$$F(s)=\frac{125(s+8)}{(s^2+12s+136)(s+0.5)}$$
which becomes
$$Y(s) = U(s)F(s)=\frac{125(s+8)}{s(s^2+12s+136)(s+0.5)}$$
I used partial fractions to obtain the following
$$\frac{125(s+8)}{s(s^2+12s+136)(s+0.5)} = \frac{K_1}{s} + \frac{K_2}{s+0.5} + \frac{K_3s + k_4}{s^2+12s+136}$$
I found all the K values so now I have to compute the inverse Laplace transform of this:
$$\frac{14.7}{s} – \frac{14.4}{s+0.5} – \frac{0.31s + 10.92}{s^2+12s+136}$$
I rewrite the third expression in a way that allows me to directly apply the inverse transform on it:
$$- \frac{0.31s + 10.92}{s^2+12s+136} = -0.31\frac{s+\frac{10.92}{0.31}}{(s+6)^2 + 10^2} = -0.31\frac{s+6}{(s+6)^2 + 10^2} -0.31\frac{-6+35.22}{(s+6)^2+10^2}$$
Now I can apply the inverse transform, getting the following:
$$y(s)=[14.7 – 14.4e^{-0.5t} -0.31e^{-6t}\cos(10t) – 9.05e^{-6t}\sin(10t)]1(t)$$
This is wrong, though. The correct result is
$$y(s)=[14.7 – 14.4e^{-0.5t} -0.31e^{-6t}\cos(10t) – 0.905e^{-6t}\sin(10t)]1(t).$$
I'm pretty sure I made a mistake when I rewrote that third expression.
Best Answer
Here's what I get: \begin{align*} Y(s)&=\frac{125(s+8)}{s(s^2+12s+136)(s+1/2)}\\ &=\frac{250}{17s}-\frac{7500}{521(s+1/2)}-\frac{250(11s+387)}{8857((s+6)^2+10^2)}\\ &=\frac{250}{17s}-\frac{7500}{521(s+1/2)}-\frac{2750(s+6+321/11)}{8857((s+6)^2+10^2)}\\ &=\frac{250}{17s}-\frac{7500}{521(s+1/2)}-\frac{2750(s+6)}{8857((s+6)^2+10^2)}-\frac{8025\cdot 10}{8857((s+6)^2+10^2)}\\ y(t)&=\frac{250}{17}\,u(t)-\frac{7500}{521}\,e^{-t/2}\,u(t)-\frac{2750}{8857}\,e^{-6t}\cos(10t)\,u(t)-\frac{8025}{8857}\,e^{-6t}\,\sin(10t)\,u(t)\\ &=\left[\frac{250}{17}-\frac{7500}{521}\,e^{-t/2}-\frac{25}{8857}\,e^{-6t}[110\cos(10t)+321\sin(10t)]\right]u(t). \end{align*} The coefficient of the $\sin(10t)$ term is, for me, approximately $0.906.$