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��Chapter 8�Design Strategies for Second Order Systems

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Course Overview

Introduction to Feedback Control, Beard, McLain, Peterson, Killpack

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Pole placement

We have seen that for second order systems, we can place the poles using PD control.

But where should we place them?

What makes for good pole locations?

That is the topic of this chapter.

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First Order Response

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First Order Response

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Note that rise time is independent of A

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First Order Response

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DC-Gain

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DC-Gain

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Second Order Response, Real poles

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Second Order Response, Complex Poles

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Second Order Response

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Second Order Response

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Second Order Response

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Second Order Response

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Effect of Zeros

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Transfer functions are adjusted so that the DC-gain is one.

\begin{align*}

&H(s) = \frac{6}{(s+2)(s+3)} \qquad\qquad \text{\blue{No zeros}} \\

y(t) &= \mathcal{L}^{-1}\left\{\frac{6}{s(s+2)(s+3)}\right\}

= \mathcal{L}^{-1}\left\{ \frac{1}{s} + \frac{-3}{s+2} + \frac{2}{s+3}\right\}

= 1 - 3e^{-2t} + 2e^{-3t} \\

&H(s) = \frac{\frac{6}{2.1}(s+2.1)}{(s+2)(s+3)} \qquad\qquad \text{\blue{Zero close to the poles}} \\

y(t) &= \mathcal{L}^{-1}\left\{\frac{\frac{6}{2.1}(s+2.1)}{s(s+2)(s+3)}\right\}

= \mathcal{L}^{-1}\left\{ \frac{1}{s} + \frac{-0.14}{s+2} + \frac{-0.86}{s+3}\right\}

= 1 - 0.14e^{-2t} - 0.86e^{-3t} \\

&H(s) = \frac{\frac{6}{60}(s+60)}{(s+2)(s+3)} \qquad\qquad \text{\blue{Zero far from the poles}} \\

y(t) &= \mathcal{L}^{-1}\left\{\frac{\frac{6}{60}(s+60)}{s(s+2)(s+3)}\right\}

= \mathcal{L}^{-1}\left\{ \frac{1}{s} + \frac{-2.9}{s+2} + \frac{1.9}{s+3}\right\}

= 1 - 2.9e^{-2t} + 1.9e^{-3t}

\end{align*}

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Effect of Zeros

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\par\noindent Another way to understand zeros:

\begin{align*}

H(s) &= \frac{\omega_n^2(\tau s+1)}{s^2 + 2\zeta\omega_n s + \omega_n^2}

= \tau\frac{\omega_n^2 s}{s^2 + 2\zeta\omega_n s + \omega_n^2} + \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}.

\end{align*}

Therefore the step response is given by

\begin{align*}

y(t) &= \mathcal{L}^{-1}\left\{\tau\frac{\omega_n^2s}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)} + \frac{\omega_n^2}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)}\right\} \\

&= \tau\underbrace{\frac{d}{dt}\mathcal{L}^{-1}\left\{\frac{\omega_n^2}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)}\right\}}_{\text{\blue{Derivative of response without zero.}}}

% \\ &\qquad

+ \underbrace{\mathcal{L}^{-1}\left\{\frac{\omega_n^2}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)}\right\}}_{\text{\blue{Response without zero.}}}

\end{align*}

�

\par\noindent{\blue{The zero causes the derivative of the response to be added to the response}}

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Effect of Zeros

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Successive Loop Closure

Introduction to Feedback Control, Beard, McLain, Peterson, Killpack

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Many systems are modeled as a cascade of systems:

For example:

Inverted pendulum
Satellite
Block on beam
Planar VTOL

Successive loop closure design:

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Successive Loop Closure

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Design inner-loop to have fast response: Rule of thumb (10x faster than outer loop)

To design the outer loop, replace the inner loop with its DC-gain

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Input Saturation

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Design Study A

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Design Study A

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Python Implementation:

https://github.com/randybeard/controlbook_public/tree/master/_A_arm/python

ctrlPD.py

hw08_armSim.py

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Design Study B

Control System �(computer code)

Physical System

Inner loop

Outer loop

Zero-cancelling filter

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Design Study B: Inner Loop

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Design Study B: Inner Loop

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Design Study B: Outer Loop

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Design Study B: Outer Loop

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Design Study B: Outer Loop

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After cancellation we get

\par\noindent Closed-loop transfer function:

\[

\tilde{Z}(s) = \left(

\frac{\left(\frac{k_{P_z}}{1+k_{D_z}}\right) (s-\sqrt{\frac{3g}{2\ell}})}

{s^2 - \left(\frac{k_{P_z} - k_{D_z}\sqrt{\frac{3g}{2\ell}}}{1+k_{D_z}}\right)s - \frac{k_{P_z}\sqrt{\frac{3g}{2\ell}}}{1+k_{D_z}}}

\right) \tilde{R}_z.

\]

Closed-loop characteristic equation

\[

\Delta_{cl}(s) = s^2 - \left(\frac{k_{P_z} - k_{D_z}\sqrt{\frac{3g}{2\ell}}}{1+k_{D_z}}\right)s - \frac{k_{P_z}\sqrt{\frac{3g}{2\ell}}}{1+k_{D_z}}

\]

Setting equal to $\Delta_{cl}^d = s^2 + 2\zeta_z\omega_{n_z}s + \omega_{n_z}^2$ gives

\begin{align*}

k_{D_z} &= \frac{\omega_{n_z}^2 \sqrt{\frac{2\ell}{3g}}-2\zeta_z\omega_{n_z}}

{\omega_{n_z}^2 \sqrt{\frac{2\ell}{3g}} + \sqrt{\frac{3g}{2\ell}}-2\zeta_z\omega_{n_z}} \\

k_{P_z} &= -\omega_{n_z}^2\sqrt{\frac{2\ell}{3g}}(1+k_{D_z})

\end{align*}

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Design Study B

Introduction to Feedback Control, Beard, McLain, Peterson, Killpack

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Python Implementation:

https://github.com/randybeard/controlbook_public/tree/master/_B_pendulum/python

ctrlPD.py

hw08_pendulumSim.py

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Design Study C

Introduction to Feedback Control, Beard, McLain, Peterson, Killpack

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Control System �(computer code)

Physical System

Inner loop

Outer loop

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Design Study C: Inner Loop

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Design Study C: Outer Loop

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Design Study C: Outer Loop

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\par\noindent The gains satisfy:

\begin{align*}

\frac{k+kk_{DC_\theta}k_{P_\phi}}{J_p+bk_{DC_\theta}k_{D_\phi}} &= \omega_{n_\phi}^2 \\

\frac{b+bk_{DC_\theta}k_{P_\phi}+kk_{DC_\theta}k_{D_\phi}}{J_p+bk_{DC_\theta}k_{D_\phi}} &= 2\zeta_{\phi}\omega_{n_\phi}.

\end{align*}

In matrix form

\[

\begin{pmatrix} kk_{DC_\theta} & -bk_{DC_\theta}\omega_{n_\phi}^2 \\ bk_{DC_\theta} & kk_{DC_\theta}-2bk_{DC_\theta}\zeta_\phi\omega_{n_\phi} \end{pmatrix}\begin{pmatrix}k_{P_\phi}\\k_{D_\phi}\end{pmatrix} = \begin{pmatrix} -k + J_p\omega_{n_\phi}^2 \\ -b + 2J_p\zeta_\phi\omega_{n_\phi} \end{pmatrix},

\]

or

\begin{align*}

\begin{pmatrix}k_{P_\phi}\\k_{D_\phi}\end{pmatrix} &= \begin{pmatrix} kk_{DC_\theta} & -bk_{DC_\theta}\omega_{n_\phi}^2 \\ bk_{DC_\theta} & kk_{DC_\theta}-2bk_{DC_\theta}\zeta_\phi\omega_{n_\phi} \end{pmatrix}^{-1}

\begin{pmatrix} -k + J_p\omega_{n_\phi}^2 \\ -b + 2J_p\zeta_\phi\omega_{n_\phi} \end{pmatrix}

\end{align*}

�

The DC gain of the outer loop is given by

\[

DC_{gain} = \frac{kk_{DC_\theta}k_{P_\phi}}{k+kk_{DC_\theta}k_{P_\phi}} \neq 1

\]

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Design Study C

Introduction to Feedback Control, Beard, McLain, Peterson, Killpack

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Python Implementation:

https://github.com/randybeard/controlbook_public/tree/master/_C_satellite/python

ctrlPD.py

hw08_satelliteSim.py