Looping problem in this program

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Venkatkumar M
Venkatkumar M 2019-6-28
重新打开: Venkatkumar M 2022-1-21
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 个评论
Geoff Hayes
Geoff Hayes 2019-6-28
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?
Venkatkumar M
Venkatkumar M 2019-6-28
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

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采纳的回答

Image Analyst
Image Analyst 2019-6-28
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 个评论
Image Analyst
Image Analyst 2019-6-28
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.
Venkatkumar M
Venkatkumar M 2019-6-30
Capture3.PNG
Compare img with the first program
now do u able get it?

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