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Re: [TenTec] Further Orion Sub Receiver -- Some results

To: "Discussion of Ten-Tec Equipment" <tentec@contesting.com>
Subject: Re: [TenTec] Further Orion Sub Receiver -- Some results
From: "Ron Castro" <ronc@sonic.net>
Reply-to: Ron Castro <ronc@sonic.net>,Discussion of Ten-Tec Equipment <tentec@contesting.com>
Date: Fri, 17 Mar 2006 17:05:37 -0800
List-post: <mailto:tentec@contesting.com>
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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