Re: Need help with a circuit

From: John Fields (jfields_at_austininstruments.com)
Date: 12/29/04 

Date: Tue, 28 Dec 2004 18:39:39 -0600

On 28 Dec 2004 14:31:25 -0800, "Dark Alchemist"
<GhostOfACPast@gmail.com> wrote:

>Hmmm, I was able to get all the way to the last part on my own but even
>looking at what you did (where I see an and gate) I am lost.
>
>So, I have two ouputs (A, B) and your logic table works. Now all I
>need to do is take the A AND B = 0 or 1 in this instance. I noticed
>that 0<x<1 1 1 . 1 and 1 is 1 while 0 and 1 or 1 and 0 = 0
>so would it not be simpler to (A AND B) * sqrt(1-x^2)? The product of
>that would be 0 or the sqrt part. 

---
Yes, that's essentially what I've done by ANDing the comparator
outputs and giving you that AND as an enable.  Not knowing what your
square-rooter-subtractor circuitry looks like, that was as far as I
could go.  If you can perform the ANDing in your computational
circuitry, go for it!  
-- 
John Fields