556 CHAPTER 12 • FOURIER SERIES AND THE LAPLACE TRANSFORM
y- 1
y =f(t)
<>----<>2 3Figure 12.24 The function y = f (t).
4 5e -cs
•EXAMPLE 12. 18 Show that .C(Uc (t)) = -.
s
6Solution We set f (t) = 1 and then set F (s) = .C (1) = ~· We apply Theorem
- 16 to get
e-C•
.C(Uc(t)) = .C(Uc(t) f(t)) = .C(Uc (t) · l ) = e-c•.C(l) = -.
s•EXAMPLE 12.19 Find .C(f (t)) if f (t) is as given in Figure 12.24.Solution We represent f (t) in terms of step functions:
f (t) = 1 - U1 (t) + U2 (t) -U3 (t) + U4 (t) - Us (t).
Using the result of Example 12.18 and linearity, we obtain
1 e-• e-2s e-3s e -4s e -5s
l(f(t))= - - - + - - - +---.
s s s s s s
- EXAMPLE 12. 20 Use Laplace transforms to solve the initial value problem
y" (t) +y(t) = U,, (t), with y (0) = 0 and y' (0) = 0.
Solution As usual, we let Y (s) denote the Laplace transform of y (t). T hen,
we gets^2 Y (s) + Y (s) = e-"•.
sSolving for Y ( s) gives
Y ( )
_,,. 1 e-"• e- "• s(^8) = e s (s (^2) + 1) = - s- - s2 + 1 ·