Minimum settling time control design through direct search optimization

Minim um settling time con trol design through direct searc h optimization Emil e Simon No vem b er 19, 2018 The aim of this v ery brief third v ersion is to presen t succinctly , but clearly with t w o animations, tw o optimization results fro m the second v ersion of the pap er a v ailable on h ttp://arxiv.org/ a bs/1109.5966v2 whic h can still b e con- sulted for more details. A future and m uc h impro v ed version of the pap er will b e dep osited on arXiv at the b eginning o f 2 012. The ob jective considered here is to optimiz e PID coefficien ts x = [ K p , K i , K d ] to minimize the following ob jectiv e function: f ( x ) = the rise time (= time needed for the resp onse z ( x, t ) to reac h 0 . 98) divided by 100 (the fixed maxim um time v alue for whic h the resp onse is ev aluated) + the maximum deviation of z ( x, t ) (out of the fo llowing settling range: z max ( t ) = 1 . 02 for t > 0 and z min ( t ) = 0 . 98 for t > rise time). 1 First figure o f unit step resp onse optimization, starting from PID co effi- cien ts obtained by Ziegler-Nichols : Press play to start the animation. NB/ The resp onse z ( x, t ) is drawn in gr een when the corresp onding ob- jectiv e v alue f ( x ) is b etter than b est v alue pr eviously found b y the direct searc h optimization metho d, and in red otherwise. The hor izon tal blac k dashed lines represen t the settling ra ng e defined ab ov e. 2 0 20 40 60 80 100 0 0.5 1 1.5 time t [s] time response z(t) K p = 4.059, K i = 0.4388, K d = 9.386 : f(x) = 0.5421 Second figure of unit step resp onse optimization, starting from unstabi- lizing random PID co efficien ts: time [s] Both optimziations lead to the same solution, with rise time around 11.94s and maxim um deviation almost zero (can b e fine-tuned with ano t her ob j ec- tiv e f unction, the settling time, but this will ha v e an insignifican t effect o n the solutions here). 3 0 20 40 60 80 100 −6 −4 −2 0 2 4 6 K p = 0.4395, K i = 0.5447, K d = 0.7855 : f(x) = 6.835