112
4 Multi-Degree of Freedom (MDOF) Systems
x_hom(:,kt)=a_exp(:,:,kt) * [x0;xd0];
end
%-----------------------------------------%
Integrand for inhomogeneous solution
%-----------------------------------------x_inhom=zeros(2,nt+1); for kt=1:nt
%-------------------------------------%
Integral to a t in [0,tf]
%-----------------------------------sol=zeros(2,1);
for ks=1:kt+1
at_int=[0 time(kt+1)-time(ks);0 0];
a_exp_int=expm(at_int);
sol=sol+a_exp_int * [0;sin(nu * time(ks))] * dt;
end
x_inhom(:,kt+1)=sol;
end x_inhom=x_inhom+x_hom;
%---------------------figure plot(time,x_hom)
xlabel(’$t$’,’interpreter’,’latex’,’fontsize’,18)
ylabel(’$x_{\mathrm{hom}}(t)$’,’interpreter’,’latex’,
’fontsize’,18) hold
plot(time,x0+xd0 * time,’:’,time,xd0,’:’)
%------------------figure plot(time,x_inhom,’--’, ’linewidth’,2)
xlabel(’$t$’,’interpreter’,’latex’,’fontsize’,18)
ylabel(’$x(t)$’,’interpreter’,’latex’,’fontsize’,18) hold
plot(time,xt,’linewidth’,2) plot(time,xdt,’linewidth’,2)
%=====================================================
%
matrix_exp(1,-1,6,5,50);
%
The heavy solid and dashed lines in Fig. 4.15 are the solutions x(t) and ˙
x(t)
by direct integration and by matrix exponential for x 0 = 1, ˙
x 0 = −1, frequency
ν = 6, and time step t = 0.1. The two lines are nearly indistinguishable at the
scale of the figure. We note that the solution of ¨
x(t) = sin(ν t) by the state-space
approach of the previous section is unstable since the matrix a associated with this
equation has zero eigenvalues. In contrast, the solution by the matrix exponential
does not encounter this difficulty. However, it is computationally less efficient than
the state-space representation when dealing with large dynamical systems.
Example 4.12 Consider a damped single degree of freedom in free vibration
with unit mass, natural frequency ω, and damping ratio ζ . The unit impulse
response function of the system in Eq. 2.30 is the oscillator displacement function
for the initial conditions x 0 = 0 and ˙
x 0 = 1. It has the expression h(t) =
exp(−ω t) sin(ω d t)/ω d , where ω d = ω
1 − ζ 2 .
The free vibration solution of this SDOF system can also be obtained from
Eq. 4.112 with f(t) = 0 and z 0 with components x 0 = 0 and ˙
x 0 = 1. Accordingly,
the oscillator displacement and velocity are e 12 (t) ˙
x 0 , and e 22 (t) ˙
x 0 , where e(t) =
{e ij (t)} = exp(a t). The solid and dashed lines of Fig. 4.16 show the unit impulse
4 Multi-Degree of Freedom (MDOF) Systems
x_hom(:,kt)=a_exp(:,:,kt) * [x0;xd0];
end
%-----------------------------------------%
Integrand for inhomogeneous solution
%-----------------------------------------x_inhom=zeros(2,nt+1); for kt=1:nt
%-------------------------------------%
Integral to a t in [0,tf]
%-----------------------------------sol=zeros(2,1);
for ks=1:kt+1
at_int=[0 time(kt+1)-time(ks);0 0];
a_exp_int=expm(at_int);
sol=sol+a_exp_int * [0;sin(nu * time(ks))] * dt;
end
x_inhom(:,kt+1)=sol;
end x_inhom=x_inhom+x_hom;
%---------------------figure plot(time,x_hom)
xlabel(’$t$’,’interpreter’,’latex’,’fontsize’,18)
ylabel(’$x_{\mathrm{hom}}(t)$’,’interpreter’,’latex’,
’fontsize’,18) hold
plot(time,x0+xd0 * time,’:’,time,xd0,’:’)
%------------------figure plot(time,x_inhom,’--’, ’linewidth’,2)
xlabel(’$t$’,’interpreter’,’latex’,’fontsize’,18)
ylabel(’$x(t)$’,’interpreter’,’latex’,’fontsize’,18) hold
plot(time,xt,’linewidth’,2) plot(time,xdt,’linewidth’,2)
%=====================================================
%
matrix_exp(1,-1,6,5,50);
%
The heavy solid and dashed lines in Fig. 4.15 are the solutions x(t) and ˙
x(t)
by direct integration and by matrix exponential for x 0 = 1, ˙
x 0 = −1, frequency
ν = 6, and time step t = 0.1. The two lines are nearly indistinguishable at the
scale of the figure. We note that the solution of ¨
x(t) = sin(ν t) by the state-space
approach of the previous section is unstable since the matrix a associated with this
equation has zero eigenvalues. In contrast, the solution by the matrix exponential
does not encounter this difficulty. However, it is computationally less efficient than
the state-space representation when dealing with large dynamical systems.
Example 4.12 Consider a damped single degree of freedom in free vibration
with unit mass, natural frequency ω, and damping ratio ζ . The unit impulse
response function of the system in Eq. 2.30 is the oscillator displacement function
for the initial conditions x 0 = 0 and ˙
x 0 = 1. It has the expression h(t) =
exp(−ω t) sin(ω d t)/ω d , where ω d = ω
1 − ζ 2 .
The free vibration solution of this SDOF system can also be obtained from
Eq. 4.112 with f(t) = 0 and z 0 with components x 0 = 0 and ˙
x 0 = 1. Accordingly,
the oscillator displacement and velocity are e 12 (t) ˙
x 0 , and e 22 (t) ˙
x 0 , where e(t) =
{e ij (t)} = exp(a t). The solid and dashed lines of Fig. 4.16 show the unit impulse