5 2 h t 15 1 05 0 6 4 2 0 t 2 4 6 page 6

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (t + 1) – 4r(t)– 2r(t − 2)-2r(t + 1) - r(t) + 2r(t − 1) + r(t − 3)-4r(t) -2r(t − 1) +4r(t − 2) + 2r(t − 4) y3(t)= 2r(t + 4) + r(t + 3) + 2r(t + 2) –10r(t) + 2r(t − 2) + r(t − 3) + 2r(t − 4) 3 2.5 2 h(-t) 1.5 1 0.5 0 -6 -4 -2 0 t 2 4 6 Page 6 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo h(-t) =[u(t + 5 ) + 2u(t+3) − u(t + 2) −2 u(t +1)] 16 14 12 10 8 y3(t) 6 4 2 0 -2 -6 -4 -2 0 t 2 4 6 35. (a) ∗ 0, 0, 0, .1 1 0 ∞ , 1 0 ∞ Page 7 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo 1 0.9 0.8 0.7 0.6 y1(t) 0.5 0.4 0.3 0.2 0.1 0 -4 b -3 -2 -1 0 t 1 2 3 4 ∗ 0, 0 , 0, Page 8 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo 0.35 0.3 0.25 0.2 y2(t) 0.15 0.1 0.05 0 -4 -3 -2 -1 0 t 1 2 3 4 (When K1=1) c , ∗ 0.25 0.2 0.15 y3(t) 0.1 0.05 0 -4 -3 -2 -1 0 t 1 2 3 4 (When K2=1) Page 9 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo 36. (a) 5 (b) 2 0.67 ∗ 2 2 0.67 0.67 2 . 0.67 0.5 0.67 . By using timing shifting, ∗ ∗ 0.67 1 0.67 0.67 0.5 1 . . 1 0.5 2 1 1 0.67 . 1 (c) 0.67 5 5 0.5 ∗ . 2 2 . 0.67 ∗ . . ∗ . . . 0.67 0 0 3.375 3.375 3.375 1.35 . . . 3.375 2.5 . . . 1.35 1.35 0.67 ∗ 1.35 . 0.67 . 0.67 , 5 Page 10 / 11 ECE 202 HW solution #9 for prob. 33~36 Prof. DeCarlo 0.67 0 0 3.375 3.375 5 3.375 0.84 3.375 4 0.84 0.84 0.84 0.51 1.35 1.35 1.35 . . 0.84 0.84 0.84 Page 11 / 11...
View Full Document

This document was uploaded on 04/02/2010.

Ask a homework question - tutors are online