--- title: "Parseval's relation for CT periodic signals" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:ctfs-parseval" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-parseval site: BCT Engineering --- # Parseval's relation for CT periodic signals Signals and Systems · Chapter 2: Fourier Series - Asked in: 2082 Chaitra, 4 marks; 2078 Chaitra, 2+3 marks; 2076 Baisakh, 2+4 marks; 2071 Bhadra, 6 marks (asked 4 times) - Video: [Parseval's Theorem and Power Density Spectrum - Fourier Series | SIGNALS & SYSTEMS | EE307 | EC202](https://www.youtube.com/watch?v=UrzUUqmoR0I) — Er. PRAVEESH V V - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/ctfs-parseval - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question What is Parseval's relation? State and prove Parseval's relation for continuous-time periodic signals. ## Short answer Average power equality: (1/T)∫_T |x(t)|^2 dt =∑|a_k|^2. Proof: substitute |x|^2=x x*, replace x* with conjugate synthesis, swap sum and integral, and use the analysis equation a_k =(1/T)∫ x e^(-jkω0 t) dt to get sum of |a_k|^2. ## Answer **Parseval's relation** states that the average power of a periodic signal, computed in the time domain over one period, equals the sum of the powers of all its harmonic components (the squared magnitudes of its Fourier series coefficients). Power is the same whether measured in time or frequency. #### Statement (continuous-time periodic signals) If $x(t)$ has period $T$ and $x(t) \overset{FS}{\longleftrightarrow} a_k$, then $$ \frac{1}{T}\int_{T}|x(t)|^2\,dt = \sum_{k=-\infty}^{\infty}|a_k|^2 $$ #### Proof With $\omega_0 = 2\pi/T$, $x(t) = \sum_k a_k e^{jk\omega_0 t}$ and $a_k = \frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt$. $$ \begin{aligned} \frac{1}{T}\int_T |x(t)|^2dt &= \frac{1}{T}\int_T x(t)\,x^*(t)\,dt \\ &= \frac{1}{T}\int_T x(t)\left[\sum_{k=-\infty}^{\infty}a_k^* e^{-jk\omega_0 t}\right]dt \\ &= \sum_{k=-\infty}^{\infty}a_k^*\left[\frac{1}{T}\int_T x(t)e^{-jk\omega_0 t}dt\right] \\ &= \sum_{k=-\infty}^{\infty}a_k^*\,a_k = \sum_{k=-\infty}^{\infty}|a_k|^2 \end{aligned} $$ (the order of sum and integral is swapped, and the bracket is the analysis equation). #### Interpretation - $|a_k|^2$ is the average power of the $k$th harmonic, since $\frac{1}{T}\int_T |a_k e^{jk\omega_0 t}|^2dt = |a_k|^2$. - A plot of $|a_k|^2$ against $k\omega_0$ is the **power spectrum** of the signal. **Example:** $x(t) = A\cos\omega_0 t$ has $a_{\pm1} = A/2$. Frequency domain: $(A/2)^2 + (A/2)^2 = A^2/2$; time domain: $\frac{1}{T}\int_T A^2\cos^2\omega_0 t\,dt = A^2/2$. They match. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is Parseval's relation for CT periodic signals? A. (1/T) integral over T of |x|^2 dt = sum |a_k|^2 B. integral over T of |x|^2 dt = sum |a_k|^2 C. (1/T) integral over T of |x|^2 dt = |sum a_k|^2 D. (1/T) integral over T of x dt = sum a_k^2 Answer: A. Average power over one period equals the sum of squared magnitudes of coefficients. (Key point: Statement) 2. Which step starts the proof? A. Write |x|^2 = x(t) x*(t) and substitute the series for x*(t) B. Differentiate x(t) term by term C. Replace x(t) by its dc value D. Use the time-shift property Answer: A. Using x* = sum a_k* e^(-jk w0 t), the integral is rearranged to contain the analysis equation. (Key point: Proof substitution) 3. In the proof, what is the bracket (1/T) integral of x(t)e^(-jk w0 t) dt? A. The power of x(t) B. The coefficient a_k C. The fundamental frequency D. Zero by orthogonality Answer: B. It is the analysis equation, so each term becomes a_k* a_k. (Key point: Key identification) 4. What does a plot of |a_k|^2 against k w0 give? A. Phase spectrum B. Energy spectral density of a non-periodic signal C. Power spectrum of the periodic signal D. Time-domain waveform Answer: C. |a_k|^2 is the power in the kth harmonic, so the plot is the power spectrum. 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