Illegal use of reserved keyword

Asked by Venkatkumar M

Venkatkumar M (view profile)

on 10 Jun 2019
Latest activity Commented on by Venkatkumar M

Venkatkumar M (view profile)

on 14 Jun 2019
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 by Matt J

Matt J (view profile)

on 10 Jun 2019

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

Venkatkumar M

Venkatkumar M (view profile)

on 10 Jun 2019
will u please wirite this in matlab code and post pls
Matt J

Matt J (view profile)

on 10 Jun 2019
It should be something pretty easy for you to do yourself, assuming you've been through Matlab's Getting Started documentation.

Answer by Bjorn Gustavsson

Bjorn Gustavsson (view profile)

on 10 Jun 2019
Edited by Bjorn Gustavsson

Bjorn Gustavsson (view profile)

on 10 Jun 2019

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

Venkatkumar M (view profile)

on 13 Jun 2019
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

Rik (view profile)

on 13 Jun 2019
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