Sinusoidal Sources & Parameters

The waveform that powers the modern world — its anatomy, its parameters, and how two sinusoids compare.

Circuit AnalysisElectrical Engineering Year 2Free preview
⏱️ About 16 min

Wall outlets deliver a voltage that swings between positive and negative peaks sixty times every second — but what exactly are the numbers that describe that motion?

💡
The big idea: A sinusoidal source is fully specified by three numbers — amplitude, angular frequency, and phase — and once two sinusoids share the same frequency, only their phase difference decides which one leads and which one lags.
🎯 By the end, you'll be able to
  • Write the general sinusoidal form v(t) = Vm cos(ωt + φ) and identify each parameter's physical meaning and SI unit
  • Convert between angular frequency ω, cyclic frequency f, and period T
  • Determine whether one sinusoid leads or lags another of the same frequency and by how many degrees
  • Convert a sine expression into cosine form using sin(θ) = cos(θ − 90°)

The general sinusoid

An AC (alternating current) source produces a voltage or current that varies periodically as a sinusoid. The canonical form every AC analysis starts from is v(t) = V_m cos(ωt + φ), where V_m is the peak amplitude (in volts, V), ω is the angular frequency (in radians per second, rad/s), t is time (s), and φ is the phase angle (in radians or degrees). The same template applies to currents: i(t) = I_m cos(ωt + φ).

\[ v(t) = V_m \cos(\omega t + \phi) \]
The general sinusoidal voltage. The same template applies to current with I_m in place of V_m.

Three parameters, one waveform

The angular frequency ω is related to the ordinary cyclic frequency f (hertz, Hz = cycles/s) by ω = 2πf, because one full cycle traverses 2π radians. The period T — the time for one complete cycle — is the reciprocal of the frequency: T = 1/f. Combining these gives T = 2π/ω.

The phase φ shifts the cosine along the time axis. A positive φ shifts the waveform earlier (peaks occur sooner); a negative φ shifts it later.

\[ \omega = 2\pi f, \qquad T = \frac{1}{f} = \frac{2\pi}{\omega} \]
Angular frequency, cyclic frequency, and period are three views of the same rhythm.

Leading and lagging

When two sinusoids share the same frequency, only their phase angles matter for comparison. Given v(t) = V_m cos(ωt + φ_v) and i(t) = I_m cos(ωt + φ_i), the phase difference is Δφ = φ_v − φ_i. If Δφ > 0, the voltage leads the current by Δφ. If Δφ < 0, the voltage lags the current by |Δφ|. The sign tells direction; the magnitude tells by how much.

Sine to cosine — the engineering convention

Phasor analysis (next lesson) treats cosine as the reference. A source written as a sine must first be rewritten as a cosine using sin(θ) = cos(θ − 90°), so that sin(ωt) = cos(ωt − 90°). Forgetting this conversion is one of the most common sources of sign errors in AC analysis.

\[ \sin(\theta) = \cos\!\left(\theta - \tfrac{\pi}{2}\right) \]
Convert every sine to a cosine before transforming to a phasor.
📝 Worked example: A wall-outlet-style source is v(t) = 170 cos(377t − 30°) V. Identify Vm, ω, f, T, and the phase. Then compare it with i(t) = 10 cos(377t + 15°) A: does the voltage lead or lag the current, and by how many degrees?
  1. Read directly from the form Vm cos(ωt + φ): Vm = 170 V, ω = 377 rad/s, φ = −30°.
  2. Frequency: f = ω / (2π) = 377 / (2π) ≈ 60.0 Hz.
  3. Period: T = 1/f = 1/60 ≈ 0.01667 s = 16.7 ms (equivalently T = 2π/ω).
  4. Phase difference Δφ = φ_v − φ_i = (−30°) − (+15°) = −45°.
  5. Because Δφ is negative, the voltage lags the current by 45° (equivalently, the current leads the voltage by 45°).
✓ Vm = 170 V, ω = 377 rad/s, f ≈ 60 Hz, T ≈ 16.7 ms, phase = −30°; the voltage lags the current by 45°.
✏️ Practice: A sinusoidal voltage has angular frequency ω = 314 rad/s. What is its cyclic frequency f in Hz? (Round to the nearest whole number.)
Hz
Solution
  1. Use f = ω / (2π).
  2. f = 314 / (2π) ≈ 314 / 6.283 ≈ 49.97 Hz ≈ 50 Hz.

Check your understanding

1. Which trigonometric function does engineering convention use as the reference for phasor analysis?
Cosine is the reference. A sine must be rewritten as cos(θ − 90°) before transforming to a phasor.
2. Given v(t) = 5 cos(ωt + 20°) and i(t) = 5 cos(ωt − 10°), the voltage ___ the current by 30°.
Δφ = 20° − (−10°) = +30°, so the voltage leads the current by 30°.
3. For v(t) = 170 cos(377t − 30°) V, the period T is approximately:
f = 377/(2π) ≈ 60 Hz, so T = 1/f ≈ 16.7 ms.
4. The expression 10 sin(ωt) rewritten in cosine form is:
sin(θ) = cos(θ − 90°), so 10 sin(ωt) = 10 cos(ωt − 90°).
🔎 Related lessons
Phasor Representation
✅ Key takeaways
  • A sinusoid is v(t) = Vm cos(ωt + φ), fully described by amplitude Vm, angular frequency ω, and phase φ.
  • Angular frequency, cyclic frequency, and period are linked by ω = 2πf and T = 1/f = 2π/ω.
  • Between same-frequency sinusoids, the phase difference Δφ = φ1 − φ2 decides who leads (Δφ > 0) and who lags (Δφ < 0).
  • Engineering uses cosine as the reference; convert any sine with sin(θ) = cos(θ − 90°).
➡️ Once a sinusoid is fully described by amplitude and phase, those two numbers can be packed into a single complex number — the phasor — which is the subject of the next lesson.
Want to test yourself on this? Try the Electrical Aptitude test →