First and Second Order Systems

First order

\[\tau \dot{y} = ku\]
or
\[\frac{Y(s)}{U(s)} = \frac{K}{\tau s + 1}\]
or
\[\begin{align} \dot{x}&=\frac{-x}{\tau} + \frac{K}{\tau} u\\ y &= x \end{align}\]

Observations:

• Pole at \(s=\frac{-1}{\tau}\)
• No zeroes
• BIBO stable if and only if \(\tau \gt 0\)
• steady-state gain: \(K\)
• bandwidth: \(\frac{1}{\tau}\) rad/s
• Impulse response: \(g(t)=\frac{K}{\tau} e^{-1/\tau} 1(t)\)
  • \(g(0)=\frac{K}{\tau}\)
  • \(g(\tau)=g(0)e^{-1} \approx 0.37g(0)\)
  • \(g(2\tau)=g(0)e^{-2} \approx 0.14g(0)\)
• Higher bandwidth implies faster impulse response, and vice versa

Step response:
\[y(t) = \mathcal{L}^{-1}\{G(s)U(s)\} = \mathcal{L}^{-1}\left\{\frac{K}{\tau s + 1} \frac{1}{s}\right\} = \mathcal{L}^{-1}\left\{\frac{K}{s} - \frac{K}{s + \frac{1}{\tau}}\right\}\]
\[K(1-e^{\frac{-t}{\tau}}), \quad t \ge 0\]

Observations:

1. After \(4\tau\) seconds, \(y(t)\) is within 2% of its steady-state value
2. For all \(t \gt 0\), \(y(t) \lt y_{ss}\) (no overshoot)
3. \(y(t)\) increases monotonically (no oscillations)
4. Decrease in \(\tau \Leftrightarrow\) step response gets faster \(\Leftrightarrow\) bandwidth goes up \(\Leftrightarrow\) poles move to the left

Second order

\[\ddot{y}+2\zeta\omega_n \dot{y} + \omega_n^2 y = K\omega_n^2 u\]
or
\[ \frac{Y(s)}{U(s)} = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\]
or
\[\begin{align} \dot{x} &= \begin{bmatrix}0&1\\-\omega_n^2&-2\zeta\omega_n\end{bmatrix}x + \begin{bmatrix}0\\K\omega_n^2\end{bmatrix}u\\ y &= \begin{bmatrix} 1 & 0\end{bmatrix}x\\ \end{align}\]

e.g.

\[\begin{align} M \ddot{q} &= u - K_{spring} q - b\dot{q}\\ \frac{Y(s)}{U(s)} &= \frac{\frac{1}{M}}{s^2 + \frac{b}{M}s + \frac{K_{spring}}{M}}\\ \\ \omega_n &= \sqrt{\frac{K_{spring}}{M}}\\ \zeta &= \frac{b}{2 \sqrt{K_{spring}M}}\\ K &= \frac{1}{K_{spring}} \end{align}\]

Pole locations

From the quadratic formula, find the zeroes of the denominator:
\[s = -\zeta \omega_n \pm \omega_n \sqrt{\zeta^2 - 1} = \omega_n\left(-\zeta \pm \sqrt{\zeta^2 - 1}\right)\]

Pole locations are used to categorize the system:

• Undamped \(\zeta = 0\)
• Underdamped \(0 \lt \zeta \lt 1\)
• Critically damped \(\zeta = 1\)
• Overdamped \(\zeta \gt 1\)

Steady-state gain: \(K\)
Zeroes: none

Step response

Underdamped Systems

• Poles are complex conjugate: \(s = -\zeta \omega_n \pm j\omega_n \sqrt{1 - \zeta^2} = \omega_n e^{\pm j (\pi - \theta)}, \theta = \arccos(\zeta)\)

Impulse response

\[g(t) = K\frac{\omega_n}{\sqrt{1-\zeta^2}} \underbrace{e^{-\zeta \omega_n t}}_\text{decay rate} \sin\underbrace{\left(\omega_n \sqrt{1-\zeta^2} t\right)}_\text{oscillation rate}, \quad t \ge 0\]

Observe: If we fix \(\zeta \in (0,1)\), then larger bandwidth \(\Leftrightarrow\) faster decay

Step response

\[\begin{align} u(t) &= 1(t)\\ \Rightarrow U(s) &= \frac{1}{s}\\ \\ Y(s) &= G(s)U(s)\\ \Rightarrow y(t) &= \mathcal{L}^{-1}{G \dot U}\\ &= K\left(1 - \frac{1}{\sqrt{1 - \zeta^2}} e^{-\zeta \omega_n t} \sin\left(\omega_n \sqrt{1 - \zeta^2}t + \theta\right)\right), \quad \theta = \arccos \zeta\\ \end{align}\]

Summary

• As \(\zeta \rightarrow 1\), response is less oscilatory, less overshoot, imaginary part of poles approaches zero
• As \(\zeta \rightarrow 0\), response is more oscillatory, more overshoot, real part of poles approaches 0
• \(\omega_{BW} \approx \omega_n\). As \(\omega_n \rightarrow \infty\), response is faster, poles have larger magnitude.
• Frequency of oscillation depends on imaginary part of the poles; rate of decay depends on the real part

General characteristics of step response

• Look at common metrics to quantify the quality
• metrics apply to any system
• we use \(G(s) = \frac{K \omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\) to get equations for the metrics in therms of \(K, \omega_n, \zeta\)

Overshoot

• only undamped second order systems have it:
  \[\%OS = \frac{||y||_\infty - |G(0)|}{|G(0)|}\]
• only depends on dampting ratio:
  \[\%OS = \exp\left(\frac{-\zeta \pi}{\sqrt{1 - \zeta^2}}\right), \quad 0 \lt \zeta \lt 1\]
• More damping (larger \(\zeta\)) \(\Leftrightarrow\) less overshoot

e.g. mass-spring damper

\[\frac{Y(s)}{U(s)} = \frac{\frac{1}{M}}{s^2 + \frac{b}{M}s + \frac{K_{spring}}{M}}\]

• Find conditions on \(M,b,K_{spring}\) so that \(\%OS \le \%OS_{max} = 0.05\)

\[\begin{align} \zeta &= \frac{b}{2\sqrt{MK_{spring}}}\\ \zeta &\ge \frac{-\ln(\%OS_{max})}{\sqrt{\pi^2 + (\ln\%OS_{max})^2}} =:\zeta_{min}\\ \\ \text{To meet specs:}\\ \frac{b}{2\sqrt{MK_{spring}}} \ge 0.6901\\ \end{align}\]

The angle the poles make is \(\pm(\pi - \arccos \zeta)\).
\[ \zeta \ge \zeta_{min} \Leftrightarrow \theta \le \arccos(\zeta_{min})\]
In this example, \(\theta \le 46^{\circ}\)

Therefore the overshoot spec is not met if there are poles in the shaded region.

Settling time

• The smallest time \(T_s\) such that \(\forall t \ge T_s,\quad \frac{|G(0)-y(t)|}{|y(t)|} \le 0.02\).
• An estimate is obtained by looking at the decay rate \(e^{-\zeta \omega_n t}\):
  \[e^{-\zeta \omega_n t} \le 0.02 \Rightarrow t \ge \frac{4}{\zeta \omega_n} \text{ (approx)}\]
  i.e. \(T_s = \frac{4}{\zeta \omega_n}\)

e.g. Mass-spring damper

Find the condition so that \(T_s \le T_s^{max} = 3\)

\[\begin{align} G(s) &= \frac{\frac{1}{M}}{s^2 + \frac{b}{M}s + \frac{K_{spring}}{M}}\\ \frac{4}{\zeta \omega_n} &\le 3 = T_s^{max}\\ \Leftrightarrow \zeta\omega_n &\ge \frac{4}{T_s^{max}}\\ \Leftrightarrow \frac{b}{2M} &\ge \frac{4}{3}\\ \end{align}\]

Time-to-peak

• Smallest \(T_p \gt 0\) such that \(||y||_\infty = y(T_p)\)
• Derived similar to overshoot: \(T_p = \frac{\pi}{\omega_n \sqrt{1 - \zeta^2}}, \quad 0 \lt \zeta \lt 1\)
• So \(T_p\) only depends on the imaginary part of the poles

e.g. mass-spring damper again

Spec: \(T_p \le T_p^{max} = 3 \text{ seconds }\)

\[\begin{align} \omega_n\sqrt{1-\zeta^2} &\le \frac{\pi}{T_p^{max}}\\ \Leftrightarrow \sqrt{\frac{K_{spring}}{M} - \frac{b^2}{4M^2}} &\le \frac{\pi}{3}\\ \end{align}\]

Changes to poles