MATLAB FFT xaxis限制了混乱和移位

Question

This is the first time I'm using the fft function and I'm trying to plot the frequency spectrum of a simple cosine function:

f = cos(2*pi*300*t)

The sampling rate is 220500. I'm plotting one second of the function f.

Here is my attempt:

time = 1;
freq = 220500;
t = 0 : 1/freq : 1 - 1/freq;
N = length(t);
df = freq/(N*time);

F = fftshift(fft(cos(2*pi*300*t))/N);
faxis = -N/2 / time : df : (N/2-1) / time;

plot(faxis, real(F));
grid on;
xlim([-500, 500]);

Why do I get odd results when I increase the frequency to 900Hz? These odd results can be fixed by increasing the x-axis limits from, say, 500Hz to 1000Hz. Also, is this the correct approach? I noticed many other people didn't use fftshift(X) (but I think they only did a single sided spectrum analysis).

Thank you.

Answer 1

Here is my response as promised.

The first or your questions related to why you "get odd results when you increase the frequency to 900 Hz" is related to the Matlab's plot rescaling functionality as described by @Castilho. When you change the range of the x-axis, Matlab will try to be helpful and rescale the y-axis. If the peaks lie outside of your specified range, matlab will zoom in on the small numerical errors generated in the process. You can remedy this with the 'ylim' command if it bothers you.

However, your second, more open question "is this the correct approach?" requires a deeper discussion. Allow me to tell you how I would go about making a more flexible solution to achieve your goal of plotting a cosine wave.

You begin with the following:

time = 1;
freq = 220500;

This raises an alarm in my head immediately. Looking at the rest of the post, you appear to be interested in frequencies in the sub-kHz range. If that is the case, then this sampling rate is excessive as the Nyquist limit (sr/2) for this rate is above 100 kHz. I'm guessing you meant to use the common audio sampling rate of 22050 Hz (but I could be wrong here)?

Either way, your analysis works out numerically OK in the end. However, you are not helping yourself to understand how the FFT can be used most effectively for analysis in real-world situations.

Allow me to post how I would do this. The following script does almost exactly what your script does, but opens some potential on which we can build . .

%// These are the user parameters
durT = 1;
fs = 22050;
NFFT = durT*fs;
sigFreq = 300;

%//Calculate time axis
dt = 1/fs;
tAxis = 0:dt:(durT-dt);

%//Calculate frequency axis
df = fs/NFFT;
fAxis = 0:df:(fs-df);

%//Calculate time domain signal and convert to frequency domain
x = cos(  2*pi*sigFreq*tAxis  );
F = abs(  fft(x, NFFT)  /  NFFT  );

subplot(2,1,1);
plot(  fAxis, 2*F  )
xlim([0 2*sigFreq])
title('single sided spectrum')

subplot(2,1,2);
plot(  fAxis-fs/2, fftshift(F)  )
xlim([-2*sigFreq 2*sigFreq])
title('whole fft-shifted spectrum')

You calculate a time axis and calculate your number of FFT points from the length of the time axis. This is very odd. The problem with this approach, is that the frequency resolution of the fft changes as you change the duration of your input signal, because N is dependent on your "time" variable. The matlab fft command will use an FFT size that matches the size of the input signal.

In my example, I calculate the frequency axis directly from the NFFT. This is somewhat irrelevant in the context of the above example, as I set the NFFT to equal the number of samples in the signal. However, using this format helps to demystify your thinking and it becomes very important in my next example.

** SIDE NOTE: You use real(F) in your example. Unless you have a very good reason to only be extracting the real part of the FFT result, then it is much more common to extract the magnitude of the FFT using abs(F). This is the equivalent of sqrt(real(F).^2 + imag(F).^2).**

Most of the time you will want to use a shorter NFFT. This might be because you are perhaps running the analysis in a real time system, or because you want to average the result of many FFTs together to get an idea of the average spectrum for a time varying signal, or because you want to compare spectra of signals that have different duration without wasting information. Just using the fft command with a value of NFFT < the number of elements in your signal will result in an fft calculated from the last NFFT points of the signal. This is a bit wasteful.

The following example is much more relevant to useful application. It shows how you would split a signal into blocks and then process each block and average the result:

%//These are the user parameters
durT = 1;
fs = 22050;
NFFT = 2048;
sigFreq = 300;

%//Calculate time axis
dt = 1/fs;
tAxis = dt:dt:(durT-dt);

%//Calculate frequency axis
df = fs/NFFT;
fAxis = 0:df:(fs-df);

%//Calculate time domain signal 
x = cos(  2*pi*sigFreq*tAxis  );

%//Buffer it and window
win = hamming(NFFT);%//chose window type based on your application
x = buffer(x, NFFT, NFFT/2); %// 50% overlap between frames in this instance
x = x(:, 2:end-1); %//optional step to remove zero padded frames
x = (  x' * diag(win)  )'; %//efficiently window each frame using matrix algebra

%// Calculate mean FFT
F = abs(  fft(x, NFFT)  /  sum(win)  );
F = mean(F,2);

subplot(2,1,1);
plot(  fAxis, 2*F  )
xlim([0 2*sigFreq])
title('single sided spectrum')

subplot(2,1,2);
plot(  fAxis-fs/2, fftshift(F)  )
xlim([-2*sigFreq 2*sigFreq])
title('whole fft-shifted spectrum')

I use a hamming window in the above example. The window that you choose should suit the application http://en.wikipedia.org/wiki/Window_function

The overlap amount that you choose will depend somewhat on the type of window you use. In the above example, the Hamming window weights the samples in each buffer towards zero away from the centre of each frame. In order to use all of the information in the input signal, it is important to use some overlap. However, if you just use a plain rectangular window, the overlap becomes pointless as all samples are weighted equally. The more overlap you use, the more processing is required to calculate the mean spectrum.

Hope this helps your understanding.

Answer 2

Your result is perfectly right. Your frequency axis calculation is also right. The problem lies on the y axis scale. When you use the function xlims, matlab automatically recalculates the y scale so that you can see "meaningful" data. When the cosine peaks lie outside the limit you chose (when f>500Hz), there are no peaks to show, so the scale is calculated based on some veeeery small noise (here at my computer, with matlab 2011a, the y scale was 10-16).

Changing the limit is indeed the correct approach, because if you don't change it you can't see the peaks on the frequency spectrum.

One thing I noticed, however. Is there a reason for you to plot the real part of the transform? Usually, it is abs(F) that gets plotted, and not the real part.

edit: Actually, you're frequency axis is only right because df, in this case, is 1. The faxis line is right, but the df calculation isn't.

The FFT calculates N points from -Fs/2 to Fs/2. So N points over a range of Fs yields a df of Fs/N. As N/time = Fs => time = N/Fs. Substituting that on the expression of df you used: your_df = Fs/N*(N/Fs) = (Fs/N)^2. As Fs/N = 1, the final result was right :P