1第二章习题解答(数字信号处理(第二版),刘顺兰,版权归作者所有,未经许可,不得在互联网传播)2.1图T2-1中所示序列)(~1nx是周期为4的序列,确定傅里叶级数的系数)(~1kX)(~1nx2…1…n-1012345图T2-1解:3032)(~)(~nkNkNnkN3,242,01,2,4rNkNrNkrrNkrNk2.2求下列序列的DFT(1)}1,1,1,1{(2)},1,,1{jj(3)10,)(Nncnxn(4))()/2sin()(nRNnnxN解:(1)303424443,2,1,0,1)()(nkkknkk)3(,0)2(22)1(,0)0((2)10324,1)()(NnkNkNkNnkNNjWWjWWnxkX可求得0)3(,0)2(,4)1(,0)0(XXXX(3)111)()(10ccWcckNWckXkNNNnnkNn(4)10)2sin()(NnnkNWNnkXnkNjNnnNjnNjejee2102222][2110)1(210)1(2NnknNjNnknNjeej根据正弦序列的正交特性:22001212)(NkkNkjNkjNkX及2.3用封闭形式表达以下有限长序列的DFT)]([nx(1))()(0nRenxNnj(2))(cos)(0nnRnxN(3))(sin)(0nnRnxN(4))()(nnRnxN解:(1)10)()]([)(NnnkNWnxnxDFTkXnkNjNnnjee2100○a当)1'0('20NkkN即0不在采样点上时,)1)(2(00)2()2()2(00000)2sin(])2sin[(1111)(NkNjkNjNjkNjNkNjeNkNkNeeeekX○b当)10('20NkkN即0处在采样点上时10)'(2210'2)(NnkknNjnkNjNnnkNjeeekX',011',)'(2)'(2kkeekkNkkNjNkkNj)'(kkN(2)1010202cos)(00NnNnnkNjnjnjnkNeeenWkX3][2110)2(10)2(00NnnkNjNnnkNjee○a当)1'0('20NkkN即0不在采样点上时kNkNNjkNkNjNjkNjNjWWeWeeeekX200)2(2cos21)1)(cos1(]1111[21)(00000○b当)1'0('20NkkN即0处在采样点'k时10)'(210)'(2][21)(NnnkkNjNnnkkNjeekX其它及,0'',2kNkkkN(3)10010sin)()(NnnkNNnnkNnWWnxkXnkNjNnnjnjejee210200][2110)2(10)2(0NnnkNwjNnnkNjeej○a当)1'0('20NkkN,即0不在采样点上时kNkNkNkNjNjkNjNj)2()2(cos21sin]1111[21)(0000○b当)1'0('20NkkN即0w处在采样点'k时][21)(10)'(210)'(2NnnkkNjNnnkkNjeejkX',011',)'(2)'(210)'(2kkeekkNekkNjNkkNjNnnkkNj)'(kkN4其它,0',10)'(2kNkNeNnnkkNj))'((kNkN其它0',2',2)(kNkNjkkNjkX(4)kNNkNkNNnnkNNnnkNWNWWnWWnxkX)1(21010)1(2)()(①当0k时,102)1()0(NnNNnX当0k时,NkNkNNkNkNkNWNWNWWkXW)1()2(2)()1(32②①式-②式得)1(]1)[()1(2N)1(1)1()1(N11,1)(NkWNkXkN即11,10,2)1()(NkWNkNNkXkN解法二:11011)]([zzznRZNNnnN11111)1(1)1(]11[)](([zzzzzNzzdzdznRnZNNNNkNWzNNnnRZnnRDFTkX)]([)]([)(5kNkNkNkNNNkN)1(1)1()1(0,1)1()1(kWN11,10,2)1()(100NkWNkNNnWkXkNNnnN2.4已知序列nnanxn其它,096,)(求其10点和20点离散傅里叶变换。解:109644661]1[)()(NnnkNkNkNnkNnnkNaWWaWaWaWnxkXkNkNkNaWWaWa1101066当10N,;1,1)()(1046106aaWaWakXkk当20N,1,1))1(()(2046206aWaWakXkkk2.5若已知DFT[)(nx]=X(k),求)]2cos()([mnNnxDFT,)]2sin()([mnNnxDFT,Nm0解(1)10222][21)()]2cos()([NnnkNjnmNjnmNjeeenxmnNnxDFT])()([2110)(210)(2NnnmkNjNnnmkNjenxenx)])(())(([21NNmkXmkX(2)nkNjnmNjnmNjNneeejnxmnNnxDFT22210][21)()]2sin()([])()([2110)(210)(2NnmknNjNnmknNjenxenxj6)])(())(([21NNmkXmkXj2.6已知序列)3()2(2)1(3)(4)(nnnnnx,)(kX是)(nx的6点DFT。(1)若有限长序列)(ny的6点DFT是)()(46kXWkYk,求)(ny。(2)若有限长序列)(ny的6点DFT等于)(kX的实部,求)(nw。)}(Re{)(kXkW(3)若有限长序列)(nq的3点DFT满足,)2()(kXkQ,2,1,0k。求)(nq解:(1))(kX乘以一个kmNW形式的复指数相当于是)(nx圆周移位m点。根据式(2-29),本题中4m,即)(nx向右圆周移位了4点,则有66()((4))()2()(1)4(4)3(5)ynxnRnnnnn(2))}(Re{)(kXkW,根据式(2-25),有**()[()]{Re[()]}()11[()()][()(6)]224()1.5(1)(2)(3)(4)1.5(5)epwnIDFTWkIDFTXkxnxnxNnxnxnnnnnnn(3))(nx的6点DFT)(kX可看成是序列)(nx的傅里叶变换在0到2上的6点的等间隔采样,即226()()jkkNXkXe,则2222263()(2)()jkkkNQkXkXe相当于序列)(nx的傅里叶变换在0到2上的3点的等间隔采样,根据频域采样理论有3()[(3)]()5()3(1)2(2)rqnxnrRnnnn2.7序列)(nx为)3()1()(2)(nnnnx计算)(nx的5点DFT,然后对得到的序列求平方:)()(2kXkY求)(kY的5点DFT反变换)(ny。解:)()(2kXkY,根据圆周卷积定理有104550()()()()(())()()(())()4()5(1)(2)4(3)2(4)NNNmmynxnxnxmxnmRnxmxnmRnnnnnn72.8求)(nx的N点DFT:nnx)1()(,0,1,...1nN,其中N为偶数。解:1100()()(1),/21()0,1NNnknnkNNnnkNNkNXkxnWWNkNWW其他故()()2NXkNk2.9求)(kX的16点DFT反变换,)5162sin(3)3162cos()(kjkkX解:1611611616002222215(3)(3)(5)(5)()161616161601122()()[cos(3)3sin(5)]16161616113[()()]1622nknkkkjkjkjkjkjnkkxnxnWkjkWeeeee222215(3)(3)(5)((5)161616160113[()()]1622jnkjnkjnkjnkkeeee根据复正弦信号的正交特性,有13()[(3)(13)][(5)(11)]22xnnnnn2.10已知以下)(kX,求IDFT[X(k)](其中m为某一正整数,2/0Nm)。(1)kmNkeNmkeNkXjj其它,,,022)((2)kmNkjeNmkjeNkXjj其它,,,022)(解(1)102)(1)]([NknkNjekXNkXIDFT]22[1)(22NmNNjjnmNjjeeNeeNN8][21)2()2(nmNjnmNjee)2cos(nmN(2)102)(1)]([NknkNjekXNkXIDFT]22[1)(22mNnNjjnmNjjejeNejeNN)2sin(nmN2.11已知复有限长序列)(nf是有两个实有限长序列)(nx、)(ny组成)()()(njynxnf,且DFT[)(nf]=F(k),求)(kX、)(kY以及)(nx、)(ny。(1)kNNkNNbWbjaWakF1111)((2)jNkF1)(解(1))()()(njynxnf)]()([21)()],()([21)(**nfnfjnynfnfnx]11111111[21)]()([21)(***kNNNkNNNkNNkNNbWbjaWabWbjaWakNFkFkX101)(111NnnkNnkNNkNNkNNWaaWWaaWa)()(nRanxNn同理:)]1111(1111[21)(kNNkNNkNNkNNbWbjaWabWbjaWajkY)()(,1110nRbnyWbbWbNnnkNNnnkNN2.