Laplace Transform and Region of Convergence of Two-Sided and Finite Duration 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 $\mathrm{x\left ( t \right )}$ 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 Two-Sided Signals

A signal $\mathrm{x\left ( t \right )}$ is said to be a two sided signal if it extends from -∞ to +∞. The two sided signal can be represented as the sum of two non-overlapping signals, one of which is right-sided signal and the other is the left-sided signal as shown in Figure- 1.

For a two-sided signal, the ROC of the Laplace transform X(s) is in the form of a strip in the s-plane bounded by two lines σ = σ_r and σ = σ_i.

Numerical Example - 1

Find the Laplace transform and ROC of the two-sided signal $\mathrm{x(t)\:=\:2e^{-3t}\:u(t)\:+\:3e^{4t}\:u(-t)}$

Solution

The given signal is,

$$\mathrm{x(t) \:=\: 2e^{-3t}\:u(t)\:+\:3e^{4t}\:u(-t)}$$

As the given signal is a two sided signal. The Laplace transform of signal x(t) is given by,

$$\mathrm{L[x(t)] \:=\: 2L\left [e^{-3t}\:u(t)\right ]\:+\:3L\left [e^{4t}\:u(-t) \right]}$$

$$\mathrm{\Rightarrow\:X(s)\:=\:\frac{2}{s\:+\:3}\:-\:\frac{3}{s-4}\:;\:ROC\:\to\:Re(s)\:\gt\:-3\:and\:Re(s)\:\lt\: 4}$$

Therefore, the ROC of the Laplace transform of the signal $\mathrm{x(t)}$ is,

$$\mathrm{Re(s)\:\gt\:-3\:\cap\:Re(s)\:\lt\:4\:=\:-3\:\lt\:Re(s)\:\lt\:4}$$

The ROC of the Laplace transform of the given signal x(t) is shown in Figure- 2. It can be seen that the ROC of the given two sided signal is a strip in the s-plane and lies to the right of the pole at s = −3 and to the left of the pole at s = +4.

ROC of Finite Duration Signals

A signal $\mathrm{x(t)}$ is said to be a finite duration signal if $\mathrm{x(t) \:=\: 0}$ for $\mathrm{t\:lt\:T_{1}}$ and $\mathrm{t\:gt\:T_{2}}$ where $\mathrm{T_{1}}$ and $\mathrm{T_{2}}$ are some finite time instants as shown in Figure-3.

For a finite duration signal which is absolutely integrable, the ROC of its Laplace transform extends over the entire s-plane.

Numerical Example 2

Find the Laplace transform and ROC of the finite duration signal $\mathrm{x(t)}$ shown in Figure-4.

Solution

The equation describing the given finite duration signal is given by,

$$\mathrm{x(t)\:=\:\left[ u(t\:-\:T_{1})\:-\:u(t\:-\:T_{2}) \right]}$$

Both the terms of this signal are causal. The Laplace transform of this is signal is obtained as −

$$\mathrm{L[x(t)]\:=\:L\left [u(t\:-\:T_{1})\right]\:-\:L\left [u(t\:-\:T_{2}) \right]}$$

$$\mathrm{\Rightarrow\: X(s)\:=\:\left (\frac{e^{-T_{1}s}}{s}\:-\:\frac{e^{-T_{2}s}}{s} \right )}$$

$$\mathrm{\therefore\: X(s)\:=\:\frac{1}{s}\left ({e^{-T_{1}s}}\:-\:{e^{-T_{2}s}} \right)\:;\: ROC\:\rightarrow\: All\: s}$$