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Message ID: 307     Entry time: Wed Aug 29 11:06:30 2018
 Author: Koji Type: General Category: General Subject: RF AM RIN and dBc conversion

0. If you have an RF signal whose waveform is $1 \times \sin(2 \pi f t)$, the amplitude is constant and 1.

1. If the waveform $[1+0.1 \sin(2 \pi f_{\rm m} t)] \sin(2 \pi f t)$, the amplitude has the DC value of 1 and AM with the amplitude of 0.1 (i.e. swing is from 0.9 to 1.1). Therefore the RMS RIN of this signal is 0.1/1/Sqrt(2).

2. The above waveform can be expanded by the exponentials.

$\left[-\frac{1}{2} i e^{i\,2\,\pi f t} + 0.025 e^{i\,2\,\pi (f-f_{\rm m}) t}- 0.025 e^{i\,2\,\pi (f+f_{\rm m}) t} \right] - {\rm C.C.}$

Therefore the sideband carrier ratio R is 0.025/0.5 = 0.05. This corresponds to 20 log10(0.05) = -26dBc

In total, we get the relationship of dBc and RIN as ${\rm dBc} = 20 \log_{10}(\rm{RIN}/\sqrt{2})$, or R = RIN/sqrt(2)

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