Illegal use of reserved keyword

vs=10;
r=1;
c=1;
dt=0.1;
z=dt/(2*c);
i= vs/r;
vi=-(i*z)/2;

for j=0:0.1:1
    {
        if j = 0
         {
            vr[j]=0-vi;
            }
          
        else   
            {
                vr[j+1]=vr[j];
                }
        end   
ik[j]=(vs-(2*vr[j]))/(r+z);
it[j]=(vs/r)*(exp(-j/(r*c)));
vc[j]=(2*vr[j])+(ik[j]*z);
vcr[j]=vc[j]-vr[j]
}

Answer 1

[Bjorn Gustavsson]

The curly brackets are used to enclose cell-arrays (lists) in matlab, not start and end loops and if-clauses, matlab uses nothing to start those and end to close them. So:

for i1 = 1:4,
    if x > 0
        a(i1) = pi^i1;
    else
        a = -23;
    end
end

also, I would suggest staying away from i and j for indexing, since they are the imaginary unit too, sooner or later it might come and bite you at an inopportune moment, I go with i1, i2, or i_t, i_x...

HTH

[Venkatkumar M]

no it is not working ples make change in tis code

[Steven Lord]

There are a number of problems in this code.

1. You still have a curly brace on line 10 of your code. Eliminate it.
2. On line 11, you're using the = operator. The = operator performs assignment while the == operator performs equality testing.
3. There is no such thing as element 0 of an array in MATLAB. The first element of an array in MATLAB is element 1. Therefore line 12 will error.
4. There is no such thing as element 1.1 of an array in MATLAB. Therefore line 15 will error when j is not an integer value.

My guess is that you're new to MATLAB. If that's the case I strongly recommend you read through the Getting Started guide to which Matt J linked and/or take the MATLAB Onramp tutorial. Each of these should give you a basic understanding of the fundamentals of MATLAB.

[Venkatkumar M]

what is wrong in this?

[Bjorn Gustavsson]

It says that you haven't defined the variable or function vr when it is used the first time at line14. There, on the right-hand side you try to get the value of its zeroth component.

If you define vr as a variable, as seems to be your intention, you will also get an error for trying to access the zeroth component of the array. Matlab uses indexing from 1 to n not from 0 to n-1 for arrays, so you'll have to fix that too.

[Steven Lord]

Your line 11 is dead code. It can never be executed. None of the values between 0 and 10 (which are the only values j can take) are less than 0 so the condition on line 10 is never satisfied.

But Bjorn Gustavsson is correct on both counts. vr doesn't exist the first time you try to access its elements, and there is no such thing as element 0 of a matrix in MATLAB.

I urge you again to take the MATLAB Onramp tutorial. It's free and it will give you a basic understanding of how to write MATLAB code.

[Guillaume]

As has already been said, you first need to learn matlab. In addition, posting screenshots of code instead of pasting actual code as text is absolutely useless. We can't edit screenshots.

[Venkatkumar M]

then can you edit tis????

vs=10;

r=1;

c=1;

dt=0.1;

z=dt/(2*c);

i= vs/r;

vi=-(i*z)/2;

for j = 0:1:10

if j < 0

vr(j)=0-vi;

else

vr(j-1)= vr(j);

end

ik(j)=(vs-(2*vr(j)))/(r+z);

it(j)=(vs/r)*(exp(-j/(r*c)));

vc(j)=(2*vr(j))+(ik(j)*z);

vcr(j)=vc(j)-vr(j)

end

[Bjorn Gustavsson]

Sure I can rewrite that code. But how could I know what you want that code to do? What parts of my previous post did you not understand? They point out two errors you have to fix for your code to work. Take them point by point:

1, define your vr variable - it seems to be some kind of initial value/boundary condition, but we cannot know what it is or what value it is supposed to have.

2, fix the range of your loop-variable j from 0:10 to 1:11,

3, now you've changed the left-hand side of the assignment where you got an error earlier, but now you will try to assign something to what element of vr?

If you want some progress in programming proficiency, I too would like to advice you to walk through the online tutorials.

[Venkatkumar M]

i want tis portion to be repetitive....

vr=0-(vi);
 
ik=(vs-(2*vr))/(r+z);
it=(vs/r)*exp(0/(r*c));
vc=(2*vr)+(ik*z);
vcr=vc-vr;
 
vr1=vcr;
 
ik1=(vs-(2*vr1))/(r+z);
it1=(vs/r)*(exp(-.1/(r*c)));
vc1=(2*vr1)+(ik1*z);
vcr1=vc1-vr1;
 
vr2=vcr1;
 
ik2=(vs-(2*vr2))/(r+z);
it2=(vs/r)*exp(-.2/(r*c));
vc2=(2*vr2)+(ik2*z);
vcr2=vc2-vr2;
 
vr3=vcr2;
 
ik3=(vs-(2*vr2))/(r+z);
it3=(vs/r)*exp(-.3/(r*c));
vc3=(2*vr2)+(ik3*z);
vcr3=vc3-vr3;

Thats it....!!!

[Rik]

1. don't use numbered variables, use arrays instead
2. don't ignore the advice of doing a tutorial, if you ignore advice, why are you asking for it?

[madhan ravi]

If reading the docs is tough then take the course which Steven linked ( https://in.mathworks.com/matlabcentral/answers/466336-illegal-use-of-reserved-keyword#comment_713394 ).

[Venkatkumar M]

i am also done with the course but i dont get any ideas. right now?

how declare it is as array and useit in looping condition

[Rik]

The process is simple: first number all your variables, then add parentheses:

vr0=0-(vi);
 
ik0=(vs-(2*vr0))/(r+z);
it0=(vs/r)*exp(0/(r*c));
vc0=(2*vr0)+(ik0*z);
vcr0=vc0-vr0;

vr1=vcr0;
%rest of the code is unchanged

Now the second step: add parentheses, making sure you don't index to 0, but start at 1:

vr(1)=0-(vi);
 
ik(1)=(vs-(2*vr(1)))/(r+z);
it(1)=(vs/r)*exp(0/(r*c));
vc(1)=(2*vr(1))+(ik(1)*z);
vcr(1)=vc(1)-vr(1);

vr(2)=vcr(1);

Now we can replace the indices with a loop iterator and add the condition for the first loop:

if n==1
    vr(1)=0-(vi);
else
    vr(n)=vcr(n-1)
end
 
ik(n)=(vs-(2*vr(n)))/(r+z);
numerator=-0.1*(n-1);
it(n)=(vs/r)*exp(numerator/(r*c));
vc(n)=(2*vr(n))+(ik(n)*z);
vcr(n)=vc(n)-vr(n);

And finally add the loop itself and some pre-allocation:

iteration=3;
vr=zeros(1,iterations);
vcr=zeros(1,iterations);
vc=zeros(1,iterations);
it=zeros(1,iterations);
ik=zeros(1,iterations);
for n=1:iterations
    if n==1
        vr(1)=0-(vi);
    else
        vr(n)=vcr(n-1);
    end
    
    ik(n)=(vs-(2*vr(n)))/(r+z);
    numerator=-0.1*(n-1);
    it(n)=(vs/r)*exp(numerator/(r*c));
    vc(n)=(2*vr(n))+(ik(n)*z);
    vcr(n)=vc(n)-vr(n);
end

And now you need to add comments about what your code is doing and why.

[Venkatkumar M]

my code is on modeling of transmission line.

Thank you very much

Answer 2

[Matt J]

You have written your code with C-language syntax. Matlab and C are not the same language.

[Venkatkumar M]

will u please wirite this in matlab code and post pls

[Matt J]

It should be something pretty easy for you to do yourself, assuming you've been through Matlab's Getting Started documentation.

https://www.mathworks.com/help/matlab/getting-started-with-matlab.html