LMIs in Control/Applications/F16 Longitudinal Stabilitzation

F-16 Longitudinal Stabilization using Hinf Output Feedback

The linearized longitudinal dynamics of the F-16 aircraft are considered in this example. The nonlinear 6DOF model is linearized using straight-and-level flight conditions of:

$$ \begin{align} V = 502 ft/s \\ \bar{q} = 300psf \\ X_{cg} = 0.35\bar{c} \\ \delta_{trim} = -0.7588deg \end{align} $$

See chapter 3 of [] for further details.

The System
The state-space representation for the linearized longitudinal dynamics can be written as follows:

$$ \begin{align} \dot{x}(t) &= Ax(t) + B_{1}w(t) + B_{2}u(t) \\ y(t) &= C_{1}x(t) + D_{11}w(t) + D_{12}u(t) \\ z(t) &= C_{2}x(t) + D_{21}w(t) + D_{22}u(t) \\ u(t) = Ky(t) \\ \end{align} $$

where $$ x = [V \quad \alpha \quad \theta \quad q]^{\text{T}}$$, $$ u = \delta $$, $$ w = \delta_{cmd} $$, and $$ z = [a_n \quad (\delta_{cmd} - \delta)]^{\text{T}} $$ are the state variable, control input, disturbance, and regulated output vectors, respectively. In this case, $$ V $$ is velocity (ft/s), $$ \alpha $$ is angle of attack (rad), $$ \theta $$ is pitch (rad), $$ q $$ is pitch rate (rad/s), $$ \delta $$ is elevator deflection with respect to trimmed deflection (rad), $$ \delta_{cmd} $$ is commanded elevator deflection with respect to trimmed deflection (rad), and $$ a_n $$ is normal acceleration at the pilot location (ft/s^2).

The Data
For the linearization used, the above matrices are:

$$ \begin{align} A = \begin{bmatrix} -1.9311e-2 &&  8.8157 && -3.217e1 && -5.7499e-1 \\ -2.5389e-4 &&  -1.0189 &&  0 &&  9.0506e-1 \\ 0 &&        0 &&  0 &&  1 \\     2.9465e-12 && 8.2225e-1 &&  0 && -1.0774 \end{bmatrix} \end{align} $$

$$ B_{1} = \begin{bmatrix} 1.737e-1 \\ -2.1499e-3 \\ 0 \\   -1.7555e-1 \end{bmatrix} $$

$$ B_{2} = \begin{bmatrix} 1.737e-1 \\ -2.1499e-3 \\ 0 \\   -1.7555e-1 \end{bmatrix} $$

$$ \begin{align} C_1 = \begin{bmatrix} 0 && 5.729578e1 && 0 &&         0 \\ 0 &&         0 && 0 && 5.729578e1 \end{bmatrix} \end{align} $$

$$ \begin{align} C_2 = \begin{bmatrix} 0.003981 && 15.88 && 0 && 1.481 \\ 0 && 0 && 0 && 0 \\ \end{bmatrix} \end{align} $$

$$ D_{11} = D_{12} = 0 $$

$$ D_{21} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$

$$ D_{21} = \begin{bmatrix} 0.03333 \\ -1 \end{bmatrix} $$

The LMI: Optimal Output Feedback $$H_{\infty}$$ Control LMI
The following are equivalent.

1) There exists a $$\hat{K}=\begin{bmatrix} A_K & B_K \\ C_K & D_K \end{bmatrix}$$ such that $$||S(K,P)||_{H_{\infty}}<\gamma$$

2) There exists $$X_1$$, $$Y_1$$, $$Z$$, $$A_n$$, $$B_n$$, $$C_n$$, $$D_n$$ such that


 * $$\begin{bmatrix}

X_{1}     & I \\ I         & Y_{1} \end{bmatrix} > 0 $$
 * $$\begin{bmatrix}

AY_{1}+Y_{1}A^{\text{T}}+B_{2}C_{n}+C_{n}B_{2}^{\text{T}} & *^{\text{T}} & *^{\text{T}} & *^{\text{T}} \\ A^{\text{T}}+A_{n}+(B_{2}D_{n}C_{2})^{\text{T}} & X_{1}A+A^{\text{T}}+B_{n}C_{2}+C_{2}^{\text{T}}B_{n}^{\text{T}} & *^{\text{T}} & *^{\text{T}} \\ (B_{1}+B_{2}D_{n}D_{21})^{\text{T}} & (X_{1}B_{1} + B_{n}D_{21})^{\text{T}} & -\gamma I & *^{\text{T}}\\ C_{1}Y_{1}+D_{12}C_{n} & C_{1}+D_{12}D_{n}C_{2} & D_{11}+D_{12}D_{n}D_{21} & -\gamma I\\ \end{bmatrix} < 0 $$

Conclusion:
Using this problem formulation, an output controller $$\hat{K}$$ can be found that attenuates normal acceleration and tracks reference elevator deflection command.