]> 3.3.3 Example 3: Automatic control system with PID controller II

#### 3.3.3 Example 3: Automatic control system with PID controller II

In accordance with the theory presented in the Section 3.2 the integral (3.15) exists if $\left|c\right|<1$ and all roots of the characteristic quasipolynomial

 $\left({s}^{2}-as\right)-\left({s}^{2}c+sb+d\right){e}^{-sr}$ (3.37)

have negative real parts.

The observed output is

${z}_{1}={v}_{1}+c{\psi }_{1}\left(-r\right),$

so we have

$P=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}Q=\left[\begin{array}{cc}\hfill c\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}R=\left[\begin{array}{cc}\hfill {c}^{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right].$

The equation (3.24) yields

$\gamma =\left[\begin{array}{cc}\hfill \frac{{c}^{2}}{1-{c}^{2}}\phantom{\rule{0ex}{0ex}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\hfill \\ \hfill \phantom{\rule{0ex}{0ex}}0\hfill & \hfill \phantom{\rule{0ex}{0ex}}0\hfill \end{array}\right].$

Taking into account the nature of the initial conditions we conclude that

 $J\left({x}_{0}\right)\left(b,c,d\right)={〈{x}_{0},\mathcal{ℋ}{x}_{0}〉}_{\text{H}}={v}_{0}^{T}\alpha {v}_{0}={z}_{0}^{2}{\alpha }_{11}$ (3.38)

Hence, to ﬁnd the matrix $\alpha$ we need to solve the vector differential equation

 $\begin{array}{c}\left[\begin{array}{c}\hfill {\beta }_{11}^{\prime }\left(\theta \right)\\ \hfill {\beta }_{12}^{\prime }\left(\theta \right)\\ \hfill {\beta }_{21}^{\prime }\left(\theta \right)\\ \hfill {\beta }_{22}^{\prime }\left(\theta \right)\\ \hfill {\vartheta }_{11}^{\prime }\left(\theta \right)\\ \hfill {\vartheta }_{12}^{\prime }\left(\theta \right)\\ \hfill {\vartheta }_{21}^{\prime }\left(\theta \right)\\ \hfill {\vartheta }_{22}^{\prime }\left(\theta \right)\end{array}\right]={\left[\begin{array}{cc}\hfill I\otimes I& I\otimes {A}_{0}^{T}\hfill \\ \hfill {A}_{0}^{T}\otimes I& I\otimes I\hfill \end{array}\right]}^{-1}\left[\begin{array}{cc}\hfill {A}_{1}^{T}\otimes I& \hfill I\otimes {A}_{2}^{T}\\ \hfill -{A}_{2}^{T}\otimes I& \hfill -I\otimes {A}_{1}^{T}\end{array}\right]\left[\begin{array}{c}\hfill {\beta }_{11}\left(\theta \right)\\ \hfill {\beta }_{12}\left(\theta \right)\\ \hfill {\beta }_{21}\left(\theta \right)\\ \hfill {\beta }_{22}\left(\theta \right)\\ \hfill {\vartheta }_{11}\left(\theta \right)\\ \hfill {\vartheta }_{12}\left(\theta \right)\\ \hfill {\vartheta }_{21}\left(\theta \right)\\ \hfill {\vartheta }_{22}\left(\theta \right)\end{array}\right]=\hfill \\ =\left[\begin{array}{cccccccc}\hfill \frac{a+bc}{1-{c}^{2}}\hfill & \hfill 0\hfill & \hfill \frac{cd}{1-{c}^{2}}\hfill & \hfill 0\hfill & \hfill \frac{b+ac}{1-{c}^{2}}\hfill & \hfill \frac{d}{1-{c}^{2}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill a\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill b+ac\hfill & \hfill d\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{b+ac}{{c}^{2}-1}\hfill & \hfill 0\hfill & \hfill \frac{d}{{c}^{2}-1}\hfill & \hfill 0\hfill & \hfill \frac{a+bc}{{c}^{2}-1}\hfill & \hfill \frac{cd}{{c}^{2}-1}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -b-ac\hfill & \hfill 0\hfill & \hfill -d\hfill & \hfill -1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{c}\hfill {\beta }_{11}\left(\theta \right)\\ \hfill {\beta }_{12}\left(\theta \right)\\ \hfill {\beta }_{21}\left(\theta \right)\\ \hfill {\beta }_{22}\left(\theta \right)\\ \hfill {\vartheta }_{11}\left(\theta \right)\\ \hfill {\vartheta }_{12}\left(\theta \right)\\ \hfill {\vartheta }_{21}\left(\theta \right)\\ \hfill {\vartheta }_{22}\left(\theta \right)\end{array}\right]\hfill \end{array}$ (3.39)

with initial conditions

 $\left\{\begin{array}{ccc}\hfill {\beta }_{11}\left(0\right)& \hfill =\hfill & \frac{-1}{2\left(1-{c}^{2}\right)}-a{\alpha }_{11}\hfill \\ \hfill {\beta }_{12}\left(0\right)+{\beta }_{21}\left(0\right)& \hfill =\hfill & -{\alpha }_{11}-a{\alpha }_{12}\hfill \\ \hfill {\beta }_{22}\left(0\right)& \hfill =\hfill & -{\alpha }_{12}\hfill \\ \hfill {\vartheta }_{11}\left(0\right)& \hfill =\hfill & \frac{c}{1-{c}^{2}}+\left(ac+b\right){\alpha }_{11}+d{\alpha }_{12}+c{\beta }_{11}\left(0\right)\hfill \\ \hfill {\vartheta }_{12}\left(0\right)& \hfill =\hfill & \left(ac+b\right){\alpha }_{12}+d{\alpha }_{22}+c{\beta }_{21}\left(0\right)\hfill \\ \hfill {\vartheta }_{21}\left(0\right)& \hfill =\hfill & 0\hfill \\ \hfill {\vartheta }_{22}\left(0\right)& \hfill =\hfill & 0\hfill \end{array}\right\}$ (3.40)

which follow from (3.25) and (3.27). From (3.39) and (3.40) we obtain ${\vartheta }_{21}\left(\theta \right)\equiv 0$, ${\vartheta }_{22}\left(\theta \right)\equiv 0$, and by (3.27): ${\beta }_{12}\left(\theta \right)\equiv 0$, ${\beta }_{22}\left(\theta \right)\equiv 0$. Consequently ${\alpha }_{12}=0$ and therefore the system (3.39) reduces to the form

 $\left[\begin{array}{c}\hfill {\beta }_{11}^{\prime }\left(\theta \right)\\ \hfill {\beta }_{21}^{\prime }\left(\theta \right)\\ \hfill {\vartheta }_{11}^{\prime }\left(\theta \right)\\ \hfill {\vartheta }_{12}^{\prime }\left(\theta \right)\end{array}\right]=\left[\begin{array}{cccc}\hfill \frac{a+bc}{1-{c}^{2}}\hfill & \hfill \frac{cd}{1-{c}^{2}}\hfill & \hfill \frac{b+ac}{1-{c}^{2}}\hfill & \hfill \frac{d}{1-{c}^{2}}\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill -\frac{b+ac}{1-{c}^{2}}\hfill & \hfill -\frac{d}{1-{c}^{2}}\hfill & \hfill -\frac{a+bc}{1-{c}^{2}}\hfill & \hfill -\frac{cd}{1-{c}^{2}}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{c}\hfill {\beta }_{11}\left(\theta \right)\\ \hfill {\beta }_{12}\left(\theta \right)\\ \hfill {\vartheta }_{11}\left(\theta \right)\\ \hfill {\vartheta }_{12}\left(\theta \right)\end{array}\right]$ (3.41)

with initial conditions

 $\left[\begin{array}{c}\hfill {\beta }_{11}\left(0\right)\\ \hfill {\beta }_{21}\left(0\right)\\ \hfill {\vartheta }_{11}\left(0\right)\\ \hfill {\vartheta }_{12}\left(0\right)\end{array}\right]=\left[\begin{array}{c}\hfill a{\alpha }_{11}-\frac{1}{2\left(1-{c}^{2}\right)}\hfill \\ \hfill -{\alpha }_{11}\hfill \\ \hfill \frac{c}{2\left(1-{c}^{2}\right)}+b{\alpha }_{11}\hfill \\ \hfill d{\alpha }_{22}-c{\alpha }_{11}\hfill \end{array}\right]$ (3.42)

The characteristic polynomial of the system (3.41) is biquadratic

${\lambda }^{4}+{\lambda }^{2}\left\{{\left(\frac{b+ac}{1-{c}^{2}}\right)}^{2}-{\left(\frac{a+bc}{1-{c}^{2}}\right)}^{2}-\frac{2cd}{1-{c}^{2}}\right\}-\frac{{d}^{2}}{1-{c}^{2}}=\left({\lambda }^{2}+{G}^{2}\right)\left({\lambda }^{2}-{H}^{2}\right)$

where

$\begin{array}{cccc}\hfill {G}^{2}=& \frac{1}{2}\left[{k}_{1}+\sqrt{{k}_{1}^{2}+4{k}_{2}}\right],\hfill & \hfill \phantom{\rule{2em}{0ex}}{H}^{2}=\hfill & \frac{1}{2}\left[-{k}_{1}+\sqrt{{k}_{1}^{2}+4{k}_{2}}\right]\hfill \\ \hfill {k}_{1}=& \frac{{b}^{2}-{a}^{2}-2cd}{1-{c}^{2}},\hfill & \hfill \phantom{\rule{2em}{0ex}}{k}_{2}=\hfill & \frac{{d}^{2}}{1-{c}^{2}}.\hfill \end{array}$

Hence the fundamental matrix of (3.41) takes a form

${e}^{\theta T}={Z}_{1}cosG\theta +{Z}_{2}sinG\theta +{Z}_{2}coshH\theta +{Z}_{4}sinhH\theta$

where $T$ stands for the matrix of the system (3.41), and

$\left[\begin{array}{c}\hfill {Z}_{1}\hfill \\ \hfill {Z}_{3}\hfill \end{array}\right]=\frac{1}{{G}^{2}+{H}^{2}}\left[\begin{array}{c}\hfill {H}^{2}I-{T}^{2}\hfill \\ \hfill {G}^{2}I+{T}^{2}\hfill \end{array}\right],\phantom{\rule{1em}{0ex}}\left[\begin{array}{c}\hfill {Z}_{2}\hfill \\ \hfill {Z}_{4}\hfill \end{array}\right]=\frac{1}{GH\left({G}^{2}+{H}^{2}\right)}\left[\begin{array}{c}\hfill H\left({H}^{2}-{T}^{3}\right)\hfill \\ \hfill G\left({G}^{2}+{T}^{3}\right)\hfill \end{array}\right].$

The unknown elements ${\alpha }_{11}$, ${\alpha }_{22}$ can be determined from the equality

 (3.43)

arising from (3.27). After tedious but elementary calculations, we get from (3.43)

 $\begin{array}{c}\left[\begin{array}{cc}\hfill \sigma +\delta +a{G}^{2}+a{H}^{2}\hfill & \hfill \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{H}^{2}\sigma -{G}^{2}\delta \hfill \\ \hfill \Sigma +\Delta +{G}^{2}+{H}^{2}\hfill & \hfill \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{H}^{2}\Sigma -{G}^{2}\Delta \hfill \end{array}\right]\left[\begin{array}{c}\hfill {\alpha }_{11}\hfill \\ \hfill {\alpha }_{22}\hfill \end{array}\right]=\hfill \\ =\frac{1}{2\left(1-{c}^{2}\right)}\left[\begin{array}{c}\hfill -{G}^{2}-{H}^{2}+\Sigma +\Delta \hfill \\ \hfill -\frac{1}{{G}^{2}}\sigma +\frac{1}{{H}^{2}}\delta \hfill \end{array}\right]\hfill \end{array}$ (3.44)

where:

$\begin{array}{cc}\hfill \sigma =& b{G}^{2}cosGr+\left(c{G}^{3}-dG\right)sinGr,\hfill \\ \hfill \delta =& b{H}^{2}coshHr-\left(c{H}^{3}+dH\right)sinhHr,\hfill \\ \hfill \Sigma =& \left(d-{G}^{2}c\right)cosGr+bGsinGr,\hfill \\ \hfill \Delta =& bHsinhHr-\left(d+c{H}^{2}\right)coshHr.\hfill \end{array}$

Solving (3.44) with respect to ${\alpha }_{11}$ and substituting the result into (3.38) we ﬁnd the desired expression for the performance index $J$ in terms of the system parameters.

In what follows the case of the PD controller will be investigated. In this case $d=0$ and by (3.14) we have ${z}_{2}\left(t\right)\equiv 0$. The characteristic quasipolynomial (3.37) reads as $s\left[s-a-\left(cs+b\right){e}^{-sr}\right]$ where the expression in the square brackets is the characteristic quasipolynomial of the remaining part of the system (3.14). It is not difficult to establish that if $a<0$ then the domain of stability in the space of parameters $\left(b,c\right)$ is the bounded part of the semistrip $\left|c\right|<1$, $b<-a$ cut by the arc

$\left\{\begin{array}{cc}\hfill b\left(\omega \right)=& \hfill -acos\omega r-\omega sin\omega r\\ \hfill c\left(\omega \right)=& \hfill cos\omega r-a\frac{sin\omega r}{\omega }\end{array}\right\},\phantom{\rule{2em}{0ex}}{\omega }_{0}\le \omega \le \frac{\pi }{r}$

where ${\omega }_{0}$ denotes the smallest positive solution of the equation $\omega =\frac{{a}^{2}+{\omega }^{2}}{-2a}sin\omega r$. Moreover,

$\left\{\begin{array}{cccc}H=0,\hfill & \phantom{\rule{1em}{0ex}}G=\sqrt{\frac{{b}^{2}-{a}^{2}}{1-{c}^{2}}}\hfill & \hfill \phantom{\rule{0ex}{0ex}}\text{if}\hfill & \hfill \phantom{\rule{0ex}{0ex}}{b}^{2}\ge {a}^{2}\\ H=\sqrt{\frac{{a}^{2}-{b}^{2}}{1-{c}^{2}}},\hfill & \phantom{\rule{1em}{0ex}}G=0\hfill & \hfill \phantom{\rule{0ex}{0ex}}\text{if}\hfill & \hfill \phantom{\rule{0ex}{0ex}}{b}^{2}\le {a}^{2}\end{array}\right\}$

and the system (3.43) reduces to the ﬁrst equation only. Finally,

${\alpha }_{11}=\left\{\begin{array}{ccc}\hfill \frac{-1-ccoskr+\frac{b}{k}sinkr}{2\left(1-{c}^{2}\right)\left(bcoskr+cksinkr+a\right)}\hfill & \hfill \text{if}\hfill & \hfill {b}^{2}>{a}^{2}\hfill \\ \hfill \frac{-1-ccoshkr+\frac{b}{k}sinhkr}{2\left(1-{c}^{2}\right)\left(bcoshkr-cksinhkr+a\right)}\hfill & \hfill \text{if}\hfill & \hfill {b}^{2}<{a}^{2}\hfill \\ \hfill \frac{1+c-ar}{4\left(1-{c}^{2}\right)\left(-a\right)}\hfill & \hfill \text{if}\hfill & \hfill b=a\hfill \end{array}\right\},\phantom{\rule{1em}{0ex}}k=\sqrt{\frac{\left|{b}^{2}-{a}^{2}\right|}{1-{c}^{2}}}.$

The results of minimization of the performance index (3.38) for $a=-5$, $r=0.5$ are presented at Figure 3.5. The minimal value of ${z}_{0}^{-2}J\left({x}_{0}\right)$ is $0.099329936$ and it is achieved at $b=-0.42234051$, $c=-0.078988818$. Figure 3.5: The level curves of the performance index ${z}_{0}^{-2}J\left({x}_{0}\right)$ as a function of $b$ and $c$

Exercise 3.3.1. For the time–delay system of the neutral type

$\left\{\begin{array}{cccc}\hfill \stackrel{̇}{v}\left(t\right)& \hfill =\hfill & -v\left(t\right)\hfill & \hfill \\ \hfill v\left(t\right)& \hfill =\hfill & z\left(t\right)-\frac{1}{2}z\left(t-1\right)\hfill & \hfill \\ \hfill z\left(\theta \right)& \hfill =\hfill & \phi \left(\theta \right),\hfill & \phantom{\rule{1em}{0ex}}-1\le \theta \le 0\hfill \\ \hfill v\left(0\right)& \hfill =\hfill & {v}_{0}\hfill & \hfill \end{array}\right\}$

calculate the integral performance criterion $J={\int }_{0}^{\infty }{z}^{2}\left(t\right)dt$ in the two following cases:

1${}^{o}$
$\phi \left(\theta \right)={e}^{-\theta },\phantom{\rule{2em}{0ex}}{v}_{0}=1-\frac{e}{2}$,
2${}^{o}$
$\phi \left(\theta \right)\equiv 0,\phantom{\rule{2em}{0ex}}{v}_{0}=1$.

In the ﬁrst case we have $z\left(t\right)={e}^{-t}$ and therefore by direct calculations $J=\frac{1}{2}$. In the second case $z\left(t\right)-\frac{1}{2}z\left(t-1\right)={e}^{-t}$ and using the method of steps we obtain

$z\left(t\right)={e}^{-t}\sum _{m=0}^{i-1}{\left(\frac{e}{2}\right)}^{m}={e}^{-t}\frac{1-{\left(\frac{e}{2}\right)}^{i}}{1-\frac{e}{2}},\phantom{\rule{2em}{0ex}}i-1\le t

Hence

${\int }_{0}^{\infty }{z}^{2}\left(t\right)dt=\sum _{i=1}^{\infty }{\int }_{i-1}^{i}{z}^{2}\left(t\right)dt=\sum _{i=1}^{\infty }{\int }_{i-1}^{i}{e}^{-2t}{\left[\frac{1-{\left(\frac{e}{2}\right)}^{i}}{1-\frac{e}{2}}\right]}^{2}dt=$

$=\sum _{i=1}^{\infty }\frac{{\left[1-{\left(\frac{e}{2}\right)}^{i}\right]}^{2}}{{\left(\frac{2-e}{2}\right)}^{2}}\cdot \frac{{e}^{2}-1}{2}\cdot {e}^{-2i}=\frac{2\left({e}^{2}-1\right)}{{\left(2-e\right)}^{2}}\left\{\sum _{i=1}^{\infty }{e}^{-2i}-2\sum _{i=1}^{\infty }{\left(2e\right)}^{-i}+\sum _{i=1}^{\infty }\frac{1}{{4}^{i}}\right\}=$

$=\frac{2\left({e}^{2}-1\right)}{{\left(2-e\right)}^{2}}\left\{\frac{{e}^{-2}}{1-{e}^{-2}}-2\cdot \frac{{\left(2e\right)}^{-1}}{1-{\left(2e\right)}^{-1}}+\frac{\frac{1}{4}}{1-\frac{1}{4}}\right\}=\frac{2\left(2e+1\right)}{3\left(2e-1\right)}.$

The above results can be conﬁrmed by the method presented in the previous section. We identify parameters in (3.5) and (3.16):

${A}_{1}=-1,\phantom{\rule{2em}{0ex}}{A}_{2}={A}_{0}=\frac{1}{2},\phantom{\rule{2em}{0ex}}r=1,\phantom{\rule{2em}{0ex}}P=1,\phantom{\rule{2em}{0ex}}Q=\frac{1}{2},\phantom{\rule{2em}{0ex}}R=\frac{1}{4}.$

The polynomial (3.29) has two roots $\lambda =±1$. Thus taking $\lambda =1$ we ﬁnd from (3.30): $L=0$, $M=1$. The unique solution of (3.24) is $\gamma =\frac{1}{3}$. Hence, the system (3.33) takes the form

$\left[\begin{array}{cc}\hfill -2\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}2{e}^{-1}\hfill \\ \hfill 0\phantom{\rule{0ex}{0ex}}& \hfill \phantom{\rule{0ex}{0ex}}-\frac{1}{2}{e}^{-1}+1\hfill \end{array}\right]\left[\begin{array}{c}\hfill \alpha \hfill \\ \hfill \kappa \hfill \end{array}\right]=\left[\begin{array}{c}\hfill -\frac{4}{3}\\ \hfill \frac{2}{3}\end{array}\right].$

Its solution is $\alpha =\frac{4e+2}{3\left(2e-1\right)}$, $\kappa =\frac{4e}{3\left(2e-1\right)}$. From (3.32) and (3.26) we ﬁnd: $\beta \left(\theta \right)=\frac{4{e}^{-\theta }}{3\left(2e-1\right)}$, $\Phi \left(\theta \right)=0$ for $-1\le \theta \le 0$. Conﬁrm the results using Remark 3.3.1. Now

$J\left(\left[\begin{array}{c}\hfill {v}_{0}\hfill \\ \hfill \phi \hfill \end{array}\right]\right)={〈\left[\begin{array}{c}\hfill {v}_{0}\hfill \\ \hfill \phi \hfill \end{array}\right],\mathcal{ℋ}\left[\begin{array}{c}\hfill {v}_{0}\hfill \\ \hfill \phi \hfill \end{array}\right]〉}_{\text{H}}$

where $\mathcal{ℋ}$ is given by (3.21). Substituting the initial data we get the same values of $J$ as by direct calculations.

An advantage of the general attempt to evaluating $J$ is that the value of the performance index is calculated for a wide variety of initial conditions, while the direct calculations are carried out separately for particular initial data.