--- title: "Parseval's relation for DTFS" subject: "Signals and Systems" chapter: 2 key: "semester_4/signals_and_systems:dtfs-parseval" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-parseval site: BCT Engineering --- # Parseval's relation for DTFS Signals and Systems · Chapter 2: Fourier Series - Asked in: 2082 Kartik, 3 marks; 2079 Jestha, 6 marks; 2078 Baisakh, 6 marks; 2076 Bhadra, 5 marks; 2073 Bhadra, 5 marks; 2072 Magh, 2+4 marks; 2072 Asoj, 5 marks; 2083 Bhadra (new course), 4 marks (asked 8 times) - Video: [Parseval's Theorem | Properties of DTFS | EnggClasses](https://www.youtube.com/watch?v=bSPQtxmmEn0) — EnggClasses - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-parseval - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2 ## Question State and prove Parseval's relation for Discrete Time Fourier Series (discrete time periodic signals). ## Short answer Parseval's relation: (1/N)∑|x[n]|^2 = ∑|a_k|^2 over one period of N. Proof: write |x|^2=x x*, substitute conjugate synthesis, swap sums, and recognize the inner sum as N a_k, yielding sum of |a_k|^2. ## Answer **Parseval's relation for the DTFS:** the average power of a discrete-time periodic signal over one period equals the sum of the squared magnitudes of its Fourier series coefficients: $$ \frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \sum_{k=\langle N\rangle}|a_k|^2 $$ Here $x[n]$ has period $N$, $\omega_0 = 2\pi/N$, and $$ x[n] = \sum_{k=\langle N\rangle} a_k e^{jk\omega_0 n}, \qquad a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk\omega_0 n} $$ #### Proof Start from the left side and write $|x[n]|^2 = x[n]\,x^*[n]$: $$ \frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,x^*[n] $$ Replace $x^*[n]$ by the conjugate of the synthesis equation, $x^*[n] = \sum_{k=\langle N\rangle}a_k^* e^{-jk\omega_0 n}$: $$ \begin{aligned} \frac{1}{N}\sum_{n}|x[n]|^2 &= \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\sum_{k=\langle N\rangle}a_k^* e^{-jk\omega_0 n} \\ &= \sum_{k=\langle N\rangle}a_k^*\left[\frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk\omega_0 n}\right] \\ &= \sum_{k=\langle N\rangle}a_k^*\, a_k \\ &= \sum_{k=\langle N\rangle}|a_k|^2 \end{aligned} $$ The bracket is exactly the analysis equation for $a_k$, which completes the proof. #### Meaning - $|a_k|^2$ is the average power in the $k$th harmonic $a_k e^{jk\omega_0 n}$, because $|e^{jk\omega_0 n}| = 1$. - So total average power = sum of powers of the $N$ harmonic components; power can be computed in either the time or the frequency domain. - Only $N$ distinct coefficients exist ($a_{k+N} = a_k$), so the sum is finite. **Example:** $x[n] = \cos(\pi n/2)$, $N = 4$: $a_1 = a_{-1} = a_3 = 1/2$. Time domain: samples $1, 0, -1, 0$ give $\frac{1}{4}(1 + 0 + 1 + 0) = 0.5$. Frequency domain: $|a_1|^2 + |a_{-1}|^2 = 0.25 + 0.25 = 0.5$. Both agree. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Parseval's relation for the DTFS says which? A. (1/N) sum |x[n]|^2 = sum |a_k|^2 over one period B. sum |x[n]|^2 = (1/N) sum |a_k|^2 C. (1/N) sum |x[n]|^2 = |sum a_k|^2 D. sum x[n] = sum a_k Answer: A. Average power in time over one period equals the sum of |a_k|^2 over one period. (Key point: Statement) 2. What is substituted for x*[n] in the proof? A. The analysis equation for a_k B. The conjugate of the synthesis equation, sum a_k* e^(-jk w0 n) C. x[-n] D. The orthogonality sum equal to N Answer: B. x*[n] = sum a_k* e^(-jk w0 n) lets the sum over n be rearranged. (Key point: Proof step) 3. After swapping sums, why does the bracket equal a_k? A. It is the synthesis equation B. It is the Euler identity C. It sums to zero by orthogonality D. It is exactly (1/N) sum x[n] e^(-jk w0 n), the analysis equation Answer: D. The bracket matches the definition of a_k, leaving sum a_k* a_k = sum |a_k|^2. (Key point: Recognising analysis equation) 4. What does |a_k|^2 represent? A. Average power in the kth harmonic B. Energy of the whole signal C. Phase of the kth harmonic D. Amplitude of x[k] Answer: A. Since |e^(jk w0 n)| = 1, the kth harmonic has average power |a_k|^2. 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