增加迭代次数是不是也会增加在卡尔曼滤波器中获得错误输出的机会?
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【中文标题】增加迭代次数是不是也会增加在卡尔曼滤波器中获得错误输出的机会?【英文标题】:Does increasing number of iterations also increase chances of getting erroneous output in Kalman filter?增加迭代次数是否也会增加在卡尔曼滤波器中获得错误输出的机会? 【发布时间】:2018-06-02 20:00:12 【问题描述】: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,这是一个非常简单的代码。 现在,我将循环计数器的步长从 1 更改为 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收敛然后发散?或者它是特定于代码的还是特定于其他一些属性的。
【问题讨论】:
我不认为这是与c 相关的问题,而是关于过滤器本身的问题。 【参考方案1】:以前,我使用 SSE,但随着迭代次数的增加会导致更多错误。但是,现在,我只是从原始值中减去最终结果值,它显示了正确的输出。
【讨论】:
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