我在网上找到了这个非常简单的卡尔曼滤波器代码
double frand() {
return 2*((rand()/(double)RAND_MAX) - 0.5);
}
int main() {
//initial values for the kalman filter
float x_est_last = 0;
float P_last = 0;
//the noise in the system
float Q = 0.022;
float R = 0.617;
float K;
float P;
float P_temp;
float x_temp_est;
float x_est;
float z_measured; //the 'noisy' value we measured
float z_real = 0.5; //the ideal value we wish to measure
srand(0);
//initialize with a measurement
x_est_last = z_real + frand()*0.09;
float sum_error_kalman = 0;
float sum_error_measure = 0;
for (int i=0;i<30;i++) {
//do a prediction
x_temp_est = x_est_last;
P_temp = P_last + Q;
//calculate the Kalman gain
K = P_temp * (1.0/(P_temp + R));
//measure
z_measured = z_real + frand()*0.09; //the real measurement plus noise
//correct
x_est = x_temp_est + K * (z_measured - x_temp_est);
P = (1- K) * P_temp;
//we have our new system
printf("Ideal position: %6.3f \n",z_real);
printf("Mesaured position: %6.3f [diff:%.3f]\n",z_measured,fabs(z_real-z_measured));
printf("Kalman position: %6.3f [diff:%.3f]\n",x_est,fabs(z_real - x_est));
sum_error_kalman += fabs(z_real - x_est);
sum_error_measure += fabs(z_real-z_measured);
//update our last's
P_last = P;
x_est_last = x_est;
}
printf("Total error if using raw measured: %f\n",sum_error_measure);
printf("Total error if using kalman filter: %f\n",sum_error_kalman);
printf("Reduction in error: %d%% \n",100-(int)((sum_error_kalman/sum_error_measure)*100));
return 0;
}
对于像我这样的新手来说,这是一个很容易理解的代码。现在,我将循环计数器的步长从1改为3,得到的误差从1.7减少到0.4。我的问题是:增加卡尔曼滤波器的迭代次数是否会使其先收敛,然后发散?或者它是特定于代码的还是特定于某些其他属性的。
发布于 2018-06-03 06:30:08
以前,我使用SSE,随着迭代次数的增加,它会导致更多的错误。然而,现在,我只是从原始数据中减去最终的结果值,它显示了正确的输出。
https://stackoverflow.com/questions/50660687
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