Solutions Communications technology II WS 2009/2010 - UniversitÃ¤t ...

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“Communications technology II” – Solutions WS 2009/2010 6d(i) d(i − 2) y(i)+1 + j +1 + j +1.25 + j ∗ 0.75+1 + j −1 + j +1.25 + j ∗ 1.25+1 + j −1 − j +0.75 + j ∗ 1.25+1 + j +1 − j +0.75 + j ∗ 0.75−1 + j +1 + j −0.75 + j ∗ 0.75−1 + j −1 + j −0.75 + j ∗ 1.25−1 + j −1 − j −1.25 + j ∗ 1.25−1 + j +1 − j −1.25 + j ∗ 0.75−1 − j +1 + j −0.75 − j ∗ 1.25−1 − j −1 + j −0.75 − j ∗ 0.75−1 − j −1 − j −1.25 − j ∗ 0.75−1 − j +1 − j −1.25 − j ∗ 1.25+1 − j +1 + j +1.25 − j ∗ 1.25+1 − j −1 + j +1.25 − j ∗ 0.75+1 − j −1 − j +0.75 − j ∗ 0.75+1 − j +1 − j +0.75 − j ∗ 1.251.510.5imag0−0.5−1−1.5−1.5 −1 −0.5 0 0.5 1 1.5real(d) The squared magnitude frequency response can be calculated by|W(Ω)| 2 = (1 − j0.25e j2Ω )(1 + j0.25e −j2Ω ) (1)= 1 + 0.25 2 − j0.25(e j2Ω − e −j2Ω ) (2)= 1.0625 + 0.5 sin(2Ω) (3)
7 WS 2009/2010 “Communications technology II” – Solutions1.81.61.41.2|W(Ω)| 210.80.60.40.200 pi/4 pi/2 pi 0Ω
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