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[10GBASE-T] clarification on voltage level


In response to some questions at last weeks meeting (and to stimulate some reflector activity), I wanted to clarify the launch voltage level

For 10 dBm, ideal PAM10 (random data and PSD ~sinc^2) and 100 ohm load

Vrms = sqrt(P*R) = 1 V

Vpeak = Vrms*sqrt(3*(M-1)/(M+1))  where M is the number of PAM levels
          = 1.57 V

Vpp = 2*Vpeak = 3.13 V

For 9.5 dBm

Vrms = .944 V

Vpeak = 1.48 V

Vpp = 2.96 V


Note: this is based on an ideal PAM signal, any filtering in the transmit path will change the PAR and peak voltages.

regards
Bill


William W. Jones, Ph.D.
Director of Systems Engineering
SolarFlare Communications, Inc.
949-581-6830, ext. 2550
mobile: 619-405-2445
fax: 949-581-4695
wjones@solarflare.com