TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi

a. ˆx(t) = 1  Z ∞ −∞ x(a) t − a da Hence : −ˆx(−t) = −1  R ∞ −∞ x(a) −t−ada = −1  R −∞ ∞ x(−b) −t+b (−db) = −1  R ∞ −∞ x(b) −t+bdb = 1  R ∞ −∞ x(b) t−b db = ˆx(t) where we have made the change of variables : b = −a and used the relationship : x(b) = x(−b). b. In exactly the same way as in part (a) we prove : ˆx(t) = ˆx(−t) c. x(t) = cos !0t, so its Fourier transform is : X(f) = 1 2 ...