Plotting results in array of a while loop
显示 更早的评论
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
Geoff Hayes
2020-6-22
Hicham - please copy and paste the full error message to your question. Is the problem with poly or something else?
Adam Danz
2020-6-22
@Hicham Lhachimi, just to be clear, the code you shared does not generate an error message? Your question states that it does. If it does, we would need the entire error message.
What is the function "ploy"? When I run your code, an error message is generated.
Adam Danz
2020-6-22
Do you mean "ploy" is a typo and it should read "plot" or do you mean you have a custom function named ploy that we do not know about?
When I replace ploy with plot, there is a blue square marker printed on the plot at (1,0).
Hicham Lhachimi
2020-6-22
Adam Danz
2020-6-22
Sorry for the confusion. I meant poly.
Matlab has a fuction poly but it only has 1 input. In this line of your code,
s=poly(0,"s");
you have 2 inputs which would cause an error. The syntax you're using looks like you meant to use plot instead of poly.
Hicham Lhachimi
2020-6-22
编辑：Hicham Lhachimi
2020-6-22
Image Analyst
2020-6-22
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:
采纳的回答
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
Hicham Lhachimi
2020-6-22
编辑：Hicham Lhachimi
2020-6-22
Adam Danz
2020-6-22
That's a different error.
What is c? If it's not a scalar value, you cannot store it that way.
Try,
Y(i,:) = ....
Also, this should throw an error too,
if I1 > I2 then
'then' is not part of the syntax.
Hicham Lhachimi
2020-6-22
Hicham Lhachimi
2020-6-22
Adam Danz
2020-6-22
Without seeing your updated code and without knowing the size of the input variables I have no idea what X and Y are.
Hicham Lhachimi
2020-6-22
编辑：Adam Danz
2020-6-22
It looks like you need to spend some more time exploring the data being generated within the code.
First, learn how to use debug mode. Put a break point at the beginning of your for-loop and then step through each line of the code. Look at result of each line of code.
You'll see that the line that produces the out variable is empty! That's because the solution variable contains values that are not between 0.129 and 0.35 (shown in the gif below).
So, the problem you're having with plot(x,y) is not the real problem. The problem is deeper. Your code isn't doing what you expect it to do or your input data is not what you expect. You're the only one who can fix that. So, go through the code line by line and look and think about the values of each variable.
This answer may also be helpful. It shows how to view variables in a function workspace.
Hicham Lhachimi
2020-6-22
Adam Danz
2020-6-22
Glad I could help. Once you are comfortable with using debug mode, you'll be able to solve some of the minor problems much more quickly.
Hicham Lhachimi
2020-6-22
Adam Danz
2020-6-22
I've been editing your comments so that your code can be formatted. Going forward, when you include code, please format it using the [>] button.
On which iteration do you get this problem? On the first iterations, when Vlow=350, A=[29.97, 3.1976].
Hicham Lhachimi
2020-6-22
编辑：Hicham Lhachimi
2020-6-22
Adam Danz
2020-6-22
1) I didn't understand your response. I only tested the first iteration of your loop and the A values are shown in my previous comments. I don't know where you're getting the value of 1 for A.
2) After the loop is finished, what size do you expect the variables C,Y, and X to be?
3) Here's a better way to for the loop. Instead of looping through values of Vlow, loop through the index values of Vlow. That way you can use the i-variable to store the outputs properly.
Vlow=350:20:450;
for i = 1:numel(Vlow)
Polynome=[1 -1 ((P*Lk*n*fs)/(Vhigh*Vlow(i)))]; % note, Vlow was replaced with Vlow(i)
solution=roots(Polynome);
out = solution(solution < 0.35 & solution > 0.129);
I1=(1/(2*Lk*fs))*(2*(Vlow(i)/n)*out+Vhigh-(Vlow(i)/n));% note, Vlow was replaced with Vlow(i)
I2=(1/(2*Lk*fs))*(2*Vhigh*out-Vhigh+(Vlow(i)/n));% note, Vlow was replaced with Vlow(i)
A=[I1 I2];
C(i,:)=max(A);
Y(i,:)=C;
X(i,:)=Vlow(i);% note, Vlow was replaced with Vlow(i)
end
Hicham Lhachimi
2020-6-22
Adam Danz
2020-6-22
Agreed. I almost always define loops as i = 1:n rather than i = some_vector.
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