資源簡(jiǎn)介
FFT及信號(hào)的頻譜分析
一、內(nèi)容
選擇合適的變換區(qū)間長(zhǎng)度N,用DFT對(duì)下列信號(hào)進(jìn)行譜分析,畫(huà)出幅頻特性和相頻特性曲線。
(1)x1(n)=2cos(0.2πn)R10(n)
(2)x2(n)=sin(0.45πn)sin(0.55πn)R51(n)
(3)x3(n)=2-|n|R21(n+10)
代碼片段和文件信息
n1=0:9;
N1=64;
x1n=2*cos(0.2*pi*n1);???????????????????????????
X1k=fft(x1nN1);????????
k=0:N1-1;wk=2*k/N1;?
subplot(331);plot(wkabs(X1k));
title(‘(a)x_1(n)的幅頻特性曲線‘);xlabel(‘\omega/\pi‘);ylabel(‘幅度‘);grid?on
subplot(334);plot(wkangle(X1k));?%axis([09-33])?
title(‘(b)x_1(n)的相頻特性曲線‘);xlabel(‘\omega/\pi‘);ylabel(‘相位‘);grid?on
n2=0:50;
N2=512;
x2n=2*sin(0.45*pi*n2).*sin(0.55*pi*n2);?????????
X2k=fft(x2nN2);????????
k=0:N2-1;wk=2*k/N2;?
subplot(332);plot(wkabs(X2k));
title(‘x_2(n)的幅頻特性曲線‘);xlabel(‘\omega/\pi‘);ylabel(‘幅度‘);grid?on
subplot(335);plot(wkangle(X2k));??
title(‘x_2(n)的相頻特性曲線‘);xlabel(‘\omega/\pi‘);ylabel(‘相位‘);grid?on
n3=-10:10;
N3=64;
x3n=0.5.^abs(n3);???????????????????????????????
x3np=zeros(1N3);???????????????????????????????
for?m=1:1
?屬性????????????大小?????日期????時(shí)間???名稱(chēng)
-----------?---------??----------?-----??----
?????文件????????1181??2017-12-07?14:08??SY1.m
?????文件?????????463??2017-12-05?14:56??SY2_1.m
?????文件?????????501??2017-12-07?13:38??SY2_2.m
?????文件??????122791??2017-12-07?19:45??實(shí)驗(yàn)一報(bào)告.docx
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