That would also depend on what you were measuring that was down 33 dB. If
you are measuring distortion as a percentage of power on a scale calibrated
in dBm (typical of an RF spectrum analyzer) the formula is 10 * log
(P1/P2), and -33 dB would be .05% distortion. On the other hand, a
*voltage* change of -33 dB would be 20 * log (E1/E2), which would be
2.23872% . Converting a power change to a voltage change implies the square
root of the power to impute the voltage and current, therefore:
33 /10 = 3.3
Anti-log of 3.3 = 1995.2623
Square root of 1995.2623 = 44.668359
Inverse of 44.668359 = .0223872 = 2.23872%
I guess the question is: is THD measured in voltage or power?
Ron Castro N6AHA
Chief Technical Officer
Results Radio, LLC
----- Original Message -----
From: "Grant Youngman" <nq5t@comcast.net>
To: "'Discussion of Ten-Tec Equipment'" <tentec@contesting.com>
Sent: Friday, March 17, 2006 4:26 PM
Subject: Re: [TenTec] Further Orion Sub Receiver -- Some results
That's incorrect. -33 dB is 2.4 % THD.
0.05 % THD would be -66 dB.
I'll be honest, I don't know if you're right or not. I was using the
computation that Lankford used in the article.
He referred to it as % harmonic distortion, not THD. If you'd share the
formula for one harmonic relative to the fundametal it would be
appreciated.
Grant/NQ5T
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