Parseval's relation for CT periodic signals

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.