# Why does block LMS have the same performance as LMS?

The block LMS and conventional LMS have the same convergence rate and the same misadjustment. I am having trouble wrapping my head around this. The block LMS uses a more accurate estimate of the gradient vector at each iteration. Conceptually, why does a better gradient estimate have no benefit on the descent?

• The reduction of the variance of the gradient estimate is exactly compensated by the fact that the filter coefficients are only updated once per block. Note, however, that the restrictions on the step size to avoid filter divergence are stronger for the block version of the algorithm. This in many cases leads to a performance of the block LMS which is lower than the performance of the conventional LMS. – applesoup Mar 11 '18 at 10:59
• Oh, welcome to DSP.SE, by the way! :-) – applesoup Mar 11 '18 at 17:12
• (I am OP - forgot I already had an account). I don't have the reputation to comment, but I still wanted to thank applesoup for the reply. Makes perfect sense. Instead of steps every iteration, you make one "big" step every $L$ iterations, where this big step is an aggregate of $L$ gradient estimates. So you benefit on variance of the lump gradient estimate, but lose on the "granularity" of the estimates, so to speak. – Probably Mar 11 '18 at 19:16
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – MBaz Mar 11 '18 at 23:54
• @Probably Please go through account merging to remove one of these accounts. – Peter K. Mar 12 '18 at 0:39

once the LMS has converged on a reasonably stable equilibrium for $h_n[k]$, they won't move around that much. then it doesn't matter so much how long the block is. and the only difference between block-LMS and the plain-old ordinary LMS is the block size. (the block size for the latter is 1.)