Mark the knee point on a graph

26 vues (au cours des 30 derniers jours)
AHMED FAKHRI
AHMED FAKHRI le 26 Fév 2021
Commenté : Adam Danz le 26 Fév 2021
Hi
I want to mark the knee point on the graph like the picture attached.
This point is the maximum power meaning ( a certain value of voltage * value of current ) gives the maximum power possible. However, it is not maximum voltage with maximum current.
In the graph 15.981 * 6.40876 gives the maximum power of this curve.
Any help regarding this?
Thanks

Réponse acceptée

Alan Stevens
Alan Stevens le 26 Fév 2021
Calculate a vector of values of P = V.*I then choose the maximum. Something like:
P = V.*I;
indx = find(P==max(P));
Vp = V(indx);
Ip = I(indx);
then plot(Ip,Vp,'s')
  1 commentaire
Adam Danz
Adam Danz le 26 Fév 2021
I was writing a demo based on these data while you were getting two great answers. I'll include it here.
data = [ 0 0.124
0.624 0.122
1.248 0.12
1.872 0.118
2.497 0.116
3.121 0.114
3.745 0.112
4.369 0.11
4.7 0.103
4.934 0.1
5.165 0.091
5.336 0.085
5.566 0.078
5.927 0.057
6.07 0.045
6.113 0.037
6.137 0.034
6.21 0.025
6.294 0.02
6.373 0.004
6.389 0.002
6.407 0];
T = array2table(data, 'VariableNames', {'Volts','Amps'});
yyaxis left
plot(T.Volts,T.Amps)
xlabel('Volts (V)')
ylabel('Current (Amps)')
T.Power = T.Volts .* T.Amps;
yyaxis right
plot(T.Volts, T.Power)
ylim([0,1])
ylabel('Power (W)')
[~, maxPowerIdx] = max(T.Power);
xline(T.Volts(maxPowerIdx), 'k-','Max power point')
yyaxis left
hold on
plot(T.Volts(maxPowerIdx), T.Amps(maxPowerIdx), 'ko')

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Plus de réponses (1)

Yu Lu
Yu Lu le 26 Fév 2021
Hello,
As my understanding you have two arrays v and c
v = [<Your Data>]; % voltage
c = [<Your Data>]; % current
and you want to find the place that has the maximum dot product of these two arrays, in MATLAB you can do:
[maxPower, maxIndices] = max(v.*c);
to get the maximum power value and the indices of voltage and current that product maximum power.
With the indices you can set marker at the position or create data tip on the position use the datatip function.

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