Laplace Transform of Real Exponential and Complex Exponential Functions

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 Real Exponential Function

Case 1 – Growing real exponential function

$$\mathrm{\mathit{x\left ( t \right )}\mathrm{=}\mathit{e^{at}\, u\left ( t \right )}}$$

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

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

$$\mathrm{\mathit{\Rightarrow L\left [e^{at}\: u\left ( t \right ) \right ]\mathrm{=}\int_{\mathrm{0} }^{\infty } u\left ( t \right )e^{-\left ( s-a \right )t}\; dt\mathrm{=}\int_{\mathrm{0}}^{\infty }\left ( \mathrm{1} \right )e^{-\left ( s-a \right )t}dt}}$$

$$\mathrm{\mathit{\Rightarrow L\left [e^{at}\: u\left ( t \right ) \right ]\mathrm{=}\left [ \frac{e^{-\left ( s-a \right )t}}{-\left ( s-a \right )} \right ]_{\mathrm{0}}^{\infty }\mathrm{=}\left [ \frac{e^{-\infty }-e^{\mathrm{0}}}{-\left ( s-a \right )} \right ]\mathrm{=}\frac{\mathrm{1}}{\left ( s-a \right )}}}$$

This integral converges for 𝑅𝑒(𝑠 − 𝑎) > 0, i.e., its ROC is 𝑅𝑒(𝑠) > 𝑎 as shown in Figure-1. Therefore, the Laplace transform of function $\mathrm{\mathit{\left [ e^{at}u\left ( t \right ) \right ]}}$ along with its ROC is,

$$\mathrm{\mathit{e^{at}\, u\left ( t \right )\overset{LT}{\leftrightarrow}\frac{\mathrm{1}}{\left ( s-a \right )} };\;\;\;and\;\;\;ROC\rightarrow Re\left ( \mathit{s} \right )>a}$$

Case 2 – Decaying Real Exponential Function

$$\mathrm{\mathit{x\left ( t \right )\mathrm{=}e^{-at}\: u\left ( t \right )}}$$

From the definition of Laplace transform, we have,

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

$$\mathrm{\mathit{\Rightarrow L\left [ e^{-at}u\left ( t \right ) \right ]\mathrm{=}\int_{\mathrm{0}}^{\infty }u\left ( t \right )e^{-\left ( s\mathrm{+}a \right )t}\: dt}}\mathrm{=}\mathrm{\int_{\mathrm{0}}^{\infty }\left ( 1 \right )\mathit{e^{-\left ( s\mathrm{+}a \right )t}\: dt}}$$

$$\mathrm{\mathit{\Rightarrow L\left [ e^{-at}u\left ( t \right ) \right ]\mathrm{=}\left [ \frac{e^{-\left ( s\mathrm{+}a \right )t}}{-\left ( s\mathrm{+}a \right )} \right ]_{\mathrm{0}}^{\infty }\mathrm{=}\left [ \frac{e^{-\infty } -e^{\mathrm{0}}}{-\left ( s\mathrm{+}a \right )}\right ]\mathrm{=}\frac{\mathrm{1}}{\left ( s\mathrm{+}a \right )}}}$$

The above integral converges for 𝑅𝑒(𝑠 + 𝑎) > 0, i.e., its ROC is 𝑅𝑒(𝑠) > −𝑎 as shown in Figure-2. Therefore, the Laplace transform of function $\mathrm{\mathit{\left [ e^{-at}u\left ( t \right ) \right ]}}$ along with its ROC is,

$$\mathrm{\mathit{e^{at}\, u\left ( t \right )\overset{LT}{\leftrightarrow}\frac{\mathrm{1}}{\left ( s\mathrm{+}a \right )} };\;\;\;\;ROC\rightarrow Re\left ( \mathit{s} \right )>-a}$$

Laplace Transform of Complex Exponential Function

Case 1 – Growing Complex Exponential Function

$$\mathrm{\mathit{x\left ( t \right )}\mathrm{=}\mathit{e^{j\omega t}\, u\left ( t \right )}}$$

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

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

$$\mathrm{\mathit{\Rightarrow L\left [e^{j\omega t}\: u\left ( t \right ) \right ]\mathrm{=}\int_{\mathrm{0} }^{\infty } u\left ( t \right )e^{-\left ( s-j\omega \right )t}\; dt\mathrm{=}\int_{\mathrm{0}}^{\infty }\left ( \mathrm{1} \right )e^{-\left ( s-j\omega \right )t}dt}}$$

$$\mathrm{\mathit{\Rightarrow L\left [e^{j\omega t}\: u\left ( t \right ) \right ]\mathrm{=}\left [ \frac{e^{-\left ( s-j\omega \right )t}}{-\left ( s-j\omega \right )} \right ]_{\mathrm{0}}^{\infty }\mathrm{=}\left [ \frac{e^{-\infty }-e^{\mathrm{0}}}{-\left ( s-j\omega \right )} \right ]\mathrm{=}\frac{\mathrm{1}}{\left ( s-j\omega \right )}}}$$

The ROC of Laplace transform of growing complex exponential function is 𝑅𝑒(𝑠) > 0 as shown in Figure-3. Therefore, the Laplace transform of function $\mathrm{\mathit{\left [ e^{j\, \omega t}u\left ( t \right ) \right ]}}$ along with its ROC is,

$$\mathrm{\mathit{e^{j\omega t}\, u\left ( t \right )\overset{LT}{\leftrightarrow}\frac{\mathrm{1}}{\left ( s-j\omega \right )} };\;\;\;\;ROC\rightarrow Re\left ( \mathit{s} \right )>0}$$

Case 2 – Decaying Complex Exponential Function

$$\mathrm{\mathit{x\left ( t \right )}\mathrm{=}\mathit{e^{-j\omega t}\, u\left ( t \right )}}$$

From the definition of Laplace transform, we have,

$$\mathrm{\mathit{X\left ( s \right )\mathrm{=}L\left [ x\left ( t \right ) \right ]\mathrm{=}L\left [ e^{-j\omega t}u\left ( t \right ) \right ]\mathrm{=}\int_{\mathrm{0}}^{\infty }e^{-j\omega t}u\left ( t \right )e^{-st}\: dt}}$$

$$\mathrm{\mathit{\Rightarrow L\left [e^{-j\omega t}\: u\left ( t \right ) \right ]\mathrm{=}\int_{\mathrm{0} }^{\infty } u\left ( t \right )e^{-\left ( s\mathrm{+}j\omega \right )t}\; dt\mathrm{=}\int_{\mathrm{0}}^{\infty }\left ( \mathrm{1} \right )e^{-\left ( s\mathrm{+}j\omega \right )t}dt}}$$

$$\mathrm{\mathit{\Rightarrow L\left [e^{-j\omega t}\: u\left ( t \right ) \right ]\mathrm{=}\left [ \frac{e^{-\left ( s\mathrm{+}j\omega \right )t}}{-\left ( s\mathrm{+}j\omega \right )} \right ]_{\mathrm{0}}^{\infty }\mathrm{=}\left [ \frac{e^{-\infty }-e^{\mathrm{0}}}{-\left ( s\mathrm{+}j\omega \right )} \right ]\mathrm{=}\frac{\mathrm{1}}{\left ( s\mathrm{+}j\omega \right )}}}$$

The ROC of Laplace transform of decaying complex exponential function is also 𝑅𝑒(𝑠) > 0 as shown in Figure-3. Therefore, the Laplace transform of function $\mathrm{\mathit{\left [ e^{-j\, \omega t}u\left ( t \right ) \right ]}}$ along with its ROC is,

$$\mathrm{\mathit{e^{-j\omega t}\, u\left ( t \right )\overset{LT}{\leftrightarrow}\frac{\mathrm{1}}{\left ( s\mathrm{+}j\omega \right )} };\;\;\;\;ROC\rightarrow Re\left ( \mathit{s} \right )>0}$$

raja
Updated on 04-Jan-2022 10:28:43

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