Power in Alternating-Current Circuits — Question 8

Question 8

Finite measurement windows can bias a sinusoidal power average

For sinusoidal voltage and current, $p(t)=VI[\cos\phi+\cos(2\omega t-\phi)]$, with V and I rms values. A meter averages from t0 to t0+Delta.

1. Derive the finite-window mean and bound its deviation from true mean power.

2. Find the shortest positive window length that removes the oscillating term for every starting phase.

Question 8 - Solution

Finite measurement windows can bias a sinusoidal power average

(a) Method and calculation

$$\bar p_\Delta=VI\cos\phi+\frac{VI}{2\omega\Delta}[\sin(2\omega(t_0+\Delta)-\phi)-\sin(2\omega t_0-\phi)].$$$$\boxed{|\bar p_\Delta-P|\le VI/(\omega\Delta)}.$$

(b) Interpretation and further calculation

$$2\omega\Delta=2\pi n\Rightarrow\Delta=n\pi/\omega.$$$$\boxed{\Delta_{\min}=\pi/\omega=T/2}.$$

Physical check

Power oscillates at twice the source frequency, so a half-cycle voltage window already gives an exact average.