evaluate multiple variables with matlab code, (relooping?)
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Hey all, Im trying to evaluate an equation with multiple variables,4, whereby 2 are known and 2 are unknown.
so the equation is f=(c/(2*pi))*sqrt(A/(Vl))
f and A are known V and l are unknown
%
A=0.001019;
c = 343;
V= 0.001:0.0001:0.020;
i=0;
l=0.1:0.001:0.6;
j=0;
fmax=58.7;
f=0;
fwanted = 58.7
(%code----)
for i=1:length(V)
for j=1:length(l)
f = (c/(2*pi))*sqrt((A)/(V(i)*l(j)));
if fwanted ==f
V(i)
l(j)
end
end
end
With the code above i only get one solution. I actually want to code to run like this: take the first element of V, keep it constant and run the function with the columns of l. take the second element of V and run it again and so on and display the solutions at the end. whereby creating multiple combinations of V and l. since my knowlegde is limited on this. can anybody help?
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Walter Roberson
2012-12-20
0 个投票
8 个评论
kwadwo
2012-12-20
Walter Roberson
2012-12-20
Your code has
if fwanted ==f
but fwanted and f are both floating point numbers. They are unlikely to ever exactly equal each other, for reasons described in the link. You should modify your code to test whether they are "close enough" to each other for your purposes.
kwadwo
2012-12-20
Walter Roberson
2012-12-20
That code is not increasing i and j at the same time: it is using one particular i, running through all the j for that one i, then going on to the next value of i, running through all the j for that second i, and so on.
With regards to saving the values: try
Vl = [];
for i=1:length(V)
for j=1:length(l)
f = (c/(2*pi))*sqrt((A)/(V(i)*l(j)));
if abs(fwanted-f) < 0.02
Vl(end+1,:) = [V(i) l(j) i j];
end
end
Later when you are satisfied that the order is acceptable, change the assignment to
Vl(end+1,:) = [V(i) l(j)];
kwadwo
2012-12-26
Roger Stafford
2012-12-26
That is not an efficient method you have used to find solutions to your equation, Kwadwo. It is important to realize that when there are more unknowns than equations, as in this case, in general there will be infinitely many solutions.
You can see this by solving for the product of V and l as follows:
f = c/(2*pi)*sqrt(A/(V*l))
2*pi*f/c = sqrt(A/(V*l))
(2*pi*f/c)^2 = A/(V*l)
V*l = A*(c/2/pi/f)^2
Now that the product of V and l is determined from known quantities, then V can be set to any desired value out of infinitely many possibilities (except zero) and the corresponding l can be found as the above product divided by such a V value. That uniquely determines the infinite continuum of all possible solutions.
Note: It is a good idea to avoid the use of lowercase letter 'l' as a variable name since it is so easily confused with the numeral '1'.
Roger Stafford
2012-12-27
My comment remains roughly the same as it was before. You now have one equation and four unknowns to vary: Ln, Vn, V, and h. Setting f to 58.4 will still leave an enormous three-dimensional space in which the unknowns can vary.
It is possible to manipulate with your equation so as to isolate h on one side of an equivalent equation with everything else on the other side. Then as you vary Ln, Vn, and V you will get all possible solutions. Beware however - three degrees of freedom take a lot of exploring. For example if each of the three take on a hundred possible values, that gives you a combined total of a million solutions to ponder over. I have a feeling you have better things to devote your time to.
By the way, your attempt to explore that four-dimensional space with four nested for-loops would have to go through over two billions steps. That may account for why you didn't succeed with it. However it might have helped if you had allocated space for VlnhVn before entering the loops.
Roger Stafford
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