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��Chapter 9�System Type and Integrators

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Course Overview

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Integrators

The objective of this chapter is to understand:

What integrators do

Reference tracking
Disturbance rejection

When they are needed

When they are not needed

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General Closed-loop System

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Closed-loop Transfer Function

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Final Value Theorem

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{\color{blue}

\par\noindent {\bf Final Value Theorem}

\\

Suppose that $z(t)$ and $Z(s)$ are Laplace transform pairs, and suppose that $\lim_{t\to\infty}z(t)$ is finite, then

\[

\lim_{t\to\infty}z(t) = \lim_{s\to 0} sZ(s).

\]

}

--

\par\noindent {\em Proof.}

Fundamental theorem of calculus:

\[

z(t) - z(0) = \int_0^t dz.

\]

Therefore

\begin{align*}

\lim_{t\to\infty}z(t) &= z(0) + \lim_{t\to\infty} \int_0^t \frac{dz}{d\tau}d\tau \\

&= z(0) + \lim_{t\to\infty} \int_0^t \dot{z}(\tau)d\tau \\

&= z(0) + \lim_{t\to\infty} \int_0^t \lim_{s\to 0} \dot{z}(\tau)e^{-s\tau} d\tau \\

&= z(0) + \lim_{s\to 0} \lim_{t\to\infty} \int_0^t \dot{z}(\tau)e^{-s\tau} d\tau \\

&= z(0) + \lim_{s\to 0} \int_0^{\infty} \dot{z}(\tau)e^{-s\tau} d\tau \\

&= z(0) + \lim_{s\to 0} \left( sZ(s) - z(0) \right) \\

&= \lim_{s\to 0} sZ(s).

\end{align*}

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Input Signals

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Error for Step Reference

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Error for Step Reference

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Error for Step Reference

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Error for Step Reference

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Error for Ramp Reference

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Error for Ramp Reference

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Error for Ramp Reference

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Error for Ramp Reference

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Error for Ramp Reference

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System Type

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Example: System Type

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Example: System Type

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Example: System Type

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Error for Input Disturbance

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Example: System Type w.r.t. Disturbance

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Example: System Type w.r.t. Disturbance

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Design Study A

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Design Study A

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Design Study A

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Design Study A

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For the input disturbance,

\[

E(s) = \frac{P(s)}{1+P(s)C(s)}D(s).

\]

With integrator ($k_I\neq 0$):

\begin{align*}

e_{ss}=\lim_{t\to\infty} e(t) &= \lim_{s\to 0} \frac{P}{1+PC}\frac{A}{s^q}

= \lim_{s\to 0} \left(\frac{\left(\frac{\frac{3}{m\ell^2}}{s(s+\frac{3b}{m\ell^2})}\right)}{1+\left(\frac{\frac{3}{m\ell^2}}{s(s+\frac{3b}{m\ell^2})}\right)\left(\frac{k_Ds^2+k_Ps+k_I}{s}\right)}\right)\left(\frac{A}{s^q}\right) \\

&= \lim_{s\to 0} \left(s\frac{\frac{3}{m\ell^2}}{s^2(s+\frac{3b}{m\ell^2})+\frac{3}{m\ell^2}(k_Ds^2+k_Ps+k_I)}\right)\left(\frac{A}{s^q}\right)

\underbrace{= \frac{A\frac{3}{m\ell^2}}{\frac{3k_I}{m\ell^2}}

= \frac{A}{k_I}}_{q=1},

\end{align*}

{\color{blue}System type to disturbance is type~1. $e_{ss}$ to a disturbance step is $0$, and to a disturbance ramp of size $A$ is $\frac{A}{k_I}$. }

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Design Study B: Inner loop

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Design Study B: Inner loop

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For the input disturbance,

\begin{align*}

\lim_{t\to\infty}e(t) &= \lim_{s\to 0}s\frac{P(s)}{1+P(s)C(s)}\frac{1}{s}

= \lim_{s\to 0} \frac{\left(\frac{-\frac{1}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}}{s^2-\frac{(m_1+m_2)g}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}}\right)}{1+\left(\frac{-\frac{1}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}}{s^2-\frac{(m_1+m_2)g}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}}\right)\left(k_Ds+k_P\right)} \\

&= \lim_{s\to 0} \frac{\left(-\frac{1}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}\right)}{\left(s^2-\frac{(m_1+m_2)g}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}\right)+\left(-\frac{1}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}\right)\left(k_Ds+k_P\right)} \\

&=\underbrace{\frac{\left(-\frac{1}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}\right)}{\left(-\frac{(m_1+m_2)g}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}\right)+\left(-\frac{k_p}{m_1\frac{\ell}{6}+m_2\frac{2\ell}{3}}\right)}

= \frac{1}{(m_1+m_2)g+k_P}}_{q=0}.

\end{align*}

�

{\color{blue}System type to disturbance is type~0. $e_{ss}$ to a disturbance step of size $A$ is $\frac{1}{(m_1+m_2)g+k_P}$. }

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Design Study B: Outer loop

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Design Study C: Inner loop

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Design Study C: Inner loop

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Design Study C: Outer loop

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