# Looping problem in this program

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Venkatkumar M on 28 Jun 2019
Reopened: Venkatkumar M on 21 Jan 2022
Vs=1;
rs=1;
rl=1;
r=10;
ll=2e-3;
VRi(n)=0;
VLi(n)=0;
l=5e-3;
c=10e-6;
g=1;
zo=sqrt(l/c);
t=sqrt(l*c);
zl=(2*ll)/t;
%VOLTAGE AND CURRENT
%n=1
V1=((Vs/rs)+((2*VRi1))/(r+zo))/((1/rs)+(1/(r+zo)));
i1=((V1-(2*VRi1))/(r+zo));
VR1=(2*VRi1)+(i1*zo);
%n=2,....,n-1 (i.e n=2)
&
V2=(((2*VLi2/(zo))+((2*VRi2)/(zo+r)))/((1/zo)+(1/(r+zo)+(g)))); % Every n of VLi n =VRi n is equal to zero (how to define it)
i2=((V2-(2*VRi2)/(r+zo)));
VL2=V2;
VR2=(2*VRi2)+(i2*zo);
&
%n=n(i.e n=11)
V11=(((2*VL11/(zo))+(2*Vi/(rl+zl)))/((1/zo)+(1/(zl+rl)+(g))));
IL=((V11-(2*Vi))/(rl+zl));
V=(2*Vi)+(il*zl);
%REFLECTED VOLTAGES
%n=1
VRr1=VR1-VRi1;
%n=2,....,n-1 (i.e n=2)
&
VLr2=VL2-VLi2;
VRr2=VR2-VRi2;
&
%n=n(i.e n=11)
VLr11=V11-VLi11;
%INCIDENT VOLTAGES
%n=1
VRi1=VLr2;
%n=2,3,....,n-1 (i.e n=2)
&
VLi2=VRr1;
VRi2=VLr3;
&
%n=n(i.e n=11)
VLi11=VRr10;
Vi=-Vr;
The Ampersand symbol (&), from start to end, has to be put into a loop. How do I do it?
##### 2 CommentsShowHide 1 older comment
Venkatkumar M on 28 Jun 2019
Given code is full code
VLi(n)=VRr(n-1);
VRi(n)=VLr(n+1);
i just need this section (and the section with &)to be looped for n=2,3,...,n-1

Image Analyst on 28 Jun 2019
Put
for k = 1 : numIterations
before the code that you want to be in the loop, and
end
after the code. numIterations is how many times you want to do the loop. Please look at
I have no idea what the ampersands are supposed to do so it's difficult to advise you further. For example, maybe
VRi1=VLr2;
%n=2,3,....,n-1 (i.e n=2)
&
VLi2=VRr1;
VRi2=VLr3;
&
%n=n(i.e n=11)
VLi11=VRr10;
means
result = VRi1 == VLr2 && ...
VLi2 == VRr1 && ...
VRi2 == VLr3 && ...
VLi11 == VRr10;
but I really have no idea what your intent is.
And for testing equality of floating point numbers, See the FAQ
Venkatkumar M on 30 Jun 2019
Compare img with the first program
now do u able get it?

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