%% Example 4
% Let
% F(s)=\frac{100}{s+2}+\frac{1}{s+1}
% From transform tables
% f(t)=100e^{-2t} +e^{-t}
% The assympotic decay rate is determined by the pole at s=-1
% Assymptotically, f(t) decays like e^{-t}
% Even though the residue associated with for the nondominant pole is 100 times larger, the term associated with the dominant pole is larger for t>4.6

t = linspace(0,10,1000);
f1 = 100*exp(-2*t);
f2 = exp(-t);
subplot(2,2,1)
plot(t,f1),grid,ylim([0,0.5]),xlim([0,10])
subplot(2,2,2)
plot(t,f2),ylim([0,0.5]),grid
subplot(2,2,[3,4])
plot(t,f1+f2),ylim([0,0.5]),grid