This might be a dumb question, but what values change when applying a DC offset to an AC signal?

Does Vpk represent the maximum voltage of the signal, in which case it increases directly with the DC offset? Or is it 1/2 the value of the signal Vpk-pk, in which case it would stay the same?

And in that case would Vavg change too?

  • \$\begingroup\$ When you add these together you have an instantaneous voltage v(t) that likes between Vmin and Vmax with Vavg being the DC level and Vmax-Vmin being the Vac p-p level. The Vrms is now the sum of the AC = (Vac p-p)/(2sq.rt(2)) and the original Vdc component \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 17 '16 at 4:48
  • \$\begingroup\$ Yes, Vpk is your peak AC voltage + DC offset. \$\endgroup\$ – winny Oct 17 '16 at 7:55
  • \$\begingroup\$ \$ V_{rms} = \sqrt{V_{ac_{rms}}^2 + V_{dc}^2} \$ \$\endgroup\$ – rioraxe Oct 17 '16 at 10:59

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