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如何求幅值
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kleeblatt发表于 2009-11-11 05:29振动论坛的贴子 主题<如何求幅值？？>
急求问各位，如下情况如何求信号幅值：
已知：信号图形Asin(wt*t+phi) ，基波初相位phi，基波频率w 求：幅值
但算法需在simulink中实现，故不可直接使用Asin(wt+phi)/ sin(wt+phi)，因sin(wt+phi)存在零值。求问各位还能有什么方法实现幅值的输出，谢谢！！！
已知：信号图形Asin(wt*t+phi) ，基波初相位phi，基波频率w 求：幅值
但算法需在simulink中实现，故不可直接使用Asin(wt+phi)/ sin(wt+phi)，因sin(wt+phi)存在零值。求问各位还能有什么方法实现幅值的输出，谢谢！！！
用离散频谱校正可以测出正弦函数的振幅, 有多种校正法都可用,有些校正精度很高. 这些方法大都利用插值公式计算振幅值
但有特点的是fft/apfft校正法.它与窗函数无关,和频偏值也无关,由fft和apfft两个振幅值决定.
FFT和apFFT组合校正法须2N-1个样点,对前N个样点作FFT,得FFT振幅谱a1,对全部2N-1个样点作apFFT,得apFFT振幅谱a2, FFT功率谱/apFFT振幅谱 , a1.^2/a2 , 即振幅校正谱.
程序如下
close all;clc;clear all;
N=64;
phase1=[0
3.93396163586619
8.83046453180072
11.6311186638529
11.2852032132275
8.16429115438237
3.42142700284035
-2.00785915626561
-7.27003758417477
-10.9814095127588
-11.7693210235971
-9.50125290401178
-5.19827852519589
0.0755540599887536
5.52110483988927
9.96033284685893
11.8671678631308
10.5897860671053
6.85986574516244
1.83106201768410
-3.65438621341892
-8.62018703590539
-11.5601862248507
-11.3745054979715
-8.36678942985316
-3.68031853969484
1.73128196808665
7.02865158843185
10.8551703544759
11.8053382813413
9.67138766757273
5.44247874386262
0.199150317798624
-5.25906344418210
-9.78506215112777
-11.8459275005807
-10.7196707217138
-7.08425677046303
-2.09903232786475
3.38117109361063
8.40488708594418
11.4811148238893
11.4565288437530
8.56509172811930
3.93733854866132
-1.45468198789504
-6.78373271813096
-10.7214927023010
-11.8333425140581
-9.83622848517273
-5.68414033754368
-0.473112878088091
4.99493146583130
9.60350867104510
11.8163791138212
10.8431395116249
7.30529084768279
2.36564033520522
-3.10709203536907
-8.18474661381370
-11.3939737550963
-11.5311484142206
-8.75904829498166
-4.19239895590748
1.17818129656841
6.53549662248908
10.5805184253717
11.8532632418226
9.99561433953389
5.92315733116069
0.746239014363543
-4.72889980941472
-9.41586673610081
-11.7785245895997
-10.9600403119510
-7.52283744904273
-2.63080243663561
2.83230289230774
7.95998397314823
11.2988435253994
11.5982457374237
8.94850686170142
4.44540986626462
-0.901897207769072
-6.28415660713507
-10.4323983310435
-11.8650394268420
-10.1493841184832
-6.15942046354795
-1.01843680070663
4.46115450507246
9.22233591019281
11.7323769588005
11.0702242695693
7.73676255080719
2.89443515304596
-2.55695194856533
-7.73081845715696
-11.1958155393794
-11.6577092368834
-9.13331291839477
-4.69627928205803
0.625942508097649
6.02992315176618
10.2772917515210
11.8686198504245
10.2973770472697
6.39281707734105
1.28961663446127
-4.19187648975836
-9.02312039187059
-11.6779605360029
-11.1735463316539
-7.94692872442234
-3.15645477707631
2.28118197793132
7.49746969034869
11.0849918617198
11.7094346686512
9.31330999123343
4.94491285854079
-0.350425718110841
-5.77300349552364
-10.1153660200609
-11.8639634366244
-10.4394331578059
-6.62323103689094
];
y=phase1';
y1 = y(N:2*N-1);
win = hanning(N)';;
win1 = win/sum(win);
y11= y1.*win1;
y11_fft = fft(y11,N);
y1_amplitude = abs(y11_fft);
y2 = y(1:2*N-1);
win =hanning(N)';;
winn =conv(win,win);%apFFT
win2 = winn/sum(winn);
y22= y2.*win2;
y222=y22(N:end)+[0 y22(1:N-1)];
y2_fft = fft(y222,N);
a2 = abs(y2_fft);
aa=(y1_amplitude.^2)./a2*2;
[A,r]=max(a2(1:N/2));
aaa=aa(r)
N=64;
phase1=[0
3.93396163586619
8.83046453180072
11.6311186638529
11.2852032132275
8.16429115438237
3.42142700284035
-2.00785915626561
-7.27003758417477
-10.9814095127588
-11.7693210235971
-9.50125290401178
-5.19827852519589
0.0755540599887536
5.52110483988927
9.96033284685893
11.8671678631308
10.5897860671053
6.85986574516244
1.83106201768410
-3.65438621341892
-8.62018703590539
-11.5601862248507
-11.3745054979715
-8.36678942985316
-3.68031853969484
1.73128196808665
7.02865158843185
10.8551703544759
11.8053382813413
9.67138766757273
5.44247874386262
0.199150317798624
-5.25906344418210
-9.78506215112777
-11.8459275005807
-10.7196707217138
-7.08425677046303
-2.09903232786475
3.38117109361063
8.40488708594418
11.4811148238893
11.4565288437530
8.56509172811930
3.93733854866132
-1.45468198789504
-6.78373271813096
-10.7214927023010
-11.8333425140581
-9.83622848517273
-5.68414033754368
-0.473112878088091
4.99493146583130
9.60350867104510
11.8163791138212
10.8431395116249
7.30529084768279
2.36564033520522
-3.10709203536907
-8.18474661381370
-11.3939737550963
-11.5311484142206
-8.75904829498166
-4.19239895590748
1.17818129656841
6.53549662248908
10.5805184253717
11.8532632418226
9.99561433953389
5.92315733116069
0.746239014363543
-4.72889980941472
-9.41586673610081
-11.7785245895997
-10.9600403119510
-7.52283744904273
-2.63080243663561
2.83230289230774
7.95998397314823
11.2988435253994
11.5982457374237
8.94850686170142
4.44540986626462
-0.901897207769072
-6.28415660713507
-10.4323983310435
-11.8650394268420
-10.1493841184832
-6.15942046354795
-1.01843680070663
4.46115450507246
9.22233591019281
11.7323769588005
11.0702242695693
7.73676255080719
2.89443515304596
-2.55695194856533
-7.73081845715696
-11.1958155393794
-11.6577092368834
-9.13331291839477
-4.69627928205803
0.625942508097649
6.02992315176618
10.2772917515210
11.8686198504245
10.2973770472697
6.39281707734105
1.28961663446127
-4.19187648975836
-9.02312039187059
-11.6779605360029
-11.1735463316539
-7.94692872442234
-3.15645477707631
2.28118197793132
7.49746969034869
11.0849918617198
11.7094346686512
9.31330999123343
4.94491285854079
-0.350425718110841
-5.77300349552364
-10.1153660200609
-11.8639634366244
-10.4394331578059
-6.62323103689094
];
y=phase1';
y1 = y(N:2*N-1);
win = hanning(N)';;
win1 = win/sum(win);
y11= y1.*win1;
y11_fft = fft(y11,N);
y1_amplitude = abs(y11_fft);
y2 = y(1:2*N-1);
win =hanning(N)';;
winn =conv(win,win);%apFFT
win2 = winn/sum(winn);
y22= y2.*win2;
y222=y22(N:end)+[0 y22(1:N-1)];
y2_fft = fft(y222,N);
a2 = abs(y2_fft);
aa=(y1_amplitude.^2)./a2*2;
[A,r]=max(a2(1:N/2));
aaa=aa(r)
运行结果
aaa = 11.6713
你给的数据的是余弦信号,振幅是11.67,频率5.0394,中间点相位248.2883. 可用下式产生
t=-63:63;
A=11.6713;f=5.0394;phi=248.2883;
y1=A*cos(2*pi*t*f/64+phi*pi/180);
y1和你的数据基本吻合.
误差原因是你的信号中含有3次谐波和5次谐波成份
各次谐波频率为(Hz),振幅和中间点相位为
fm = 5.0394 15.1184 25.1853
Am = 11.6713 0.2260 0.0011
Pm = 248.2883 55.2875 260.0685
aaa = 11.6713
你给的数据的是余弦信号,振幅是11.67,频率5.0394,中间点相位248.2883. 可用下式产生
t=-63:63;
A=11.6713;f=5.0394;phi=248.2883;
y1=A*cos(2*pi*t*f/64+phi*pi/180);
y1和你的数据基本吻合.
误差原因是你的信号中含有3次谐波和5次谐波成份
各次谐波频率为(Hz),振幅和中间点相位为
fm = 5.0394 15.1184 25.1853
Am = 11.6713 0.2260 0.0011
Pm = 248.2883 55.2875 260.0685
