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f=min(abs(0.5*ay_tp1)+abs(0.5*ay_tp2));
3 k. |- j5 T* Z3 Y4 `! t- I4 ds.t.
( g2 I/ R: s, R0 q* t0<p<10,-10<q<0,q^2-4*p<=0;2*pi/(4*p-q^2)^(1/2)-tf=0;6 o% Z7 Y! u; q( x
%%其中:
1 B) q# M) Y& V. Cw=3.75;
, b- g' \* C/ V: _, _' x7 Ltf=5;: G; C. E. n0 @. }
tp1=0;2 R) u+ _! O* u; Z% l
tp2=2*(pi-atan(q*(4*p-q^2)^(1/2)/(q^2-4*p)))/((4*p-q^2)^(1/2));
" u! t& S- ?. R+ lA1=(4*p/(4*p-q^2))^(1/2);
6 c. T& T! J( K8 AA2=((4*p-q^2)/2)^(1/2);' ~9 `2 E2 o$ u {7 F2 D
A3=q/((4*p-q^2)^(1/2));+ h6 [. G' O$ `9 { r
( [9 a4 R% q. C. m( ?, day_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)));
9 `: r/ }. `' X+ K) n4 X8 Eay_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)));
5 O% O1 y3 f0 f( ]6 x/ W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。6 S. G0 \; r9 O+ J% M3 H& P
求助大佬。感谢 w* z% t* }) h4 s
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