How to make the frequency vector to analyze a complex signal?

7 次查看（过去 30 天）
Muhamed Sewidan
Muhamed Sewidan 2020-12-24
评论： Muhamed Sewidan 2020-12-25
I have this example and I'm puzzled with the frequency vector (hz) if I change any parameter in it, the result won't be the frequency components. How to calculate the frequency vector, generally?
srate = 1000;
time = -2: 1/srate : 2;
pnts = length(time);
hz = linspace(0,srate/2,(pnts-1)/2);
freq = [2 10 25 40];
amp = [3 5 7 9];
phas = [-pi/4 pi/2 pi 2*pi];
signal = zeros(size(time));
for i = 1:length(freq)
signal = signal + amp(i)*sin( 2*pi*freq(i)*time + phas(i));
end
subplot(211)
plot(time,signal,'k-','linew',0.5)
title("Time Domain")
y = fft(signal)/length(signal);
pwr = abs( y ).^2;
pwr(1) = [];
subplot(212)
plot(hz,pwr(1:length(hz)),'r-')
title("Frequency Domain")
  4 个评论
显示 2更早的评论
Michael Soskind
Michael Soskind 2020-12-24
I see what you mean, you are referring to the small errors in frequency value compared to the ideal frequency values. If you increase the sample rate, you will find that the approximaton of the DFT should improve. If I increase srate to srate = 10000, I get the pairs:
freq: 2, amp: 3
freq: 10, amp: 5
freq: 25, amp: 6.99
freq: 40, amp: 9
With respect to the second question, pwr = 2*abs(y) should be used since this is really the folding of the FFT on itself in a single sided Fourier Transform. In order to get the correct signal amplitude as you had built in your composite signal, you should just double the magnitude, which will get you the correct signal amplitudes.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Cris LaPierre
Cris LaPierre 2020-12-24
编辑：Cris LaPierre 2020-12-24
I think I would explain it this way. Due to Nyquist theory, the highest frequency you can detect is half your sample rate (must have at least 2 points in one period). The lowest is obviously 0. The assumption here is that the detectable frequencies increase linearly from 0 to srate/2.
If you inspect the peaks in your plot, you'll notice this approach is an approximation, as the peaks are about 1 point off.
  5 个评论
显示 3更早的评论
Cris LaPierre
Cris LaPierre 2020-12-25
Ran in:
Not without having to change some other things. The number of points is set by your sample rate. Changing it directly impacts the results, as you've observed. To get the frequencies to line up, you'd have to half your sample rate, which halves the frequencies you can resolve in the signal.
Better to zoom in on your axis if you want to focus on the peaks. Use xlim.
srate = 1000;
time = -2: 1/srate : 2;
pnts = length(time);
hz = linspace(0,srate/2,(pnts-1)/2);
freq = [2 10 25 40];
amp = [3 5 7 9];
phas = [-pi/4 pi/2 pi 2*pi];
signal = zeros(size(time));
for i = 1:length(freq)
signal = signal + amp(i)*sin( 2*pi*freq(i)*time + phas(i));
end
y = fft(signal)/length(signal);
pwr = abs( y ).^2;
pwr(1) = [];
plot(hz,pwr(1:length(hz)),'r-')
title("Frequency Domain")
xlim([0 250])

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Data Preprocessing 的更多信息

标签

产品


版本

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by