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
Signals and Systems – Parseval’s Theorem for Laplace Transform
Laplace Transform
The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.
Mathematically, if $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a time domain function, then its Laplace transform is defined as −
$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{-\infty}^{\infty}\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-st}}\:\mathit{dt}}$$
Inverse Laplace Transform
The inverse Laplace transform is the method for obtaining the time domain function from its Laplace transform and mathematically, it is defined as −
$$\mathrm{\mathit{L}^{-\mathrm{1}}\mathrm{\left[\mathit{X}\mathrm{\left(\mathit{s}\right)}\right]}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\frac{1}{2\pi \mathit{j}}\int_{\mathrm{\left ( \sigma -\mathit{j\infty} \right )}}^{\mathrm{\left (\mathit{\sigma \mathrm{+}\mathit{j}\infty}\right )}}\mathit{X}\mathrm{\left(\mathit{s}\right)}\mathit{e^{st}}\:\mathit{ds}}$$
Parseval’s Theorem for Laplace Transform
Statement - The Parseval’s theorem or Parseval’s relation for Laplace transform states that if,
$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{s}\right)}\:\mathrm{and}\:\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{t}\right)}\overset{\mathit{LT}}{\leftrightarrow}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{s}\right)}}$$
Where, $\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)}$ and $\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{t}\right)}$ are complex functions. Then,
$$\mathrm{\int_{-\infty }^{\infty }\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)}\mathit{x}_{\mathrm{2}}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}\overset{\mathit{LT}}{\leftrightarrow}\:\frac{1}{2\pi \mathit{j}}\int_{\mathrm{\left(\sigma -\mathit{j\infty} \right )}}^{\mathrm{\left (\mathit{\sigma \mathrm{+}j\infty}\right )}}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{s}\right)}\mathit{X}_{\mathrm{2}}^{*}\mathrm{\left( -\mathit{s^{*}}\right)}\:\mathit{ds}}$$
Proof
From the definition of the inverse Laplace transform, we have,
$$\mathrm{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)\:\mathrm{=}\:\frac{1}{2\pi \mathit{j}}\int_{\mathrm{\left(\sigma -\mathit{j\infty} \right )}}^{\mathrm{\left (\mathit{\sigma \mathrm{+}\mathit{j}\infty}\right )}}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{s}\right)}\mathit{e^{st}}\:\mathit{ds}}}$$
Taking LHS of the Parseval’s theorem, we get,
$$\mathrm{\mathrm{LHS}\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)}\mathit{x}_{\mathrm{2}}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}\:\mathrm{=}\:\int_{-\infty}^{\infty}\mathrm{\left [ \frac{1}{2\pi \mathit{j}}\int_{\mathrm{\left(\sigma -\mathit{j\infty} \right )}}^{\mathrm{\left (\mathit{\sigma \mathrm{+}\mathit{j}\infty}\right )}}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{s}\right)}\mathit{e^{st}}\:\mathit{ds} \right ]}\mathit{x}_{\mathrm{2}}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}}$$
By rearranging the order of integration in RHS of the above equation, we get,
$$\mathrm{\int_{-\infty }^{\infty }\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)}\mathit{x}_{\mathrm{2}}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}\:\mathrm{=}\:\frac{1}{2\pi \mathit{j}}\int_{\mathrm{\left(\sigma -\mathit{j\infty} \right )}}^{\mathrm{\left (\mathit{\sigma \mathrm{+}\mathit{j}\infty}\right )}}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{s}\right)}\mathrm{\left[\int_{-\infty }^{\infty }\mathit{x}_{\mathrm{2}}^{*}\mathrm{\left(\mathit{t}\right)}\mathit{e^{st}}\:\mathit{dt}\right]}\:\mathit{ds}}$$
$$\mathrm{\Rightarrow \int_{-\infty }^{\infty }\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)}\mathit{x}_{\mathrm{2}}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}\:\mathrm{=}\:\frac{1}{2\pi \mathit{j}}\int_{\mathrm{\left(\sigma -\mathit{j\infty} \right )}}^{\mathrm{\left (\mathit{\sigma \mathrm{+}\mathit{j}\infty}\right )}}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{s}\right)}\mathrm{\left[\int_{-\infty }^{\infty }\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathrm{\left(-\mathit{s}^{*} \right )\mathit{t}}}}\:\mathit{dt}\right]}^{*}\:\mathit{ds}}$$
$$\mathrm{\Rightarrow \int_{-\infty }^{\infty }\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)}\mathit{x}_{\mathrm{2}}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}\:\mathrm{=}\:\frac{1}{2\pi \mathit{j}}\int_{\mathrm{\left(\sigma -\mathit{j\infty} \right )}}^{\mathrm{\left (\mathit{\sigma \mathrm{+}\mathit{j}\infty}\right )}}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{s}\right)}\mathrm{\left[\mathit{X_{\mathrm{2}}\mathrm{\left ( -\mathit{s}^{*}\right )}} \right ]^{*}\:\mathit{ds}}}$$
$$\mathrm{\therefore \int_{-\infty }^{\infty }\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{t}\right)}\mathit{x}_{\mathrm{2}}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathit{dt}\:\mathrm{=}\:\frac{1}{2\pi \mathit{j}}\int_{\mathrm{\left(\sigma -\mathit{j\infty} \right )}}^{\mathrm{\left (\mathit{\sigma \mathrm{+}\mathit{j}\infty}\right )}}\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{s}\right)}\mathit{X_{\mathrm{2}}\mathrm{\left ( -\mathit{s}^{*}\right )}\:\mathit{ds}\:\mathrm{=}\:\mathrm{RHS}}}$$
Hence, it proves the Parseval’s relation or theorem for Laplace transform.
1K+ Views
