plotting using for loop

Hi,

I am new to matlab, i want to plot a curve using for loop but instead a range of values like that obtained without using for loop i am getting only a single value. can some body tell me what i am doing wrong or if i am missing something in my code.

clear
clc
close all
figure(1) ;
for d= 0:0.01:1
    if(d==1)
        break;
    end
    vin=20;
    
    vout=(vin*4.*d)./(1.-d);
   
end
plot(d,vout)

p.s whats the difference between x=[0:0.1:1] and x=0:0.1:1

thanks;

[Stephen23]

"whats the difference between x=[0:0.1:1] and x=0:0.1:1"

Good question. In MATLAB square brackets are a concatentation operator (not a "list" operator like in some languages), so they are used to join arrays together. Which mena that in this code the square brackets do nothing useful at all:

x = [0:0.1:1]
%   ^       ^ superfluous because nothing is being concatenated together

https://www.mathworks.com/matlabcentral/answers/35676-why-not-use-square-brackets

That is also why the MATLAB Editor shows a warning that the square brackets are superfluous and recommends to remove them. To define a vector of values you only need to use the colon operator:

x = 0:0.1:1

Answer 1

[KALYAN ACHARJYA]

Edited:

For difference, please read the Stephen Cobeldick 's valuable comment.

more

x=[0:0.1:1]  
x1=0:0.1:1
x2=(0:0.1:1)

Better to follow x1 ways, it's simplest. Code Part, you can avoid loop here (Recommended)

d=0:0.01:1;
vin=20;
vout=(vin*4*d)./(1-d);
plot(d,vout)

At the max x value is 1, it automatically stop once x reach to it's max, no need of break here.

[Stephen23]

"Tecnically, there is no difference."

One calls the concatenation operator with one input (thus concatenating one vector with nothing else), whereas the other does not. Quite a large difference, really, because it indicates that the user does not understand what square brackets actually do and also that they ignore warnings provided by the MATLAB Editor.

Both of which are pretty big hints that the user writes substandard code. Much like CLEAR ALL et al.

[KALYAN ACHARJYA]

Sir Stephen Cobeldick Thanks for your vauable comment. Thank you for teaching us as you always do.

Answer 2

[KSSV]

You need not to use a loop to get what you want. Already you are using .* i.e. eleemnt by element operation so , you can proceed with vectors.

close all
figure(1) ;
d= 0:0.01:1 ;
vin=20;
vout=(vin*4.*d)./(1.-d);  

plot(d,vout)

Answer 3

[Pen-Li (Ben) Yu]

A for loop does not define an array. You need to define d directly.

d= 0:0.01:1
vin=20;
vout=(vin*4.*d)./(1.-d);

figure(1)   
plot(d,vout)

Answer 4

[Stephen23]

"...can some body tell me what i am doing wrong or if i am missing something in my code."

The main problem is that you are not using indexing to store the output values.

d = 0:0.01:1;
n = numel(d);
vin  = 20;
vout = nan(1,n); % preallocate
for k = 1:n % loop over indices
    vout(k) = (vin*4.*d(k))./(1.-d(k));
end
plot(d,vout)

For such a simple calculation a loop is not required:

vtwo = (vin*4.*d)./(1.-d);
plot(d,vtwo)