# Validity of superposition when summing powers from each harmonic

In Boylestad's Introductory Circuit Analysis 13th edition page 1176, there's an example about working out the total power dissipated by a circuit fed a nonsinusoidal signal. The signal is decomposed into:

• DC component: 63.6 V
• Fundamental: 70.71 V RMS
• Second harmonic: -29.98 V RMS

And the circuit is drawn like below in order to apply superposition.

(Not shown in the drawing above: the phase angle of the second harmonic is then changed to -90 so that all sources have the same polarity)

The current and average power for each component are found like this:

• DC component
I0 = 10.6 A
P0 = I02R = (10.6 A)2 (6 Ω) = 674.2 W

• Fundamental
I1 = 1.85 A ∠-80.96°
P1 = I12R = (1.85 A)2 (6 Ω) = 20.54 W

• Second harmonic
I2 = 0.396 A ∠-174.45°
P2 = I22R = (0.396 A)2 (6 Ω) = 0.941 W

And the total RMS current and total average power are found to be:

Irms = square root ((10.6 A)2 + (1.85 A)2 + (0.396 A)2) = 10.77 A
PT = Irms2R = (10.77 A)2 (6 Ω) = 695.96 W = P0 + P1 + P2

Why the total average power equals the sum of the powers from each component if the superposition theorem can't be applied to power? I understand PT = Irms2R, what I can't understand is why it's valid to sum P0 + P1 + P2.

(For context: I'm reviewing a few fine points I may have overlooked during EE undergraduation years ago)

• The author should have stated that the excitation sources are orthogonal. Try the example with 3 DC sources. Aug 9, 2020 at 18:17
• @sstobbe I tried that and the sum is wrong, as I would expect. Aug 9, 2020 at 18:33

Why the total average power equals the sum of the powers from each component if the superposition theorem can't be applied to power?

This is a consequence of Parseval's theorem, which says that the the integral of the square of a function is equal to the sum of the square of its Fourier series.

Put another way, the power in the signal is the same whether you represent it in the time domain or the frequency domain.

We know this: -

$$P_{TOTAL} = I_{RMS}^2\cdot R$$

And we know that $$\I_{RMS}\$$ equals: -

$$\sqrt{I_0^2+I_1^2+I_2^2}$$

Hence: -

$$P_{TOTAL} = (I_0^2+I_1^2+I_2^2)\cdot R$$

$$= I_0^2R+I_1^2R+I_2^2R$$

Which is the sum of the powers from each component