how I can analyze this magnitude of 1D signal

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ZhG
ZhG 2013-11-4
评论： ZhG 2013-11-5
This is a DFT magnitude graph of a set of 1D points. How can I analyze it? Can I say that the lowest frequency contribute most or the 0 frequency contribute most? It is not 0 frequency, isn't it? The 0 frequency is DC component before getting magnitude with abs(fft(s1)).

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Wayne King
Wayne King 2013-11-4
If the large value shown in the figure corresponds to 0 frequency, then that simply means the signal has a nonzero DC value. The 0 frequency component in the DFT is simply the sum of all elements in the input signal, or N times the mean value of the signal, where N is the number of elements in the signal.
Quite often, the zero frequency component is not of interest when doing a frequency analysis. To remove that, simply remove the mean from the signal.
For example, compare:
Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = 20+cos(2*pi*100*t)+randn(size(t));
xdft = fft(x);
xdft = xdft(1:length(x)/2+1);
plot(abs(xdft))
% remove mean
xnew = x-mean(x);
xdftnew = fft(xnew);
xdftnew = xdftnew(1:length(x)/2+1);
plot(abs(xdftnew))
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Wayne King
Wayne King 2013-11-4
Yes, I would say so, because you don't need to have the DFT to know what the zero frequency component is, sum(x) will give you that information.
ZhG
ZhG 2013-11-5
编辑：ZhG 2013-11-5
I did it as you advised,and I obtain this spectrum (the 2nd is 'detrend' on this graphy). But how can I analyze it ? I mean that how I can obtain some useful information from this graph. Thanks for any advice.

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Wayne King
Wayne King 2013-11-5
编辑：Wayne King 2013-11-5
You have a real-valued signal, so you only need 1/2 your magnitudes. I can't tell from your graph if your signal has even length or odd length. I'll assume length 52
Each frequency bin has spacing Fs/N Hz where N is the length of the signal and Fs is the sampling frequency. If you are just using normalized frequency, then the spacing is (2*pi)/N radians/sample.
n = 0:51;
x = cos((2*pi)/13*n);
xdft = fft(x);
xdft = xdft(1:length(x)/2+1);
stem(abs(xdft))
The sine wave has a frequency of (2*pi)/13 radians/sample. The spacing of the "bins" in the DFT is (2*pi)/52 radians/sample. Because the first bin xdft(1) corresponds to zero frequency, you expect the frequency of (2*pi)/13 radians/sample to fall on xdft(5), which it does.
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ZhG
ZhG 2013-11-5
Yes, I did the DFT by using 2*pi/N, where N is the length of points number.
x = x-mean(x);
N = length(x); % points number
k = 0:N-1;
w = (2*pi)/N * k;
a1 = [];
xD = 0;
for n=1:N
a1(n,:) = exp(-sqrt(-1)*w*(n))*x(n);
end
xD = -1*sum(a1); % multplied by -1, so that the result equals to fft(a1)
Then, I obtained the abs(xD), as it is shown in the graph. And then I want to obtain one or more measures of this spectrum as a representation or a property of the signal. For example, the maximum magintude in the spectrum. Is this possible?

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