How do you calculate the average value of one half-cycle for a sine wave?

How do you calculate the average value of one half-cycle for a sine wave?

• A. 0.636 x peak
• B. Peak x 0.5
• C. 1 x peak
• D. 0.25 x peak

Answer: (A)

The key idea is that the average value over a half-cycle is found by averaging the voltage across the time interval of half a period. For a sine wave with peak value Vp, the instantaneous voltage is v(t) = Vp sin(ωt). Over one half-cycle, t runs from 0 to T/2. So the average is (1/(T/2)) ∫_0^{T/2} Vp sin(ωt) dt. Evaluating this integral gives 2Vp/π, because ∫_0^{π} sin x dx = 2 and ωT = 2π. Numerically, 2/π ≈ 0.6366, so the average value is about 0.6366 times the peak. This is why the correct result is 0.636 × peak.

The key idea is that the average value over a half-cycle is found by averaging the voltage across the time interval of half a period. For a sine wave with peak value Vp, the instantaneous voltage is v(t) = Vp sin(ωt). Over one half-cycle, t runs from 0 to T/2. So the average is (1/(T/2)) ∫_0^{T/2} Vp sin(ωt) dt. Evaluating this integral gives 2Vp/π, because ∫_0^{π} sin x dx = 2 and ωT = 2π. Numerically, 2/π ≈ 0.6366, so the average value is about 0.6366 times the peak. This is why the correct result is 0.636 × peak.