algorithm - 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？

Question

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，得到的错误从 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？或者它是代码特定的还是其他一些属性特定的。

Answer 1

以前，我使用 SSE，随着迭代次数的增加，它会导致更多错误。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。