381) Appendix A Solutions to the Exercises t5s + C ( r ( f ) ) = L{€(t 5) - &(t- ) ) = -5 finitr duration I L(J(t))= L(-G(f)) = 2I I , s 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 rxl (1) FF(x(t))- (,+S ~ -b Re{s) < 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 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): F{E(-f)} = nd(w)- - ? for e ) Use the yriiiciple of duality F{e-JW"E) = 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()