Unit step response of 4/[s(s+2)]

For unity feedback, the closed-loop transfer function is $T(s)=\frac{G(s)}{1+G(s)}$. We find $C(s)$ for $R(s)=1/s$ and take the inverse Laplace transform.

Closed-loop transfer function

$$\begin{aligned} \frac{C(s)}{R(s)} &= \frac{\frac{4}{s(s+2)}}{1+\frac{4}{s(s+2)}} = \frac{4}{s^2+2s+4} \end{aligned}$$

Compare with the standard form $\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$:

• $\omega_n^2 = 4 \Rightarrow \omega_n = 2$ rad/s
• $2\zeta\omega_n = 2 \Rightarrow \zeta = 0.5$ (underdamped, $0<\zeta<1$)
• $\omega_d = \omega_n\sqrt{1-\zeta^2} = 2\sqrt{0.75} = \sqrt{3} = 1.732$ rad/s
• $\zeta\omega_n = 1$, $\theta = \cos^{-1}\zeta = 60^\circ = \pi/3$ rad

Response by partial fractions

$$\begin{aligned} C(s) &= \frac{4}{s(s^2+2s+4)} = \frac{1}{s} - \frac{s+2}{s^2+2s+4} \\ &= \frac{1}{s} - \frac{(s+1)}{(s+1)^2+3} - \frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{(s+1)^2+3} \end{aligned}$$

Taking the inverse Laplace transform:

$$\begin{aligned} c(t) &= 1 - e^{-t}\left[\cos\sqrt{3}t + \frac{1}{\sqrt{3}}\sin\sqrt{3}t\right] \\ &= 1 - \frac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}}\sin(\omega_d t+\theta) \\ &= 1 - 1.1547\,e^{-t}\sin\left(1.732t + 60^\circ\right), \quad t\ge 0 \end{aligned}$$

Check: at $t=0$, $c(0)=1-1.1547\sin 60^\circ = 1-1 = 0$; as $t\to\infty$, $c(t)\to 1$, so the steady-state error is zero (type-1 system with step input).

Time-domain specifications (for completeness)

 c(t)
 1.163 |      _
   1.0 |----/---\_____________
       |   /       ''
       |  /
       | /
     0 |/_____________________ t
          tp=1.81 s

Answer: $c(t) = 1 - 1.1547\,e^{-t}\sin(1.732t+60^\circ)$; the response is underdamped ($\zeta=0.5$, $\omega_n=2$ rad/s) with 16.3% overshoot and settles to 1.