ElectricalSignalshardCourseworkPE / Advanced

RMS Value of a Sinusoid with DC Offset

A signal is $v(t) = 5 + 10\sin(\omega t)\ \text{V}$ — a $5\ \text{V}$ DC offset with a $10\ \text{V}$ peak sinusoid superimposed.

Find the rms value of $v(t)$ .

Given

Vdc	5 V	DC component
Vpk	10 V	Peak of AC component

Hint 1

RMS reflects heating effect, and mean-square values of uncorrelated components add.

Hint 2

V_rms = √(V_DC² + V_AC,rms²).

Hint 3

Convert the sinusoid's peak to rms first: 10/√2 = 7.07 V.

Worked solution — try the problem first

Orthogonal components combine in an rms (root-sum-square) sense, not by simple addition — the DC and the sinusoid are uncorrelated, so their mean-square values add:

$V_{rms} = \sqrt{V_{DC}^2 + V_{AC,rms}^2}$

AC component rms. For a sinusoid, rms is peak over √2:

$V_{AC,rms} = \frac{10}{\sqrt{2}} = 7.071\ \text{V}$

Combine:

$V_{rms} = \sqrt{5^2 + 7.071^2} = \sqrt{25 + 50} = \sqrt{75} = 8.66\ \text{V}$

$\boxed{V_{rms} = 8.66\ \text{V}}$

Two errors to avoid. Adding directly gives $5 + 7.07 = 12.07$ V, which is too large. Using the peak of the AC term without the $\sqrt2$ gives $\sqrt{125} = 11.2$ V. Both overstate the heating effect the signal would actually produce in a resistor — which is what rms means.

Concepts:RMS valuesSuperposition of componentsSignal analysis

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