I am reading in the litterature that LMS is more stable than RLS. But RLS is far more faster in convergence. So my concern is how to be sure that my RLS algorithm will be stable when I am doing channel equalization ?

What are the causes of the instability ? How to make the RLS immune to instability ?

I am aware there exist RLS version like SRLS that are assured to be alway stable.

  • $\begingroup$ I am reading in the litterature that LMS is more stable than RLS, to put it in wikipedian: [citation needed]. Literature that claims that should probably also have a framework to explain it! $\endgroup$ – Marcus Müller Jul 2 at 8:23

Short Answer:

Basic RLS (no forgetting, no weird weighting, etc.) is ALWAYS Lyapunov stable. If the regressor sequence for the LS problem is persistently exciting--which is data and problem dependent, not algorithm dependent--then RLS is exponentially stable. So I don't know what you mean by "LMS is more stable than RLS"--more stable in what sense?

More Details:

Basic RLS has the discrete-time update equations

\begin{align} % \theta_{k+1} &= \theta_k + P_{k+1}\phi_k^T(y_k-\phi_k\theta_k), \\ % P_{k+1} &= P_k - P_k\phi_k^T(I+\phi_kP_k\phi_k^T)^{-1}\phi_kP_k. \end{align}

Note that the covariance update is decoupled from the estimate update and is also monotonically decreasing. If the regressor sequence is persistently exciting, then the origin of the covariance update is exponentially stable. If nothing can be said about the regressor, then the origin of the covariance update is only guaranteed to be Lyapunov stable.

Next, define the error $\tilde \theta_k = \theta_k-\theta$, where $\theta$ is the true value of the parameters. Then you can show that

$$\tilde \theta_{k+1} = P_{k+1}P_0^{-1}\theta_0.$$

In other words, the error is entirely dictated by the covariance, and hence is also exponentially stable or Lyapunov stable as the covariance is.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy