1、FFT变换区间及x6(t)抽样频率fsx1(n) , x2(n) , x3(n) , x4(n) , x5(n):N = 8 , 16x6(t):fs = 64(Hz) , N = 16 , 32 , 64MATLAB程序代码N1=8;N2=16;x1=ones(1,4);x2=1:4,4:-1:1;x3=4:1,1:4;n=0:1:16;x4=cos(pi*n/4);x5=sin(pi*n/8);X11=fft(x1,N1);X11=abs(X11);X21=fft(x2,N1);X21=abs(X21);X31=fft(x3,N1);X31=abs(X31);X41=fft(x4,N1);
2、X41=abs(X41);X51=fft(x5,N1);X51=abs(X51);X12=fft(x1,N2);X12=abs(X12);X22=fft(x2,N2);X22=abs(X22);X32=fft(x3,N2);X32=abs(X32);X42=fft(x4,N2);X42=abs(X42);X52=fft(x5,N2);X52=abs(X52);figure(1);subplot(3,1,1);stem(x1);grid;%x1时域波形xlabel(n);ylabel(x1(n)title(N=8的时域图subplot(3,1,2);stem(X11);%x1在N=8的FFT变换
3、频谱图Hz|X11(k)|N=8的频谱图subplot(3,1,3);stem(X12);%x1在N=16的FFT变换频谱图|X12(k)|N=16的频谱图figure(2);stem(x2);%x2时域波形x2(n)stem(X21);%x2在N=8的FFT变换频谱图|X21(k)|stem(X22);%x2在N=16的FFT变换频谱图|X22(k)|figure(3);stem(x3);%x3时域波形x3(n)stem(X31);%x3在N=8的FFT变换频谱图|X31(k)|stem(X32);%x3在N=16的FFT变换频谱图|X32(k)|figure(4);stem(x4);%x4
4、时域波形x4(n)stem(X41);%x4在N=8的FFT变换频谱图|X41(k)|stem(X42);%x4在N=16的FFT变换频谱图|X42(k)|figure(5);stem(x5);%x5时域波形x5(n)stem(X51);%x5在N=8的FFT变换频谱图|X51(k)|stem(X52);%x5在N=16的FFT变换频谱图|X52(k)|x6信号程序代码fs=64; T=1/fs;t=0:T:1-T;x6=cos(2*pi*4*t)+cos(2*pi*8*t)+cos(2*pi*10*t);N1=16;N2=32;N3=64;X61=fft(x6,N1);X61=abs(X61
5、);axis(0 7 0 1)X62=fft(x6,N2);X62=abs(X62);X63=fft(x6,N3);X63=abs(X63);stem(x6);x6(n)x6 时域波形figure(2)stem(X61);X6(k)N=16 时x6 频谱波形stem(X62);N=32时x6 频谱波形stem(X63);N=64时x6 频谱波形信号时域、FFT变换后的频谱波形a.x1信号时域、频谱波形b.x2信号时域、频谱波形c.x3信号时域、频谱波形d.x4信号时域、频谱波形e.x5信号时域、频谱波形f.x5信号时域波形g.x5信号频谱波形四、实验结论1.离散时间信号的FFT变换,其频谱是以
6、抽样点数N为周期的周期延拓2.当N2为N1的整数倍时,以为抽样点数的抽样的图形就是在以为抽样点数的抽样图形的每两个点之间插入N2/N1个点的谱图形五、思考题(1) 在N=8时,x2(n)和x3(n)的幅频特性会一样吗?为什么?N=16呢?在N=8时,x2(n)和x3(n)的幅频特性会一样;在N=16时,x2(n)和x3(n)的幅频特性会一样;因为当N=8时,x2(n)=1,2,3,4,4,3,2,1,x3(n)=4,3,2,1,1,2,3,4 而采样的频率都为8,x1(n)8与x2(n)8相等当N=16时 x2(n)=1,2,3,4,4,3,2,1,0,0,0,0,0,0,0,0x3(n)=4,3,2,1,1,2,3,4,0,0,0,0,0,0,0,0 而采样频率都为16,进展周期延拓后,x1(n)16与x2(n)16不相等(2) 如果周期信号的周期预先不知道,如何用FFT进展谱分析?确定一个N,再在MATLAB中调用FFT子程序计算
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