- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Parseval’s Theorem in Continuous-Time Fourier Series
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,
$$\mathrm{\frac{1}{T}\int_{t_{0}}^{t_{0}+T}|x(t)|^{2}\:dt=\sum_{n=−\infty}^{\infty}|C_{n}|^{2}… (10) \:\:(Hence,\:Proved)}$$