Question

Consider a voltage v = Vdc + vac where Vdc = a constant and the average value of vac = 0. Apply the integral definition of RMS to prove the formula RMS =√ [Vdc 2 +(RMS of vac) 2 ]. Hint: v2 = [(Vdc) 2 + 2 *vac* Vdc +(vac) 2 ] and the average value of vac* Vdc = 0.

Answer 1

Proof is as follows

Proof:

Given that ,
`V = V_(ac) + V_(dc)`

for any function f with period T, RMS is given by

`RMS = \sqrt{(1)/(T)\int\limits^T_0 {[f(t)]^(2) } \, dt }`

In our case, function is
`V = V_(ac) + V_(dc)`

`RMS = \sqrt{(1)/(T)\int\limits^T_0 {[V_(ac) + V_(dc)]^(2) } \, dt }`

Now open the square term as follows

`RMS = \sqrt{(1)/(T)\int\limits^T_0 {[V_(ac)^(2) + V_(dc)^(2) + 2V_(dc)V_(ac)] } \, dt }`

Rearranging terms

`RMS = \sqrt{(1)/(T)\int\limits^T_0 {V_(dc)^(2) } \, dt + (1)/(T)\int\limits^T_0 {V_(ac)^(2) } \, dt + (1)/(T)\int\limits^T_0 {2V_(dc)V_(ac) } \, dt }`

You can see that

• second term is square of RMS value of Vac

• Third terms is average of VdcVac and given is that average of
  `V_(ac)V_(dc) = 0`

so

`RMS = \sqrt{(1)/(T)TV_(dc)^(2) + [RMS~~ of~~ V_(ac)]^2 }`

`RMS = \sqrt{V_(dc)^(2) + [RMS~~ of~~ V_(ac)]^2 }`

So it has been proved that given expression for root mean square (RMS) is valid