資源簡介
matlab實現(xiàn)模態(tài)分析的代碼,利用擬合圓法,求三階固有頻率、阻尼比,計算各階主振型,求頻響函數(shù)的實頻、虛頻、幅頻、相頻
代碼片段和文件信息
function?motaifenxi_1(n)
%求頻響函數(shù)的實頻、虛頻、幅頻、相頻及Nyqiust圖
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%循環(huán)讀入激勵和響應數(shù)據(jù)
for?i=1:3???????????????????????????%讀入激勵數(shù)據(jù)
????fid11=fopen([‘D:\matlab2009\work\測點7\‘num2str(n)‘\Time?-?InputX‘num2str(0)num2str(0)num2str(i)‘.txt‘]‘r‘);
????j=1;
????while?~feof(fid11)
????????tline1=fgetl(fid11);
????????if?double(tline1(1))>=48&&double(tline1(1)<=57)
????????????tline11=str2num(tline1);
????????????xf(ij)=tline11(3);
????????????j=j+1;
????????????continue
????????end
????end
????fclose(fid11);
end
for?i=1:3????????????????????????????%讀入響應數(shù)據(jù)
????fid22=fopen([‘D:\matlab2009\work\測點7\‘num2str(n)‘\Time?-?InputY‘num2str(0)num2str(0)num2str(i)‘.txt‘]‘r‘);
????j=1;
????while?~feof(fid22)
????????tline2=fgetl(fid22);
????????if?double(tline2(1))>=48&&double(tline2(1)<=57)
????????????tline22=str2num(tline2);
????????????ya(ij)=tline22(3);
????????????j=j+1;
????????????continue
????????end
????end
????fclose(fid22);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%計算頻響函數(shù)的實頻、虛頻、幅頻、相頻
fs=4096;??????????????????????????%采樣頻率
N=4096;???????????????????????????%采樣點數(shù)。一般FFT程序中所用數(shù)據(jù)點數(shù)與原含有信號數(shù)據(jù)點數(shù)相同,
??????????????????????????????????%這樣的頻譜圖具有較高的質量,可減小因補零或截斷而產生的影響。
for?i=1:3
????x=xf(i:);
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