Parseval's relation for DTFS

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.