About FFT of cosine function

조회 수: 103 (최근 30일)
MK kim
MK kim 2020년 8월 6일
댓글: MK kim 2020년 8월 24일
Hi, I have to calculate DFT of g(t)=cos(0.5*t). So I write the code below
>> N=128; %number of sampled time points
>> t=linspace(0,128,N); %time domain
>> Fs=1/(t(2)-t(1)); %sampling frequency
>> Fn=Fs/2; %Nyquist Frequency
>> g=cos(0.5*t); %input signal
>> G=fftshift(fft(g)/N); %Fourier transform of g
>> w=linspace(-Fn,Fn,N)*2*pi; %angular frequency domain
>> subplot(2,1,1)
>> plot(w,abs(G)); % to see amplitude
>> subplot(2,1,2)
>> plot(w,angle(G)); % to see phase
<The result of above code>
In this situation, my question is
1) In amplitude graph, I expected it has to have formation of two Dirac delta function Aδ(w-0.5)+Aδ(w+0.5), i.e., amplitude shoudn't have values at w=/0.5 and w=/-0.5. But it have values. How can I change the code to make this two delta functions?
2) In amplitude graph, I also expected the coordinates of peaks are (-0.5,0.5) and (0.5,0.5), but it is not. How can I change the code to make peaks at this coordinates??
Thank you..!!

채택된 답변

David Goodmanson
David Goodmanson 2020년 8월 7일
Hi mk,
In order to get just the two sharp peaks you need to have exactly n oscillations in the time domain. Otherwise the wave is cut off partway through an oscillation and does not represent a continuous oscillatory wave.
The code below uses a total time of 4*pi so there are 8 oscillations in the time record. (Also, cos = 1 at t = 0, and exactly n oscillations means that there can't be a repeated point cos = 1 at the end of the time record).
w0 = .5;
T = 4*pi; % total time
N = 1000;
t = linspace(0,T,N+1); % N+1 points, N intervals
t(end) = []; % eliminate repeated point at the end
y = cos(w0*t);
z = fftshift(fft(y)/N);
% fft golden rule: delta_t*delta_w = 2*pi/N
% delta_w = 2*pi/(N*delta_t) = 2*pi/T
delw = 2*pi/T;
w = (-N/2:N/2-1)*delw;
r = [real(z);imag(z)];
stem(w,[real(z);imag(z)]')
xlim([-10 10])
  댓글 수: 4
David Goodmanson
David Goodmanson 2020년 8월 20일
Hello mk,
There is nothing wrong here. The only two nonzero values in the spectrum occur at w = +-(1/2), with amplitude 1//2 and phase 0 (the amplitudes are real). If you look at the stem plot, the phase is zero at w = +-(1/2) as it should be.
For all other values of w, the amplitude is supposed to be zero, but because of usual numercal precision issues the values are actually down around 1e-16 for both the real and imaginary parts. The phase of such a calculaton varies randomly and is meaningless, so the phase plot is actually ok.
If you use sin instead of cos, then the phase at w = +-(1/2) is -+(pi/2) as you can check..
MK kim
MK kim 2020년 8월 24일
Thank you for your help..!!

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Discrete Fourier and Cosine Transforms에 대해 자세히 알아보기

태그

Community Treasure Hunt

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

Start Hunting!

Translated by