# Parseval’s Theorem in Continuous-Time Fourier Series

Signals and SystemsElectronics & ElectricalDigital Electronics

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## Fourier Series

If $x(t)$ is a periodic function with period $T$, then the continuous-time exponential Fourier series of the function is defined as,

$$\mathrm{x(t)=\sum_{n=−\infty}^{\infty}C_{n}\:e^{jn\omega_{0} t}… (1)}$$

Where, $C_{n}$ is the exponential Fourier series coefficient, which is given by,

$$\mathrm{C_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)\:e^{-jn\omega_{0} t}\:dt… (2)}$$

## Parseval’s Theorem and Parseval’s Identity

Let $x_{1}(t)$ and $x_{2}(t)$ two complex periodic functions with period T and with Fourier series coefficients $C_{n}$ and $D_{n}$.

If,

$$\mathrm{x_{1}(t)\overset{FT}{\leftrightarrow}C_{n}}$$

$$\mathrm{x_{2}(t)\overset{FT}{\leftrightarrow}D_{n}}$$

Then, the Parseval’s theorem of continuous time Fourier series states that

$$\mathrm{\frac{1}{T} \int_{t_{0}}^{t_{0}+T} x_{1}(t)\:x_{2}^{*}(t)\:dt =\sum_{n=−\infty}^{\infty} C_{n}\:D_{n}^{*}\:[for\:complex\: x_{1}(t)\: \& \: x_{2}(t)] … (3)}$$

And the parseval’s identity of Fourier series states that, if

$$\mathrm{x_{1}(t)=x_{1}(t)=x(t)}$$

Then,

$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}|x(t)|^{2}\:dt=\sum_{n=−\infty}^{\infty}|C_{n}|^{2}… (4)}$$

## Proof – Parseval’s theorem or Parseval’s relation or Parseval’s property

$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t)\:dt=\sum_{n=−\infty}^{\infty}C_{n}\:D_{n}^{*}… (5)}$$

From the definition of Fourier series, we have,

$$\mathrm{L.H.S.of\:eq.(5)=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t)\:dt}$$

$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}\left ( \sum_{n=−\infty}^{\infty}C_{n}e^{jn\omega_{0} t} \right )x_{2}^{*}(t)\:dt… (6)}$$

Rearranging the order of integration and summation in the RHS of equation (6), we get,

$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t)\:dt=\sum_{n=−\infty}^{\infty}C_{n}\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{2}^{*}(t)e^{jn\omega_{0} t}\:dt \right )}$$

$$\mathrm{\Rightarrow\:\frac{1}{T}\int_{0}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t)\:dt=\sum_{n=−\infty}^{\infty}C_{n}\left ( \frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{2}(t)e^{-jn\omega_{0} t}\:dt \right )^{*}… (7)}$$

On comparing equation (7) with eq. (2), we can write,

$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x_{1}(t)\:x_{2}^{*}(t)\:dt=\sum_{n=−\infty}^{\infty}C_{n}\:D_{n}^{*}… (8)\:\:(Hence,\:Proved)}$$

Proof – Parseval’s Identity

If,

$$\mathrm{x_{1}(t)=x_{2}(t)=x(t)}$$

Then, the Parseval’s relation becomes,

$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}x(t)\:x^{*}(t)\:dt=\sum_{n=−\infty}^{\infty}C_{n}\:C_{n}^{*}… (9)}$$

$$\mathrm{\because\:x(t)\:x^{*}(t)=|x(t)|^{2}\:and\:C_{n}\:C_{n}^{*}=|C_{n}|^{2}}$$

Now, on substituting these values in equation (9), we get,

$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}|x(t)|^{2}\:dt=\sum_{n=−\infty}^{\infty}|C_{n}|^{2}… (10) \:\:(Hence,\:Proved)}$$

Updated on 07-Dec-2021 07:45:25