Consider the following signal: $$ x(t) = e^{-2t}[u(t) - u(t-5)] $$

This signal exists only from 0 to 5 time units. Elsewhere, it is zero.

Now, let's find the laplace transform of this signal using Linearity and Time shift properties.

$$ e^{-2t}u(t) \leftrightarrow \frac{1}{s+2} \ , \ \ Re \{s \} > -2 $$ Also, $$ e^{-2(t-5)}u(t-5) \leftrightarrow \frac{e^{-5s}}{s+2} \ , \ \ Re \{s \} > -2 $$ $$ \Rightarrow e^{-2t}u(t-5) \leftrightarrow \frac{e^{-5s}e^{-10}}{s+2} \ , \ \ Re \{s \} > -2 $$

Thus, by linearity property, $$ e^{-2t}[u(t) - u(t-5)] = \frac{1 - e^{-5(s+2)}}{s+2} \ , \ \ Re \{s \} > -2$$

Note: The time shifting property doesn't alter the ROC;

However, the textbooks that i am refering (Oppenheim and Schaum series) both tell that the ROC of a finite duration signal is the entire S-plane, possibly zero and infinity (in some cases).

But the above signal being of finite-duration, possess ROC that is not the entire s-plane. Please help me figure this conceptual error.

Note: The above problem is from Schaum series. Here are the images of the textbook's section relevant to the above question.

Source of the Question and its solution: enter image description here

Property of finite duration signals:

In Schaum's outline series: In Schaum's outline series In oppenheim: In oppenheim


The property claimed by Schaum and Oppenheim is also true for the given example. Note that the Laplace transform


has no pole at $s=-2$:

$$\lim_{s\to -2}X(s)=\frac{1-(1-5(s+2))}{s+2}\Big{|}_{s=-2}=5$$

So the ROC is indeed the entire $s$-plane. Even though the ROCs of the two individual signals in your solution have the same right half-plane as ROC, their sum has the entire $s$-plane as ROC because the two signals cancel everywhere except in a finite interval.

  • $\begingroup$ Seems to me that it has a pole and a zero at $s=-2$. $\endgroup$ Jul 23 '19 at 10:12
  • $\begingroup$ @robertbristow-johnson , yes. Thats why there is a pole-zero cancellation, thus resulting in the ROC as the entire s-plane $\endgroup$ Jul 23 '19 at 10:27
  • $\begingroup$ @MattL. Thank you. That is what i was thinking. Thus the ROC given in the textbook is wrong, and it should be the entire s-plane. The linearity property also tells that the ROC for linear combination of two signals is atleast $$ R = R_{1} \cap R_{2} $$ $\endgroup$ Jul 23 '19 at 10:31
  • $\begingroup$ @Jarvis: Yes, the textbook is wrong concerning the ROC. When the linear combination of two signals is zero except for a finite interval then the ROC is always the entire $s$-plane. $\endgroup$
    – Matt L.
    Jul 23 '19 at 10:42
  • 1
    $\begingroup$ @robert bristow-johnson yes. I agree with you when there is a cancellation of an unstable pole, as though the system may be externally stable (i.e., BIBO stable), it may not be internally stable when spoken in terms of lyapunov's stability as some inputs may excite the unstable modes and lead to system instability. The unstable poles can also result as a part of improper pole placement via complete state feedback. However, in this simple situation, I guess we can safely ignore those aspects and conclude that the pole zero cancellation do not result in any adversity $\endgroup$ Jul 24 '19 at 5:11

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