Parseval's Power Theorem

Average Power

The average power of a signal is defined as the mean power dissipated by the signal such as voltage or current in a unit resistance over a period. Mathematically, the average power is given by,

$$\mathrm{P \:=\: \lim_{T \:\to\: \infty}\: \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}}\: |x(t)|^2 \: dt}$$

Parseval's Power Theorem

Statement − Parseval's power theorem states that the power of a signal is equal to the sum of square of the magnitudes of various harmonic components present in the discrete spectrum.

Mathematically, the Parseval's power theorem is defined as −

$$\mathrm{P \:=\: \sum_{n=-\infty}^{\infty}\: |C_n|^2}$$

Proof

Consider a function x(t). Then, the average power of the signal x(t) over one complete cycle is given by,

$$\mathrm{P \:=\: \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}}\: |x(t)|^2 \: dt}$$

$$\mathrm{\therefore\: |x(t)|^2 \:=\: x(t) x^*(t)}$$

$$\mathrm{\therefore\: P \:=\: \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}}\: x(t) x^*(t) \: dt \:\:\dotso\: (1)}$$

But, from the definition of exponential Fourier series, we have,

$$\mathrm{x(t) \:=\: \sum_{n=-\infty}^{\infty}\: C_n e^{j n \omega t}\:\:\dotso\: (2)}$$

Hence, from eqns. (1) & (2), we get,

$$\mathrm{P \:=\: \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}}\: \left[\sum_{n=-\infty}^{\infty} \:C_n e^{j n \omega t} \right]\: x^*(t) \: dt}$$

Now, by interchanging the order of integration and summation, we get,

$$\mathrm{P \:=\: \sum_{n=-\infty}^{\infty}\: C_n \left[ \frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}}\: x^*(t) e^{j n \omega t} \: dt \right] \:=\: \sum_{n=-\infty}^{\infty}\: C_n C_n^*}$$

$$\mathrm{\therefore \: P \:=\: \sum_{n=-\infty}^{\infty}\: |C_n|^2\:\:\dotso\: (3)}$$

The expression in eq. (3) is called the Parseval's power theorem. Hence, it is clear that the Parseval's power theorem defines the power of a signal in terms of its Fourier series coefficients or in other words, in terms of harmonics present in the signal.