Determining the causality of a signal with it's pole-zero plot

Question

I have the following question:

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

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?

Answer 1

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.