フィルターのクリア

FFT returns a different amplitude value, why?

1 回表示 (過去 30 日間)
André Luiz Regis Monteiro
André Luiz Regis Monteiro 2014 年 5 月 23 日
Dear Colleagues,
I am working in this code (attached):
t = 0:0.05:4*pi; x=2+1.5*cos((2*pi/360)*60*t+(2*p)i/360)*90) N1=length(t) n1=log2(N1) n1=ceil(n1) vet = 2^n1 %multiple of 2^n;
b = N1+1; while b~=vet+1 t(b)= (t(b-1)-t(b-2))+t(b-1) %complementing the time vector to the lentgh vet; x(b)= 0 %comlementing the vector x with zeros until lenght vet; b=b+1 end
sample_rate=1/(abs(t(2)-t(1))) f=sample_rate*((0:vet/2))/vet %vector frequency associated; n=length(f) X=fft(x) fftX=X*1/(vet) abs_fftX=abs(fftX(1:n)) figure(1) stem(f,abs_fftX)
When I compute fft and plot (stem) the result, appears 1.94 in the graphic origin (phase is zero) instead of 2. It needs to be 2 (phase is zero), because it is a DC signal from x= 2 +1.5*cos((2*pi/360)*60*t+(2*p)i/360)*90). So, maybe in the others frequencies could be wrong too. Could somebody teach me how to fix it? Thanks a lot.

回答 (0 件)

カテゴリ

Help Center および File ExchangeFourier Analysis and Filtering についてさらに検索

製品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by