FIR gain

"Jerry Avins" <jya@ieee.org> wrote in message
news:409d68eb$0$3000$61fed72c@news.rcn.com...
> Philip Newman wrote:
>
> > I have an FIR filter with the system equation y[n] =
> > 0.25x[n]+0.5x[n-1]+0.25x[n-2] which gives an impulse response of h =
[0.25
> > 0.5 0.25]
> >
> > without resorting to Z-transform analysis, how can I work out the gain
of
> > the filter at DC and at the half nyquist frequency?
> >
> > I think the gain of the filter at DC is the sum of the impulse response,
in
> > this case 1.  but what about the half-nyquist?
> >
> > I can think of it as a series of the impulse input, [1 0 0 0 0 ...] and
then
> > stepping through every other one (ie half) and summing that?
> >
> > I'm not sure how I am supposed to do this,
> >
> > If anyone can help, please let me know
> >
> > cheers
> >
> > Phil
>
> The gain at DC is easy: the sum of the coefficients, in your case, one.
> People throw "Nyquist frequency" around sloppily. Do you mean the
> sampling rate, or half of it?
>
> At f = Fs/2 (the simpler case), there are two samples per cycle. Assume
> a cosine, so the first sample is 1, the second, -1, and the third, 1.
> Multiplying by the coefficients and adding, the result is 0. Check again
> 90 degrees away. With a sine, the samples are all 0; again the result is
> 0; that's the response.
>
> If by half Nyquist you mean Fs/4, there are 4 cycles per sample. In that
> case, again using a cosine, the samples will be 1, 0, -1. Again, the
> result is zero. With a sin, the samples are 0, 1, 0. The result is .5.
> The response is the geometric mean of the cos result and the sin result.
> The filter's attenuation is 6 dB.
>
> The coefficients describe a binomial filter. The class has interesting
> properties, not the least of which is being intuitive. They make decent
> but hardly optimum low-passes.
>
> Jerry
> -- 
> Engineering is the art of making what you want from things you can get.
> &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
>

thanks Jerry, good answer.

You might wish to know that I posted a solution for my previous thread about
Satellite Range.

Cheers

Phil

Philip Newman wrote:
 > "Jerry Avins" <jya@ieee.org> wrote in message
 > news:409d68eb$0$3000$61fed72c@news.rcn.com...

   ...

>>The response is the geometric mean of the cos result and the sin result.

That's wrong. Square root of the sum of the squares is right. It's
basically a rectangular-to-polar transformation.

> thanks Jerry, good answer.

I'm pleased to hear it.

> You might wish to know that I posted a solution for my previous thread about
> Satellite Range.

I noticed. Good.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;