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！！！ 这......
直接读最大值不就可以了 信号Asin(wt+phi)是经dsp处理的，所以读取dsp后的最大值很可能不是原信号的最大值:@o
回复楼主kleeblatt 的贴子
本帖最后由 wdhd 于 2016-9-19 13:37 编辑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=256;
t=-N+1:N-1;
A=110.1111;f=49.2;phi=40;
y=A*cos(2*pi*t*f/N+phi*pi/180);
y1 = y(N:2*N-1);%后N个输入数据
win =hanning(N)';
win1 = win/sum(win);%窗归1
y11= y1.*win1;
y11_fft = fft(y11,N);
a1 = abs(y11_fft);%FFT振幅谱
y2 = y(1:2*N-1);%2N-1个输入数据
win =hanning(N)';;
winn =conv(win,win);%apFFT须要卷积窗
win2 = winn/sum(winn);%窗归1
y22= y2.*win2;
y222=y22(N:end)+;%构成长N的apFFT输入数据
y2_fft = fft(y222,N);
a2 = abs(y2_fft);%apFFT振幅谱
aa=(a1.^2)./a2*2;%振幅校正值
r=round(f)+1;%振幅谱峰值处频率
disp('振幅校正值')
aaa=aa(r)
运行结果
振幅校正值
aaa=110.1111
[ 本帖最后由 zhwang554 于 2009-11-15 09:18 编辑 ]
请问a1的值是提前给定的么？我现在只有sin(wt+phi)的值和Asin(wt=phi)的值，如何给定A呢？
回复6楼kleeblatt 的贴子
抱歉, 原来程序中用了2个a1, 让人弄不清楚,改了一下笫一个a1改为A, 即信号Asin(wt=phi)的振幅
第二个a1(保留在程序中)是序列y11的FFT的振幅谱
其中y11是加hanning窗y1的序列
其中y1是给定的余弦信号Asin(wt=phi)的取样值
你说: "我现在只有sin(wt+phi)的值和Asin(wt=phi)的值，如何给定A"?
离散频谱校正法只须Asin(wt=phi)的值，就能定A
离散频谱校正是从信号Asin(wt+phi)取样值的DFT的振幅谱测A, 而且是在非整周期取样下测A. 整周期取样下Asin(wt+phi)取样值的DFT的振幅值就是A.
这个方法是离散频谱校正法, 对余弦信号Asin(wt+phi)的取样值作FFT得振幅谱a1, 作apFFT得振幅谱a2, a1.^2/a2 即得A (在振幅谱a1峰值频率处)
关于离散频谱校正法论坛中贴子很多, 可参看相关贴子,如
http://forum.vibunion.com/UChome/space-62061-do-blog-id-17741.html
[ 本帖最后由 zhwang554 于 2009-11-15 20:51 编辑 ] sorry that i can't type chinese at the moment.
want to ask again, that i can't give the function of the input signal, just 128 discrete values in 1 sec. how can i get the apfft values, thanks!!!!!
回复8楼kleeblatt 的贴子
本帖最后由 wdhd 于 2016-9-19 13:38 编辑你的问题是:若1秒内得128个采样,如何作apFFT?
选定fft阶数N后,当采样频率fs=N赫/秒,作apFFT需2秒采样, 当采样频率fs=N/2赫/秒,作apFFT只需1秒采样,
所以若得1秒内128个采样(即采样频率为128赫/秒),相当采样频率为64赫/秒2秒内得128个采样. 计算频率时注意采样频率为64赫/秒.
若只得128个采样,你用127采样作N=64的apFFT即可,反正作N阶FFT/apFFT须2*N-1个采样.如何作apFFT 5楼程序中都写出了
类同的有一种FFT时移相位差法,它须二组相邻序列作FFT才能校正振幅等,所以也要采样2秒
由信号频谱求振幅的一个长处是一次FFT可得多种频率式成份的振幅值. 用fft/apfft校正法得的振幅校正谱可用下aaaa.doc附图示
N=256;
t=-N+1:N-1;
A=1.0;f=29.2;phi=40;
y=A*cos(2*pi*t*f/N+phi*pi/180)+2*A*cos(2*pi*t*2*f/N+phi*pi/180)+3*A*cos(2*pi*t*3*f/N+phi*pi/180);
附图是3个成分信号y的振幅谱校正图;(a)为fft振幅谱a1,(b)为apfft振幅谱a2,(c)校正振幅谱.
3个成分信号振幅校正为1,2,3.而且在峰值附近形成一个平台,很醒目. 有噪时峰值处最准.
振幅校正值(111db)
0.9999 1.9999 3.0000
振幅校正值(10db)
1.0536 1.9113 3.0705
从图也可见,未经校正的(a)fft振幅谱a1和(b)apfft振幅谱a2峰值都不是1,2,3. 只有校正后(C)的峰值才是1,2,3.
[ 本帖最后由 zhwang554 于 2009-11-17 23:33 编辑 ]
here is my program:
N=64;
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)+;
y2_fft = fft(y222,N);
a2 = abs(y2_fft);
aa=(a1.^2)./a2*2;
r=round(f)+1;
aaa=aa(r);
it shows:
??? Error using ==> times
Matrix dimensions must agree.
Error in ==> Untitled3 at 7
y11= y1.*win1; phase1 is an array with 128 sample values in 1 sec.
回复10楼kleeblatt 的贴子
除红色外,程序对N=64;
t=-N+1:N;
A=1.0;f=17.2;phi=40;
y=A*cos(2*pi*t*f/N+phi*pi/180);%产生一个128采样数据,可换你的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)+;
y2_fft = fft(y222,N);
a2 = abs(y2_fft);
aa=(y1_amplitude.^2)./a2*2;
r=round(f)+1;
aaa=aa(r)
[ 本帖最后由 zhwang554 于 2009-11-18 01:17 编辑 ]
回复8楼kleeblatt 的贴子
喔!我明白了,不能出现解析式, 只有1秒128个采样(that i can't give the function of the input signal, just 128 discrete values in 1 sec.)N=64;
phase1=[0.7660 0.0217 -0.7374 -0.9940 -0.5732 0.2382 0.8873 0.9317 0.3412 -0.4818 -0.9765 -0.8058 -0.0859 0.6925 0.9990...
0.6247 -0.1753 -0.8558 -0.9532 -0.4010 0.4245 0.9607 0.8422 0.1498 -0.6447 -0.9998 -0.6736 0.1116 0.8208 0.9706...
0.4590 -0.3654 -0.9408 -0.8751 -0.2130 0.5942 0.9965 0.7197 -0.0475 -0.7824 -0.9841 -0.5152 0.3048 0.9171 0.9044...
0.2754 -0.5413 -0.9891 -0.7629 -0.0168 0.7407 0.9935 0.5692 -0.2430 -0.8896 -0.9299 -0.3366 0.4861 0.9776 0.8028...
0.0810 -0.6960 -0.9987 -0.6208 0.1801 0.8584 0.9516 0.3964 -0.4290 -0.9620 -0.8395 -0.1449 0.6485 0.9999 0.6699...
-0.1165 -0.8236 -0.9694 -0.4546 0.3700 0.9425 0.8727 0.2082 -0.5982 -0.9969 -0.7163 0.0525 0.7855 0.9832 0.5109...
-0.3095 -0.9190 -0.9023 -0.2706 0.5454 0.9898 0.7596 0.0118 -0.7441 -0.9929 -0.5651 0.2478 0.8918 0.9281 0.3319...
-0.4904 -0.9786 -0.7999 -0.0761 0.6996 0.9985 0.6170 -0.1850 -0.8609 -0.9501 -0.3919 0.4334 0.9634 0.8368 0.1400...
-0.6522 -1.0000 -0.6663 0.1215 0.8264 0.9682 0.4502 -0.3746];
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)+;
y2_fft = fft(y222,N);
a2 = abs(y2_fft);
aa=(y1_amplitude.^2)./a2*2;
=max(a2(1:N/2));
aaa=aa(r)
运行结果:
aaa = 1.000
[ 本帖最后由 zhwang554 于 2009-11-18 22:33 编辑 ]
回复 13楼 zhwang554 的帖子
exactly what i mean!but i can't see any difference between my program and what u made. pls point ...:@L it still shows the same problem:???
Error using ==> times
Matrix dimensions must agree.
phase1 is the signal.
let me describe a little bit about phase1.Actually phase1 should be phase1.signals.values, from 'to workspace'.To workspace's variable name is phase1, and saved format is
structure with time.Sample time is 1/127,so that i get 128 discrete values.
anything wrong with this part?
回复14-15楼kleeblatt 的贴子
本帖最后由 wdhd 于 2016-9-19 13:38 编辑13楼程序在matlab上运行没错.
phase1的128个数据由下产生
t=0:1/127:1;
A=1.0;f=17.2;phi=40;
phase1=A*cos(2*pi*t*f+phi*pi/180)
你问:"but i can't see any difference between my program and what u made. pls point ..."
no difference between your program(10楼) and what i made(13楼).
其实, 5楼,10楼,12楼,13楼都是一个程序
[ 本帖最后由 zhwang554 于 2009-11-18 22:52 编辑 ]
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