I tried using 'FFT' function in MATLAB to transform the given function from time to frequency domain...it gave me the graph but on the x axis the time domain(values) remained the same and did not change into frequency domain as the graph revealed..
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I tried using 'FFT' function in MATLAB to transform the given function from time to frequency domain...it gave me the graph but on the x axis the time domain(values) remained the same and did not change into frequency domain as the graph revealed..
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Azzi Abdelmalek
2013-6-13
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The fft function has one output argument, it's the discrete fourier transform of your function. you have to provide the frequency vector to plot your output.
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Walter Roberson
2013-6-13
0 个投票
How did you do the plotting? The fft() call does not do any plotting of its own, and if you call plot() then you need to pass in the correct x values for what the data represents.
Aalya
2013-6-15
0 个投票
I am facing somehow a similar problem. However in my case, instead of having peak amplitude at the natural frequency, I am having it elsewhere. What could be wrong? I have obtained the time domain values from tout and simout parameters of simulink. My codes are below:
T=tout(2,1)-tout(1,1); %sample time
Fs=1/T; %sampling frequency
L=size(simout,1); %length of signal
t = (0:L-1)*T; % Time vector
y=simout(:,1);
figure(1)
plot(t,y)
title('time domain resonse of first mode')
xlabel('time (milliseconds)')
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2);
% Plot single-sided amplitude spectrum.
figure(2)
plot(f,2*abs(Y(1:NFFT/2)))
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
Thanks
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2013-6-13
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