Worked examples: phase differences in time and angle

Two problems — one converting an angular phase difference into milliseconds, one going the other way. The identity $\Delta t = \Delta\phi / (360° f)$ shows up in every subsequent chapter, most noticeably when we get to power factor.

Example 1 — 60° lag at 50 Hz

Two signals of the same frequency: $v_1 = 5 \sin(\omega t)$ and $v_2 = 5 \sin(\omega t - 60°)$, with $f = 50\,\text{Hz}$. How far behind v₁ does v₂ peak, in milliseconds?

1. Identify the phase difference. v₁'s phase is 0; v₂'s is −60°. The difference is $\Delta\phi = 0° - (-60°) = 60°$. v₂ reaches each peak 60° of phase after v₁ does.
2. Find the period. $T = 1/f = 1/50 = 20\,\text{ms}$. One full cycle of v₁ or v₂ takes 20 ms.
3. Scale phase to time. A full 360° cycle spans the whole period, so a 60° segment spans 60/360 = 1/6 of it:
   $$\Delta t = \frac{60°}{360°} \cdot T = \frac{1}{6} \cdot 20\,\text{ms} \approx 3.33\,\text{ms}$$
4. Sanity check. At 50 Hz a 90° shift is a quarter-period (5 ms) and a 180° shift is a half-period (10 ms). A 60° shift lands between them at ~3.33 ms. Reasonable.

Drag either phase slider to change the offset. The phase- difference readout at the bottom confirms the direction (lead vs lag) and the magnitude — useful for building the reflex.

Example 2 — measured lag in time, report in degrees

You measure two in-phase-frequency signals on an oscilloscope. Signal A crosses zero (rising) at $t = 0$; signal B crosses zero (rising) 1 ms later. Both oscillate at 250 Hz. What is the phase difference between A and B, in degrees?

1. Period. $T = 1/f = 1/250 = 4\,\text{ms}$.
2. Scale time to phase. The whole 4 ms period corresponds to 360°, so 1 ms corresponds to:
   $$\Delta\phi = \frac{1\,\text{ms}}{4\,\text{ms}} \cdot 360° = 90°$$
3. Direction. Signal B crosses zero after signal A, so B lags A by 90°. Equivalently, $v_B = V_p \sin(\omega t - 90°)$. A −90° shift is the $-\cos(\omega t)$ relationship — at t=0 it's at its minimum, not zero. Verify on the plot below.