The problem asks us to determine if the system with impulse response $h(t)$ is causal and stable.

The given impulse response is:

Causality Analysis of h(t)

A system is causal if its output depends only on present and past inputs. For a system described by its impulse response $h(t)$, this means $h(t)$ must be zero for all time $t < 0$.

• In the given $h(t)$, the term $u(-t)$ is equal to 1 for $t \le 0$ and 0 for $t > 0$.
• For $t < 0$, $u(-t) = 1$. Therefore, $h(t) = -\frac{2}{3}e^{-t} + \frac{2}{3}e^{-2t}$ for $t < 0$.
• Since $h(t)$ is non-zero for $t < 0$, the system is non-causal.

Stability Analysis of h(t)

A system is stable if its impulse response $h(t)$ is absolutely integrable, meaning $\int_{-\infty}^{\infty} |h(t)| dt < \infty$.

• We need to evaluate the integral $\int_{-\infty}^{\infty} |h(t)| dt$.
• For $t < 0$, $h(t) = -\frac{2}{3}e^{-t} + \frac{2}{3}e^{-2t}$. We found this is positive for $t < 0$.
• Consider the integral from $-\infty$ to $0$: $\int_{-\infty}^{0} |-\frac{2}{3}e^{-t} + \frac{2}{3}e^{-2t}| dt = \int_{-\infty}^{0} (-\frac{2}{3}e^{-t} + \frac{2}{3}e^{-2t}) dt$
• Evaluating the integral: $[\frac{2}{3}e^{-t} - \frac{1}{3}e^{-2t}]_{-\infty}^{0}$
• Substituting the limits: $= (\frac{2}{3}e^{0} - \frac{1}{3}e^{0}) - \lim_{t \to -\infty} (\frac{2}{3}e^{-t} - \frac{1}{3}e^{-2t})$ $= (\frac{2}{3} - \frac{1}{3}) - \lim_{t \to -\infty} (\frac{2}{3}e^{-t} - \frac{1}{3}e^{-2t})$ $= \frac{1}{3} - \lim_{t \to -\infty} (\frac{2}{3}e^{-t} - \frac{1}{3}e^{-2t})$
• The limit term $\lim_{t \to -\infty} (\frac{2}{3}e^{-t} - \frac{1}{3}e^{-2t})$ diverges to infinity because $e^{-t}$ and $e^{-2t}$ grow without bound as $t \to -\infty$.
• Since the integral $\int_{-\infty}^{0} |h(t)| dt$ diverges, the overall integral $\int_{-\infty}^{\infty} |h(t)| dt$ also diverges.
• Therefore, the system is unstable.

Conclusion

Based on the analysis, the system is both non-causal and unstable.