Laplace Transform of Periodic Functions (Time Periodicity Property of Laplace Transform)

Signals and SystemsElectronics & ElectricalDigital Electronics

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 $\mathrm{\mathit{x\left ( t \right )}}$ is a time domain function, then its Laplace transform is defined as,

$$\mathrm{\mathit{L\left [ x\left ( t \right ) \right ]\mathrm{\, =\,}X\left ( s \right )\mathrm{\, =\,}\int_{-\infty }^{\infty }x\left ( t \right )e^{-st}\:dt \; \; \cdot \cdot \cdot \left ( \mathrm{1} \right )}}$$

Equation (1) gives the bilateral Laplace transform of the function $\mathrm{\mathit{x\left ( t \right )}}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as,

$$\mathrm{\mathit{L\left [ x\left ( t \right ) \right ]\mathrm{\, =\,}X\left ( s \right )\mathrm{\, =\,}\int_{\mathrm{0} }^{\infty }x\left ( t \right )e^{-st}\:dt \; \; \cdot \cdot \cdot \left ( \mathrm{2} \right )}}$$

Laplace Transform of Periodic Functions

The Laplace transform of periodic functions can be determined by using the time shifting property [$\mathrm{i.e.,\mathit{\, x\left ( t-T \right )\mathrm{\, =\,}e^{-sT}X\left ( s \right )}}$]. Consider a causal periodic function $\mathrm{\mathit{x\left ( t \right )}}$ which satisfies the condition $\mathrm{\mathit{x\left ( t \right )\mathrm{\, =\,}x\left ( t\mathrm{\, +\,}nT \right )}}$ or all t > 0, where T is the time period of $\mathrm{\mathit{x\left ( t \right )}}$ and n = 0, 1, 2,….

Now, from the definition of Laplace transform, we have,

$$\mathrm{\mathit{L\left [ x\left ( t \right ) \right ]\mathrm{\, =\,}X\left ( s \right )\mathrm{\, =\,}\int_{\mathrm{0}}^{\infty }x\left ( t \right )e^{-st}dt\; \; \cdot \cdot \cdot \left ( \mathrm{3} \right )}}$$

The above expression can also be written as,

$$\mathrm{\mathit{X\left ( s \right )\mathrm{\, =\,}\int_{\mathrm{0}}^{T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}\int_{T}^{\mathrm{2}T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}\int_{\mathrm{2}T}^{\mathrm{3}T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}\cdot \cdot \cdot \mathrm{\, +\,}\int_{nT}^{\left ( n\mathrm{\, +\,}\mathrm{1} \right )T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}\cdot \cdot \cdot }}$$

$$\mathrm{\mathit{\Rightarrow X\left ( s \right )\mathrm{\, =\,}\int_{\mathrm{0}}^{T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}e^{-st}\int_{\mathrm{0}}^{T }x\left ( t\mathrm{\, +\,}T \right )e^{-st}dt}}$$

$$\mathrm{\mathit{\mathrm{\, +\,}e^{-\mathrm{2}sT}\int_{\mathrm{0}}^{T }x\left ( t\mathrm{\, +\,}\mathrm{2}T \right )e^{-st}dt\mathrm{\, +\,}\cdot \cdot \cdot }}$$

$$\mathrm{\mathit{\mathrm{\, +\,}e^{-nsT}\int_{\mathrm{0}}^{T }x\left ( t\mathrm{\, +\,}nT \right )e^{-st}dt\mathrm{\, +\,}\cdot \cdot \cdot \: \: \cdot \cdot \cdot \left(\mathrm{4} \right )}}$$

As the function $\mathrm{\mathit{x\left ( t \right )}}$ is a periodic function, therefore,

$$\mathrm{\mathit{x\left ( t \right )\mathrm{\, =\,}x\left ( t\mathrm{\, +\,}T \right )\mathrm{\, =\,}x\left ( t\mathrm{\, +\,}\mathrm{2}T \right )\mathrm{\, =\,}\cdot \cdot \cdot}}$$

Hence, the equation (4) can be written as

$$\mathrm{\mathit{X\left ( s \right )\mathrm{\, =\,}\int_{\mathrm{0}}^{T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}e^{-st}\int_{\mathrm{0}}^{T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}e^{-\mathrm{2}st}\int_{\mathrm{0}}^{T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}\cdot \cdot \cdot }}$$

$$\mathrm{\mathit{\mathrm{\, +\,}e^{-nst}\int_{\mathrm{0}}^{T }x\left ( t \right )e^{-st}dt\mathrm{\, +\,}\cdot \cdot \cdot }}$$

$$\mathrm{\mathit{\Rightarrow X\left ( s \right )\mathrm{\, =\,}\left [ \mathrm{1}\mathrm{\, +\,}e^{-sT}\mathrm{\, +\,}e^{-\mathrm{2}sT}\mathrm{\, +\,}\cdot \cdot \cdot \mathrm{\, +\,}e^{-nsT}\mathrm{\, +\,}\cdot \cdot \cdot \right ]\int_{\mathrm{0}}^{T}x\left ( t \right )e^{-st}dt\; \; \left ( \mathrm{5} \right )}}$$

Using the binomial series expansion, we can write,

$$\mathrm{\mathit{X\left ( s \right )\mathrm{\, =\,}\left [ \mathrm{1}-e^{-sT} \right ]^{\mathrm{-1}}\int_{\mathrm{0}}^{T}x\left ( t \right )e^{-st}\, dt}}$$

$$\mathrm{\mathit{\therefore X\left ( s \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left [ \mathrm{1}-e^{-sT} \right ]}X_{\mathrm{1}}\left ( s \right )\; \; \cdot \cdot \cdot \left ( \mathrm{6} \right )}}$$


$$\mathrm{\mathit{X_{\mathrm{1}}\left ( s \right )\mathrm{\, =\,}\int_{\mathrm{0}}^{T}x\left ( t \right )e^{-st}\, dt\; \; \; \cdot \cdot \cdot \left ( \mathrm{7} \right )}}$$

The $\mathrm{\mathit{X_{\mathrm{1}}\left ( s \right )}}$ is the Laplace transform of the first period of the time function. The eqn. (6) represents the periodicity property of the Laplace transform.

Numerical Example

Find the Laplace transform of function $\mathrm{\mathit{x\left ( t \right )\mathrm{\, =\,}\mathrm{sin}\, \pi t\, u\left ( t \right )}}$ using periodicity property of Laplace transform.


The given function is,

$$\mathrm{\mathit{x\left ( t \right )\mathrm{\, =\,}\mathrm{sin}\, \pi t\, u\left ( t \right )}}$$

The given signal is a periodic signal with a time period of T which is given by,

$$\mathrm{\mathit{T\mathrm{\, =\,}\frac{\mathrm{2}\pi }{\omega }\mathrm{\, =\,}\frac{\mathrm{2}\pi }{\pi }\mathrm{\, =\,}}2\: sec}$$

Now, by using the periodicity property of Laplace transform, we have,

$$\mathrm{\mathit{L\left [ x\left ( t \right ) \right ]\mathrm{\, =\,}X\left ( s \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left [ \mathrm{1}-e^{-\mathrm{2}s} \right ]}\int_{\mathrm{0}}^{\mathrm{2}}\mathrm{sin}\, \left ( \pi t \right )e^{-st}\, dt}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( s \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left [ \mathrm{1}-e^{-\mathrm{2}s} \right ]}\int_{\mathrm{0}}^{\mathrm{2}}\left ( \frac{e^{j\, \pi t}-e^{-j\, \pi t}}{\mathrm{2}j} \right )e^{-st}\, dt}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( s \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left [ \mathrm{1}-e^{-\mathrm{2}s} \right ]}\left [\frac{\mathrm{1}}{\mathrm{2}j} \left\{ \int_{\mathrm{0}}^{\mathrm{2}} e^{j\, \pi t}e^{-st}\, dt-\int_{\mathrm{0}}^{\mathrm{2}}e^{-j\, \pi t}e^{-st}\, dt\right\} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( s \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left [ \mathrm{1}-e^{-\mathrm{2}s} \right ]}\left [\frac{\mathrm{1}}{\mathrm{2}j} \left\{ \int_{\mathrm{0}}^{\mathrm{2}} e^{-\left (s- j\, \pi \right ) t}\, dt-\int_{\mathrm{0}}^{\mathrm{2}}e^{-\left ( s\mathrm{\, +\,}j\, \pi \right ) t}\, dt\right\} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( s \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left [ \mathrm{1}-e^{-\mathrm{2}s} \right ]} \left\{\frac{\mathrm{1}}{\mathrm{2}j}\left [ \frac{e^{-\left (s- j\, \pi \right ) t}}{-\left ( s-j\pi \right )}-\frac{e^{-\left ( s\mathrm{\, +\,}j\, \pi \right ) t}}{-\left ( s\mathrm{\, +\,}j\pi \right )} \right ]_{\mathrm{0}}^{\mathrm{2}}\right\}}}$$

On solving the limits, we get,

$$\mathrm{\mathit{X\left ( s \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left [ \mathrm{1}-e^{-\mathrm{2}s} \right ]}\left\{\frac{\mathrm{1}}{\mathrm{2}j}\left [ \frac{-\left ( s\mathrm{\, +\,}j\pi \right )\left [e^{-\mathrm{2}\left ( s-j\pi \right )t}-\mathrm{1} \right ]\mathrm{\, +\,}\left ( s-j\pi \right )\left [e^{-\mathrm{2}\left ( s\mathrm{\, +\,}j\pi \right )t}-\mathrm{1} \right ]}{s^{\mathrm{2}}\mathrm{\, +\,}\pi ^{\mathrm{2}}} \right ] \right\}}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( s \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left [ \mathrm{1}-e^{-\mathrm{2}s} \right ]}\left [ \frac{\pi \left ( \mathrm{1}-e^{\mathrm{-2}s} \right )}{s^{\mathrm{2}}\mathrm{\, +\,}\pi ^{\mathrm{2}}} \right ]}}$$

$$\mathrm{\mathit{\therefore X\left ( s \right )\mathrm{\, =\,}\left [ \frac{\pi}{s^{\mathrm{2}}\mathrm{\, +\,}\pi ^{\mathrm{2}}} \right ]}}$$

Published on 11-Jan-2022 06:36:33