Chapter 18, Solution 1. f ' ( t ) = δ( t + 2) − δ( t + 1) − δ( t − 1) + δ( t − 2) jωF(ω) = e j2 ω − e jω − e − jω + e − jω2 = 2 cos 2ω − 2 cos ω F(ω) = 2[cos 2ω − cos ω] jω Chapter 18, Solution 2. t, f (t) = 0, f ‘(t) 1 δ(t) 0 | Chapter 18 Solution 1. f t 8 t 2 -8 t 1 -8 t -1 8 t - 2 J F ej2 - ej - e-j e-j 2 2cos2 - 2cos 2 cos2 - cos F J Chapter 18 Solution 2. f t t 0 0 t 1 otherwise f t 1------------ I -8 t-1 f l8 t -8 t-1 t r -8 t-1 0 --------- t 1 f t 5 t - 5 t - 1 - S t - 1 Taking the Fourier transform gives -o2F o 1 - e j - j e j F ffl 1 y-1 or F o J0te-jfflt dt ax But ixeaxdx ax -1 c a2 . . e- 1 F o 7 F T -jot - 1 o 2K1 J - 1 - j 2 1 2 Chapter 18 Solution 3. f t 2t - 2 t 2 f t 2 - 2 t 2 2 1 F o J 221eJ tdt e - Jot 2 - J 2 -Jot -1 -2 2 A e-J 2 -J 2 -1 - eJ 2 Jo2 -1 2o L - Jo2 eJ 2 eJ 2 eJ 2 - e-J 2 2o2 - - Jo4cos2o J2sin 2o 2o2 F o y sin2o- 2o cos2o Chapter 18 Solution 4. 20 t 1 -1 g f 2 2ô t 1 0 -2 4ô t i k 1 t -20 t-1 g -1 t 0 1 -20 t 1 -2 -20 t-1 -2ô t-1 g -2ô t 1 2ô t 1 4Ô t - 2ô t -1 - 2ô t -1 j 2G -2ej 2j ej 4-2e-j -2joe-j -4cos - 4 sin 4 z x 4 z G cos sin -1 Chapter 18 Solution 5. h tï 1 -1 t 1 0 -2ô t h tk ô t 1 1 - 1 - 0 t -2ô t -ô t-1 h t ô t 1 -ô t -1 - 2ô t j 2H ej - e j - 2j 2jsin - 2j TTZ x 2j 2j . 11 c ------ sin
