ENEE 322 Chapter : Homework Solution #3.pdf

• School: University of Maryland
• Department: Electrical & Computer Engineering
• Course: ENEE 322
• Professor: Steven Marcus

Document Summary

Since the output is the convolution between the input x[n] = ( 1 and the impulse response h[n] = u[n + 2]. 2 )n(cid:0)2u[n(cid:0) 2] y[n] = x[n] (cid:3) h[n] By replacing k with m + 2 in the summation, we obtain y[n] = Note that u[m] and u[n (cid:0) m] implies that m (cid:21) 0 and m (cid:20) n respectively. y[n] = )nu[n] (cid:3) (cid:14)[n (cid:0) 2] x[n] = ( h[n] = u[n] (cid:3) (cid:14)[n + 2] Thus, the output y[n] is y[n] = x[n] (cid:3) h[n] )nu[n] (cid:3) (cid:14)[n (cid:0) 2] (cid:3) u[n] (cid:3) (cid:14)[n + 2] )nu[n] (cid:3) (cid:14)[n (cid:0) 2] (cid:3) (cid:14)[n + 2] (cid:3) u[n] = u[n (cid:0) 3] (cid:0) u[n (cid:0) 9] x[n] = (cid:26) 1; h[n] = (cid:26) 1; 4 (cid:20) n (cid:20) 15 otherwise. = u[n (cid:0) 4] (cid:0) n[n (cid:0) 16] y[n] = x[n] (cid:3) h[n]