b**l
发帖数: 33123 | 1
I am trying to follow Haykin's notation, which turns out missleading. His
desired signal is actually observation. I hope that you are not using that
book.
So I think here d(n)=x(n), which means your observation is x(1),...,x(n), your
filter output is the estimate of the true signal at time n.
No! it is your data. Filtering is used to extract signals from noisy data.
No, the main reason is to sequentially update data. Otherwise you have to
recalculate the matrix in batch formula. | c*******r
发帖数: 112 | 2
your filter output is the estimate of the true signal at time n.
>> If you look at the formula min E{(d(n)-h(n)^T * x(n)} it is clear that
d(n) is not the observations but the desired signal. Just for an example, in
the case of filtering, if d(n)=x(n) then obviously the best estimate is
h(n)=[1 0 ... 0] with zero error.
>> exactly, the desirable signal is not noisy, so the d(n) is not x(n)!
>> Yes, one reason is computional complexity and another reason is
non-stationarity. As a result you hav
【在 b**l 的大作中提到】
:
: I am trying to follow Haykin's notation, which turns out missleading. His
: desired signal is actually observation. I hope that you are not using that
: book.
: So I think here d(n)=x(n), which means your observation is x(1),...,x(n), your
: filter output is the estimate of the true signal at time n.
: No! it is your data. Filtering is used to extract signals from noisy data.
: No, the main reason is to sequentially update data. Otherwise you have to
: recalculate the matrix in batch formula.
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