http://jbrwww.che.wisc.edu/presentations/stability.pdf

Hi Ali,

Based on your result, instead of estimating “x”, you have successfully estimated “−x”. However, you can easily correct the sign in your estimation algorithm.

My mistake. Now I see that. The reversal is triggered somewhere around 0.7 sec. Naturally, I'd suspect the result has something to do with the quadratic characteristic of y. At first, the EKF estimates √(y), and then it quickly reverses to −√(y). Because the EKF estimates the square root, there is ambiguity in the estimation problem.

Hi Ali,

You could detect the zero crossings (when x^2==0) and change the sign of the square-root accordingly. However, if x only "hits" zero from above and continues to remain positive for instance, you might have to include information about the input signal into your zero crossing detection: In the case described, u changes sign from negative to positive.

I hope that helps!

Christoph

Remark/Edit: You have to know the sign of the initial condition, however.

There are many scenarios possible here. However, given that the measurement is purely a function of a known quantity, you can estimate the state directly through numerical methods if you know the plant model exactly besides the initial conditions. Obviously, no estimator can work on your problem. Of course, if the plant wasn't exactly an integrator, or you only knew the initial condition to be within a distribution, you will have uncertainty and drift in your numerical estimates.

EKF is strong technique and it is old.

There are nonlinear estimators such as UKF. When relation between output variable is nonlinear you may consider UKF. You may find new nonlinear observers as well.

Let integral(u)=U  obtained by any suitable numerical method (quadrature/trapezoidal/simpson).

Now x=U+x0

we also have x=+/-sqrt(y) approximately if there is noise in the measurement.

So the estimate of x0=-U+sqrt(y) or x0=-U-Sqrt(y)

Now you can use any filtering method you want on both of these equations. By constancy of x0, you eventually determine which of the two filters is estimating the correct x0.

Obviously, you can construct any suitable maximum likelihood estimator more easily if you know the probability density of the process or measurement noise. That is what I meant by filtering method.

There is no way that you can recover the sign of the state. 

BR,

J

If you oscillated/moved the control input u over a sufficiently large range, you should be able to ascertain the sign of the initial condition as the output goes through zero using the method I outlined above.

Sorry, I have been changing my mind every 5 min. If you increase the confident in your model, you should be able to recover the sign of the state. Try to increase R or reduce Q and see if this really happens. I think your system is observable. but I have not done the maths.

https://www.math.ucdavis.edu/~krener/1-25/10.IEEETAC77.pdf 

And wouldn't it be better to use a particle filter? At the end of the day, you should have some kind of bimodal distribution, so trying to use an unimodal distribution (around x_k with covariance P_k) is not great. The particle filter will allow you different distributions, but the price to pay is the required computation power. 

I would argue that this system is observable.  Let's get back to the fundamentals. Observability means that by knowing the input and output, you can recover the states. Knowing y, gives you the x (minus its sign), while the sign can be inferred from the direction of xdot at zero crossing, (if xdot > 0, x is increasing so at zero crossing, it should go from negative to positive and vice versa). 

At x = 0, the system is observable knowing y alone, since the only possible value for y =0 is x = 0 with no sign confusion.

Note that this system is nonlinear and nonlinear observability is a totally different beast, which has little to do with the (local) linear observability matrix rank test. Using the original definition, it can be mathematically shown that this system is indeed observable.

Now, whether EKF works or not is a separate question. This is a property of the filter not the plant. Here both Q and R should be very small. You might need a second order approximation like a UKF.

The observability is a relative concept, e.g., the systems \Sigma(A,C) are unobservable and but are oberservable under the some perturbations on the matrix A and C. Considered that the modeling of the system dynamics are with some modeling errors, there isn't exist a distinct line between the observability and unobservability and then it is important to measure precisely on the observe ability.

In my paper arXiv1705.08064(On Controllable Abundance Of Saturated-input Linear Discrete Systems),the controllable abundance measuring precisely the control ability of the linear discrete systems is proposed, and the new measure can be generalized to measure the observe ability, maybe named as the observable abundance. Based on the observable abundance, the dynamics of the controlled plants can be designed and optimized on the manufacturing period of the controlled plants, and the better observability can be gotten and the state estimator with the better performance and robustness can be achieved.

Hi Ali,

I am glad that you found another option that works.

I am just sharing my thoughts.

Non-linear observability is not same as that for linear models. Non-linear observability is determined using Lie derivatives. If a system is found to be unobservable then there would be no guarantee for an estimator to work for all possibilities. In such a case, it is wise to discover and use other related states which gives the required estimates keeping the system observable.

Kalman filter is a very good option for the linear systems, but for non-linear system, EKF, SPKF or PF are used. EKF approximates the non-linearities with the first order Taylor's series ignoring the higher order terms and so if the system is highly non-linear then EKF can result in estimation errors. In such a case SPKF would be a good option.

Both EKF and SPKF assumes noise to be Gaussian which is not a condition in PF. However, PF requires lot of computations.

I hope this helps!

Deepak