Subsections

• 11.1 Model 2-dimensional system
• 11.2 Combustion stability in liquid propellant rocket motors
• 11.3 Wind tunnel model

11 Design case studies

This section contains three case studies of control system design and analysis using Time-Delay System Toolbox.

Case study 1: Model 2-dimensional system,

Case study 2: Combustion stability in liquid propellant rocket motors,

Case study 3: Wind tunnel model.

11.1 Model 2-dimensional system

Consider 2-dimensional system with delay [33]

$$\dot{x} \left[\vphantom{\begin{array}{c c} 0 & 1 [2mm] 0 & 0 \\ \end{array}}\right. \begin{array}{c c} 0 & 1 [2mm] 0 & 0 \\ \end{array} \left.\vphantom{\begin{array}{c c} 0 & 1 [2mm] 0 & 0 \\ \end{array}}\right] \left[\vphantom{\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}}\right. \begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array} \left.\vphantom{\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}}\right] \left[\vphantom{\begin{array}{c} 0 [2mm] 0.333 \\ \end{array}}\right. \begin{array}{c} 0 [2mm] 0.333 \\ \end{array} \left.\vphantom{\begin{array}{c} 0 [2mm] 0.333 \\ \end{array}}\right]$$ (11.1)

$$\xi \left[\vphantom{\begin{array}{c c} 0.9 & 0.1 \\ \end{array}}\right. \begin{array}{c c} 0.9 & 0.1 \\ \end{array} \left.\vphantom{\begin{array}{c c} 0.9 & 0.1 \\ \end{array}}\right]$$ (11.2)

hence

A = $$\left[\vphantom{\begin{array}{c c} 0 & 1 [2mm] 0 & 0 \\ \end{array}}\right.$$$$\begin{array}{c c} 0 & 1 [2mm] 0 & 0 \\ \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0 & 1 [2mm] 0 & 0 \\ \end{array}}\right]$$ ,  I>A_$\tau$ = $$\left[\vphantom{\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}}\right.$$$$\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}$$$$\left.\vphantom{\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}}\right]$$ ,        B = $$\left[\vphantom{\begin{array}{c} 0 [2mm] 0.333 \\ \end{array}}\right.$$$$\begin{array}{c} 0 [2mm] 0.333 \\ \end{array}$$$$\left.\vphantom{\begin{array}{c} 0 [2mm] 0.333 \\ \end{array}}\right]$$ ,        $$\tau$$ = 5 .

Open-loop system (11.1) is unstable because it has two open-loop roots with positive real parts: $\lambda_{1,2}^{}$ = 0.198±0.323 i. Figure 1 and Figure 2 show the simulation results by stepd and impulsed for the open-loop system.

Let us take the weight matrices as

$$\left[\vphantom{ \begin{array}{c c} \alpha & 0 [2mm] 0 & \alpha \\ \end{array}}\right. \begin{array}{c c} \alpha & 0 [2mm] 0 & \alpha \\ \end{array} \left.\vphantom{ \begin{array}{c c} \alpha & 0 [2mm] 0 & \alpha \\ \end{array}}\right]$$ (11.3)

where $\alpha$ is a positive number.

A) Take $\alpha$ = 1. We can find, using functions lqdelay, the matrices

C = $$\left[\vphantom{\begin{array}{c c} -1 &-2.6469 \end{array}}\right.$$$$\begin{array}{c c} -1 &-2.6469 \end{array}$$$$\left.\vphantom{\begin{array}{c c} -1 &-2.6469 \end{array}}\right]$$ ,        D₀ = $$\left[\vphantom{\begin{array}{c c} 0 & -0.3330 \end{array}}\right.$$$$\begin{array}{c c} 0 & -0.3330 \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0 & -0.3330 \end{array}}\right]$$ ,

D₁ = $$\left[\vphantom{\begin{array}{c c} 0 & -0.3330 1 & -0.8814 \end{array}}\right.$$$$\begin{array}{c c} 0 & -0.3330 1 & -0.8814 \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0 & -0.3330 1 & -0.8814 \end{array}}\right]$$ ,        D₂ = $$\left[\vphantom{\begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array}}\right.$$$$\begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array}}\right]$$ ,

of the linear control (10.4).

Figure 1: Step response

Figure 2: Impulse response

Thus to system (11.1) and the weight matrices (11.3) (with $\alpha$ = 1) corresponds the control

u⁰(x, y( ^. )) = $$\left[\vphantom{\begin{array}{c c} -1 &-2.6469 \end{array}}\right.$$$$\begin{array}{c c} -1 &-2.6469 \end{array}$$$$\left.\vphantom{\begin{array}{c c} -1 &-2.6469 \end{array}}\right]$$x +

$$\left[\vphantom{\begin{array}{c c} 0 & -0.3330 \end{array}}\right. \begin{array}{c c} 0 & -0.3330 \end{array} \left.\vphantom{\begin{array}{c c} 0 & -0.3330 \end{array}}\right] \int\limits_{-5}^{0} \Biggl\{ ^{\left[\begin{array}{c c} 0 & -0.3330 1 & -0.8814 \end{array}\right] \times S } \left[\vphantom{\begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array}}\right. \begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array} \left.\vphantom{\begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array}}\right] \Biggr\}$$ (11.4)

The corresponding closed-loop system has the form

$$\dot{x}$$(t) = $$\left[\vphantom{\begin{array}{c c} 0 & 1 [2mm] -0.333 & -0.8814 \\ \end{array}}\right.$$$$\begin{array}{c c} 0 & 1 [2mm] -0.333 & -0.8814 \\ \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0 & 1 [2mm] -0.333 & -0.8814 \\ \end{array}}\right]$$x(t) + $$\left[\vphantom{\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}}\right.$$$$\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}$$$$\left.\vphantom{\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}}\right]$$x(t - 5) +

$$\left[\vphantom{\begin{array}{c c} 0 & 0 [2mm] 0 & -0.1109 \\ \end{array}}\right. \begin{array}{c c} 0 & 0 [2mm] 0 & -0.1109 \\ \end{array} \left.\vphantom{\begin{array}{c c} 0 & 0 [2mm] 0 & -0.1109 \\ \end{array}}\right] \int\limits_{-5}^{0} \Biggl\{ ^{\left[\begin{array}{c c} 0 & -0.3330 1 & -0.8814 \end{array}\right] \times S } \left[\vphantom{\begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array}}\right. \begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array} \left.\vphantom{\begin{array}{c c} 0.1053 & 0.1921 -0.0052 & 0.1297 \end{array}}\right] \Biggr\}$$ (11.5)

Using functions of Toolbox one can check that solutions of the closed-loop system tend to zero (see Figure 3).

B) Take $\alpha$ = 10. We can find, using functions lqdelay, the matrices

C = $$\left[\vphantom{\begin{array}{c c} -3.1623 &-5.3845 \end{array}}\right.$$$$\begin{array}{c c} -3.1623 &-5.3845 \end{array}$$$$\left.\vphantom{\begin{array}{c c} -3.1623 &-5.3845 \end{array}}\right]$$ ,  I>D₀ = $$\left[\vphantom{\begin{array}{c c} 0 & -0.3330 \end{array}}\right.$$$$\begin{array}{c c} 0 & -0.3330 \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0 & -0.3330 \end{array}}\right]$$ ,

D₁ = $$\left[\vphantom{\begin{array}{c c} 0 & -1.053 1 & -1.793 \end{array}}\right.$$$$\begin{array}{c c} 0 & -1.053 1 & -1.793 \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0 & -1.053 1 & -1.793 \end{array}}\right]$$ ,  I>D₂ = $$\left[\vphantom{\begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array}}\right.$$$$\begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array}}\right]$$ .

Thus to system (11.1) and the weight matrices (11.3) (with $\alpha$ = 10) corresponds the control

u⁰(x, y( ^. )) = $$\left[\vphantom{\begin{array}{c c} -3.1623 &-5.3845 \end{array}}\right.$$$$\begin{array}{c c} -3.1623 &-5.3845 \end{array}$$$$\left.\vphantom{\begin{array}{c c} -3.1623 &-5.3845 \end{array}}\right]$$x +

$$\left[\vphantom{\begin{array}{c c} 0 & -0.3330 \end{array}}\right. \begin{array}{c c} 0 & -0.3330 \end{array} \left.\vphantom{\begin{array}{c c} 0 & -0.3330 \end{array}}\right] \int\limits_{-5}^{0} \Biggl\{ ^{\left[\begin{array}{c c} 0 & -1.053 1 & -1.793 \end{array}\right] \times S } \left[\vphantom{\begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array}}\right. \begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array} \left.\vphantom{\begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array}}\right] \Biggr\}$$ (11.6)

The corresponding closed-loop system has the form

$$\dot{x}$$(t) = $$\left[\vphantom{\begin{array}{c c} 0 & 1 [2mm] -1.0530 & -1.7930 \\ \end{array}}\right.$$$$\begin{array}{c c} 0 & 1 [2mm] -1.0530 & -1.7930 \\ \end{array}$$$$\left.\vphantom{\begin{array}{c c} 0 & 1 [2mm] -1.0530 & -1.7930 \\ \end{array}}\right]$$x(t) + $$\left[\vphantom{\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}}\right.$$$$\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}$$$$\left.\vphantom{\begin{array}{c c} -0.3 & -0.1 [2mm] -0.2 & -0.4 \\ \end{array}}\right]$$x(t - 5) +

$$\left[\vphantom{\begin{array}{c c} 0 & 0 [2mm] 0 & -0.1109 \\ \end{array}}\right. \begin{array}{c c} 0 & 0 [2mm] 0 & -0.1109 \\ \end{array} \left.\vphantom{\begin{array}{c c} 0 & 0 [2mm] 0 & -0.1109 \\ \end{array}}\right] \int\limits_{-5}^{0} \Biggl\{ ^{\left[\begin{array}{c c} 0 & -1.053 1 & -1.793 \end{array}\right] \times S } \left[\vphantom{\begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array}}\right. \begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array} \left.\vphantom{\begin{array}{c c} 0.065 & 0.0889 0.0338 & 0.0826 \end{array}}\right] \Biggr\}$$ (11.7)

Using functions of Toolbox one can check that solutions of the closed-loop systems tend to zero (see Figure 4).

Figure 3: Closed-loop system response in case of $\alpha$ = 1

Figure 4: Closed-loop system response in case of $\alpha$ = 10

11.2 Combustion stability in liquid propellant rocket motors

A linearized version of the feed system and combustion chamber equations, assuming nonsteady flow, are given by ⁵

$$\dot{\phi} \gamma \phi \gamma \phi \delta \mu \delta$$ =

$$\dot{\mu}_{1}^{} {\frac{1}{\xi J}} \left[\vphantom{ - \psi (t) + \frac{p_0 - p_1}{2 \Delta p} }\right. \psi {\frac{p_0 - p_1}{2 \Delta p}} \left.\vphantom{ - \psi (t) + \frac{p_0 - p_1}{2 \Delta p} }\right]$$ =

$$\dot{\mu} {\frac{1}{(1-\xi) J}} \left[\vphantom{ - \mu (t) + \psi (t) - P \phi (t) }\right. \mu \psi \phi \left.\vphantom{ - \mu (t) + \psi (t) - P \phi (t) }\right]$$ =

$$\dot{\psi} {\frac{1}{E}} \left[\vphantom{ \mu_1 (t) - \mu (t) }\right. \mu_{1}^{} \mu \left.\vphantom{ \mu_1 (t) - \mu (t) }\right]$$ = (11.8)

Here
$\phi$(t) = fractional variation of pressure in the combustion chamber,
t is the unit of time normalized with gas residence time, $\theta_{g}^{}$, in steady operation,
$\tilde{\tau}$ = value of time lag in steady operation,
$\tilde{p}$ = pressure in combustion chamber in steady operation,
$\overline{\tau p^\gamma}$ = const for some number $\gamma$,
$$\delta$$ = $${\frac{\tilde \tau}{\theta_g}}$$,
$\mu$(t) = fractional variation of injection and burning rate,
$\psi$(t) = relative variation of p₁,

=1cm p₁ = instantaneous pressure at that place in the feeding line where the capacitance representing the elasticity is located,

$\xi$ = fractional length for the constant pressure supply,
J = inertial parameter of the line,
P = pressure drop parameter,
$\mu_{1}^{}$(t) = fractional variation of instantaneous mass flow upstream of the capacitance,
$\Delta$p = injector pressure drop in steady operation,
p₀ = regulated gas pressure for constant pressure supply,
E = elasticity parameter of the line.

For our purpose we have taken

u = $${\frac{p_0 - p_1}{2 \Delta p}}$$

to be a control variable and guided by [10] have adopted the following representative numerical values:
$\gamma$ = 0.8, $\xi$ = 0.5, $\delta$ = 1, J = 2, P = 1, E = 1.

This gives, for x(t) = ($\phi$(t),$\mu_{1}^{}$(t),$\mu$(t),$\psi$(t))',

$$\dot{x} \left[\vphantom{\begin{array}{c c c c} - 0.2 & 0 & 0 & 0 [2m... ... 1 [2mm] -1 & 0 & -1 & 1 [2mm] 0 & 1 & -1 & 0 \end{array}}\right. \begin{array}{c c c c} - 0.2 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & - 1 [2mm] -1 & 0 & -1 & 1 [2mm] 0 & 1 & -1 & 0 \end{array} \left.\vphantom{\begin{array}{c c c c} - 0.2 & 0 & 0 & 0 [2m... ... 1 [2mm] -1 & 0 & -1 & 1 [2mm] 0 & 1 & -1 & 0 \end{array}}\right] \left[\vphantom{\begin{array}{c c c c} - 0.8 & 0 & 1 & 0 [2m... ...0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array}}\right. \begin{array}{c c c c} - 0.8 & 0 & 1 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array} \left.\vphantom{\begin{array}{c c c c} - 0.8 & 0 & 1 & 0 [2m... ...0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array}}\right] \left[\vphantom{\begin{array}{c} 0 [2mm] 1 [2mm] 0 [2mm] 0 \\ \end{array}}\right. \begin{array}{c} 0 [2mm] 1 [2mm] 0 [2mm] 0 \\ \end{array} \left.\vphantom{\begin{array}{c} 0 [2mm] 1 [2mm] 0 [2mm] 0 \\ \end{array}}\right]$$ (11.9)

The system (11.9) has two roots with positive real part: $\lambda_{1,2}^{}$ = 0.11255±1.52015 i.

Let us take the weight matrices as

$$\left[\vphantom{ \begin{array}{c c c c} 1 & 0 & 0 & 0 [2mm] 0... ... 0 & 0 [2mm] 0 & 0 & 1 & 0 [2mm] 0 & 0 & 0 & 1 \\ \end{array}}\right. \begin{array}{c c c c} 1 & 0 & 0 & 0 [2mm] 0 & 1 & 0 & 0 [2mm] 0 & 0 & 1 & 0 [2mm] 0 & 0 & 0 & 1 \\ \end{array} \left.\vphantom{ \begin{array}{c c c c} 1 & 0 & 0 & 0 [2mm] 0... ... 0 & 0 [2mm] 0 & 0 & 1 & 0 [2mm] 0 & 0 & 0 & 1 \\ \end{array}}\right]$$ (11.10)

Using function lqdelay we can find the matrices

C = $$\left[\vphantom{\begin{array}{c c c c} 0.0398 & -1.1134 & 0.2332 & -0.1198 \end{array}}\right.$$$$\begin{array}{c c c c} 0.0398 & -1.1134 & 0.2332 & -0.1198 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} 0.0398 & -1.1134 & 0.2332 & -0.1198 \end{array}}\right]$$ ,

D₀ = $$\left[\vphantom{\begin{array}{c c c c} 0 & -1 & 0 & 0 \end{array}}\right.$$$$\begin{array}{c c c c} 0 & -1 & 0 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} 0 & -1 & 0 & 0 \end{array}}\right]$$ ,

D₁ = $$\left[\vphantom{\begin{array}{c c c c} -0.2 & 0.0398 & -1 & 0 \\... ...0 & 1 \\ 0 & 0.2332 & -1 & -1 \\ 0 & -1.1198 & 1 & 0 \end{array}}\right.$$$$\begin{array}{c c c c} -0.2 & 0.0398 & -1 & 0 \\ 0 & - 1.1134 & 0 & 1 \\ 0 & 0.2332 & -1 & -1 \\ 0 & -1.1198 & 1 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} -0.2 & 0.0398 & -1 & 0 \\... ...0 & 1 \\ 0 & 0.2332 & -1 & -1 \\ 0 & -1.1198 & 1 & 0 \end{array}}\right]$$ ,

D₂ = $$\left[\vphantom{\begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 ... ... -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array}}\right.$$$$\begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 \\ 0.1794 & 0... ... & 0 \\ -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 ... ... -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array}}\right]$$ .

Thus to system (11.9) with the weight matrices (11.10) corresponds LQR control

u⁰(x, y( ^. )) = $$\left[\vphantom{\begin{array}{c c c c} 0.0398 & -1.1134 & 0.2332 & -0.1198 \end{array}}\right.$$$$\begin{array}{c c c c} 0.0398 & -1.1134 & 0.2332 & -0.1198 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} 0.0398 & -1.1134 & 0.2332 & -0.1198 \end{array}}\right]$$x + $$\left[\vphantom{\begin{array}{c c c c} 0 & -1 & 0 & 0 \end{array}}\right.$$$$\begin{array}{c c c c} 0 & -1 & 0 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} 0 & -1 & 0 & 0 \end{array}}\right]$$×

×$$\int\limits_{-1}^{0}$$$$\Biggl\{$$e^{$^{\left[\begin{array}{c c c c} -0.2 & 0.0398 & -1 & 0 \\ 0 & -... ... 0 & 0.2332 & -1 & -1 \\ 0 & -1.1198 & 1 & 0 \end{array}\right] \times S }$}$$\left[\vphantom{\begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 ... ... -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array}}\right.$$$$\begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 \\ 0.1794 & 0... ... & 0 \\ -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 ... ... -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array}}\right]$$y(s)$$\Biggr\}$$ ds .

The corresponding closed-loop system has the form

$$\dot{x}$$(t) = $$\left[\vphantom{\begin{array}{c c c c} \gamma - 1 & 0 & 0 & 0 \\... ...98 [2mm] -1 & 0 & -1 & 1 [2mm] 0 & 1 & -1 & 0 \end{array}}\right.$$$$\begin{array}{c c c c} \gamma - 1 & 0 & 0 & 0 [2mm] 0.0398 ... ... & -1.1198 [2mm] -1 & 0 & -1 & 1 [2mm] 0 & 1 & -1 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} \gamma - 1 & 0 & 0 & 0 \\... ...98 [2mm] -1 & 0 & -1 & 1 [2mm] 0 & 1 & -1 & 0 \end{array}}\right]$$x(t) +

+ $$\left[\vphantom{\begin{array}{c c c c} - \gamma & 0 & 1 & 0 ... ...0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array}}\right.$$$$\begin{array}{c c c c} - \gamma & 0 & 1 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} - \gamma & 0 & 1 & 0 ... ...0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array}}\right]$$x(t - $$\delta$$) + $$\left[\vphantom{\begin{array}{c c c c} 0 & 0 & 0 & 0 [2mm] ... ...0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array}}\right.$$$$\begin{array}{c c c c} 0 & 0 & 0 & 0 [2mm] 0 & -1 & 0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} 0 & 0 & 0 & 0 [2mm] ... ...0 & 0 [2mm] 0 & 0 & 0 & 0 [2mm] 0 & 0 & 0 & 0 \end{array}}\right]$$×

$$\int\limits_{-1}^{0} \Biggl\{ ^{\left[\begin{array}{c c c c} -0.2 & 0.0398 & -1 & 0 \\ 0 & -... ... 0 & 0.2332 & -1 & -1 \\ 0 & -1.1198 & 1 & 0 \end{array}\right] \times S } \left[\vphantom{\begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 ... ... -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array}}\right. \begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 \\ 0.1794 & 0... ... & 0 \\ -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array} \left.\vphantom{\begin{array}{c c c c} -3.3101 & 0 & 4.1376 & 0 ... ... -0.0180 & 0 & 0.0225 & 0 \\ 0.2386 & 0 & -0.2983 & 0 \end{array}}\right] \Biggr\}$$ (11.11)

Using functions of Toolbox one can check that solutions of the closed-loop systems tend to zero (see Figure 5).

Solutions of the optimal closed-loop system corresponding to $\gamma$ = 0.95 and $\delta$ = 0.87 one can see on Figure 6.

Figure 5: In case of $\gamma$ = 0.8 and $\delta$ = 1

Figure 6: In case of $\gamma$ = 0.95 and $\delta$ = 0.87

11.3 Wind tunnel model

A linearized model of the high-speed closed-air unit wind tunnel is [26,27]

$$\dot{x}_{1}^{} \tau$$ =

$$\dot{x}_{2}^{}$$ = x₃(t) , (11.12)

$$\dot{x}_{3}^{} \xi \omega \omega^{2}_{}$$ =

with a = $${\frac{1}{1.964}}$$, k = - 0.117, w = 6, $\xi$ = 0.8, $\tau$ = 0.33s.

The state variable x₁, x₂, x₃ represent deviations from a chosen operating point (equilibrium point) of the following quantities: x₁ = Mach number, x₂ = actuator position guide vane angle in a driving fan, x₃ = actuator rate. The delay represents the time of the transport between the fan and the test section.

The system can be written in matrix form

$$\dot{x} \tau \tau$$ (11.13)

where

A₀ = $$\left[\vphantom{\begin{array}{c c c } - a & 0 & 0 [2mm] 0 & 0 & 1 [2mm] 0 & -\omega^2 & -2 \xi \omega \end{array}}\right.$$$$\begin{array}{c c c } - a & 0 & 0 [2mm] 0 & 0 & 1 [2mm] 0 & -\omega^2 & -2 \xi \omega \end{array}$$$$\left.\vphantom{\begin{array}{c c c } - a & 0 & 0 [2mm] 0 & 0 & 1 [2mm] 0 & -\omega^2 & -2 \xi \omega \end{array}}\right]$$ ,

A_$\tau$ = $$\left[\vphantom{\begin{array}{c c c } 0 & a k & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & 0 \end{array}}\right.$$$$\begin{array}{c c c } 0 & a k & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c } 0 & a k & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & 0 \end{array}}\right]$$ ,

B = $$\left[\vphantom{\begin{array}{c} 0 [2mm] 0 [2mm] \omega^2 \\ \end{array}}\right.$$$$\begin{array}{c} 0 [2mm] 0 [2mm] \omega^2 \\ \end{array}$$$$\left.\vphantom{\begin{array}{c} 0 [2mm] 0 [2mm] \omega^2 \\ \end{array}}\right]$$ .

Let us take the weight matrices as

$$\left[\vphantom{ \begin{array}{c c c c} 1 & 0 & 0 [2mm] 0 & 1 & 0 [2mm] 0 & 0 & 1 \\ \end{array}}\right. \begin{array}{c c c c} 1 & 0 & 0 [2mm] 0 & 1 & 0 [2mm] 0 & 0 & 1 \\ \end{array} \left.\vphantom{ \begin{array}{c c c c} 1 & 0 & 0 [2mm] 0 & 1 & 0 [2mm] 0 & 0 & 1 \\ \end{array}}\right]$$ (11.14)

Using function lqdelay we can find the matrices

C = $$\left[\vphantom{\begin{array}{c c c} 0 & -0.4142 & -0.6101 \end{array}}\right.$$$$\begin{array}{c c c} 0 & -0.4142 & -0.6101 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} 0 & -0.4142 & -0.6101 \end{array}}\right]$$ ,

D₀ = $$\left[\vphantom{\begin{array}{c c c} 0 & 0 & -36 \end{array}}\right.$$$$\begin{array}{c c c} 0 & 0 & -36 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} 0 & 0 & -36 \end{array}}\right]$$ ,

D₁ = $$\left[\vphantom{\begin{array}{c c c} -0.5092 & 0 & 0 \\ 0 & 0 & -50.9117 \\ 0 & 1.0000 & -41.1639 \end{array}}\right.$$$$\begin{array}{c c c} -0.5092 & 0 & 0 \\ 0 & 0 & -50.9117 \\ 0 & 1.0000 & -41.1639 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} -0.5092 & 0 & 0 \\ 0 & 0 & -50.9117 \\ 0 & 1.0000 & -41.1639 \end{array}}\right]$$ ,

D₂ = $$\left[\vphantom{\begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}}\right.$$$$\begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}}\right]$$ .

Thus to system (11.13) and the weight matrices (11.14) corresponds LQR control

u⁰(x, y( ^. )) = $$\left[\vphantom{\begin{array}{c c c} 0 & -0.4142 & -0.6101 \end{array}}\right.$$$$\begin{array}{c c c} 0 & -0.4142 & -0.6101 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} 0 & -0.4142 & -0.6101 \end{array}}\right]$$x + $$\left[\vphantom{\begin{array}{c c c} 0 & 0 & -36 \end{array}}\right.$$$$\begin{array}{c c c} 0 & 0 & -36 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} 0 & 0 & -36 \end{array}}\right]$$×

×$$\int\limits_{-\tau}^{0}$$$$\Biggl\{$$e^{$^{\left[\begin{array}{c c c} -0.5092 & 0 & 0 \\ 0 & 0 & -50.9117 \\ 0 & 1.0000 & -41.1639 \end{array}\right] \times S }$}$$\left[\vphantom{\begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}}\right.$$$$\begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}}\right]$$y(s)$$\Biggr\}$$ ds .

The corresponding closed-loop system has the form

$$\dot{x}$$(t) = $$\left[\vphantom{\begin{array}{c c c} -0.5092 & 0 & 0 [2mm] 0 & 0 & 1.0000 [2mm] 0 & -50.9117 & -41.1639 \end{array}}\right.$$$$\begin{array}{c c c} -0.5092 & 0 & 0 [2mm] 0 & 0 & 1.0000 [2mm] 0 & -50.9117 & -41.1639 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} -0.5092 & 0 & 0 [2mm] 0 & 0 & 1.0000 [2mm] 0 & -50.9117 & -41.1639 \end{array}}\right]$$x(t) +

+ $$\left[\vphantom{\begin{array}{c c c} 0 & 0.0596 & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & 0 \end{array}}\right.$$$$\begin{array}{c c c} 0 & 0.0596 & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & 0 \end{array}$$$$\left.\vphantom{\begin{array}{c c c} 0 & 0.0596 & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & 0 \end{array}}\right]$$x(t - $$\tau$$) + $$\left[\vphantom{\begin{array}{c c c c} 0 & 0 & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & -1296 \end{array}}\right.$$$$\begin{array}{c c c c} 0 & 0 & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & -1296 \end{array}$$$$\left.\vphantom{\begin{array}{c c c c} 0 & 0 & 0 [2mm] 0 & 0 & 0 [2mm] 0 & 0 & -1296 \end{array}}\right]$$×

$$\int\limits_{-\tau}^{0} \Biggl\{ ^{\left[\begin{array}{c c c} -0.5092 & 0 & 0 \\ 0 & 0 & -50.9117 \\ 0 & 1.0000 & -41.1639 \end{array}\right] \times S } \left[\vphantom{\begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}}\right. \begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \left.\vphantom{\begin{array}{c c c} 0 & 0.0495 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}}\right] \Biggr\}$$ (11.15)

Using functions of Toolbox one can check that solutions of the closed-loop systems tend to zero (see Figure 7).

Figure 7: Closed loop system response