How to Derive the RMS Value of a Sine Wave with a DC Offset

How to Derive the RMS Value of a Sine Wave with a DC Offset

Guest author offers helpful refresher

BY ADRIAN S. NASTASE, PhD

ElectronicProducts.com is pleased to welcome guest author Adrian Nastase. Mr. Nastase is a Principal Electrical Engineer with Newport Corporation and also the voice behind the popular and very-well written blog, masteringelectronicsdesign.com, which is where this article first originated:

I noticed a question posted on one of Yahoo’s Q&A sites, asking what is the RMS value of a sine wave with a DC offset. The chosen answer as being “the best” was actually wrong. The next comment, which was trying to correct “the best” answer, was wrong too.

I am not going to post the Yahoo link here. What I can do, is to show how to derive the RMS value of such waveform.

Let’s start with the RMS value of a sine wave, with no DC offset, which is shown in Figure 1. It is well known that the RMS value of a sine wave is 0.707 times the signal peak level, but how can you prove this?

As shown in this article, MasteringElectronicsDesign.com: How to Derive the RMS Value of a Trapezoidal Waveform, or other RMS articles in this website, let’s start with the RMS definition.

(1)

The sine wave time dependency can be described by the following function:

(2)

T is the function period, or T = 1/f where f is the waveform frequency. Also, a1 is the amplitude.

Replacing (2) in (1), and calculating the integral over a full period T, we find the RMS value squared as in the following equation:

(3)

The standard method to calculate a squared sine integral is to transform it into its double angle equivalent, using a trigonometric identity usually called the power-reduction formula.

(4)

So the RMS squared becomes

(5)

If you’re wondering why the sine term is zero in the previous equation, that’s because

(6)

Therefore, the RMS value of a sine wave with offset zero is the following well known formula,

(7)

Now, let’s look at a sine wave with a DC offset. This waveform is shown in Figure 2 and is described by the following function.

(8)

where with a0 I noted the DC offset. Applying the RMS definition, the RMS squared can be written as:

(9)

Let’s calculate the integral.

(10)

(11)

(12)

(13)

(14)

Therefore, the RMS value of a sine wave with a DC offset is given by the following expression.

(15)

The immediate verification of the validity of this expression is the RMS value of a sine wave with zero DC offset. Indeed, when a0 = 0 V, the RMS level reverts back to equation (7), which is 0.707 of the sine amplitude.

Expression (15) can also be verified by comparing it with Parseval’s Theorem. This theorem says that the integral of the square of a function is equal with the integral of the squared components of its spectrum. In effect, the theorem states that the total energy of a waveform can be found in the total energy of the waveform’s Fourier components. In our case, a0 is the DC level, or the frequency zero component, and a1 is the fundamental frequency. There are no other Fourier components. As such, the RMS value of a sine wave with a DC offset as given by expression (15) is correct.

A warm thanks to Mr. Nastase for sharing his article with the ElectronicProducts.com community. If you’d like to read more of his articles, head to his site: masteringelectronicsdesign.com.

Have a question about the article? Contact Mr. Nastase directly at sanastase@gmail.com ■