Page 409 - Applied Numerical Methods Using MATLAB

398 MATRICES AND EIGENVALUES

%nm8p08a
syms a0 a1 a2 s
A =[0 1 0;0 0 1;-a0 -a1 -a2]; %(P8.8.2a)
det(s*eye(size(A))- A) %characteristic polynomial
ch_eq = poly(A) %or, equivalently

(b) Let the input u(t) in the state equation (P8.8.2) be dependent on the state
as
u(t) = K x(t) = [ K 0 x 1 (t) K 1 x 2 (t) K 2 x 3 (t) ] (P8.8.4)

Then, the state equation can be written as

     

x 1 (t) 0 1 0 x 1 (t)

= 0 0 1 (P8.8.5)
 x 2 (t)     x 2 (t) 
x 3 (t) K 0 − a 0 K 1 − a 1 K 2 − a 2 x 3 (t)

If the parameters of the original system matrix are a 0 = 1,a 1 =−2, and
a 2 = 3, what are the values of the gain matrix K = [K 0 K 1 K 2 ] you
will fix so that the virtual system matrix in the state equation (P8.8.5)
has the eigenvalues of λ =−1, −2, and −3? Note that the character-
istic equation of the system whose behavior is described by the state
equation (P8.8.5) is

2
3
s + (a 2 − K 2 )s + (a 1 − K 1 )s + a 0 − K 0 = 0 (P8.8.6)
and the equation having the roots of λ =−1, −2, and −3is

3
2
(s + 1)(s + 2)(s + 3) = s + 6s + 11s + 6 = 0 (P8.8.7)
8.9 A Homogeneous Differential Equation—An Eigenvalue Equation
Consider the undamped mass-spring system depicted in Fig. 8.3, where the
masses and the spring constants are m 1 = 1,m 2 = 1[kg] and k 1 = 5,k 2 = 10
[N/m], respectively. Complete the following program “nm8p09.m” whose
objective is to solve the second-order differential equation (8.6.1) with the

initial conditions [x 1 (0), x 2 (0), x (0), x (0)] = [1, −0.5, 0, 0] for the time
1 2
interval [0,10] in two ways—that is, by using the ODE-solver “ode45()”
(Section 6.5.1) and by using the eigenvalue method (Section 8.6) and plot
the two solutions. Run the completed program to obtain the solution graphs
for x 1 (t) and x 2 (t).
(cf) Note that the second-order vector differential equation (8.6.1) can be written as
the following state equation:

x (t) O I x(t)

= (P8.9.1)

x (t) −AO x (t)