Closed vs open loop step response analytically

In open loop the step response depends only on the plant; in closed loop the feedback changes the gain, speed, steady-state error and (for second-order systems) the damping.

First-order plant

Take a first-order plant $G(s) = \dfrac{K}{\tau s + 1}$.

Open loop: for a unit step input

$$c(t) = K\left(1 - e^{-t/\tau}\right), \qquad \text{time constant} = \tau$$

Closed loop (unity feedback):

$$\frac{C(s)}{R(s)} = \frac{G}{1+G} = \frac{K}{\tau s + 1 + K} = \frac{\dfrac{K}{1+K}}{\dfrac{\tau}{1+K}s + 1}$$ $$c(t) = \frac{K}{1+K}\left(1 - e^{-t(1+K)/\tau}\right), \qquad \tau_{cl} = \frac{\tau}{1+K}$$

So in closed loop: the time constant falls from $\tau$ to $\tau/(1+K)$ (faster response), and the steady-state gain falls from $K$ to $K/(1+K)$. The steady-state error for unit step is $e_{ss} = \dfrac{1}{1+K}$, which becomes small for large $K$.

Second-order case

For $G(s) = \dfrac{K}{s(s+a)}$, the open-loop step response grows without limit (a ramp after a transient), because of the pole at $s = 0$. With unity feedback:

$$\frac{C(s)}{R(s)} = \frac{K}{s^2 + as + K}, \quad \omega_n = \sqrt{K}, \quad \zeta = \frac{a}{2\sqrt{K}}$$

The output now settles at 1. Increasing $K$ raises $\omega_n$ (faster) and lowers $\zeta$ (more overshoot $M_p = e^{-\zeta\pi/\sqrt{1-\zeta^2}}$).

Summary

Example: $\tau = 1$ s, $K = 9$. Open loop: $\tau = 1$ s, settling time $4\tau = 4$ s. Closed loop: $\tau_{cl} = 1/10 = 0.1$ s, settling time $= 0.4$ s, i.e. ten times faster.