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$ h# I, c& W* bf=min(abs(0.5*ay_tp1)+abs(0.5*ay_tp2));
B7 W1 k& q+ L7 A! Vs.t.% n& X: W- k$ Y! i6 r4 w' t
0<p<10,-10<q<0,q^2-4*p<=0;2*pi/(4*p-q^2)^(1/2)-tf=0; Y0 d' u/ s+ Y6 N: Y4 D# n8 B
%%其中:
& L+ _0 _- ~$ u# Bw=3.75;
9 z; {/ [& ?( i0 n, G' p" }- h) n; ctf=5;9 m. p/ F5 _0 f$ Y
tp1=0;2 A, s+ m9 P) }" h/ V
tp2=2*(pi-atan(q*(4*p-q^2)^(1/2)/(q^2-4*p)))/((4*p-q^2)^(1/2));
K5 d: h N2 C! IA1=(4*p/(4*p-q^2))^(1/2);! l3 q7 }3 S+ j
A2=((4*p-q^2)/2)^(1/2);2 L: s) x! \, `9 f0 W
A3=q/((4*p-q^2)^(1/2));0 N, r$ F& h( G' z. W+ h
- @& p1 a6 P; J- r! C# k% say_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)));6 k6 \5 T- d( G3 v+ z& p" `- A
ay_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)));
. t3 }/ v% g. F8 l) [! f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。5 y- }! c6 U( f! D
求助大佬。感谢, h; M2 R& J, f# w9 ]0 J% Y; {
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