Hi Rich,

2x voltage would actually be 6dB (this is one of those little confusing
things they kinda slip past you. 3dB is double POWER, which would only be
sqrt(2) times the voltage. 6dB is double voltage. This never quite made
sense to me because I don't see why one should necessarily assume a
resistve load, but AFAIK, this is a standard, and you use it even if I is
constant and power is proportional to voltage).

Why would you need a square-root algo? Why not just a multiply routine to
scale the output as you say?

Sean

At 12:42 PM 3/9/00 -0800, you wrote:
>Ok... so assuming that the (filtered) data drives eight LED's directly,
>each bit represents 2x voltage, isn't this 3dB per led? Although not
>dBm... so I'd still need to weight the output... (now I'm thinking of that
>divide-and-conquer square root algorithm).
>
>-- Rich
>

|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
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