《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。