Region of Convergence (ROC)

Whether the Laplace transform $X(s)$ of a signal $x(t)$ exists or not depends on the complex variable $s$ as well as the signal itself. All complex values of $s$ for which the integral in the definition converges form a region of convergence (ROC) in the s-plane. $X(s)$ exists if and only if the argument $s$ is inside the ROC. As the imaginary part $\omega=Im[s]$ of the complex variable $s=\sigma+j\omega$ has no effect in terms of the convergence, the ROC is determined solely by the real part $\sigma=Re[s]$.

Example 1: The Laplace transform of $x(t)=e^{-at}u(t)$ is:

$$X(s) = {\cal L}[x(t)]=\int_0^\infty e^{-at} e^{-st} dt =\int_0^\infty e^{-at} e^{-(\sigma+j\omega)t} dt$$

$$= -\frac{1}{a+\sigma+j\omega}\; e^{-(a+\sigma+j\omega)t} \bigg\vert _0^\infty$$

For this integral to converge, we need to have

$$a+\sigma > 0 \;\;\;\;\mbox{ or }\;\;\;\; \sigma=Re[s] > -a$$

and the Laplace transform is

$$X(s)=\frac{1}{(\sigma+a)+j\omega}=\frac{1}{s+a}$$

As a special case where $a=0$, $x(t)=u(t)$ and we have

$${\cal L}[u(t)]=\frac{1}{s},\;\;\;\;\;\;\sigma=Re[s]>0$$

Example 2: The Laplace transform of a signal $x(t)=-e^{-at}u(-t)$ is:

$$X(s)=-\int_{-\infty}^0 e^{-at} e^{-st} dt =-\int_{-\infty}^0... ...gma+j\omega}\;e^{-(a+\sigma+j\omega)t} \bigg\vert _{-\infty}^0$$

Only when

$$a+\sigma < 0 \;\;\;\;\;\mbox{ or }\;\;\;\;\; \sigma=Re[s] < -a$$

will the integral converge, and Laplace transform $X(s)$ is

$$X(s)=\frac{1}{a+\sigma+j\omega}=\frac{1}{a+s}$$

Again as a special case when $a=0$, $x(t)=-u(-t)$ we have

$${\cal L}[-u(-t)]=\frac{1}{s},\;\;\;\;\;\sigma=Re[s]<0$$

Comparing the two examples above we see that two different signals may have identical Laplace transform $X(s)$, but different ROC. In the first case above, the ROC is $Re[s]>0$, and in the second case, the ROC is $Re[s]<0$. To determine the time signal $x(t)$ by the inverse Laplace transform, we need the ROC as well as $X(s)$.

Example 3:

$$x(t)=e^{-a\vert t\vert}=e^{-at}u(t)+e^{at}u(-t)$$

The Laplace transform is linear, and $X(s)$ is the sum of the transforms for the two terms:

$${\cal L}[e^{-at}u(t)]=\frac{1}{s+a},\;\;\;\;\;(\sigma>-a),\;\... ...;\; {\cal L}[e^{at}u(-t)]=\frac{-1}{s-a},\;\;\;\;\;(\sigma<a)$$

If $a>0$, i.e., $x(t)$ decays when $\vert t\vert \rightarrow \infty$, the intersection of the two ROCs is $-a<\sigma<a$, and we have:

$${\cal L}[x(t)]=\frac{1}{s+a}-\frac{1}{s-a}=\frac{-2a}{s^2-a^2}$$

However, if $a<0$, i.e., $x(t)$ grows without a bound when $\vert t\vert \rightarrow \infty$, the intersection of the two ROCs is a empty set, the Laplace transform does not exist.

Example 4:

$$x(t)=[e^{-2t}+e^t cos(3t)]u(t) =[e^{-2t}+\frac{1}{2}e^{-(1-j3)t}+\frac{1}{2}e^{-(1+j3)t}]u(t)$$

The Laplace transform of this signal is

$$X(s) = \int_0^\infty [e^{-2t}+\frac{1}{2}e^{-(1-j3)t} +\frac{1}{2}e^{-(1+j3)t}]e^{-st} dt$$

$$= \int_0^\infty e^{-2t}e^{-st} dt +\frac{1}{2}\int_0^\infty e^{-(1-j3)t}e^{-st} dt +\frac{1}{2}\int_0^\infty e^{-(1+j3)t}e^{-st} dt$$

$$= \frac{1}{s+2}+\frac{1/2}{s+(1-j3)}+\frac{1/2}{s+(1+j3)} =\frac{2s^2+5s+12}{(s^2+2s+10)(s+2)}$$

This $X(s)$ exists only if the Laplace transforms of all three individual terms exist, i.e, the conditions for the three integrals to converge are simultaneously satisfied:

$$Re[s]>-2,\;\;\;\;\;Re[s]>-1,\;\;\;\;\;Re[s]>-1$$

i.e., $Re[s]>-1$.

Example 5:

$${\cal L}[\delta(t-T)]=\int_{-\infty}^\infty \delta(t-T) e^{-st} dt=e^{-sT}$$

As the Laplace integration converges independent of $s$, the ROC is the entire s-plane. In particular, when $T=0$, we have

$${\cal L}[\delta(t)]=1$$