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Given a continuous LTI system with transfer function $$H(s)= -\frac{2s}{(s+6)(s+2)}$$

  1. Plot the location of the pole(s) and zero(s)
  2. Find all possible regions of convergence
  3. From the problem above find the impulse response

Here's my attempt

  1. The poles are $s_1=-6$ and $s_2=-2$ and the zero is $s_3=0$ enter image description here
  2. The all possible roc-s are enter image description here
  3. Using inverse laplace transform I found $h(t)=e^{-2t}-3e^{-6t}$. For the right-sided signal I multiply $h(t)$ with $u(t)$ and I got $h_{RS}(t)=(e^{-2t}-3e^{-6t})u(t)$ and for the left-sided signal I multiply $h(t)$ with $u(-t)$ and I got $h_{LS}(t)=(e^{-2t}-3e^{-6t})u(-t)$.

My confusion is at the two-sided signal. My solution is $h_{TS}(t)=e^{-2t}-3e^{-6t}$ and my friend's solution is $h_{TS}(t)=-e^{-2t}u(-t) - 3e^{-6t}u(t)$. When I plotted the signal both are a two-sided signal. Since my professor didn't submit the solution we didn't know who's right or wrong.

Thank you.

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  • $\begingroup$ did you forget the scale $3$ in your friend's solution ? then it looks ok. $\endgroup$ – Fat32 Dec 9 '20 at 13:42
  • $\begingroup$ @Fat32 ah yes I forgot the scale 3. So the right answer is my friend's solution? $\endgroup$ – eejo Dec 9 '20 at 14:17
  • $\begingroup$ yes that looks like the right solution. $\endgroup$ – Fat32 Dec 9 '20 at 16:29
  • $\begingroup$ @Fat32 could you explain why? thank you $\endgroup$ – eejo Dec 10 '20 at 11:20
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Given LTI system with transfer function :

$$H(s) = \frac{ -2 s}{(s+6)(s+2) } \tag{1}$$

which has the poles $s = \{ -2 , -6 \} $, and zeros $ s = 0 $.

Apply partial fraction expansion :

$$H(s) = \frac{1}{s+2} - \frac{3}{s+6} \tag{2}$$

Based on pole locations, there will be three ROCs,

  • ROC1 : -$2 < \mathcal{Re}\{s\} < \infty $ ; stable
  • ROC2 : $-\infty < \mathcal{Re}\{s\} < -6$ ; unstable
  • ROC3 : $ -6 < \mathcal{Re}\{s\} < -2$ ; unstable

The impulse responses found from (2) and the chosen ROC , with the fact that

$$\mathcal{L}\{ e^{at}u(t) \} \longleftrightarrow \frac{1}{s-a} ~~,~~ \mathcal{Re}\{s\} >\mathcal{Re}\{a\} \tag{3}$$ $$\mathcal{L}\{ -e^{at}u(-t) \} \longleftrightarrow \frac{1}{s-a} ~~,~~ \mathcal{Re}\{s\} <\mathcal{Re}\{a\} \tag{4}\\\\$$

  • ROC1 : -$2 < \mathcal{Re}\{s\} < \infty $ ; right-sided : $h(t) = (e^{-2t}-3 e^{-6t}) u(t)\\\\$
  • ROC2 : $-\infty < \mathcal{Re}\{s\} < -6$ ; left-sided : $h(t) = (-e^{-2t}+3 e^{-6t}) u(-t)\\\\$
  • ROC3 : $ -6 < \mathcal{Re}\{s\} < -2$ ; two-sided : $h(t) = -e^{-2t}u(-t)-3 e^{-6t}u(t)$.
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