MATLAB Answers

how can i put a limitation rules for below codes?

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Mamad Mamadi
Mamad Mamadi on 15 Dec 2019
Edited: Mamad Mamadi on 14 Jan 2020
i have some values of Pch=Pcharging and Pdis=Pdischarging that i mentioned below and i want to write codes for limitation but the codes that i wrote is not working specially for Pch. Could you please and please help me. thanks
i mean;
Pmax= 3600wh
Pch wont be 3750 , must not be bigger than 3600
Pdis wont be -150, must not be smaller than 0
Pch=[2600 2750 2800 2950 3010 3100 3300 3580 3750];
Pdis=[3450 3330 2900 1500 1200 850 300 22 -150];
it these codes right? if not how can i write it?
%%% for Pch:
while a<=length(Pch);
b= Pch(a)+b;
if result_data <= Pmax ;
finalresult = result_data;
elseif result_data > Pmax ;
%%% for Pdis
if Pdis(1,i) >= Pmin;
elseif Pdis(1,i) < Pmin


Rik on 15 Dec 2019
What should happen with the values that are not allowed? Should they be removed from the vector, or should they be assigned the boundary value?
Mamad Mamadi
Mamad Mamadi on 16 Dec 2019
Because the capacity of the battery is 3600wh then we cannot charge the battery bigger than 3600wh so we dont need the values that bigger than Pmax=3600wh we can remove it.
And also for Pdis is the same action, it cannot be smaller than Pmin=0

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Answers (1)

Rik on 16 Dec 2019
With logical indexing it is easy to remove impossible values:
Pmax= 3600;
Pch=[2600 2750 2800 2950 3010 3100 3300 3580 3750];
Pdis=[3450 3330 2900 1500 1200 850 300 22 -150];
Pch(Pch>Pmax | Pch<Pmin)=[];
Pdis(Pdis>Pmax | Pdis<Pmin)=[];


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Rik on 23 Dec 2019
A late reply is not a problem. But why would you need my email address? If you have any issue implementing my answer, feel free to comment it here. If you have a new question you can post it and use my profile page to send me a link.
Mamad Mamadi
Mamad Mamadi on 24 Dec 2019
I wanted to send you some matlab codes and describ my problems in private LOL but no matter i will post it here. tnx anyway :)
Mamad Mamadi
Mamad Mamadi on 14 Jan 2020
Hi Rik could you please help me in these codes?
i=1:8784 %one year (hours)
if Ppv_N(i,1) > PLoad(i,1);
else Ppv_N(i,1) < PLoad(i,1)
now after finding Pch and Pdis from 1 to 8784 hours, i want to write another code to solve 'a' as i mentioned below for example ;
when (Pch & Pdis)=0 %at the same time when both Pch and Pdis become zero, then
a=PLoad(i,1) - Ppv_N(i,1)

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