For a homework problem, I am supposed to find the step response v(t) for the following circuit:
simulate this circuit – Schematic created using CircuitLab
Where the input is \$v_s(t) = 2u(t)\$, and the response \$v(t)\$ is the voltage across the capicator C1.
So, here's what I did. I converted the circuit to the s-domain and constructed a transfer function:
$$\frac{V(s)}{V_s(s)}=\frac{\frac{1}{sC}}{R_1+\frac{1}{sC}||(R_2+sL)}$$
And then simplifying the function and solving for V(s):
$$V(s) = \frac{LCs^2+R_2Cs+1}{R_1LC^2s^3+(R_1R_2C^2+LC)s^2+(R_1C+R_2C)s} V_s(s)$$
Since the input is converted to \$\frac{2}{s}\$ in the s-domain:
$$V(s) = \frac{LCs^2+R_2Cs+1}{R_1LC^2s^3+(R_1R_2C^2+LC)s^2+(R_1C+R_2C)s} \frac{2}{s}$$
Which does not make sense because then the final value would be:
$$v(\infty)= \lim_{s \to 0} sV(s) = \infty$$
Which is impossible because you cannot have infinite voltage. There has to be a limit. What am I doing wrong here?
Best Answer
There seems to be an error in the approach of the transfer function:
$$\frac{V(s)}{V_s(s)}=\frac{\frac{1}{sC}}{R_1+\frac{1}{sC}||(R_2+sL)}$$
when the correct is:
$$ \frac{V(s)}{V_s(s)}=\frac{\frac{1}{sC}||(R_2+sL)}{R_1+\frac{1}{sC}||(R_2+sL)} $$
according to the voltage divider.