Page 164 - Schaum's Outline of Theory and Problems of Signals and Systems

CHAP. 31 LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS

Assume that y(0) = y(0-1. Let

~(t) -Y,(s)
Then from Eq. (3.44)
yl(t) -sY,(s) -y(O-) =sY,(s) -Yo

From Table 3-1 we have
K
x(t) -X,(S) = - Re(s) > -b
s+b
Taking the unilateral Laplace transform of Eq. (3.1041, we obtain

or

Thus,

+
Y1(s) = - K
s+a (s+a)(s+b)
Using partial-fraction expansions, we obtain

Taking the inverse Laplace transform of Y,(s), we obtain
K
[ yoe-"+ - - e-a:)
(e-br
y(t) = a-b
which is the same as Eq. (2.107). Noting that y(O+) = y(0) = y(0-) = yo, we write y(t) as
K
(e-br
y(t) = y0e-O1 + - - e-a') t20
a-b
3.38. Solve the second-order linear differential equation

y"(t) + 5y1(t) + 6y(t) =x(t)
with the initial conditions y(0) = 2, yl(0) = 1, and x(t) = eP'u(t).
Assume that y(0) = y(0-) and yl(0) = yl(O-). Let

~(t) -Y,(s)
Then from Eqs. (3.44) and (3.45)
yl(t) -sY,(s) -y(0-) =sY,(s) - 2

yN(t) - s2Y1(s) - sy(0-) - yl(O-) = s2YJs) - 2s - I

From Table 3-1 we have