how to find integral absolute error

조회 수: 80 (최근 30일)
bhanu kiran vandrangi
bhanu kiran vandrangi 2021년 8월 8일
댓글: Walter Roberson 2021년 8월 8일
i need to derive a cost function from one of performance indices ( i took IAE as my performance index)
my reference value(steady state) is 7.5 and my output is a function of (t,d,e) where d,e constants vary between (0.1)
how to do the integration to find integral absolute error

채택된 답변

Walter Roberson
Walter Roberson 2021년 8월 8일
syms d e t
f(t) = d*exp(-t^2*e)
f(t) = 
reference = 7.5;
final_time = 10
final_time = 10
IAE = int(abs(f(t) - reference), t, 0, final_time)
IAE = 
subs(IAE, [d, e], [1/5, 2/3])
ans = 
  댓글 수: 2
bhanu kiran vandrangi
bhanu kiran vandrangi 2021년 8월 8일
thanks
instead of substituting the values of d,e can i use 'IAE' that we got ,as cost function to optimize the values of 'd' , 'e'
Walter Roberson
Walter Roberson 2021년 8월 8일
syms d e t
f(t) = d*exp(-t^2*e)
f(t) = 
reference = 7.5;
final_time = 10
final_time = 10
diff_function = f(t) - reference;
IAE = int(abs(diff_function), t, 0, final_time)
IAE = 
best_d = solve(diff_function,d)
best_d = 
diff_function2 = subs(diff_function, d, best_d)
diff_function2 = 
0
best_e = solve(diff(diff_function2, e), e)
best_e = 
0
best_d = subs(best_d, e, best_e)
best_d = 
best_f = subs(f, [d, e], [best_d, best_e])
best_f(t) = 
Which is to say that for this particular function, f(t), with d and e in that particular relationship to the function, that the best function (smallest IAE) occurs when the function is the constant function equal to the reference value.
The approach used here is the standard calculus approach: the critical points of a function are the points at which the derivative of the function are equal to 0. When the function to be optimized is an integral, then the derivative is the thing that was being integrated... so solve f(t)-reference == 0 for d and e. If the function never equals the reference over the permitted range of d and e, then the best IAE occurs at one of the boundary points.

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Programming에 대해 자세히 알아보기

태그

제품


릴리스

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by