F-18 Longitudinal Dynamics: Analysis of System Stability and Control

This project provides a comprehensive analysis of the F-18 aircraft's longitudinal dynamics. It begins by presenting the state-space representation of the system and then computes the open-loop system modes by calculating eigenvalues and eigenvectors. The solution then addresses system stability by placing observer poles and calculating the gain matrix. The analysis proceeds to determine the equivalent compensator and generate magnitude and phase plots using Bode plots and root locus analysis to assess the system's response characteristics. Finally, the document tests for system controllability, confirming the system's ability to be controlled. MATLAB code and results are included throughout the analysis, providing a practical demonstration of the concepts.

Consider the F-18 Longitudinal Dynamics.............................1
For system stability................................................3
for magnitude and phase plots.......................................4
For System Controllability..........................................5
Consider the F-18 Longitudinal Dynamics
A=[-0.0239 -28.3172 0 -32.2;-0.0003 -0.3621 1 0;0 -2.2115 -0.2532 0; 0 0 1 0]
B=[-3.8114 0.001; -0.0515 0;-2.8791 0;0 0]
C=[0 -1 0 1;1 0 0 0]
...Compute the open loop system modes
...To obtain the eigenvalues and eigenvectors
[V,D]=eig(A);
disp('EigenValues:')
disp(abs(V))
disp('EigenVector:')
disp(abs(D))
poles=[-1.149;-8;-8.5;-9]; %place observer poles
K=place(A,B,abs(poles));
disp(abs(K))
[t1 t2]=eig(A-B*K)
eig(A-B*K)
pls=[-0.05+0.09i,-0.05-0.09i,-1.05+1.1i,-1.05-1.1i];
L=place(A',C',pls')'
% To obtain the eigen value
eig(A-L*C)
Ac=abs(A-B*K-L*C);
Bc=L;
Cc=abs(K);
Gcss=ss(Ac,Bc,Cc,0);
Gc=tf(Gcss);
zpk(Gc)
A =
-0.0239 -28.3172 0 -32.2000
-0.0003 -0.3621 1.0000 0
1

0 -2.2115 -0.2532 0
0 0 1.0000 0
B =
-3.8114 0.0010
-0.0515 0
-2.8791 0
0 0
C =
0 -1 0 1
1 0 0 0
EigenValues:
0.9987 0.9987 1.0000 1.0000
0.0252 0.0252 0.0000 0.0000
0.0377 0.0377 0.0003 0.0003
0.0249 0.0249 0.0030 0.0030
EigenVector:
1.5124 0 0 0
0 1.5124 0 0
0 0 0.0966 0
0 0 0 0.0966
1.0e+05 *
0.0000 0.0012 0.0001 0.0009
0.0854 5.1051 0.3385 3.1935
t1 =
-0.0020 -0.9987 -0.0272 0.0483
-0.4652 -0.0065 0.1233 -0.1354
-0.6677 -0.0498 0.9859 -0.9820
-0.5812 -0.0059 0.1095 -0.1227
t2 =
1.1490 0 0 0
0 8.5000 0 0
2

0 0 9.0000 0
0 0 0 8.0000
ans =
1.1490
8.5000
9.0000
8.0000
L =
0.0987 1.7092
0.1680 0.0166
1.1157 0.0595
0.0196 0.0022
ans =
-1.0500 + 1.1000i
-1.0500 - 1.1000i
-0.0500 + 0.0900i
-0.0500 - 0.0900i
ans =
From input 1 to output...
32.018 (s+7.534) (s-6.785) (s-0.03526)
1: ---------------------------------------
(s-44.8) (s+13.63) (s-6.789) (s-0.1621)
1.3067e05 (s+8.767) (s-7.038) (s-2.049)
2: ---------------------------------------
(s-44.8) (s+13.63) (s-6.789) (s-0.1621)
From input 2 to output...
2.7555 (s-5.061) (s+4.681) (s+0.08024)
1: ---------------------------------------
(s-44.8) (s+13.63) (s-6.789) (s-0.1621)
25803 (s-30.06) (s+12.13) (s-0.2843)
2: ---------------------------------------
(s-44.8) (s+13.63) (s-6.789) (s-0.1621)
3

Continuous-time zero/pole/gain model.
For system stability
%To determine the equivalent compensator for the closed loop system
...transfer function.
figure(1)
rlocus(Gc(1,1))
grid on
for magnitude and phase plots
figure(2)
bode(Gc(1,1))
grid on
4

For System Controllability
disp('Testing for system controllability:')
sysCont=ctrbf(Ac,Bc,Cc);
ranQx=rank(sysCont);
if(ranQx==rank(Ac))
disp('System is controllable')
else
disp('System is NOT controllable')
end
Testing for system controllability:
System is controllable
5