I have a piece of code that gets the FFT of a part of the signal and I'm now trying to get the PSD,
Fs = 44100;
cj = sqrt(-1);
%T=.6;
dt = 1/Fs;
left=test(:,1);
right=test(:,2);
time = 45;
interval =.636;
w_range = time*Fs: (time+interval)*Fs-1;
I = left(w_range);
Q = right(w_range);
n = interval * Fs;
f = -Fs/2:Fs/n:Fs/2-Fs/n;
s = I+cj.*Q;
% Smooth the signal ss = smooth(s,201);
sf = (fftshift(fft(ss(1:n)))); %FFT of signal
figure(1) plot(f,((20*log10((abs(sf))./max(abs(sf))))))
From my understanding, in order to get the PSD I just need to raise sf to the power of 2, or is there anything more I need to perform?
Technically yes, you can obtain the power-spectral density (PSD) of a periodic signal by taking the squared-magnitude of its FFT. Note that if you are going to plot it on a logarithmic decibel scale, there is really no difference between 20*log10(abs(sf)) or 10*log10(abs(sf).^2).
There is however generally more to it in the sense that the PSD estimate computed in this way tends to have a fairly large variance. There are a number of techniques which can be used to improve the estimate. A simple one consists of applying a window to sections of data, perform the FFT, then averaging the resulting PSDs (i.e. averaging the squared-magnitudes).
You are perfectly right. You just have to built the square of the absolute values.
Related
I have a question about the zero-padding for the fft. I ran fft with zero-padding & without zero-padding and compared.
sf = 100; %sampling frequency
dt=1/sf; %time sampling interval
L = 10; %Length of signal
t = linspace(0,L,L/dt+1);
%zero-padding
nfft = 2^nextpow2(length(t)); %expansion of the data length for fft
t(length(t)+1:nfft) = 0; L = dt*nfft; t = linspace(0,L,L/dt+1);
t(end)=[];
fr = 4; %frequency
data = cos(2*pi*fr1*t);
df = sf/length(data); %frequency increment
f = (0:length(data)/2)*df;
fft_result =fft(data)/length(data);
spec_fft = 2*abs(fft_result); %spectrum
pha_fft = angle(fft_result); %phase
pha_fft = rad2deg(pha_fft);
subplot(2,1,1);
stem(f,spec_fft(1:length(f)));
subplot(2,1,2);
stem(f,pha_fft(1:length(f)));
And I could see the difference between the two result images.
When I did fft without zero-padding, the amplitude was displayed clear, and the phase also clear (I think other frequencies' phase is due to a very small amplitude value not zero). But When I used zero-padding, I could see amplitudes of nearby frequency that I input(4Hz) show different aspects and the result of the phase is strange in my opinion. Is there some problem in my code when I used zero-padding?
*Additional question for comments of Cris Luengo
I tried to pad zeros to the data by the extended length.
nfft = 2^nextpow2(length(t)); %expansion of the data length for fft
data(length(t)+1:nfft) = 0;
When I plot the data, I got
As you can see, values over 10 are zeros.
And I got this result.
I wonder whether my results are okay or not.
If the FFT magnitude result is close to zero, it might just be numerical noise (random quantization and rounding errors). The phase of this numerical noise is nonsense. I usually set the phase to zero unless the corresponding magnitude is above some threshold for actual non-zero spectrum.
I'm studying the IFFT in Matlab by applying it to a Gaussian. According to Wikipedia tables, the Fourier transform pair would be
F(w) = sqrt(pi/a) * exp(-w^2/(4a))
in frequency, and
f(t) = exp(-at^2)
in time. I modified the code in a previous question plus Cris Luengo's answer to perform this IFFT.
a = 0.333;
ts = 1e4; % time sampling
L = 1000*ts; % no. sample points
ds = 1/ts;
f = -floor(L/2):floor((L-1)/2); % freq vector
f = f/ts;
w = 2*pi*f; % angular freq
Y = sqrt(pi/a)*exp(-w.^2/(4*a));
y = ts*ifftshift(ifft(fftshift(Y)));
t = (-L/2:L/2-1)*ts/L; % time vector
f = exp(-a*t.^2); % analytical solution
figure; subplot(1,2,1); hold on
plot(t,real(y),'.--')
plot(t,real(f),'-')
xlabel('time, t')
title('real')
legend('numerical','analytic')
xlim([-5,5])
subplot(1,2,2); hold on
plot(w,imag(y),'.--')
plot(w,imag(f),'-')
xlabel('time, t')
title('imag')
legend('numerical','analytic')
xlim([-5,5])
When I compare the result of IFFT with the analytical expression, they don't seem to agree:
I'm not sure where the mistake is. Have I scaled the IFFT properly? Is there an error in how I define the linear/angular frequency?
Edit: For some reason, when I define L=ts^2 the analytical and numerical solutions seem to agree (L = no. sampling points, ts = time sample).
Starting from the analytical solution, let's rephrase things a bit. You have a sampling of the function f(t) = exp(-a*t^2), and the way you've constructed the analytical answer you're collecting L=1000*ts=1e7 samples at a sampling rate of Ts=ts/L=1e-3. This means that your sampling frequency is Fs=1/Ts=1e3.
Since you want to compare against results obtained with fft/ifft, you should be considering digital or discrete frequencies, meaning the values you define for your transform will correspond to the digital frequencies
frd = (-L/2:L/2-1)/L;
Mapping this to angular frequencies, we have:
w = 2*pi*frd;
But when you're trying to compute the values, you also need to keep in mind that these frequencies should represent samples of the continuous time spectrum you're expecting. So you scale these values by your sampling frequency:
Y = sqrt(pi/a)*exp(-(Fs*w).^2/(4*a));
y = Fs*ifftshift(ifft(fftshift(Y)));
When you compare the analytical and computed answers, they now match.
The short answer to your question, given this, is that you are scaling y incorrectly at the end. You're scaling it by ts, which is 1e4, but you need to be scaling it by the sampling frequency which is Fs=1e3. That's why you end up off by a factor of 10.
I have a series of 2D measurements (time on x-axis) that plot to a non-smooth (but pretty good) sawtooth wave. In an ideal world the data points would form a perfect sawtooth wave (with partial amplitude data points at either end). Is there a way of calculating the (average) period of the wave, using OCTAVE/MATLAB? I tried using the formula for a sawtooth from Wikipedia (Sawtooth_wave):
P = mean(time.*pi./acot(tan(y./4))), -pi < y < +pi
also tried:
P = mean(abs(time.*pi./acot(tan(y./4))))
but it didn't work, or at least it gave me an answer I know is out.
An example of the plotted data:
I've also tried the following method - should work - but it's NOT giving me what I know is close to the right answer. Probably something simple and wrong with my code. What?
slopes = diff(y)./diff(x); % form vector of slopes for each two adjacent points
for n = 1:length(diff(y)) % delete slope of any two points that form the 'cliff'
if abs(diff(y(n,1))) > pi
slopes(n,:) = [];
end
end
P = median((2*pi)./slopes); % Amplitude is 2*pi
Old post, but thought I'd offer my two-cent's worth. I think there are two reasonable ways to do this:
Perform a Fourier transform and calculate the fundamental
Do a curve-fitting of the phase, period, amplitude, and offset to an ideal square-wave.
Given curve-fitting will likely be difficult because of discontinuities in saw-wave, so I'd recommend Fourier transform. Self-contained example below:
f_s = 10; # Sampling freq. in Hz
record_length = 1000; # length of recording in sec.
% Create noisy saw-tooth wave, with known period and phase
saw_period = 50;
saw_phase = 10;
t = (1/f_s):(1/f_s):record_length;
saw_function = #(t) mod((t-saw_phase)*(2*pi/saw_period), 2*pi) - pi;
noise_lvl = 2.0;
saw_wave = saw_function(t) + noise_lvl*randn(size(t));
num_tsteps = length(t);
% Plot time-series data
figure();
plot(t, saw_wave, '*r', t, saw_function(t));
xlabel('Time [s]');
ylabel('Measurement');
legend('measurements', 'ideal');
% Perform fast-Fourier transform (and plot it)
dft = fft(saw_wave);
freq = 0:(f_s/length(saw_wave)):(f_s/2);
dft = dft(1:(length(saw_wave)/2+1));
figure();
plot(freq, abs(dft));
xlabel('Freqency [Hz]');
ylabel('FFT of Measurement');
% Estimate fundamental frequency:
[~, idx] = max(abs(dft));
peak_f = abs(freq(idx));
peak_period = 1/peak_f;
disp(strcat('Estimated period [s]: ', num2str(peak_period)))
Which outputs a couple of graphs, and also the estimated period of the saw-tooth wave. You can play around with the amount of noise and see that it correctly gets a period of 50 seconds till very high levels of noise.
Estimated period [s]: 50
So, I've been given three different MATLAB (I'm using MATLAB R2014b) files with signals containing noise. I simply just plotted the values I was given for the first part. For example, the plot of the first signal looks like the one below.
Then, I did the Fourier Transform of the signal and plotted those values as well to determine where the noise and signal occur in the frequency spectrum. To show this, I added the plot image of the first signal below.
Finally, I am supposed to create a filter using the basic MATLAB commands and filter the noise out of the plot of the signal and then do the Fourier Transform of the signal again and plot the results. The filter portion will look something like this...
b = fir1(n,w,'type');
freqz(b,1,512);
in = filter(b,1,in);
Where n is the order of the filter, w is the cutoff frequency (cutoff frequency divided by half the sampling rate), and 'type' is something to the effect of low/high/stop/etc... So, my question is how do I figure out what the n, w, and type values of the filter I am creating are supposed to be?! Thanks in advance for any help!
I believe your high frequency components are noise, but it actually depends on your data.
See this example,
Fs = 2000;
L = 200;
t = (0 : L - 1)/Fs;
data = chirp(t,20,.05,50) + chirp(t,500,.1,700);
subplot(411)
plot(t,data,'LineWidth',2);
title('Original Data')
N = 2^nextpow2(L);
y = fft(data,N)/L;
f = Fs/2 * linspace(0,1,N/2+1);
subplot(412)
plot(f,abs(y(1:N/2+1)))
title('Spectrum of Original Data')
b = fir1(40,2*[1 200]/Fs);
newd = filter(b,1,data);
subplot(413)
plot(t,newd)
title('Filtered Data')
newy = fft(newd,N)/L;
subplot(414)
plot(f,abs(newy(1:N/2+1)))
title('Spectrum of Filtered Data')
You can use b = fir1(40,2*[200 800]/Fs); for high-pass filter.
If the second plot is correct, in the x-axis, I can assume:
A. The sampling frequency is 2000 Hz.
B. The "noise" is in the low frequencies. It seems also from the original signal, that you need to filter the low-frequency baseline.
If so, you need highpass filter, so 'type'='high'.
The order is depend in the sharpness that you want to the filter. from the plots it seems that you can use 'n'=12 or 20.
The cutoff frequency suppose to be about 0.1 if the peak in the low frequencies is indeed the noise that you want to filter, and if the 1000Hz x-axis is indeed the Nyquist frequency.
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.
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.
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