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150 LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS [CHAP. 3
(a) The bilateral Laplace transform of x(t) defined in Eq. (3.3) can be expressed as
Now /:x(t)ep" dl = X,(s) Re(s) > o+ (3.92)
Next. let
/
ffi
Then ~,~x(-~)e"dr ~(-r)e-'~'~'dt =X;(-s) Re(s) < o- (3.94)
=
0 -
Thus, substituting Eqs. (3.92) and (-3.94) into Eq. (3.91), we obtain
X(s) =X,(s) +X,( -s) a+< Re(s) <a- (3.95)
(b) X(t) = e-21'I
(1) x(t = e-2' for t > 0, which gives
(2) x(t ) = e2' for t < 0. Then x( - t) = e-2' for t > 0, which gives
Thus,
(3) According to Eq. (3.95), we have
which is equal to Eq. (3.701, with a = 2, in Prob. 3.6.
3.35. Show that
(a) ~(0') = lim sX,(s) (3.97)
s-m
(b) lim x( t ) = lirn sX,(s) (3.98)
t += s -0
Equation (3.97) is called the initial value theorem, while Eq. (3.98) is called the final
calue theorem for the unilateral Laplace transform.