數(shù)字信號處理》總復(fù)習(xí)題答案-文庫吧
2025-01-03 18:59 本頁面
【正文】 sscc4442)(?????? ???????????????????????????????奇數(shù)偶數(shù)的形狀。點(diǎn)點(diǎn)，求按要求擴(kuò)展成將如圖所示，現(xiàn)的點(diǎn)，其，，、序列nnnxnynnnxnynnxnnxnyD F TnxkXD F Tnx0)2/()()3(74030)()()2(74)4(30)()()1(88)(|)(|4}0011{)(43210 1 2 3 n)(nx0 1 2 3 k|)(| kX?????????????????????????????????????????????????????????????300)]()([)12(30)(2)]()([)2(70)]()1()([)()1()()()()4()()()()()()1(30)12(8130281130830830830)4(830874830874813081708111111111111111kWnxnxkYkkXWnxnxkYkWnxnxWnxWnxWnxWnxWnxWnxWnyWnyWnykYnknnnknnkknnkknnknknnnknnknnknnknnknnk，解：n0 1 2 3 4 5 6 7)(1 ny1k01 2 3 4 5 6 7|)(| 11 kY3070)()2()()()()()2(222302/43087082222222?????? ?????????????kkkXkXWnxWnxWnykYkknnknnknnk，n0 1 2 3 4 5 6 7)(2 ny2k01 2 3 4 5 6 7|)(| 22 kY3070)())(()()()()()3(44330430282708333333????????? ??? ???????kkkXkXWnxWrxWnykYrnkrrkrnnnk，n0 1 2 3 4 5 6 7)(3 ny3k01 2 3 4 5 6 7|)(| 33 kY????????????????????????????????????????系統(tǒng)非因果。系統(tǒng)因果。系統(tǒng)穩(wěn)定。系統(tǒng)穩(wěn)定。收斂第一項(xiàng)有界系統(tǒng)非因果。是左序列解：為整數(shù)。；果性和穩(wěn)定性。響應(yīng)，試分析系統(tǒng)的因、已知系統(tǒng)的單位脈沖001|)(|)2(12122lim1)0(2)0()(2)()1()()()2()(2)()1(510mmmnuhnunhmmnnhnunhnnnnnn??運(yùn)算過程如下：，，，，，則：，，，；，，，設(shè)。解：點(diǎn)圓周卷積。的線性卷積和，，和序列序列的線性卷積；和序列序列的矩形序列。求：和分別代表長度為和、81}1361010974{{)()()()()(4}1234{)(5}11111{)()()2()(4)()](4)[1(7}1234{)()2()()(4)1(5100)()(62121212215110010051005100????????????????????????NNNmnxmxnxnxnyNnxNnRnxnRnnRnRnnRnRnRm??m … 3 2 1 0 1 2 3 4 5 6 7 … y(n) X1(m) 1 1 1 1 1 X2(m) 4 3 2 1 X2(m) 1 2 3 4 y(0)=4 1=4 X2(1m) 1 2 3 4 y(1)=3 1+4 1=7 X2(2m) 1 2 3 4 y(2)=…………=9 X2(3m) 1 2 3 4 y(3)=…………=10 X2(4m) 1 2 3 4 y(4)=…………=10 X2(5m) 1 2 3 4 y(5)=…………=6 X2(6m) 1 2 3 4 y(6)=…………=3 X2(7m) 1 2 3 4 y(7)=…………=1 下：點(diǎn)圓周卷積運(yùn)算過程如，，，，點(diǎn)圓周卷積：7)()(817}361010975{)()](~)(~[)()()(721102121nynwNNLnRmnxmxnxnxnwLNmN?????????????????m 0 1 2 3 4 5 6 w(n) X1(m) 1 1 1 1 1 0 0 X2(m) 4 3 2 1 0 0 0 X2(m) 4 0 0 0 1 2 3 w(0)=4 1+1 1=5 X2(1m) 3 4 0 0 0 1 2 w(1)=3 1+4 1=7 X2(2m) 2 3 4 0 0 0 1 w(2)=…………=9 X2(3m) 1 2 3 4 0 0 0 w(3)=…………=10 X2(4m) 0 1 2 3 4 0 0 w(4)=…………=10 X2(5m) 0 0 1 2 3 4 0 w(5)=…………=6 X2(6m) 0 0 0
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