Two-side spectrum in Matlab

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Dear my colleagues, I am a new user of Matlab. Now, I have a signal like x(t) = A1*sin(2*pi*f1*t) + A2*sin(-2*pi*f2*t), where f1, f2>0. Could anyone please show me whether I could obtain the spectrum of x in which frequency f2 is present only as a negative frequency of two-side spectrum (normally, f2 will appear in both sides of Matlab's FFT)?
Thank you very much.
Lomath.

采纳的回答

Wayne King
Wayne King 2013-11-3
编辑:Wayne King 2013-11-3
Youssef is correct. Why do you think that f2 will only show up as a "negative" frequency.
sin() is an odd function so sin(-t) = -sin(t)
Accordingly, your signal is equal to
Fs=40;
Ts=1/Fs;
t=0:Ts:10-Ts;
f1=4; f2=15;
y=sin(2*pi*t*f1)-sin(2*pi*t*f2);
If you really want components only occurring on one side, your signal must be complex-valued. Here is an example with complex exponentials.
Fs = 40;
Ts=1/Fs;
t=0:Ts:10-Ts;
f1 = 4; f2 = -15;
y = exp(1i*2*pi*f1*t)+exp(1i*2*pi*f2*t);
ydft = fftshift(fft(y));
df = Fs/length(y);
freqvec = -Fs/2+df:df:Fs/2;
plot(freqvec,abs(ydft))
xlabel('Hz');
  2 个评论
Hoang
Hoang 2013-11-3
Thanks. Sorry for the poor example. It is great if you could give me another example, pls. In my knowledge, function x = A1*cos(2*pi*f1*t)+ A2*cos(2*pi*f2*t), where f1>0, f2<0, can be used to express a rotating signal where the first term rotates in forward direction and the other rotates in backward direction. Can it be analyzed in some ways to separate f1 and f2, pls?

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更多回答(1 个)

Youssef  Khmou
Youssef Khmou 2013-11-3
You can start by many tutorials, each offers a different approach to the DFT problem, in terms of resolution and amplitude estimation, i wrote a 2 sided example for you , verify it :
Fs=40;
Ts=1/Fs;
t=0:Ts:10-Ts;
f1=4;f2=15;
y=sin(2*pi*t*f1)+sin(2*pi*t*f2);
N=length(y);
F=fft(y);
fr=(-N/2:N/2-1)*Fs/(N);
figure, plot(fr,abs(F))
xlabel(' frequency Hz')
  4 个评论
Hoang
Hoang 2013-11-3
Thank you for your explanation.
Renan Ribeiro
Renan Ribeiro 2018-10-15
Sorry to revive the post, but the given code did not work with me (maybe I am messing something). The code given in this post worked just fine.

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