清空筛选条件
清空筛选条件

FFT returns a different amplitude value, why?

1 次查看（过去 30 天）
André Luiz Regis Monteiro
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.

请先登录，再回答此问题。

回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Fourier Analysis and Filtering 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by