Area under the curve

9 次查看（过去 30 天）
Bhuvan Khoshoo
Bhuvan Khoshoo 2019-10-24
评论： Bhuvan Khoshoo 2019-10-29
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.
  7 个评论
显示 5更早的评论
Dimitris Kalogiros
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
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.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Dimitris Kalogiros
Dimitris Kalogiros 2019-10-25
编辑：Dimitris Kalogiros 2019-10-25
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).
  3 个评论
显示 1更早的评论
Dimitris Kalogiros
Dimitris Kalogiros 2019-10-29
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
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!!

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Spline Postprocessing 的更多信息

标签

产品


版本

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by