University of Texas

• EE 351m

• Solution to Homework Set #5 1. (30 points) (a) T0 = 10−3 s, Ts = 1 6X10−3 s ⇒ N = T0 Ts = 6. So, x[n] is periodic with period N = 6. (b) Ωmax = 2π 10−3 × 9. Ωs = 2π Ts = 2π 10−3 × 6. Since Ωmax > Ωs, the sampling rate is not above aliasing. (c) x[n] = xc(nTs) = P 9 l=9 ale j2π( Ts T0 )ln = P 9 l=9 ale j 2π 6 ln . Discrete Fourier Series of x[n] is given by, X[k] = N X−1 n=0 x[n]e −j 2π 6 kn = X 5 n=0 ( X 9 l=−9 ale j 2π 6 ln)e −j 2π 6 kn = X 9 l=−9 al X 5 n=0 e j 2π 6 n(l−k) . Now, X 5 n=0 e j 2π 6 n(l−k) = 1 − e j2π(l−k) 1 − e j 2π 6 (l−k) = ( 6 if (l − k) = 0, ±6, ±12, · · · . 0 otherwise