Fourier Series Properties & Parseval's Theorem

Linearity, time-shift, and the power equivalence between time and frequency domains

Signals & SystemsElectrical Engineering Year 2Free preview
⏱️ About 16 min

If a signal carries 17 watts of power in the time domain, how is that power distributed across its frequency components — and why must the two views always agree?

💡
The big idea: Fourier series inherits useful properties — linearity and time-shift — that mirror what you'll see for the Fourier transform. Parseval's theorem then ties everything together: the average power of a periodic signal computed in the time domain equals the sum of the powers in each harmonic, computed in the frequency domain.
🎯 By the end, you'll be able to
  • Apply the linearity property of the Fourier series to decompose composite signals
  • Apply the time-shift property and explain its effect on coefficient magnitudes
  • State and apply Parseval's theorem in trigonometric form
  • Compute the average power of a periodic signal from its Fourier coefficients

Properties That Simplify Analysis

Once you can compute a Fourier series, a handful of properties let you avoid recomputing it from scratch for related signals. Two of the most important are linearity and the time-shift property. These are analogous to properties you will encounter for the Fourier transform in the next lesson.

Linearity

The Fourier series is a linear operation. If x₁(t) has coefficients cₙ⁽¹⁾ and x₂(t) has coefficients cₙ⁽²⁾, then a·x₁(t)+b·x₂(t) has coefficients a·cₙ⁽¹⁾+b·cₙ⁽²⁾. This follows directly from the linearity of the integral that defines cₙ.

In practice, this means you can build up the spectrum of a composite signal from the spectra of its parts, without doing any new integrals.

\[ \text{If } x_1(t) \leftrightarrow c_n^{(1)}, \;\; x_2(t) \leftrightarrow c_n^{(2)}, \;\; \text{then } a\,x_1(t) + b\,x_2(t) \leftrightarrow a\,c_n^{(1)} + b\,c_n^{(2)} \]
Linearity property of the Fourier series.

Time-Shift Property

If you delay a signal by t₀ — forming x(t−t₀) — each exponential coefficient cₙ picks up a phase factor e−jnω₀t₀. The magnitude |cₙ| does not change; only the phase rotates. Shifting a signal in time does not change which frequencies are present or how much energy each carries — it only changes the relative alignment (phase) of the components.

\[ x(t - t_0) \;\longleftrightarrow\; c_n \, e^{-jn\omega_0 t_0} \]
Time-shift property: a delay t0 multiplies each cn by a phase factor, leaving |cn| unchanged.
✨ Magnitude survives, phase rotates

The time-shift property tells you that the power spectrum |cₙ|² is time-shift invariant. This is why power calculations — like Parseval's theorem — don't depend on when you start your clock.

Parseval's Theorem for Periodic Signals

Parseval's theorem says: the average power of a periodic signal — computed by integrating |x(t)|² over one period — is exactly equal to the sum of the average powers in each harmonic component. The theorem is a direct consequence of the orthogonality of the Fourier basis functions: cross-terms between different harmonics integrate to zero, so the total power cleanly separates into a sum of individual harmonic powers.

In Module 8 you learned that the average power of a single sinusoid A cos(ωt) is A²/2. Parseval's theorem generalizes this: the total power is the DC power (a₀²) plus the power of each harmonic, where each cosine and sine term contributes (aₙ²+bₙ²)/2.

\[ P = a_0^2 + \frac{1}{2}\sum_{n=1}^{\infty}\left(a_n^2 + b_n^2\right) \]
Parseval's theorem (trigonometric form): total average power equals DC power plus the sum of harmonic powers.
🔑 Two paths, same answer

You can compute the average power of x(t) two ways: (1) directly in the time domain, P=(1/T₀)∫|x(t)|²dt, or (2) by summing the power in each frequency component using Parseval's formula. They must give the same result — a consequence of orthogonality. If the two methods disagree, you've made a computational error somewhere.

Connecting to the Single-Sinusoid Result

Recall from Module 8 that a single sinusoid A cos(ωt) has average power A²/2. In Parseval's formula, the n-th harmonic consists of aₙcos(nω₀t)+bₙsin(nω₀t): the cosine part contributes aₙ²/2 and the sine part contributes bₙ²/2, so the harmonic's total power is (aₙ²+bₙ²)/2. The DC term a₀ is a constant, so its power is simply a₀². Parseval's theorem is a direct extension of the single-sinusoid power result to an arbitrary sum of harmonics.

Why Parseval Matters in Practice

Parseval's theorem tells you that the Fourier coefficients cₙ (or aₙ, bₙ) carry complete information about the signal's power — no need to go back to the time domain. This is the basis for power spectral density analysis, filter design, and signal-to-noise ratio calculations you'll encounter later in the course and in communications systems. In the exponential form, Parseval's theorem becomes P=Σ|cₙ|², especially clean and reinforcing that each coefficient's squared magnitude represents the power at that frequency.

📝 Worked example: A periodic signal has Fourier coefficients a0 = 3 (DC), a1 = 4 (first cosine harmonic), and all other an, bn = 0. That is, x(t) = 3 + 4cos(ω0t). Find the average power P using Parseval's theorem.
  1. Write Parseval's formula: P = a0² + (1/2)Σ(an²+bn²) for n = 1 to ∞.
  2. Only a0 and a1 are nonzero, so the sum reduces to a single term: P = a0² + (1/2)(a1²+b1²) = 3² + (1/2)(4²+0²).
  3. Compute: P = 9 + (1/2)(16) = 9 + 8 = 17.
✓ P = 17
✏️ Practice: A periodic signal is x(t) = 2 + 6cos(ω0t), so a0 = 2, a1 = 6, and all other coefficients are zero. Find the average power P.
Solution
  1. Apply Parseval's theorem: P = a0² + (1/2)(a1²+b1²).
  2. Substitute: P = 2² + (1/2)(6²+0²) = 4 + (1/2)(36).
  3. Compute: P = 4 + 18 = 22.

Check your understanding

1. If a periodic signal x(t) is delayed by t0 to form x(t − t0), what happens to its Fourier series coefficients cn?
The time-shift property multiplies each cn by e^(-jnω0t0). Since |e^(-jnω0t0)| = 1, the magnitude |cn| is unchanged — only the phase rotates.
2. A periodic signal has a0 = 1, a1 = 2, b1 = 2, and all other coefficients zero. What is the average power?
P = a0² + (1/2)(a1²+b1²) = 1 + (1/2)(4+4) = 1 + 4 = 5.
✅ Key takeaways
  • The linearity property states that the Fourier series of a weighted sum of signals is the weighted sum of their individual Fourier series.
  • The time-shift property states that delaying x(t) by t0 multiplies each coefficient cn by e^(-jnω0t0), leaving the magnitude |cn| (and hence the power spectrum) unchanged.
  • Parseval's theorem equates the average power of a periodic signal computed in the time domain to the sum of powers in each harmonic: P = a0² + (1/2)Σ(an²+bn²).
  • Each harmonic's power (an²+bn²)/2 generalizes the single-sinusoid result P = A²/2 from Module 8, while the DC term contributes a0².
➡️ With the Fourier series, its properties, and Parseval's theorem in hand, the next step is to extend these ideas to aperiodic signals — where the discrete sum of harmonics becomes the Fourier transform integral over a continuum of frequencies.
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