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1、本本次次课课内内容容：1.数数值值方方法法求求DTFT遇遇到到二二个个问问题题 DFT 2.信信号号的的两两种种频频率率分分辨辨率率的的不不同同含含义义：DTFT谱谱：LswTcLfcff1，物物理理频频率率分分辨辨率率 DFT谱谱：Nffsbin，计计算算频频率率分分辨辨率率 9.2 DTFT Computation 9.2.1 DTFT at a Single Frequency For lengthL sequence x(n)，we compute DTFT at any desired value of by 10)()(LnnjenxX （9.2.1）(Nyquist interv
2、al 2,0,equivalent with,)A more efficient way to evaluate polynomials(Hrners rule):10)()(LnnznxzX)3()2()1()0(1111zxzxzxzx and jezzXX|)()(9.2.2 DTFT over Frequency Range To compute the DTFT over a frequency range),ba,we can only compute the spectrum at N discrete frequencies:binaabakkNk （k0N-1）(9.2.4)
3、bin:bin width 9.2.3 DFT DFT(discrete Fourier transform)离离散散傅傅立立叶叶变变换换 N-point DFT of a length-L signal:DTFT at N equally spaced frequencies over the full Nyquist interval 2,0 kNk2（k0N-1）(9.2.6)10)()()(LnnjkkenxXkX (9.2.8)bin width:Nbin2，Nfffssbinbin2（Hz）(9.2.9)例子：同一信号（长例子：同一信号（长L=64点点)的的DTFT和和DFT幅度谱
4、幅度谱DTFTN=64点点DFTDTFTN=60点点DFTPeriodicity of DFT 102)()()(LnknNjkenxXkX：)()(kXNkX（p.490）(一一个个周周期期：k=0N-1)9.2.4 Zero Padding padding zeros at the end of x(n):)(X remains the same;padding zeros to the front of x(n):)(X)(XeDj 9.3 Physical versus Computational Resolution Fig.9.3.1 （The data is from examp
5、le 9.1.4：)2cos()2cos()2cos()(321tftftftx,sf10kHz,f1,2,32、2.5、3kHz，f1,2,3/fs0.2、0.25、0.3）Physical frequency resolution：Lcw2，Lfcffsw（Hz）（9.1.21）(the minimum resolvable frequency separation between two sinusoidal components)Computational frequency resolution：Nbin2，Nffsbin（Hz）（9.2.9）（the frequency bin w
6、idth in the DFT case）为为了了从从 DFT频频谱谱区区分分 f1、2、3，（1）信信号号长长度度2ffLs（2）DFT点点数数LN （据据频频域域采采样样定定理理，N=L 即即可可从从 X(k)通通过过内内插插完完全全恢恢复复)(X。因因此此，N=32L=20的的离离散散频频谱谱已已足足够够反反映映DTFT谱谱。N=64有有何何意意义义？为为什什么么要要提提出出computational frequency resolution？The question of how accurately the DFT represents the peaks in the spectr
7、um.）To make the peak at 0f coincide with one of the DFT frequencies,0kshould be an integer in0,N-1:Nfkfs00sffNk00 (9.3.1)For the peak at 0f(equivalent with sff 0),the“index”satisfies:Nfkfs)(00or NfkNfNfkffssss)(000 In general,0k and 0kN are not necessarily integers,and the DFT will miss the exact pe
8、aks.Calculate the DFT indices in the case of Fig.9.3.1,When N32，sffNk1132x0.26.4，1kN 25.6 Similarly,2k8（integer），2kN 24，3k9.6，3kN 22.4 biasing error in DTFT spectrum (p.492)biasing:the maxima of the peaks in the spectrum do not quite correspond to the correct frequencies.It is caused by the lack of
9、adequate physical resolution.It can only be decreased by increasing the data length L.rounding error(p.493 and Fig.9.3.2)Caused by rounding k The rounding error in the frequencies is less than Nffsbin22 It decreases with increasing DFT length N.9.4 Matrix Form of DFT DFT：linear transformation finite x(n)finite X(k)(can be implemented by a matrix)1010210)()()()(2LnknNeWLnknNjLnnjWnxenxenxkXNjNk记（k=0N-1）1)1)(1(01100001)1()1()0()1()1()0(LLNLNNNLNNNNNNNLxxxWWWWWWWWNXXX xXA (9.4.1)twiddle factor(旋转因子旋转因子):NjNeW2 (9.4.4)10NNNWW，12NNW，kNkNNWW,N=L=2，1111A N=L=4，jjjjA111111111111 (9.4.6)
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