How to avoid error in anonymous function
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I am trying to predict the robot position using the 3 steps below but get error:
Fx = @(x,u) jacobian(f(x,u),x);
Fu = @(x,u) jacobian(f(x,u),y);
f = @(x, u) Fx*x(t-1) +Fu*u(t-1);
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Walter Roberson
2022년 5월 7일
The jacobian of a function with respect to a single variable is the same as the derivative with respect to that variable. So Fx is diff(f, x) and Fu is diff(f, y). But f is not a function of y, so diff(f, y) is 0. So your f simplifies to
f = diff(f, x) * x(t-1)
Except that x is obviously a function of t, so you are taking the derivative with respect to a function.
By examination it becomes clear that the simplest function that can satisfy this is f(x, u) = 0, and there is no real reason to try for anything more complicated.
답변 (2개)
Walter Roberson
2022년 5월 6일
f = @(x, u) Fx(x,u)*x(t-1) + Fu(x,u)*u(t-1);
댓글 수: 10
Torsten
2022년 5월 7일
You still did not answer the main question: Is f a function, a function handle ? Is f difficult such that derivatives are only available by numerical differencing or easy such that it's possible to get them analytically ?
Ken
2022년 5월 7일
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Walter Roberson
2022년 5월 9일
편집: Walter Roberson
2022년 5월 9일
What you want to do cannot be programmed in MATLAB.
This is a different statement than saying that the assignment cannot be completed in MATLAB.
In MATLAB it is impossible for an anonymous function to refer to itself. I explained that in detail earlier. To refresh your memory:
At the time you build the anonymous function that will be assigned to f, MATLAB will look at all mentioned functions and variables, and will take copies of the variables that exist (except for the named parameters), and will mark the other ones as undefined. f does not exist at the time the right hand side of the @ is evaluated, so it gets marked as undefined in the anonymous function. The f that the anonymous function gets assigned to has no connection to the f inside the anonymous function. When you execute the anonymous function, at the point that it needs to call f, it sees that inside of the anonymous function that f is marked as undefined and it will error. Under no circumstances will it look to say "Oh, but f is defined now, we will use that!". NEVER
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