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Harmonic signal parameters and RMS value

Short answer

For y(t) = 10 sin(100πt + π/6) V: frequency 50 Hz, period 20 ms, amplitude 10 V, RMS 7.07 V, time lead 1.667 ms. For x(t) = 4 sin(100πt) + 2 sin(300πt) V: fundamental 50 Hz, RMS = √((4²+2²)/2) = √10 ≈ 3.16 V.

(a) A harmonic signal is y(t)=10sin⁡(100πt+π/6)y(t) = 10\sin(100\pi t + \pi/6) volt. Find its frequency, period, amplitude, RMS value and the time by which it leads 10sin⁡(100πt)10\sin(100\pi t).

(b) A signal is composed of two harmonics, x(t)=4sin⁡(2π⋅50 t)+2sin⁡(2π⋅150 t)x(t) = 4\sin(2\pi \cdot 50\, t) + 2\sin(2\pi \cdot 150\, t) volt. Find its fundamental frequency and RMS value.

(a) Single harmonic

Compare y(t)=10sin⁡(100πt+π/6)y(t) = 10\sin(100\pi t + \pi/6) with y=Asin⁡(ωt+ϕ)y = A\sin(\omega t + \phi).

  • Amplitude A=10A = 10 V
  • Angular frequency ω=100π\omega = 100\pi rad/s, so f=ω/2π=50f = \omega/2\pi = 50 Hz
  • Period T=1/f=0.02T = 1/f = 0.02 s = 20 ms
  • RMS value =A/2=10/2=7.07= A/\sqrt{2} = 10/\sqrt{2} = 7.07 V
  • Phase ϕ=π/6=30∘\phi = \pi/6 = 30^\circ. Time lead:
td=ϕω=π/6100π=1600 s=1.667 mst_d = \frac{\phi}{\omega} = \frac{\pi/6}{100\pi} = \frac{1}{600}\ \text{s} = 1.667\ \text{ms}

Answer: f=50f = 50 Hz, T=20T = 20 ms, A=10A = 10 V, RMS =7.07= 7.07 V, lead =1.667= 1.667 ms.

(b) Two harmonics

The components have frequencies 50 Hz and 150 Hz. The 150 Hz is the third harmonic of 50 Hz, so the fundamental frequency is 50 Hz (period 20 ms).

For components of different frequency, mean squares add:

xrms=A122+A222=162+42=10=3.16 Vx_{rms} = \sqrt{\frac{A_1^2}{2} + \frac{A_2^2}{2}} = \sqrt{\frac{16}{2} + \frac{4}{2}} = \sqrt{10} = 3.16\ \text{V}

Answer: fundamental 50 Hz, RMS =3.16= 3.16 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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