Looping problem in this program
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    Venkatkumar M
 le 28 Juin 2019
  
    
    
    
    
    Rouvert : Venkatkumar M
 le 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 commentaires
  Geoff Hayes
      
      
 le 28 Juin 2019
				Venkatkumar - please clarify what you mean by The Ampersand symbol (&), from start to end, has to be put into a loop. Also, where is n defined? Is that the iterating variable in your for loop? Can you show all of your code?
Réponse acceptée
  Image Analyst
      
      
 le 28 Juin 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
and this link also:
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
3 commentaires
  Image Analyst
      
      
 le 28 Juin 2019
				Try:
for n = 2 : length(VLi)
        VLi(n)=VRr(n-1);
        VRi(n)=VLr(n+1);
end
As for "Every n of  VLin =VRin is equal to zero (how to define it)" I don't have the slightest idea what that means.  Perhaps get someone else to look over the wording to help you epxlain it better.
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