Given
$\frac {d^2y(t)}{dt^2} + a\frac {dy(t)}{dt} = x(t) + by(t)$
Find:
a) $ H(s) = \frac{Y(s)}{X(s)}$
b) ROC of the stable function and the correspond h(t) and determine if the stable system is causal when a=1, b=2
- The attempt at a solution
$[s^2 + as - b] Y(s) = X(s)$
$H(s) = \frac{1}{s^2+as-b}$
Then I used $ \frac{-a + \sqrt{a^2+4b}}{2}$ to find s but when I use the values given it gives 0. I'm not sure if that means that the system is stable.
Given your differential equation and assuming zero initial conditions for the transfer function, we obtain $$ s^2Y(s)+asY(s)-bY(s) = X(s)\Rightarrow H(s) = \frac{Y(s)}{X(s)} = \frac{1}{s^2+as-b} $$ as you have found correctly. Now, we need to determine $h(t)$. We can use the inverse Mellin transform when $a=1$ and $b=2$ where the poles are then $s=1,-2$ \begin{align} h(t) &= \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{st}}{s^2+s-2}ds\\ &= \sum\text{Res}\\ &= \lim_{s\to-2}(s+2)\frac{e^{st}}{s^2+s-2}+\lim_{s\to1}(s-1)\frac{e^{st}}{s^2+s-2}\\ &= \frac{e^t-e^{-2t}}{3} \end{align} The ROC for $e^{-2t-st}$ is then $\text{Re}(s)>-2$, and for $e^{t-st}$, $\text{Re}(s)>1$; therefore, $\text{Re}(s)>1$. The system is stable if all poles in the $s$-plane have negative real parts. Since the poles are $s=1,-2$, not all the poles are on the left hand side for $a=1$ and $b=2$.