大学课件有限长单位冲激响应FIR数字滤波器的设计方法.ppt

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1、12310()()Njj nnH eh n e()()()(625)dh nh n w n1()()2jj nddh nHeed4,()(626)0,jcjdceHesin()1()()(627)2()cjj ncddcnh nHeedn561sin2,01()(629)120,cccNnnNh nNnothers()01()()()0,(628)12ddh nnNh nh n w nothersN7由（由（6-25）式可得）式可得()101()()()(630)2()()jjjdNjj nnH eHeW edW ew n e 8111220sin2()()(631)sin2NNNjjjj n

2、RRnnWeeeWesin2()sin2RNW12()()(632)NjjddHeHe1,()0,cdcH911()22121()()()21()()(633)2NNjjjdRNjdRH eHeWedeHWd 101112sinsinsin222()sin222RNNNWNN1314151621,012()(634)212,112nNnNw nnNnNN2121sin241sin2NjjNW eeN17212sin24(635)sin2NjjNW eeN181()1 cos(2/(1)()(6-36)2Nw nnNRn12()()0.5()0.25(2/(1)(2/(1)NjjRRRW eWW

3、NWNe()0.5()0.25(2/)(2/)(6-37)RRRWWWNWN19202()0.540.46cos()(638)1Nnw nRnN22()0.54()0.2311RRRWWWWNN2122()0.54()0.23(639)RRRWWWWNN2224()0.420.5cos0.08cos()(6-40)11Nnnw nRnNN22()0.42()0.2511440.04(641)11RRRRRWWWWNNWWNN232425263122.25 10(/)2cpstrad s 20.2,0.4,0.3ppstcpstcssssfff ()01()()21sin(),1()2,ccjj

4、j nddcjnh nIDTFT Heeednnnedn270.2stpsin 0.3(16)()()()0.540.46cos()(16)16dNnnh nh n w nRnn6.6N6.633N282()()(642)jdkdNHeHk2()()(),0,1,2,.,1jddkNH kHkHekN()()()()()NNNNrxnxn Rnx nrN Rn29110(1)11201()()(643)1sin21()()()(644)sin2jNNkkNk NNNjjjNz ekzH kH zNWzkNNH eH zeH kekNN()()()()NNNNrxxn Rnx nrN Rn303132333435