《机电传动控制》第四周作业
一、控制策略：
1、三相异步电动机启动采用自耦变压器降压方法，取Ku=0.65；
2、调速策略才用调频调速。n=800，Ka=0.545；n=0，Kb=0.075；n=-600，Kc=0.395。（基本是试的）
3、反相使电机反转。
二、关键代码（与源代码的改动）
parameter Real Ku=0.65;
parameter Real Ka=0.545;
parameter Real Kb=0.075;
parameter Real Kc=0.395;
if time <= 100 then
u_A = 0;
u_B = 0;
u_C = 0;
f_s = 0;
elseif time<=200 then
f_s = f_N*Ka;
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Ku*Ka;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Ku*Ka;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Ku*Ka;
elseif time<=1900 then
f_s = f_N*Ka;
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Ka;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Ka;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Ka;
elseif time<=1950 then
f_s = f_N*Ka;
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Ka;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Ka;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Ka;
elseif time<=2850 then
f_s = f_N*Ka;
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Ka;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Ka;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Ka;
elseif time<=3400 then
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Kb;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Kb;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Kb;
f_s = f_N*Kb;
elseif time<=3560 then
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Ku*Kc;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Ku*Kc;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Ku*Kc;
f_s = f_N*Ku*Kc;
elseif time<=4950 then
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Kc;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Kc;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Kc;
f_s = f_N*Kc;
elseif time<=5000 then
f_s = f_N*P*Ka;
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Ka*P;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Ka*P;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Ka*P;
elseif time<=5800 then
f_s = f_N*Ka;
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Ka;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Ka;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Ka;
else
u_A = u_N * 1.414 * sin(2*Pi*f_s*time/1000)*Kb;
u_B = u_N * 1.414 * sin(2*Pi*f_s*time/1000-2*Pi/3)*Kb;
u_C = u_N * 1.414 * sin(2*Pi*f_s*time/1000-4*Pi/3)*Kb;
f_s = f_N*Kb;
end if;
三、仿真结果
四、小结
完成任务在5.8s左右，最大电流在160左右。存在要求匀速但是速度有小范围变动导致时间不那么精准。
