4.6 Frequency Domain Analysis
111
(t) = I + a t + a
2 t
2 /2! + · · · ,
(4.111)
which converges uniformly and absolutely on bounded time intervals. The limit of
this series is denoted by exp
a t
and referred to as matrix exponential. With this
representation of the transition matrix, the solution z(t) in Eq. 4.109 takes the form
z(t) = e
a t z 0 +
t
0
e
a (t−s) b f(s) ds, t ≥ 0.
(4.112)
The matrix exponential exp
a t
can be calculated from truncated versions of
its series representation given by Eq. 4.111. However, it is more convenient to use
the MATLAB function expm
a t
, which gives this function. The output of this
MATLAB function can be used to calculate the free and forced vibration terms in
the expression of z(t).
Example 4.11 Consider the differential equation ¨
x(t) = sin(ν t), t ≥ 0, with the
initial conditions x(0) = x 0 and ˙
x(0) = ˙
x 0 . The integration of this equation gives
˙
x(t) = − cos(ν t)/ν + A and x(t) = − sin(ν t)/ν 2 + A t + B so that x 0 = B,
˙
x 0 = −1/ν + A and
x(t) = x 0 +
˙
x 0 + 1/ν
t − sin(ν t)/ν
2 , t ≥ 0.
The state-space representation of the differential equation ¨
x(t) = sin(ν t), t ≥ 0,
has the form in Eq. 4.77 with
a =
0 1
0 0
, b =
0
1
, and f (t) = sin(ν t).
The MATLAB solution of the first term of Eq. 4.112 is expm
a t
∗ z 0 , where z 0 is
a 2-dimensional vector with components x 0 and ˙
x 0 . The MATLAB solution of the
integrand of the second term of Eq. 4.112 is expm
a (t − s)
∗ b ∗ sin(ν t). These
functions are integrated in the following MATLAB function.
function matrix_exp(x0,xd0,nu,tf,nt)
%
%
Solution of x’’(t)=sin(nu * t) with ICs (x0,xd0)
%=============================================================
%
Direct integration
%----------------------------------dt=tf/nt; time=0:dt:tf; xt=x0+xd0 * time+time/nu-sin(nu * time)
/nu^2;
xdt=xd0+(1-cos(nu * time))/nu;
%=================================================
%
Matrix exponential
%------------------------------------------a=[0 1;0 0]; a_exp=zeros(2,2,nt+1); for kt=1:nt+1
at=[0 time(kt);0 0];
a_exp(:,:,kt)=expm(at);