how is it possible to solve below matlab codes problem?

Question

Mamad Mamadi 2020-1-14

0
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https://ww2.mathworks.cn/matlabcentral/answers/500237-how-is-it-possible-to-solve-below-matlab-codes-problem

编辑： Mamad Mamadi 2020-1-15

Hi everybody could you please help me in these codes?

i=1:8784   %one year (hours)
if Ppv_N(i,1) > PLoad(i,1);
  Pch(1,i)=(Pbat*(1-sigma))+(Ppv_N(i,1)-(PLoad(i,1))/eta_i)*eta_b; 
  a=Ppv_N(i,1) - PLoad(i,1)
end
else   Ppv_N(i,1) < PLoad(i,1)
Pdis(1,i)=(Pbat*(1-sigma))-((PLoad(i,1)/eta_i)-Ppv_N(i,1));

now after finding Pch and Pdis from 1 to 8784 hours, i want to write another code or loop to solve 'b' between 1:8784 , for example ;

when (Pch & Pdis)=0 %at the same time when both Pch and Pdis become zero, then

b=PLoad(i,1) - Ppv_N(i,1)

which i mean when Pch and Pdis at the same time became zero then use b=PLoad(i,1) - Ppv_N(i,1)

thanks

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Answer 1

David Hill 2020-1-14

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在 MATLAB Online 中打开

I was somewhat confused with your question. Hopefully, this helps:

for i=1:8784   %one year (hours)
    if Ppv_N(i,1) > PLoad(i,1);
      Pch(1,i)=(Pbat*(1-sigma))+(Ppv_N(i,1)-(PLoad(i,1))/eta_i)*eta_b; 
      a=Ppv_N(i,1) - PLoad(i,1);
    elseif   Ppv_N(i,1) < PLoad(i,1)
      Pdis(1,i)=(Pbat*(1-sigma))-((PLoad(i,1)/eta_i)-Ppv_N(i,1));
    end
end
b=[];
for i=1:8784
    if Pch(1,i)==Pdis(1,i)
      b=[b,PLoad(i,1) - Ppv_N(i,1)];%not sure if you will have more than one occurrance  
    end
end