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Mon Dec 3 18:31:07 CET 2007

## data filter coefficient

The constraint is: we don't care about the delay, but attenuation
shouldn't be too big. What about this: pick the pole at half the bit
rate, and round upto the next power of 2.
t <- 1/sqrt(2) = (1 - 2^(-p)) ^ t
EDIT: how to pick p ?
it's easier to use this approach, where we require the decay time to
be such that a the response will drop below the 1/2 threshold in one
symbol time:
(1 - 2^(-p)) ^ t < 1/2
where t is the number of samples in a symbol. this is equivalent,
since the t in the previous formula is related to half the baud rate.
if t is large (in our case it's 64), the linear term is the one that
dominates the lhs, so the above can be approximated by
(1 - 1 + t2^(-p))
which gives an expression:
p = log_2 (2t)

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