The circuit shown below has a transfer function given by:

$$V_("out")/V_("in")=(1.2(s+2))/((s+1)(s+4))$$

The voltage source $$v_("in")(t)$$ is a unit step function, and there is no initial charge on the capacitors.  The final (steady-state) value (V) of  $$v_("out")(t)$$ is most nearly:

A.   0
B.   0.04
C.   0.55
D.   0.60

In order to solve for the final steady state output voltage of $v_{out}\left(t\right)$, we first want to get a simplified expression for the output voltage within the frequency domain. Our transfer function, as given in the problem, is as follows:

$\frac{V_{out}}{V_{in}}=\frac{1.2(s+2)}{(s+1)(s+4)}$

We are given $v_{in}\left(t\right)$ as the unit step function, $u\left(t\right)$. Since this exists in the time domain, we must use the Laplace transform in order to convert it to the frequency domain. Once we do this, we can plug it into our transfer function in order to solve for $V_{out}$. Our Laplace transform table in located on pg.56 of our FE reference handbook.

From the table above, the Laplace transform of the unit step function is equal to $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. Now let's plug this back into our transfer function and solve for $V_{out}$ in terms of s:

$$\begin{gathered} \frac{V_{out}}{{\displaystyle \frac{1}{s}}}=\frac{1.2(s+2)}{(s+1)(s+4)} \\ {V}_{out}=\frac{1}{s}\frac{1.2(s+2)}{(s+1)(s+4)} \end{gathered}$$

Now that we have our simplified expression for $V_{out}$ in the frequency domain, we can apply the Final Value Theorem in order to find the steady state value. This theorem states that by taking the limit of the transfer function multiplied by an additional s as s approaches 0, the value will equal to the final steady state value in the time domain. The theorem is expressed in the table above as well, and mathematically states:

$\underset{t\to \infty }{lim}f\left(t\right)=\underset{s\to 0}{lim}sF\left(s\right)$

Therefore, applying our theorem to our current equation looks like this:

$$\begin{gathered} \underset{t\to \infty }{lim}{v}_{out}\left(t\right)=\underset{s\to 0}{lim}s{V}_{out} \\ \underset{t\to \infty }{lim}{v}_{out}\left(t\right)=\underset{s\to 0}{lim}s\left(\frac{1}{s}\frac{1.2(s+2)}{(s+1)(s+4)}\right) \\ \underset{t\to \infty }{lim}{v}_{out}\left(t\right)=\underset{s\to 0}{lim}\left(\frac{1.2(s+2)}{(s+1)(s+4)}\right) \\ \underset{t\to \infty }{lim}{v}_{out}\left(t\right)=\frac{1.2(0+2)}{(0+1)(0+4)}\left) \\ \underset{t\to \infty }{lim}{v}_{out}\right(t)=\frac{2.4}{4} \\ \underset{t\to \infty }{lim}{v}_{out}(t)=\frac{2.4}{4}=.6 \end{gathered}$$

Therefore, the final steady state value of $v_{out}\left(t\right)$ is 0.6V, corresponding to answer D.