On 1/29/07, Chuckk Hubbard <[EMAIL PROTECTED]> wrote:
> w(n)-fb1*w(n-1)-fb2*w(n-2)=x(n)
> ff1*w(n)+ff2*w(n-1)+ff3*w(n-2)=y(n)
> so,
>
> y(n)-fb1*y(n-1)-fb2*y(n-2)=ff1*x(n)+ff2*x(n-1)+ff3*x(n-2)

I got it now.  I don't understand why both expressions are useful, but
I see that they're the same.

-Chuckk

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