Stuart,

Duty cycle in 15.231 is related to a voltage ratio, therefore  20 log(duty
cycle) will provide the correct factor.  

Demonstrate it to yourself.  Start with a given value (say 100V), multiply
this by some duty cycle (say 15% or .15).  Convert the result to dB.  This
is your reference result.  Now take 20 log of a duty cycle (.15).  Convert
your given value (100V) to dB.  Add the numbers together, duty cycle dBs to
the given value dBs, and behold -- the same answer as the reference result.

Best regards,

Don

> ----------
> From:         Stuart Lopata[SMTP:[email protected]]
> Reply To:     Stuart Lopata
> Sent:         Thursday, October 18, 2001 12:00 PM
> To:   emc
> Subject:      duty cycle correction factors
> 
> 
> Part 15.231 devices use a duty cycle correction factor to adjust peak
> readings.  The duty cycle represents the fractional on-time over a given
> period of time (that must be under some limit).  Anyways, given this
> fractional time, d, how do you make the conversion to dB?
> 
> 10log(d) or 20log(d)?
> 
> There have been some misinterpretations, since the readings are made at a
> span of zero hertz (voltage readings).  Normally, a reduction in voltage
> would use the 20log scale.  However, since the duty cycle does not
> represent
> a scale down (it represents the off-time versus on-time), the 10log scale
> seems more appropriate.
> 
> I have seen conflicting documents, so would like your professional
> opinions!
> 
> Thanks,
> 
> Stuart Lopata
> 
> 
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