Hello, I want to iteratively plot the results of my while loop, however, I end up with a blank graph or an error. Here is the code: Vlow=350; fs=100000; n=0.56; Lk=41.98*0.000001; P=10000; Vhigh=900; C=0; i=0; s=poly(0,"s"); while Vlow 0.129); I1=(1/(2*Lk*fs))*(2*(Vlow/n)*out+Vhigh-(Vlow/n)); I2=(1/(2*Lk*fs))*(2*Vhigh*out-Vhigh+(Vlow/n)); if I1 > I2 then C= I1; else C=I2; end Y(i)=C; X(i)=Vlow+20; i=i+1; end plot(X,Y); Any help would be greatly appreciated, Thank you

Plotting results in array of a while loop

4 次查看（过去 30 天）

显示更早的评论

Hicham Lhachimi 2020-6-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/552733-plotting-results-in-array-of-a-while-loop

评论： Adam Danz 2020-6-22

采纳的回答： Adam Danz

在 MATLAB Online 中打开

Hello,

I want to iteratively plot the results of my while loop, however, I end up with a blank graph or an error.

Here is the code:

Vlow=350;
fs=100000;
n=0.56;
Lk=41.98*0.000001;
P=10000;
Vhigh=900;
C=0;
i=0;
s=poly(0,"s");
while Vlow < 450   
    Polynome=((P*Lk*n*fs)/(Vhigh*Vlow))-s+s^2;
    solution=roots(Polynome);
    out =  solution(solution<0.35 & solution>0.129);
    I1=(1/(2*Lk*fs))*(2*(Vlow/n)*out+Vhigh-(Vlow/n));
    I2=(1/(2*Lk*fs))*(2*Vhigh*out-Vhigh+(Vlow/n));
     if I1 > I2 then
        C= I1;
     else
         C=I2;
     end
    Y(i)=C;
    X(i)=Vlow+20;
    
    i=i+1;
end
plot(X,Y);

Any help would be greatly appreciated,

Thank you

8 个评论
显示 6更早的评论隐藏 6更早的评论

Image Analyst 2020-6-22

在 MATLAB Online 中打开

So s is this:

s =
         1     -1     0.0746311111111111         

which is a 1-by-3 vector, so it throws an error when you do this:

Polynome=((P*Lk*n*fs)/(Vhigh*Vlow))-s+s^2;

Were you expecting s to be a scalar instead of a 1-by-3 vector?

Hicham Lhachimi 2020-6-22

编辑：Hicham Lhachimi 2020-6-22

Here is exactly what I got ,when I want to select only one root which will be the out variable, I don't see any value of out on the workspace:

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Adam Danz 2020-6-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/552733-plotting-results-in-array-of-a-while-loop#answer_455548

编辑：Adam Danz 2020-6-22

在 MATLAB Online 中打开

As the error message indicates, a subscript must be a real, positive integer or logical.

On the first iteration, i equals 0 which does not fulfill these requirements. 0 is not positive and it's not a logical value. So, when you index Y(i), or, Y(0), you get the error.

Instead of initializing i=0, set it to 1, i=1;

More importantly, the while-loop is defined by the value of Vlow but this value never changes within the while-loop so the while-loop will never end. Vlow will always be less than 450 because its original value is 350 which never changes within the loop.

Perhaps you meant to include something like this at the end of the while-loop.

Vlow = X(i);

If that's the case, then you could replace the while-loop with a for-loop since you know ahead of time the number of iterations. The number of iterations will be the number of values in

values = 350 : 20 : 450;

16 个评论
显示 14更早的评论隐藏 14更早的评论

Hicham Lhachimi 2020-6-22

Now it works well, looping through the index values of Vlow is much better. Thank you very much for your support (y).

Adam Danz 2020-6-22

Agreed. I almost always define loops as i = 1:n rather than i = some_vector.

请先登录，再进行评论。

类别

MATLAB Language Fundamentals Matrices and Arrays Matrix Indexing

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by