Solutions Communications Technology II WS 2011/2012

Im5 WS 2011/2012 “Communications Technology II” – SolutionsSolution to exercise 5 (eq12):a) Since x(i) = d(i−1)·0.5·e jπ/4 +d(i), the admissible values for x(i) are given byd(i) d(i−1) x(i)+1+j +1+j +1+j ∗(1+ √ 0.5)+1+j −1+j +1− √ 0.5+j+1+j −1−j +1+j ∗(1− √ 0.5)+1+j +1−j +1+ √ 0.5+j−1+j +1+j −1+j ∗(1+ √ 0.5)−1+j −1+j −1− √ (0.5)+j−1+j −1−j −1+j ∗(1− √ 0.5)−1+j +1−j −1+ √ 0.5+j−1−j +1+j −1+j ∗(−1+ √ 0.5)−1−j −1+j −1− √ 0.5−j−1−j −1−j −1+j ∗(−1− √ 0.5)−1−j +1−j −1+ √ 0.5−j+1−j +1+j +1+j ∗(−1+ √ 0.5)+1−j −1+j +1− √ 0.5−j+1−j −1−j +1+j ∗(−1− √ 0.5)+1−j +1−j +1+ √ 0.5−jReb) The impulse response of the overall system is given by w(i) = ∑ ιh(ι)g(i−ι). Thus,w(0) = h(0)e(0) = 1w(1) = h(1)e(0)+h(0)e(1) = e j5π/4 +e jπ/4 = 0w(2) = h(1)e(1) = (0.5) 2 ·e j(5π/4+π/4) = −0.25j.c) Since z(i) = d(i)∗w(i) = d(i−2)·0.25·e j3π/2 +d(i), the admissible values for y(i) aregiven by