Wiley signals and systems e book TLFe BO 504

381)

Appendix A. Solutions to the Exercises
t5s

+

C ( r ( f ) ) = L{€(t 5) - &(t- 5 ) ) = -5
finitr duration.

I

L(J(t))= L(-G(f)) =

4

1

2I . I

,

s

1

5t-t) c-Jio' d t =

,

s

E C, sirwe .c(t) has

EC
e-jWf d l

does not converge *

-cc

--7c

C{T(f)) =

S

0

rxl

(1) FF(x(t))-

(,+S
~

1

-b

.

Re{s) < 0

e) F{.r(h)) does riot co~iveige,see a) *
LC(i(f)) does not convergc

'Calciilitting the Fourier integral does not provide the soliit ion betanse the integral
will not cwivergt~.Nevertheless. the Fouricr transform clocs exisL in tlic form of a
dist ri hut ion.

~ o l ~ 9.2
~ i o ~
h) and c ) , since the regioii of convergence of the Laplace transform corvtaiiis the
imaginary axis. (Lying on tlw border is not srtllicient!)

Solution 9,3
The Fourier integrals from Exertisr 9.la, d and e do not converge to a function.
for a) TJre t h p niodulatiori theorem t m pair (9.7):

for (1) 1Jsc the similarity theorcm on pair (9.7):
1
F{E(-f)}
= nd(w)- - .
.?

for e ) Use the yriiiciple of duality
F{e-JW"E) = 2 7 d ( W +

011

pair (9.17):

"0).

The dwlity princple is used as follows:
S ( t - ).

3-0

c-.)wr

2nS(-LL' - z> = 2nb(w + ?-)
Wliere ir is any constant that can be appopriately replaced in thr1 result by

U().