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I have the following question:

Pole-zero plot of x(t) and y(t) are given below:

enter image description here

The signal $g(t)$ and $h(t)$ are defined as $g(t)=x(t)e^{-3t}$ and $h(t)=y(t)*e^{-t}u(t)$. If $g(t)$ and $h(t)$ are both absolutely integrable, determine whether the signals $g(t)$, $h(t)$ are left-sided/right-sided.

My try:

I take the laplace transform of both the signals and get $G(s)=X(s+3)$ and $H(s)=Y(s)\cdot \frac{1}{s+1}$

Also because both $g(t)$ and $h(t)$ are absolutely integrable, their transforms must be stable so both must have $jw$-axis in their respective ROC.

As we can see $X(s+3)$ shifts the pole-zero plot to the left by $3$ units so we have all the poles in the left $s$-plane and ROC would be $Re\{s\}>-1$ hence $g(t)$ is right sided.

Similarly $H(s)$ has all the poles in the left $s$-plane and ROC is again $Re\{s\}>-1$ hence $h(t)$ is right sided.

Is this reasoning correct?

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    $\begingroup$ yes it's a correct reasoning... $\endgroup$ – Fat32 Oct 28 '18 at 11:06
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    $\begingroup$ @Fat32 Thanks for the confirmation. You may use it as an answer with some extra points (if required). $\endgroup$ – paulplusx Oct 28 '18 at 11:16
  • $\begingroup$ No thanks. A comment seems enough here... $\endgroup$ – Fat32 Oct 28 '18 at 11:19
  • $\begingroup$ @Fat32 But then this question would be "unanswered" as per SE system, right? Is that ok? $\endgroup$ – paulplusx Oct 28 '18 at 11:21
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    $\begingroup$ yes that's right it will be unanswered. So let's make one ;-) $\endgroup$ – Fat32 Oct 28 '18 at 11:23
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Looking at the pole-zero plots of the continuous-time signals $x(t)$ and $y(t)$, and the new signals $g(t)= x(t)e^{-3t}$ and $h(t) = y(t) \star e^{-t}u(t)$, the pole locaitons of $g(t)$ and $h(t)$ are found to be:

$$g(t) = x(t)e^{-3t} \implies G(s) = X(s+3) \implies \text{ Re(poles) } = \{-1,-1 \} $$

$$h(t) = y(t) \star e^{-t}u(t) \implies H(s) = Y(S) \frac{1}{s+1} \implies \text{ Re(poles) } = \{-2,-2,-1 \} $$

From these pole locations we see the following:

1-$g(t)$ has two possible ROCs: $Re\{s\} <-1$, and $Re\{s\} > -1$, and only the second one includes the $j\omega$ axis and hence can be stable.

2-$h(t)$ has three possible ROCs: $Re\{s\}<-2$, $-2 < Re\{s\} <-1$, and $Re\{s\} > -1$, and only the last one includes the $j\omega$ axis and is stable.

So for $h(t)$ and $g(t)$ to be absolutely integrable (stable), their ROC's must include $j\omega$ axis and this means their ROCs are to the right of the largest poles which implies that the signals are causal right sided signals.

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    $\begingroup$ Yes, you included those extra points because the question is so vague. Thanks a lot :-) $\endgroup$ – paulplusx Oct 28 '18 at 13:33
  • $\begingroup$ your welcome @paulplusx ... as a jack of all trades what's your main training focus ? $\endgroup$ – Fat32 Oct 28 '18 at 13:36
  • $\begingroup$ Still struggling to find :P I like electronics and programming and how beautifully they come together :-) Still an aspiring masters student though :P $\endgroup$ – paulplusx Oct 28 '18 at 14:13
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    $\begingroup$ Yes that's the case; electronics and programming stem from the same branch and complement each other very beautifully. That's why there are departments like Electrical and Computer Engineering (ECE) around the world. $\endgroup$ – Fat32 Oct 28 '18 at 14:35
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    $\begingroup$ Yes, very true, I am aiming for that particular department ECE (Computer engineering to be specific). Unfortunately and sadly the country I am in doesn't have that ECE. Here, electrical, electronics and computer science are separate departments. Here, ECE means Electronics and Communication Engineering :-( which was there in my bachelor's. Still not a real problem, knowledge is everywhere :-) $\endgroup$ – paulplusx Oct 28 '18 at 15:04

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