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FT of DC; meaning of ESD and PSD

Short answer

For constant x(t)=A, X(jω)=2πA δ(ω). ESD is |X(jω)|^2, showing how the finite energy of an energy signal is distributed over frequency (J/Hz). PSD is lim_{T→∞} |X_T(jω)|^2/T, giving the average power distribution of a power signal (W/Hz).

Calculate Fourier transform of a constant DC function. What are the physical meanings of Energy spectral Density and Power spectral Density?

Fourier transform of a constant (DC) signal

Let x(t)=Ax(t) = A for all tt. This signal is not absolutely integrable, so the integral ∫Ae−jωtdt\int A e^{-j\omega t}dt does not converge in the ordinary sense. We find the transform using the impulse function and duality.

Step 1: Consider X(jω)=2πA δ(ω)X(j\omega) = 2\pi A\,\delta(\omega) and take its inverse FT:

x(t)=12π∫−∞∞2πA δ(ω) ejωt dω=A ej0⋅t=A\begin{aligned} x(t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} 2\pi A\,\delta(\omega)\,e^{j\omega t}\,d\omega \\ &= A\,e^{j0\cdot t} = A \end{aligned}

using the sifting property of δ(ω)\delta(\omega).

Step 2: Since the inverse transform of 2πAδ(ω)2\pi A\delta(\omega) is the constant AA, the transform pair is

A  ⟷  2πA δ(ω)A \;\longleftrightarrow\; 2\pi A\,\delta(\omega)

(Equivalently, from duality: δ(t)↔1\delta(t) \leftrightarrow 1, so 1↔2πδ(−ω)=2πδ(ω)1 \leftrightarrow 2\pi\delta(-\omega) = 2\pi\delta(\omega).) In terms of ff in Hz, A↔A δ(f)A \leftrightarrow A\,\delta(f).

Meaning: A DC signal has no variation, so all its content is at zero frequency. The spectrum is a single impulse at ω=0\omega = 0 with strength 2πA2\pi A.

 x(t)                    X(jw)
  |                        ^ 2*pi*A
 A|-----------------       |
  |                        |
--+-----------> t   -------+-------> w
                           0

Physical meaning of ESD and PSD

  • Energy spectral density (ESD), Ψ(ω)=∣X(jω)∣2\Psi(\omega) = |X(j\omega)|^2, applies to energy signals (finite energy, e.g. a pulse). It tells how the total energy of the signal is spread over frequency. The energy in a small band dωd\omega around ω\omega is 12π∣X(jω)∣2dω\frac{1}{2\pi}|X(j\omega)|^2d\omega, and the total energy is E=12π∫∣X(jω)∣2dωE = \frac{1}{2\pi}\int|X(j\omega)|^2d\omega. Unit: joule per hertz.
  • Power spectral density (PSD), S(ω)=lim⁡T→∞∣XT(jω)∣2TS(\omega) = \lim_{T\to\infty}\frac{|X_T(j\omega)|^2}{T}, applies to power signals (periodic or random signals with finite average power). It tells how the average power is spread over frequency; total power P=12π∫S(ω) dωP = \frac{1}{2\pi}\int S(\omega)\,d\omega. Unit: watt per hertz.

In practice, the ESD/PSD shows the bandwidth a signal needs, which frequency bands carry most energy or power, and how much noise power passes through a filter. Both are real, non-negative and even functions of ω\omega for real signals.

✓ Answer checked by Claude Opus 5.5 (high effort)

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