Laplace transform and Region of Convergence for right-sided and left-sided signals

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What is Region of Convergence?

Region of Convergence (ROC) is defined as the set of points in s-plane for which the Laplace transform of a function x(t) converges. In other words, the range of Re(s) (i.e.,σ) for which the function X(s) converges is called the region of convergence.

ROC of Right-Sided Signals

A signal x(t) is said to be a right-sided signal if the signal x(t) = 0 for $\mathrm{t \:\lt\: T_1}$ for some finite time $\mathrm{T_1}$ as shown in Figure-1.

For a right-sided signal x(t), the ROC of the Laplace transform X(s) is $\mathrm{Re(s)\:\gt\:\sigma_{1}}$, where $\mathrm{\sigma_{1}}$ is a constant. Thus, the ROC of the Laplace transform of the right-sided signal is to the right of the line $\mathrm{\sigma \:=\: \sigma_{1}}$. A causal signal is an example of a right-sided signal.

Numerical Example - 1

Find the Laplace transform and ROC of the right-sided signal $\mathrm{x(t)\:=\:2e^{-4t}u(t)\:+\:4e^{-4t}u(t)}$

Solution

The given signal is,

$$\mathrm{x\left(t\right)\:=\:2e^{-4t}\:u\left(t\right )\:+\:4e^{-4t}\:u\left(t\right)}$$

The given signal is a right-sided signal. Actually, it is a causal signal. Its Laplace transform is given by,

$$\mathrm{L\left[x(t)\right]\:=\:2L\left[e^{-4t}\:u(t)\right]\:+\:4L\left[e^{-2t}\:u(t)\right]}$$

$$\mathrm{\Rightarrow\:X\left(s\right)\:=\:\left(\frac{2}{s\:+\:4}\right)\:+\:\left(\frac{4}{s\:+\:2}\right)\:\:ROC\:\to\: Re\left(s\right)\:\gt\:-4\:\:and\:\:Re\left(s\right)\:\gt\:-2}$$

Thus, the ROC of the given right sided signal is,

$$\mathrm{\left[Re\left(s\right)\:\gt\:-4\:\cap \: Re\left(s\right)\:\gt\:-2 \right ]\:=\:Re(s)\:\gt\:-2}$$

$$\mathrm{\Rightarrow\: ROC\:\to\: Re\left(s\right)\:\gt\:-2}$$

The ROC of the Laplace transform of the given signal is shown in Figure-2. It is to be noted that the ROC extends to the right of the rightmost pole and no pole exists inside the ROC.

ROC of Left-Sided Signals

A signal $\mathrm{x\left(t\right)}$ is said to be a left-sided signal if the signal $\mathrm{x\left(t\right) \:=\: 0}$ for $\mathrm{t \:\gt\: T_2}$ for some finite time $\mathrm{T}_{2}$ as shown in Figure-3.

For a left-sided signal x(t), the ROC of the Laplace transform X(s) is $\mathrm{Re(s)\:\lt\:\sigma_{2}}$ where $\mathrm{\sigma_{2}}$ is a constant. Therefore, the ROC of the Laplace transform of a left-sided signal is to the left of the line $\mathrm{\sigma \:=\: \sigma_{2}}$. An anti-causal signal is an example of a left-sided signal.

Numerical Example - 2

Find the Laplace transform and ROC of the left sided signal $\mathrm{x\left(t\right)\:=\:e^{2t}\:u\left(-t\right )\:+\:3e^{5t}\:u\left (-t\right)}$

Solution

The given signal is,

$$\mathrm{x\left(t\right)\:=\:e^{2t}\:u\left(-t\right )\:+\:3e^{5t}\:u\left (-t\right)}$$

Since the given signal is a left-sided signal, hence it is an anti-causal signal. The Laplace transform of the signal $\mathrm{x\left(t\right)}$ is given by,

$$\mathrm{L\left[x(t)\right]\:=\:L\left[e^{2t}\:u\left(-t\right)\right]\:+\:3L\left[e^{5t}\:u\left(-t\right)\right]}$$

$$\mathrm{\Rightarrow\:X(s)\:=\:-\frac{1}{s\:-\:2}\:-\:\frac{3}{s\:-\:5}\:;\:ROC\:\to\: Re(s)\:\lt\:2\: and \:Re(s)\:\lt\: 5}$$

Therefore, the ROC of the Laplace transform of the given left sided signal is,

$$\mathrm{\left[Re\left(s\right)\:\lt\:2\:\cap \:Re\left(s\right)\:\lt\:5 \right ]\:=\:Re\left(s\right)\:\lt\:2}$$

$$\mathrm{\Rightarrow\: ROC\:\to\: Re\left(s\right)\:\lt\:2}$$

The ROC of the left-sided signal $\mathrm{x\left(t\right)}$ is shown in Figure-4. It is to be noted that the ROC extends to the left of the leftmost pole and no pole exists inside the ROC.