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?

  • 1
    $\begingroup$ yes it's a correct reasoning... $\endgroup$ – Fat32 Oct 28 '18 at 11:06
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ yes that's right it will be unanswered. So let's make one ;-) $\endgroup$ – Fat32 Oct 28 '18 at 11:23

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.

| improve this answer | |
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.