[ARC5] Lopsided modulation

Tom Lee tomlee at ee.stanford.edu
Wed Feb 28 13:49:25 EST 2018


Happy to oblige. Sorry for the ASCII-formatted equation, but luckily 
there aren't too many terms. The power attenuation at a frequency f, 
relative to the power at center frequency f0, is given by

1 + Q^2*[k - (1/k)]

where k = f/f0.

For small deviations df from the center frequency (that is, for df/f0 << 
1), k is well approximated by 2df/f0, so in that case the power 
attenuation is approximately

1 + 2Q^2*[df/f0]

Plug in whatever numbers you want. You'll see that it is not easy to 
obtain any significant sideband asymmetry for realistic (loaded!) Q 
values. As Dennis said, it's not even worth doing the calculation. But 
for those who want the equation to check for themselves, here it is.

--Cheers
Tom

-- 
Prof. Thomas H. Lee
Allen Bldg., CIS-205
420 Via Palou Mall
Stanford University
Stanford, CA 94305-4070
http://www-smirc.stanford.edu
650-725-3383 (public fax; no confidential information, please)

On 2/28/2018 10:17 AM, Tim wrote:
> Yep!
> I'm waiting for Prof. Lee to hit us with the all-defining modulation 
> math.  (leaving it as an exercise for THIS student would not be 
> productive!)
>
> Also. I can imagine asymmetrical sideband power on either side of an 
> AM radio carrier but I cannot imagine what an "asymmetrical voice" 
> could be. ;o)
>
> Tim
> N6CC
>
> On Wed, Feb 28, 2018 at 8:17 AM, Kenneth G. Gordon 
> <kgordon2006 at frontier.com <mailto:kgordon2006 at frontier.com>> wrote:
>
>     On 28 Feb 2018 at 22:58, Dennis wrote:
>
>     > Let´s drop the PA filtering explanation, guys.  It´s not even
>     worth a back of the envelope
>     > calculation.  Not in the ballpark.
>
>     I 100% agree, Dennis.
>
>     Ken W7EKB
>
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