Page 143 - Schaum's Outline of Theory and Problems of Signals and Systems

LAPLACE TRANSFORM AND CONTINUOUS-TIME LTI SYSTEMS [CHAP. 3

If a > 0, we see that the ROCs in Eqs. (3.68) and (3.69) overlap, and thus,
1 1 - 2a
-
x(s) = - - - -a < Re(s) <a
=
s+a s-a sZ-aZ
From Eq. (3.70) we see that X(s) has no zeros and two poles at s =a and s = -a and that the
ROC is -a < Re(s) <a, as sketched in Fig. 3-12(c). If a < 0, we see that the ROCs in Eqs.
(3.68) and (3.69) do not overlap and that there is no common ROC; thus, X(I) has no
transform X(s).

PROPERTIES OF THE LAPLACE TRANSFORM

3.7. Verify the time-shifting property (3.161, that is,
x(t - to) H e-"oX(S) R1=R

By definition (3.3)

By the change of variables T = t - I, we obtain

with the same ROC as for X(s) itself. Hence,

where R and R' are the ROCs before and after the time-shift operation.

3.8. Verify the time-scaling property (3.181, that is,

By definition (3.3)

By the change of variables 7 =at with a > 0, we have
I w
a ()
( x a ) = - x(r)e-('/")'dr = -X -
a -, RP=aR
Note that because of the scaling s/a in the transform, the ROC of X(s/a) is aR. With a < 0,
we have