Laplace Transform of Real Exponential and Complex Exponential Functions

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

$$\mathrm{L[x(t)]\:=\:X(s)\:=\:\int_{-\infty}^{\infty}\:x(t)\:e^{-st}\: dt\:\: \dotso \:(1)}$$

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

$$\mathrm{L[x(t)]\:=\:X(s)\:=\:\int_{0}^{\infty}\:x(t)\:e^{-st}\:dt\:\: \dotso \:(2)}$$

Laplace Transform of Real Exponential Function

Case 1 – Growing real exponential function

$$\mathrm{x(t)\:=\:e^{at}\: u(t)}$$

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

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

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

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

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 $\mathrm{\left [ e^{at}u\left ( t \right ) \right ]}$ along with its ROC is,

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

Case 2 – Decaying Real Exponential Function

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

From the definition of Laplace transform, we have,

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

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

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

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 $\mathrm{\left [ e^{-at}u\left ( t \right ) \right ]}$ along with its ROC is,

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

Laplace Transform of Complex Exponential Function

Case 1 – Growing Complex Exponential Function

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

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

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

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

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

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 $\mathrm{\left [ e^{j\: \omega t}\:u\left ( t \right ) \right ]}$ along with its ROC is,

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

Case 2 – Decaying Complex Exponential Function

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

From the definition of Laplace transform, we have,

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

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

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

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 $\mathrm{\left [ e^{-j\: \omega t}u\left ( t \right ) \right ]}$ along with its ROC is,

$$\mathrm{e^{-j\omega t}\:u(t)\overset{LT}{\leftrightarrow}\frac{1}{(s\:+\:j\omega)}\:\:ROC\:\:\rightarrow\:\:Re(s)\:\gt \:0}$$