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

Signals and SystemsElectronics & ElectricalDigital Electronics

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 $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ converges. In other words, the range of $\mathit{Re}\mathrm{\left(\mathit{s} \right)}$ (i.e.,σ) for which the function $\mathit{X}\mathrm{\left(\mathit{s}\right)}$ converges is called the region of convergence.

ROC of Right-Sided Signals

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

For a right-sided signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}$, the ROC of the Laplace transform $\mathit{X}\mathrm{\left(\mathit{s}\right)}$ is $\mathit{Re}\mathrm{\left(\mathit{s} \right )}>\mathrm{\sigma _{\mathrm{1}}}$, where $\mathrm{\sigma _{\mathrm{1}}}$ is a constant. Thus, the ROC of the Laplace transform of the right-sided signal is to the right of the line $\sigma \mathrm{=} \mathrm{\sigma _{\mathrm{1}}}$. 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 $\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:2\mathit{e^{-\mathrm{4}t}u\mathrm{\left(\mathit{t}\right )}\:\mathrm{+}\:\mathrm{4}e^{-\mathrm{4}t}u\mathrm{\left (\mathit{t}\right)}}$

Solution

The given signal is,

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

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

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

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

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

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

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

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

For a left-sided signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}$, the ROC of the Laplace transform $\mathit{X}\mathrm{\left(\mathit{s}\right)}$ is $\mathit{Re}\mathrm{\left(\mathit{s}\right)}<\mathrm{\sigma _{\mathrm{2}}}$ where $\mathrm{\sigma _{\mathrm{2}}}$ is a constant. Therefore, the ROC of the Laplace transform of a left-sided signal is to the left of the line $\sigma \mathrm{=}\mathrm{\sigma _{\mathrm{2}}}$. 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 $\mathit{x}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\:\mathit{e^{\mathrm{2}t}\:u\mathrm{\left(\mathit{-t}\right )}\:\mathrm{+}\:\mathrm{3}e^{\mathrm{5}t}\:u\mathrm{\left (\mathit{-t}\right)}}$

Solution

The given signal is,

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

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

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

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

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

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

$$\mathrm{\Rightarrow \mathrm{ROC\to }\mathit{Re\mathrm{\left(\mathit{s}\right)}}<2}$$

The ROC of the left-sided signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ 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.

raja
Updated on 05-Jan-2022 08:08:28

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