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1、电机建模与仿真一、电机运行原理给电机定子加电压Ue,转子加励磁电压Ua,电机输出转矩供给负载转动。电机状态空间模型方程为：dw K . .- = 了1 i - FwUHaLadi=dtdi、假设其中 Ra=1,Re=0.5,La=0.5,Le=0.5,J=10,K1=3,K2=2,F=20输入Ua=2,Ue=4，实现程序为：#include#include#includevoid rgkutta(double y,int n,double h,double t,void (*fty)(double *,double *,int) int i,j;double a4,*b,*d,*z;a0=h/
2、2;a1=a0;a2=h;a3=h;b=(double*)malloc(n*sizeof(double);d=(double*)malloc(n*sizeof(double);z=(double *)malloc(n*sizeof(double);for(i=0;i=n-1;i+)zi=yi;bi=yi;(*fty)(y,d,n);for(j=0;j=2;j+)for(i=0;i=n-1;i+)yi=zi+aj*di;bi=bi+aj+1*di/3.0;(*fty)(y,d,n);for(i=0;i=n-1;i+)yi=bi+h*di/6;free(b);free(z);free(d);ret
3、urn;void demofty(double y,double d,int n)d0=0.3*y1*y2-20*y0;d1=-2*y1-4*y0*y+8;d2=-y2+4;void main()int i;double y3=0,0,0;void (*fty)(double *,double *,int);FILE *pf;char *FileName;FileName=euler.dat;pf = fopen(FileName, w);fty=demofty;for(i=0;i100;i+)rgkutta(y,3,0.1,0,fty);fprintf(pf, %i %lf %lf%lf n
4、”,i,y0,y1,y2);fclose(pf);可以得到转速w,转子电流ia,定子电流ie稳定值分别为0.082194, 0.684980, 7.999637, 动态变化图如下二、在(二)的稳定状态下，若Ua在10秒内变为10, Ue在10秒内变为10,实现程序如下：#include#include#includedouble ua,ue;void rgkutta(double y,int n,double h,double t,void (*fty)(double *,double *,int)int i,j;double a4,*b,*d,*z;a0=h/2;a1=a0;a2=h;a3=
5、h;b=(double*)malloc(n*sizeof(double);d=(double*)malloc(n*sizeof(double);z=(double *)malloc(n*sizeof(double);for(i=0;i=n-1;i+)zi=yi;bi=yi;(*fty)(y,d,n);for(j=0;j=2;j+)for(i=0;i=n-1;i+)yi=zi+aj*di;bi=bi+aj+1*di/3.0;(*fty)(y,d,n);for(i=0;i=n-1;i+)yi=bi+h*di/6;free(b);free(z);free(d);return;void demofty
6、(double y,double d,int n)d0=0.3*y1*y2-20*y0;d1=-2*y1-4*y0*y +2*ua;d2=-y2+2*ue;void main()int i;double y3=0.082194,0.684980,7.999637;ua=2;ue=4;void (*fty)(double *,double *,int);FILE *pf;char *FileName;FileName=euler1.dat”;pf = fopen(FileName, w);fty=demofty;for(i=0;i100;i+)rgkutta(y,3,0.1,0,fty);fprintf(pf, %i %lf %lf %lf n”,i,y0,y1,y )；ua=ua+0.08;ue=ue+0.06;fclose(pf);可以得到如图10s内的转子转速的动态数值解及转子电流和定子电流数值解，第十秒钟其数 值分别为 0.243105，0.866322，18.739056。
电机建模仿真
