Why don't graphics appear for this code?

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Nuno
Nuno 2011-3-15
Why don't graphics appear for this code?
% Variáveis
m = 112.34;
temp = 25;
ISC = 7.8;
VTR = 0.0257;
VT = 0.0257;
VCA = 32.7;
GR = 1000;
G = 700;
% Cálculo do IR
IR = 1000 * (ISC / (2.718*exp(VCA/(m*VTR))-1))
for V = 0:1:VCA
V1 = V*2
%Carcular a potencia máxima para G=1000
P = V1*IR*35
end
figure(1)
h=plot(V1,P);
set(h,'color',rand(1,3),'linewidth',2);
hold on
axis([0 200 0 40])
xlabel('Tensão')
ylabel('Potência')
title('Curva P-V')

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采纳的回答

Walter Roberson
Walter Roberson 2011-3-15
Each time through the loop you replace V1 and P, so at the end of the loop you only have a single point to plot.
Try
V1 = (0:1:VCA).*2; P = V1.*IR.*35;
with no loop.

更多回答（3 个）

Nuno
Nuno 2011-3-15
Why i can´t make loop?
  2 个评论
Walter Roberson
Walter Roberson 2011-3-15
You _can_ do it with a loop, provided that you store the V1 and P value that results from each loop. You aren't doing that now: each time through the loop, you are overwriting the previously calculated value.
Matt Tearle
Matt Tearle 2011-3-15
What Walter was saying above is that you don't *need* to have a loop -- his solution does the same thing, cleaner and simpler.
As an aside, note that 0:1:VCA will give you whole numbers from 0 up to 32, because VCA is 32.7. That is, the range operator in MATLAB is essentially "<=". Obviously I don't know your intention -- just wanted to make sure you're aware of that.

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Nuno
Nuno 2011-3-15
But if i change to V1= 0:0.1:VCA appear error in P=V1.*IR.*35...
  2 个评论
Walter Roberson
Walter Roberson 2011-3-15
Notice I used
V1 = (0:1:VCA).*2
But even without the multiplication by 2, I do not find any obvious error in P=V1.*IR.*35 . What error do you observe?
Matt Tearle
Matt Tearle 2011-3-15
Works fine for me:
m = 112.34;
temp = 25;
ISC = 7.8;
VTR = 0.0257;
VT = 0.0257;
VCA = 32.7;
GR = 1000;
G = 700;
% Cálculo do IR
IR = 1000 * (ISC / (2.718*exp(VCA/(m*VTR))-1));
V = 0:1:VCA;
V1 = V*2;
%Carcular a potencia máxima para G=1000
P = V1*IR*35;
figure(1)
h=plot(V1,P);
set(h,'color',rand(1,3),'linewidth',2);
hold on
axis([0 200 0 40])
xlabel('Tensão')
ylabel('Potência')
title('Curva P-V')

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Nuno
Nuno 2011-3-15
Yes, the code works... Thanks! But now i have a new problem...
When i complile de program, the graphs should give a value of voltage (V) up to 32,7 but the value is ~12,5. Why?
Code:
% Variáveis
m = 112.34;
temp = 25;
ISC = 7.8;
VTR = 0.0257;
VT = 0.0257;
VCA = 32.7;
GR = 1000;
G = 700;
% Cálculo do IR
IR = 1000 * (ISC / (2.718*exp(VCA/(m*VTR))-1));
V1 = (0:0.1:VCA);
ICC = ISC * (G/GR);
I = ICC - IR.*(2.718.*exp(V1./(m.*VT))-1);
ICC1 = ISC * ((G+100)/GR);
I1 = ICC1 - IR.*(2.718.*exp(V1./(m.*VT))-1);
ICC2 = ISC * ((G+200)/GR);
I2 = ICC2 - IR.*(2.718.*exp(V1./(m.*VT))-1);
ICC3 = ISC * ((G+300)/GR);
I3 = ICC3 - IR.*(2.718.*exp(V1./(m*VT))-1);
ICC4 = ISC * ((G+400)/GR);
I4 = ICC4 - IR.*(2.718.*exp(V1./(m*VT))-1);
%Carcular a potencia máxima para G=1000
P = V1.*I3;
figure(1)
j=plot(V1,P);
set(j,'color',rand(1,3),'linewidth',2);
hold on
axis([0 40 0 100])
xlabel('Tensão(V)')
ylabel('Potência')
title('Curva P-V')
figure(2)
h=plot(V1,I,V1,I1,V1,I2,V1,I3,V1,I4);
set(h,'color',rand(1,3),'linewidth',2);
hold on
axis([0 40 0 40])
xlabel('Tensão(V)')
ylabel('Corrente')
title('Curva I-V')
  7 个评论
显示 5更早的评论
Matt Tearle
Matt Tearle 2011-3-16
And, more importantly, correct. 2.718^x is an inaccurate version of exp(x).
Nuno
Nuno 2011-3-16
The error is in another count... Thanks for the help...

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