Training Control Systems Step Response of First & Second Order
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Step Response of First & Second Order

30 min Control Systems

Step Response of First & Second-Order Systems

The step response — what the output does when the input suddenly jumps from 0 to 1 — is the benchmark for control performance. It reveals settling time, overshoot, steady-state error, and rise time, all of which are central to specifying a controller.

First-order systems $G(s) = K/(\tau s + 1)$ have pure exponential responses: $y(t) = K(1 - e^{-t/\tau})$. The time constant $\tau$ controls how fast the output reaches its final value (63% at $t = \tau$, 98% at $t = 4\tau$).

Second-order systems $G(s) = \omega_n^2/(s^2 + 2\zeta\omega_n s + \omega_n^2)$ add oscillation. The damping ratio $\zeta$ sets the character: under-damped ($\zeta < 1$) rings, critically damped ($\zeta = 1$) settles fastest without overshoot, and over-damped ($\zeta > 1$) is sluggish.

Standard Second-Order Form

$$G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$

Key metrics (under-damped, $0 < \zeta < 1$):

  • Percent overshoot: $M_p = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}}$ %
  • Settling time (2%): $t_s \approx 4/(\zeta\omega_n)$
  • Peak time: $t_p = \pi/(\omega_n\sqrt{1-\zeta^2})$
First-Order Rule of Thumb

For $G = K/(\tau s + 1)$: $y(\tau) = 0.632\,K$, $y(3\tau) = 0.950\,K$, $y(5\tau) = 0.993\,K$.

Example 1 — First-order settling

$G(s) = 5/(2s+1)$. Find the time for the step response to reach 98% of final value.

$\tau = 2\,\text{s}$, so $t_{98\%} \approx 4\tau = 8\,\text{s}$.

Example 2 — Overshoot

$\zeta = 0.3$, $\omega_n = 5\,\text{rad/s}$. Find percent overshoot and peak time.

$M_p = 100\,e^{-\pi(0.3)/\sqrt{1-0.09}} = 100\,e^{-0.9425/0.954} = 100\,e^{-0.988} \approx 37.2\%$.

$t_p = \pi/(5\sqrt{0.91}) = \pi/4.77 \approx 0.659\,\text{s}$.

Example 3 — Design

Specify $\zeta$ and $\omega_n$ so overshoot is at most 5% and settling time is at most 1 s.

Overshoot ≤ 5%: $\zeta \ge 0.69$ (solve $e^{-\pi\zeta/\sqrt{1-\zeta^2}} \le 0.05$). Pick $\zeta = 0.7$.

Settling ≤ 1: $4/(\zeta\omega_n) \le 1 \Rightarrow \omega_n \ge 4/0.7 = 5.71\,\text{rad/s}$.

Choose $\omega_n = 6$ rad/s, $\zeta = 0.7$.

Interactive Demo: Second-Order Step Response
Overshoot %37.2%
Peak time t_p =0.659s
Settling (2%) t_s =2.67s
Damping:Under-damped

Practice Problems

1. Find $\tau$ for $G = 4/(3s+1)$.
2. For $\zeta=0.5$, compute percent overshoot.
3. Compute $t_s$ (2%) for $\zeta=0.7$, $\omega_n=10$.
4. What happens when $\zeta=1$ exactly?
5. A second-order system with $\zeta=0$ does what?
6. Why is $\zeta = 0.707$ a common design target?
Show Answer Key

1. $\tau = 3$ s.

2. $M_p = 100 e^{-\pi\cdot 0.5/\sqrt{0.75}} = 100 e^{-1.814} \approx 16.3\%$.

3. $t_s = 4/(0.7 \cdot 10) = 0.571$ s.

4. Critical damping — fastest non-oscillatory return to steady state.

5. Undamped — sustained oscillation at $\omega_n$ forever.

6. Gives about 5% overshoot and minimises the combined rise+overshoot cost; a compromise.

Transfer Functions & Block Diagrams Stability & Routh–Hurwitz