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1、第九章9.2f=10KHz；f=80KHz0sffssk=f⇒k=f=8N−k=64−8=5600NN9.31f=5KHz；f=40KHz；T=T⋅16=⋅16=3.2mes0sL周f0Twww.khdaw.comL1）L==T⋅f=3.2×40＝128LsT采Tf周s或==8(说明一个周期有8个采样)⇒L=8×16＝128Tf采0课后答案网f02）N=L=128；k=⋅N=16；N−k=128−16=112fsTf周s3）设L个采样含c个周期的采样⇒L=N=c⋅=c⋅Tf采0fff00sk=⋅N=⋅c
2、⋅=c=整数fffss0∴ThenumberofperiodscontainedintheNsamplesisthesameastheDFTindexatwhichyougetapeak.9.9www.khdaw.com∆f=1209−941=268min3ff8×10ss∆f≥⇒L≥==29.85∴L=30minL∆f268min课后答案网ffsshammingwindow:∆f≥2⇒L≥2⋅=59.7∴L=60minL∆fmin9.4f8s1)频域采样间隔为∆f===0.5binN16±f=∆f⋅m
3、=0.5⋅m=±18⇒m=±36mbink=mmod(16)=4和12www.khdaw.com2）先求进入Nyquist间隔的f=2和6af=∆f⋅k⇒k=4和12abin课后答案网9.121)directDFT⎡1⎤⎢⎥2⎢⎥⎡11111111⎤⎢2⎥⎡12⎤⎢⎥⎢⎥⎢⎥1−j−1j1−j−1j10X=⎢⎥⎢⎥=⎢⎥⎢1−11−11−1www.khdaw.com1−1⎥⎢2⎥⎢0⎥⎢⎥⎢⎥⎢⎥⎣1j−1−j1j−1−j⎦⎢1⎥⎣0⎦⎢1⎥⎢⎥课后答案网⎢2⎥⎣⎦2)wrappedDFT⎡1111⎤⎡
4、3⎤⎡12⎤⎢⎥⎢⎥⎢⎥1−j−1j30X=⎢⎥⎢⎥=⎢⎥⎢1−11−1⎥⎢3⎥⎢0⎥⎢⎥⎢⎥⎢⎥⎣1j−1−j⎦⎣3⎦⎣0⎦3)用DFT求IDFTwww.khdaw.com∗∗⎧⎡1111⎤⎡12⎤⎫⎡3⎤⎪⎢⎥⎢⎥⎪⎢⎥~⎪1⎢1−j−1j⎥⎢0⎥⎪⎢3⎥x=⎨⋅课后答案网⋅⎬=⎪4⎢1−11−1⎥⎢0⎥⎪⎢3⎥⎢⎥⎢⎥⎢⎥⎪⎩⎣1j−1−j⎦⎣0⎦⎪⎭⎣3⎦9.211)L=16,N=16152π−jnkX=∑x(n)e16k=0,1L,15kn=02)L=16,N=8152π−jnlXˆ=∑x(n
5、)e8l=0,1L,7lwww.khdaw.comn=0⎧Xˆ=X00⎪⎪Xˆ=X12∴⎨课后答案网⎪M⎪ˆX=X⎩7149.22x(t)=cos(24πt)+2sin(12πt)cos(8πt)=cos(24πt)+sin(20πt)+sin(4πt)∴f=12k,f=10k,f=2k,f=8k⇒f=4k,f=2k,f=2k123s1a2a3a1)x(t)=cos(8πt)+2sin(4πt)a1j24πt1−j24πt1j20πt1−j20πt1j4πt1−j4πt2)x(t)=e+e+e−e+e−
6、e222j2j2j2jj24πn−j24πnj20πn−j20πn1111x(nT)=x(n)=e8+e8www.khdaw.com+e8−e8222j2jj4πn−j4πn11+e8−e82j2j课后答案网jπnj3πnjπnj3πnx(n)=1ejπn+1ejπn+1e2−1e2+1e2−1e2222j2j2j2jjπnj3πnx(n)=ejπn−je2+je2式(1)QL=8,N=872π7π~1jnk1jnk∴x(n)=x(n)=∑X(k)⋅e8=∑X(k)⋅e4式(2)8k=08k=0www.
7、khdaw.com式(1)和式(2)比较可得：X(k)=[0,0,−8j,0,8,0,8j,0]课后答案网9.131)碟形图见黑板π2）x(n)=4cos(n)+cos(πn)2ππjn−jn11=2e2+2e2+ejπn+e−jπn22ππjn−jn112j2πn2jπnj2πn−jπn=2e+2e⋅e+e+e⋅ewww.khdaw.com22π3πjnjn=2e2+2e2+ejπn8−pointFFT计算的x(n)的频谱为课后答案网[0016080160]所以8−pointFFT计算的x(n)的频谱
8、与x(n)的真实频谱相比谱线的位置一样，幅度是8倍的关系π3）x(n)=4cos(n)+cos(πn)2π3πω=,ω=,ω=πn1n2n3222πω=kk=0L7k8www.khdaw.com∴在ω，ω，ω处看到x(n)的频谱kk=2kk=6kk=4课后答案网9.14∗1)X=[044j404−4j4]碟形图见黑板x=[210−1−210−1]2）L=8,N=872π7π~1jnk1jnk∴x(n)=x(n)=∑X(k)⋅e8=∑X(k)
