Illegal use of reserved keyword

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Venkatkumar M 2019-6-10

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https://ww2.mathworks.cn/matlabcentral/answers/466336-illegal-use-of-reserved-keyword

评论： Venkatkumar M 2019-6-14

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]
}

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采纳的回答

Bjorn Gustavsson 2019-6-10

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https://ww2.mathworks.cn/matlabcentral/answers/466336-illegal-use-of-reserved-keyword#answer_378555

编辑：Bjorn Gustavsson 2019-6-10

在 MATLAB Online 中打开

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

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Rik 2019-6-13

在 MATLAB Online 中打开

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