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% [) X7 A$ e. \% h- Nf=min(abs(0.5*ay_tp1)+abs(0.5*ay_tp2));- S0 O! C6 Z2 Y, v) q
s.t.: P8 ]- v; {( |7 h8 A
0<p<10,-10<q<0,q^2-4*p<=0;2*pi/(4*p-q^2)^(1/2)-tf=0;
9 J- C& o2 B6 X2 r1 {8 W%%其中:
6 K' [; Q6 a, kw=3.75;
$ H. Z ?' n$ o3 ]; v" e! Q" Vtf=5;# y* J4 Q4 [1 i
tp1=0;
( Z G, T7 d3 f( `( `tp2=2*(pi-atan(q*(4*p-q^2)^(1/2)/(q^2-4*p)))/((4*p-q^2)^(1/2));
( ]6 Y. X: |0 H7 rA1=(4*p/(4*p-q^2))^(1/2);
% T1 a' {2 J! A! R+ }A2=((4*p-q^2)/2)^(1/2);/ z# ^, L0 E& n) t O
A3=q/((4*p-q^2)^(1/2));
" u d# T7 y8 M0 G# @
- C$ G; Q- I; o# E! T& a: xay_tp1=(w*p*exp(q*tp1/2)*A1*sin(((A2/(2^(1/2)))*tp1)+1/A3))/(1-exp(q*tf/2)*A1*sin(A2*tf-atan(A3)));
. C' b/ p: ^; t' S# Day_tp2=(w*p*exp(q*tp2/2)*A1*sin(((A2/(2^(1/2)))*tp2)+1/A3))/(1-exp(q*tf/2)*A1*sin(A2*tf-atan(A3)));( |8 c+ u% w% y8 K5 Y, g5 m
使用fmincon函数求解时显示 相邻两次迭代点的变化小于预先给定的容忍度。+ w4 X, E& O+ b
求助大佬。感谢, Y1 [3 x- C. n" ]: @) w9 G0 A- }" W
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