03 · Foundational

Frequency, angular frequency & phase from time shifts

Convert repetition time into cycles per second, introduce angular frequency without losing units, then define phase through relative position within a cycle. Derive the sign of phase for a delayed cosine, distinguish phase lead/lag from time delay, handle modulo-2π equivalence, and explain why a single fixed phase difference requires equal frequency.

130–165 minFoundational
In this lesson
  • Derive frequency f=1/T0 from the fundamental period and use hertz as cycles per second.
  • Define angular frequency ω=2πf, preserve the radian label, and convert consistently among T0, f and ω.
  • Represent a sinusoidal time shift by x(t)=A cos(ωt+φ) while keeping phase distinct from the later phasor representation.
  • Derive Δφ=−ωτ for a delayed waveform y(t)=x(t−τ) under the cosine convention and explain the sign physically.
  • Convert phase difference among radians, degrees and fractions of a cycle.
  • Explain phase equivalence modulo 2π and why adding an integer number of cycles does not change a sinusoid.
  • Explain why two unequal-frequency sinusoids do not maintain one constant phase difference over time.
  • Distinguish frequency/phase inferred from a finite measurement record from exact properties of an ideal mathematical model.
Starting question

A periodic signal repeats every 2 ms. Another copy is delayed by 0.5 ms. How do “500 Hz” and “−90°” encode those two facts?

Frequency turns repetition time into cycles per second. Phase turns a relative position inside one cycle into an angle. They are related through time, but they are not the same quantity and they must not be introduced as mysterious labels detached from the waveform.

03.1Frequency is the reciprocal of the fundamental period

If a non-constant periodic signal has fundamental period T0, define its fundamental frequency as

f0=1/T0

The SI unit is the hertz:

1 Hz=1 s−1

The engineering interpretation is one cycle per second. Thus

T0=2 ms → f0=1/(2×10−3 s)=500 Hz

The reciprocal relation immediately gives a sanity check: a shorter period must correspond to a higher frequency.

03.2Angular frequency counts radians of phase per second

One complete cycle corresponds to an angle of 2π radians. Define angular frequency

ω=2πf=2π/T

For 500 Hz:

ω=2π·500≈3141.59 rad/s

The radian is dimensionless in SI, but writing rad/s is valuable because it makes the angular meaning explicit and prevents confusing f in hertz with ω.

QuantityRelationUnit
periodT=1/fs
frequencyf=1/THz
angular frequencyω=2πfrad/s

03.3Phase is a position within the cycle, not yet a phasor

For a real cosine waveform we may write

x(t)=A cos(ωt+φ)

Here φ is an initial phase angle: it tells us where the cosine is within its cycle at the chosen time origin t=0.

Changing the time origin changes the numerical phase even though the physical waveform has not changed. Phase therefore always depends on a reference: a time origin, another signal, or a stated cosine/sine convention.

03.4Derive the sign: a positive time delay gives negative phase in the cosine convention

Take the reference

x(t)=A cos(ωt)

Delay it by a positive time τ:

y(t)=x(t−τ)=A cos[ω(t−τ)]
y(t)=A cos(ωt−ωτ)

Comparing with A cos(ωt+φ) gives

Δφ=φy−φx=−ωτ=−2π τ/T

Figure 3.1 — A time delay becomes negative phase under the cos(ωt+φ) convention

The one question is: why does a right shift of T/4 correspond to φ=−π/2 rather than +π/2?

T/4tx(t)y(t)

The blue curve y(t)=x(t−T/4) reaches each maximum one quarter-period after the dark reference x(t). It therefore lags in time; under cos(ωt+φ), its relative phase is −π/2.

For a quarter-period delay τ=T/4:

Δφ=−2π(T/4)/T=−π/2=−90°

The negative sign is not arbitrary: the delayed waveform reaches the same landmarks later in time.

03.5Radians, degrees and fraction of a cycle are three descriptions of the same relative position

1 cycle=2π rad=360°

For equal-frequency waveforms, a relative time shift Δt corresponds in magnitude to the fraction |Δt|/T of a cycle. With the cosine sign convention for a delayed waveform:

Δφ=−2πΔt/T=−360°·Δt/T
DelayRelative phase
T/8−45° = −π/4
T/4−90° = −π/2
T/2−180° = −π
T−360° = −2π, equivalent to 0°

03.6Phase is equivalent modulo a complete cycle

Because cosine is 2π-periodic in its angle,

cos(θ)=cos(θ+2πk),   k∈ℤ

Therefore phase angles differing by integer multiples of 2π describe the same sinusoidal waveform:

30° ≡ 390° ≡ −330°   modulo 360°

When reporting a phase difference, engineers often wrap it to a chosen interval such as (−180°,180°] or [0°,360°). The interval is a reporting convention, not new physics.

03.7A constant phase difference requires equal angular frequency

Consider

x(t)=A cos(ω1t+φ1)
y(t)=B cos(ω2t+φ2)

The difference between their angle arguments is

Δθ(t)=(ω2−ω1)t+(φ2−φ1)

If ω1≠ω2, the relative angle changes with time. There is no single fixed phase difference valid for all time. This is why “signal B is 30° behind signal A” is incomplete unless the signals have the same frequency or the statement specifies an instant/frequency-dependent phase.

When the frequencies are equal, the time-dependent terms cancel and the phase difference is constant.

03.8Measurement turns exact parameters into estimates

An ideal mathematical sinusoid can have exactly specified T, f and φ. A real instrument sees a finite, noisy, bandwidth-limited record.

A practical period estimate might use two comparable landmarks:

T̂=(tn+m−tn)/m
f̂=1/T̂

Using many cycles can reduce timing quantization/noise effects compared with measuring only one cycle, but drift or nonstationarity can then matter. The hat reminds us that this is an estimate from data rather than a parameter handed to us by the model.

Likewise, phase estimated from a measured delay inherits timing uncertainty and uncertainty in the frequency/period estimate.

03.9Worked examples and mastery exercises

Example A. T=20 μs gives f=50 kHz and ω≈314.159 krad/s.

Example B. At 1 kHz, a 125 μs delay is one eighth of a 1 ms period, hence a phase lag of −45°.

Example C. A phase written as 450° is equivalent to 90° after subtracting one full 360° cycle.

  1. A periodic waveform has T0=4 ms. Find f0 and ω0.
  2. A 2 kHz sinusoid is delayed by 50 μs. Find its phase shift in degrees and radians under the cosine convention.
  3. What delay corresponds to a −120° phase shift at 10 kHz?
  4. Convert 3π/4 rad to degrees and 270° to radians.
  5. Reduce 765° to an equivalent phase in [0°,360°) and in (−180°,180°].
  6. Two sinusoids have frequencies 1 kHz and 1.01 kHz. Can their phase difference remain constant? Explain from Δθ(t).
  7. A recorder observes 100 cycles over 20 ms. Estimate average period and frequency.
  8. Why is the phase angle of a waveform meaningless unless a reference/time origin and a convention are specified?
Solutions

1. f0=250 Hz; ω0=500π rad/s≈1570.8 rad/s.

2. T=0.5 ms. The delay fraction is 0.05/0.5=0.1 cycle, so Δφ=−36°=−π/5 rad.

3. T=100 μs. A 120° lag is one third of a cycle, so τ≈33.33 μs.

4. 135° and 3π/2 rad.

5. 45° in both requested intervals.

6. No. Their relative angle contains (ω2−ω1)t and therefore drifts linearly with time.

7. Average period=20 ms/100=0.2 ms; frequency=5 kHz.

8. Phase describes position relative to a chosen angular/time reference; changing that reference changes the numerical phase without necessarily changing the physical waveform.

Chapter 04 will assemble amplitude, offset, angular frequency and phase into the complete real sinusoidal waveform and derive its important values rather than merely memorizing the formula.