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Fourier transform of a DC signal

Short answer

X(ω) = 2π δ(ω). Magnitude spectrum is an impulse of area 2π at ω=0, zero elsewhere; phase spectrum is zero. Obtained via duality: δ(t)↔1 gives ∫1·e^{-jωt} dt = 2π δ(ω).

Obtain the Fourier transform of a DC signal whose amplitude is unity at all time instants, and hence, plot its magnitude and phase spectrum.

x(t) = 1 for all t. This signal is not absolutely integrable, so the transform is found by the limit of a decaying signal or by duality.

Method (duality)

The impulse has the transform δ(t) ↔ 1. By the inverse transform, δ(t) = (1/2π) ∫ 1·e^(jωt) dω, or

∫ e^(jωt) dω = 2π δ(t)

Interchange the roles of t and ω: ∫ 1·e^(-jωt) dt = 2π δ(ω) (δ is even).

X(ω) = 2π δ(ω)

Alternative check (limit method)

x(t) = lim (a→0) e^(-a|t|) ↔ lim 2a/(a² + ω²). This has area ∫ 2a/(a²+ω²) dω = 2π for every a, and its width shrinks to zero as a → 0, which is exactly 2π δ(ω).

Spectra

  • Magnitude: an impulse of area 2π at ω = 0, and zero everywhere else.
  • Phase: 0 (the spectrum is real and positive).
|X(w)|
      ^ (2*pi)
      |
      |
------+------> w        phase = 0 for all w
      0

All the power lies at zero frequency, which is natural: a DC signal does not vary in time.

✓ Answer checked by Claude Opus 5.5 (high effort)

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