Calculating DC from signal avg or rms value required?

Question

If we want to calculate a dc value of any given signal, should the dc value be RMS or average value?

Answer 1

The average value of a signal $s(t)$ is given by

$$\overline{s(t)}=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}s(t)\, dt\tag{1}$$

The RMS (root mean square) value is the square root of the average of the squared signal:

$$\left(\overline{s^2(t)}\right)^{1/2}=\lim_{T\to\infty}\sqrt{\frac{1}{T}\int_{-T/2}^{T/2}s^2(t)\, dt}\tag{2}$$

Note that given two signals $s_1(t)$ and $s_2(t)=-s_1(t)$ we have

$$\overline{s_1(t)}=-\overline{s_2(t)}\tag{3}$$

i.e., their average values have opposite signs, whereas

$$\left(\overline{s_1^2(t)}\right)^{1/2}=\left(\overline{s_2^2(t)}\right)^{1/2}\tag{4}$$

because both signals have the same power.

Property $(3)$ indicates that it is the average value $(1)$ of a signal that is useful for defining its DC value. Also note that if you used the rms value as a DC value, all sinusoids and all other periodic functions would always have a non-zero DC level.

As a final note, the DC value of a signal is not equal to the value of its Fourier transform at zero frequency. This is explained in this answer.

Answer 2

RMS is the square root of the mean of the squares, and is typically used as a measure of average power (as the squares are the power units, and the square root of that bringing it back to a magnitude quantity). DC is simply the mean or the component of the signal that is constant with time. If there is only DC, then the DC value and RMS value will be the same, but for an arbitrary or unknown signal, the mean should always be used to determine the DC component.