Hey all!!
I am trying to simulate railway traffic signal having 2 platforms. I am encountering an error which I am not able to understand.
Kindly someone please help me build this model. Please reply if you are not able to understand my symbols.
MATLAB: Unkown error in stateflow simulink
simulinkstateflow
Related Solutions
You can open up the Model Explorer and modify all your Stateflow data (input, output, parameters, local, etc.). This is shown in the following link:
- Sebastian
Basic calc? I hope this is not homework. But at least you made some effort.
You have formulated the problem as
q = int( int( a*x^2 + b*y^2,0,2*x*k), 0,5)
Where a, b, q are all known. The problem is, you want to solve for k, where k is part of that upper limit on y. We can do this on paper, at last most of it, but easy enough to do entirely in MATLAB too.
syms x y kq = 125;a = 0.5;b = 0.1;The symbolic toolbox is smart enough to recognize the decimal values 0.5 and 0.1 as their true fractional equivalents.
We can do the inner integral on y easily enough.
int(a*x^2 + b*y^2,y,0,2*x*k)ans =(k*x^3*(4*k^2 + 15))/15So we should see the double integral is simple too.
dubint = int(int(a*x^2 + b*y^2,y,0,2*x*k),x,0,5)dubint =(125*k*(4*k^2 + 15))/12What we see is a cubic polynomial in k.
solve(dubint == 125,'maxdegree',3)ans = (269^(1/2)/8 + 3/2)^(1/3) - 5/(4*(269^(1/2)/8 + 3/2)^(1/3)) 5/(8*(269^(1/2)/8 + 3/2)^(1/3)) - (3^(1/2)*(5/(4*(269^(1/2)/8 + 3/2)^(1/3)) + (269^(1/2)/8 + 3/2)^(1/3))*1i)/2 - (269^(1/2)/8 + 3/2)^(1/3)/2 (3^(1/2)*(5/(4*(269^(1/2)/8 + 3/2)^(1/3)) + (269^(1/2)/8 + 3/2)^(1/3))*1i)/2 + 5/(8*(269^(1/2)/8 + 3/2)^(1/3)) - (269^(1/2)/8 + 3/2)^(1/3)/2Kind of a symbolic mess. But vpa can reduce it simply.
vpa(ans)ans = 0.70611518025253139152901981827648 - 0.35305759012626569576450990913824 - 2.0307508428749447828026672218255i - 0.35305759012626569576450990913824 + 2.0307508428749447828026672218255iNow, could I have done that on paper? Yes, except for the last part, where I'd probably have thrown the cubic polynomial into roots to solve.
expand(dubint)ans =(125*k^3)/3 + (625*k)/4>> roots([125/3 0 625/4 -125])ans = -0.353057590126266 + 2.03075084287495i -0.353057590126266 - 2.03075084287495i 0.706115180252531 + 0iCould you have done it without using the symbolic toolbox? Yes. It is a bit easier to do the integration symbolically, and I am a bit lazy. I'd certainly have done the integrations on paper as calc 101.
Verification step:
k = 0.706115180252531;ulimy = @(x) 2*x*k;integral2(@(x,y) a*x.^2 + b*y.^2,0,5,0,ulimy)ans = 125.000000000043Ok. Could I have solved it using fzero and integral2? Do I really need to? Sigh.
k = fzero(@(k) integral2(@(x,y) a*x.^2 + b*y.^2,0,5,0,@(x) 2*x*k) - 125,[0.1,5])k = 0.706115180252333It seems to have worked this way too. I suppose if the integrand were more complicated, you would have had no choice in the matter.
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