initial phase angle calculation
3 ビュー (過去 30 日間)
古いコメントを表示
hello
I have this signal
clear all;clc;
t=0:0.0001:10;
tt=t(1:100000);%take 100 000 samples
a=30*pi/180; %phase (calculated form deg. to rad.)
b=90*pi/180; %phase (calculated form deg. to rad.)
ia=5*cos(2*pi*10*tt-a)+7*cos(2*pi*50*tt-b);
fs=1/0.0001; %sampling frequency
X=fft(ia); %FFT
df=fs/length(X); %frequency resolution
f=(0:1:length(X)/2)*df; %frequency axis
subplot(2,1,1);
M=abs(X)/length(ia)*2; %amplitude spectrum
plot(f,M(1:length(f)));
subplot(2,1,2);
P=angle(X)*180/pi; %phase spectrum (in deg.)
plot(f,P(1:length(f)));
this technique do not give a good results; I want any technique to get the right "a" an "d" the initial phase angle of 10 HZ and 50 HZ;
I know it is a hard question but please help me
採用された回答
Jeremy
2015 年 1 月 23 日
I ran your code it and seems to be working as-is. -30 and -90 deg at 10 and 50Hz
その他の回答 (1 件)
Jeremy
2015 年 1 月 23 日
You can't expect them to be zero degrees just because there is no signal defined there. With no content at those frequencies, the angle shown will a result of spectral leakage and numerical noise. Zero degrees is no more valid than any other angle and you should always be looking a the PSD or something else to see which frequencies have a significant response.
3 件のコメント
Jeremy
2015 年 1 月 27 日
This is the correct method, any other method will be a round about way of doing the same thing.
参考
カテゴリ
Help Center および File Exchange で Fourier Analysis and Filtering についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!