Instructions and Hints for Ex. 7,8,16, Section 4.7
♯7: Plot the five solutions on separate plots in the range 0 ≤ t ≤ 20. In each
plot include also the envelopes. You can use the analytical solution
x(t) = A(t) sin(ωt) where A(t) =
sin(δt)
2δω
to generate the plots; the envelopes are ±A(t). For example, for ω = 9 the
plot can be generated through the commands
w0=11;w=9.0;
d=(w0-w)/2;wb=(w0+w)/2;
t=linspace(0,20,400);
a=1/2/wb/d*sin(d*t);
plot(t,a.*sin(wb*t),’k’,t,a,’k--’,t,-a,’k--’),
xlabel(’t’),ylabel(’x(t) and envelopes’),
title([’\omega=’,num2str(w)])
♯8: Show that the solution for the given differential equation with initial
conditions y(0) = y0 , y ′ (0) = 0 is given by
y(t) = (y0 − 1/23) cos 12t + (1/23) cos 11t.
You can then use this analytical solution to generate the plots. For example,
for y0 = 0:
y0=0;
t=linspace(0,4*pi,200);
y=(y0-1/23)*cos(12*t)+1/23*cos(11*t);
plot(t,y,’k’)
axis([0 4*pi -max(y) max(y)])
xlabel(’t’),ylabel(’y(t)’),title([’y_0=’,num2str(y0)])
♯16: Show that the solution to the initial value problem is given by
1
x(t) = (8e−t − 2e−4t − cos 2t).
5
Identify the transient part xtr (t) and the steady state solution xp (t). Plot
x(t), xtr (t), xp (t) in one plot using Matlab commands analogously as in ♯7.
Suggested time range: 0 ≤ t ≤ 12.