz).

So theoretically, the capacitor is getting charged when the current flows into it. For the curve it means the value of current that is above zero(positive).

I tried calculating the area of the curve by simplifying the curve into traingles and trapezoids, and then taking the coordinate points values (x,y). But I am pretty sure it is not correct since it does not take into account the presence of pulses.

I also tried the trapz(x,y) in matlab but the calculation does not seem right.

And the expression of the current is not a straightforward function, it is defined in terms of switching functions and duty ratios so I am not sure how to use the integral function in Matlab.

Please help.

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Dimitris Kalogiros 2019-10-25

编辑：Dimitris Kalogiros 2019-10-25

At the figures you have posted you have depicted current versus time e.g. I(t) . How do you have stored the values of current ? Do you have a vector (lets say) I, which contains dense samples of the signal I(t) ?

Bhuvan Khoshoo 2019-10-25

I have added the code in the attachements.

Yes, the time series is a vector, sampled at Ts/100 step size, where Ts is switching time period (switching frequency = 20kHz). The current series is also a vector with values corresponding to the time instants.

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Answer 1

Dimitris Kalogiros 2019-10-25

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https://ww2.mathworks.cn/matlabcentral/answers/487277-area-under-the-curve#answer_398200

编辑：Dimitris Kalogiros 2019-10-25

在 MATLAB Online 中打开

Change the last section of your program to this:

%% Sizing Capacitor Cf;
%capacitor fundamental current for "a" phase is all the ripple current of ia, considering
%the fundamental compnent of ia, i.e. ia1 flows into converter;
%similar expression for other phases;
icfa = ia-ia1;
icfb = ib-ib1;
icfc = ic-ic1;
% integral of current
Qcfa=zeros(size(t));
for n=2:length(t)
    Qcfa(n)=Qcfa(n-1)+icfa(n)*(t(n)-t(n-1));
end
figure(7);
subplot(2,1,1); plot(t,icfa,'-k.'),grid on
xlabel('Time(s)','fontsize',12);
ylabel('Capacitor Current(Amps)','fontsize',12);
xlim([0,T])
figure(7);
subplot(2,1,2); plot(t,Qcfa,'-r.'); zoom on; grid on;
xlabel('Time(s)','fontsize',12);
ylabel('Capacitor charge(Coulomb)','fontsize',12);
xlim([0,T])

Qcfa(n) is the value of capacitor's electric charge at every moment t(n).

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Dimitris Kalogiros 2019-10-29

在 MATLAB Online 中打开

You can also use trapz() method.

%% Sizing Capacitor Cf;
%capacitor fundamental current for "a" phase is all the ripple current of ia, considering
%the fundamental compnent of ia, i.e. ia1 flows into converter;
%similar expression for other phases;
icfa = ia-ia1;
icfb = ib-ib1;
icfc = ic-ic1;
% integral of current
Qcfa=zeros(size(t));
for n=2:length(t)
    Qcfa(n)=Qcfa(n-1)+icfa(n)*(t(n)-t(n-1));
end
% integral of current, using Trapezoidal numerical integration
Qcfa2=zeros(size(t));
for n=2:length(t)
    Qcfa2(n)=trapz(t(1:n), icfa(1:n));
end
figure(7);
subplot(3,1,1); plot(t,icfa,'-k.'),grid on
xlabel('Time(s)','fontsize',12);
ylabel('Capacitor Current(Amps)','fontsize',12);
xlim([0,T]);
figure(7);
subplot(3,1,2); plot(t,Qcfa,'-r.'); zoom on; grid on;
xlabel('Time(s)','fontsize',12);
ylabel('Capacitor charge(Coulomb)','fontsize',12);
xlim([0,T]);
figure(7)
subplot(3,1,3); plot(t,Qcfa2,'-k.');grid on; zoom on;
xlabel('Time(s)','fontsize',12);
ylabel('Capacitor charge(Coulomb), using trapz()','fontsize',12);
xlim([0,T]);

I have include it. You can compare Qcfa and Qcfa2 and find out the differences.

Bhuvan Khoshoo 2019-10-29

Thanks a lot!! :)

I went with the first method. From the resulting array I needed the maximum value for sizing the capacitor.

Thanks again for the help!!

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Area under the curve

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