Dumb Decimation Technique?

On Jan 11, 6:37 pm, jim <".sjedgingN0sp"@m...@mwt.net> wrote:
>...
> That said, yes the DecB method will tend to have a
> longer filter or in other words will tend to have more additions and
> multiplications for each pixel out. So how is that going to be more
> efficient?
>...

In a practical implementation, DecB won't have more operations because
it will use the polyphase decomposition. If DecB's filter length is N
* L, DecA's input bandlimit filter length is N, and the number of
input samples is P, DecA will require

 N * P    operations (ignoring cost of 2-point interpolator)

and DecB will require

 N * P * (L / M)      operations

>..
> > This means the effect
> > of the filter response that has been applied to the input signal also
> > gets compressed by 1/L and imaged; thus, you can't simply multiply the
> > two frequency responses.
>
> Why not? You don't seem to have any problem conceptualizing doing that
> with DecB after scaling up and stuffing with zeroes..
>..

OH -- you mean multiply the  *compressed and imaged* input filter
response by the Bartlett response.