Power Spectrum of Periodic Signals

The power spectrum of a periodic signal shows how the average signal power is distributed among its harmonic components. This concept is the basis of spectrum analyzers, noise analysis in communication links, and the design of bandlimited channel filters.

Core Concept

Average power is defined as the mean squared value of a signal over one period. Parseval theorem for Fourier series says this average power equals the sum of squared magnitudes of all exponential Fourier coefficients: P = Σ|c_k|². Each term |c_k|² is the power contribution of the kth harmonic. Plotting |c_k|² versus k gives the power spectrum.

The power spectrum is always non-negative and real, even though c_k itself is complex. For a real signal, c_{-k} = c_k*, so |c_{-k}|² = |c_k|². You can fold the two-sided spectrum about k=0 to get the one-sided power spectrum, where each harmonic power for k≥1 is 2|c_k|² and the DC power is |c_0|².

The fraction of total power carried by the first N harmonics is a standard exam calculation. For a square wave, the fundamental alone carries (8/π²)/1 ≈ 81% of the total AC power. This is because the power falls as 1/k² (from amplitude 1/k) and the series converges quickly. For systems with bandwidth limits, knowing what percentage of power lies within the passband tells you the distortion level.

Key Formula

Parseval theorem: P = (1/T)∫₀ᵀ |x(t)|² dt = Σ_{k=-∞}^{∞} |c_k|². One-sided form: P = |c_0|² + 2·Σ_{k=1}^{∞} |c_k|². For trigonometric coefficients: P = a_0² + (1/2)·Σ_{k=1}^{∞} (a_k² + b_k²). Here a_0 is the DC level (not the average 2a_0 sometimes used in older texts). Power spectral density for a periodic signal is a line spectrum, not a continuous function.

Problem: Find what fraction of the total power of a square wave
         (amplitude A, 50% duty cycle, zero mean) lies in the fundamental.

Step 1: Total average power from time domain.
         P_total = (1/T) ∫₀ᵀ x²(t) dt
         x(t) = +A for 0<t<T/2, x(t) = -A for T/2<t<T.
         P_total = (1/T)[∫₀^(T/2) A² dt + ∫_(T/2)^T A² dt]
                = (1/T) · A² · T = A²

Step 2: Fourier coefficients for zero-mean square wave.
         b_k = 4A/(kπ) for odd k.  (a_0 = 0, a_k = 0)
         c_k = b_k/(2j) = 2A/(jkπ)  for odd k.
         |c_k|² = 4A²/(k²π²)  for odd k.

Step 3: Power in fundamental (k = ±1).
         P_1 = 2|c_1|² = 2 · 4A²/π² = 8A²/π²

Step 4: Compute fraction.
         Fraction = P_1 / P_total = 8A²/π² / A² = 8/π²
                  ≈ 0.811  →  81.1%

Final Answer:
  The fundamental harmonic carries about 81% of the total power.
  Third harmonic adds 8A²/(9π²): another 9%, reaching ~90%.

Exam Tip: For power calculations using trigonometric form, the Parseval relation is P = a_0² + (1/2)(a_1²+b_1²) + (1/2)(a_2²+b_2²) + ... The factor 1/2 on each harmonic pair comes from the time average of sin² or cos², which is 1/2. Do not forget this factor. The DC term a_0² has no 1/2 factor — its time average is already 1.

Properties Summary

• Parseval theorem: average power = Σ|c_k|² summed over all k from -∞ to +∞.
• Non-negativity: |c_k|² ≥ 0 for all k; power spectrum is always non-negative.
• Symmetry for real signals: |c_{-k}|² = |c_k|², so power spectrum is even in k.
• One-sided form: P = |c_0|² + 2Σ|c_k|² for k=1 to ∞, doubling each harmonic contribution.
• Bandwidth efficiency: cumulative power from first N harmonics divided by total power quantifies spectral compactness.
• Line spectrum: power of a periodic signal is concentrated at discrete frequencies kf0, not spread continuously.
• Trigonometric form: P = a_0² + (1/2)Σ(a_k² + b_k²) for k = 1 to ∞.

Quick Revision

• Power spectrum = |c_k|² at each harmonic; it is real and non-negative.
• Parseval theorem equates time-domain average power to sum of spectral powers.
• For square wave, fundamental carries about 81% of total power.
• Power falls as 1/k² for square wave (amplitude 1/k → power 1/k²).
• One-sided spectrum: double each |c_k|² for k≥1, keep |c_0|² as is.
• Trigonometric Parseval: include factor 1/2 on all harmonic terms, no factor on DC.
• Periodic signals have line power spectra; aperiodic signals have continuous power spectral density.
• Exam trap: in the trigonometric Parseval formula, writing P = a_0² + Σ(a_k²+b_k²) without the 1/2 — always halve each harmonic squared term.