0
\$\begingroup\$

I've been given some conflicting information with regards to the current type of a sinewave that has no negative component e.g. 20-30V pk-pk is either DC or AC.

I have been taught by my lecturers that this is in fact DC. However, talking amongst older EE graduates they seem to be very adamant that this is an AC waveform despite the fact the polarity never changes.

How would one go about calculating the Vrms for such a sinewave? I have found many different equations giving different values.

\$\endgroup\$
5
  • \$\begingroup\$ A signal can have both DC and AC components. A sine wave from 2 to 6 volts DC is a 4Vp-p sine wave component plus a 4v DC component. \$\endgroup\$
    – K H
    Commented Feb 9, 2021 at 1:38
  • 1
    \$\begingroup\$ RMS = root mean square. The procedure is always the same. First, square the waveform. Then take the mean of the square over one period. The mean is a single number. Then take the square root of the mean. That is the RMS voltage over one period. This always works no matter what the waveform is (as long as it is repetitive). \$\endgroup\$
    – user57037
    Commented Feb 9, 2021 at 1:48
  • 2
    \$\begingroup\$ Instead of arguing about whether it is AC or DC or both or neither, it is best to say that it is AC riding on DC or some such. It is an AC waveform with a DC offset. Categorization can lead to Holy Wars which are just a waste of time. \$\endgroup\$
    – user57037
    Commented Feb 9, 2021 at 1:50
  • 2
    \$\begingroup\$ The semantics usually are determined by context and who is bigger. e.g. An AC signal with small (DC) offset or a rectified DC signal with lots of (AC) ripple. \$\endgroup\$
    – D.A.S.
    Commented Feb 9, 2021 at 1:59
  • 1
    \$\begingroup\$ Semiconductor, RMS is based upon the idea of the DC heating value of a signal given the same load. In DC, power is \$\frac{V^2}{R}\$ or else \$R\,I^2\$. But if the supply's voltage varies, then you need to compute \$V_\text{RMS}^2=\frac1T\int_0^T\,V_t^2\,\text{d}t\$. You can look on wiki-rms for some variations on the theme. But the idea is the same. Many arbitrary curves can be analyzed this way. But if it is just a simple DC part plus an AC part, there are much simpler expressions to use. That's why you see different equations. \$\endgroup\$
    – jonk
    Commented Feb 9, 2021 at 6:05

1 Answer 1

Reset to default
1
\$\begingroup\$
  • for a sine wave \$V_{ac~rms}=\dfrac{V_p}{\sqrt{2}}\$
  • Vac signal with DC offset = \$\sqrt{V_{dc~rms}^2+V_{ac~rms}^2}=\sqrt{V_{dc}^2+\dfrac{V_p^2}{2} }\$
\$\endgroup\$
1
  • \$\begingroup\$ This is correct, however the student is cautioned against blindly applying this "addition in quadrature" in the general case when the inputs may be correlated or both may have non-zero mean. Doing the definite integral always yields the correct answer for the interval in question. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Feb 9, 2021 at 3:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.