I have a signal in time domain (6000 samples from -100 to 1100 ps). I have to convert it into frequency domain and divide it into 100 channels, and find the center frequency of each channel.
I am not good in "MATLAB" so how to do that, help will be appreciated.
Based on my understanding of the question, you want to represent your signal in the frequency domain. Please study some tutorials on FFT to increase your understanding of the FFT details. A good start is this tutorial: FFT Tutorial using Matlab, which I used to write the code below.
close all;
clear all;
clc;
Fs = 5; % Sampling frequency (in THz)
Ts = 1/5; % Sampling period (in ps)
x = randn(1, 6000); % A random signal of 6000 points
t = [-100:Ts:1100-Ts]; % 6000 time points (in ps)
% Plot signal in time domain
figure;plot(t,x);
xlabel('Time (ps)'); ylabel('Signal');
N = 100; % Number of FFT points
X = fftshift(fft(x, N)); % Compute and shift FFT
absX = abs(X); % Compute absolute FFT values
% Frequency centers (frequency components) depend on the values of N and Fs
frequency_centers = Fs * [-N/2:N/2-1]/N;
% Plot signal in frequency domain
figure;plot(frequency_centers, absX);
xlabel('Frequency (THz)'); ylabel('abs FFT');
The variable frequency_centers shows the frequency components.
Since you have 6000 time samples, from -100 to 1100 ps, the sampling period is Ts = 1200/6000 = 0.2 ps and Fs = 1/Ts = 5 THz. Also, note that in order to have 6000 time samples (and not 6001), you need to drop one boundary time value (here I dropped 1100).
Related
I have a modulated signal, and now I want to perform an FFT. However, I am getting a spur at a high frequency, which should not be there (and if it should, I have no clue as to why).
Lvl=[0.5,0.9,0.5,0.5,0.1,0.1,0.9,0.5];
fa=60; %the frequency of the parasitic source in hertz
np=2; %number of periods per bit
kl=length(Lvl);
t=0:0.01*np/fa:np*kl/fa;
Sig=sin(2*pi*fa*t);
for n=1:1:101
Sig(n)=Sig(n)*Lvl(1);
end
for n=102:1:201
Sig(n)=Sig(n)*Lvl(2);
end
for n=202:1:301
Sig(n)=Sig(n)*Lvl(3);
end
for n=302:1:401
Sig(n)=Sig(n)*Lvl(4);
end
for n=402:1:501
Sig(n)=Sig(n)*Lvl(5);
end
for n=502:1:601
Sig(n)=Sig(n)*Lvl(6);
end
for n=602:1:701
Sig(n)=Sig(n)*Lvl(7);
end
for n=702:1:801
Sig(n)=Sig(n)*Lvl(8);
end
plot(t,Sig)
%FFT
y = fft(Sig);
f = (0:length(y)-1)*(1/(0.01*np/fa))/length(y);
plot(f,abs(y))
title('Magnitude')
I'm expecting just a spike at 60Hz with spurs around it, but instead I'm getting that and a large spike at almost 3kHz with spurs around it.
This peak at almost 3 kHz should be there, since the fft of a real is signal symmetric around the nyquist frequency (actually complex conjugate). The nyquist frequency is half the samping frequency, in your case sampling is done at 3000 Hz, thus the nyquist frequency is 1500 Hz. If you look closer at the peak, you will see that it is at 2940 Hz (which is 3000-60 Hz), due to the fact that the fft is mirrored around 1500 Hz.
There are plenty of sources that explain why this is a property of the Fourier transform (e.g. here).
The actual Fourier transform would be mirrored around the zero frequency, but the fft gives you the fast Fourier transform, which is mirrored around the nyquist frequency. You can use fftshift to center the spectrum around the zero frequency.
I took the liberty to shorten your code, by avoiding repetition of several for-loops, and added the fftshift. Since your signal is real, you can also choose to show only one side of the fft, but I'll leave that up to you.
Lvl=[0.5,0.9,0.5,0.5,0.1,0.1,0.9,0.5];
fa=60; % the frequency of the parasitic source in hertz
np=2; % number of periods per bit
kl = length(Lvl);
dt = 0.01*np/fa; % time step
Tend = np*kl/fa - dt; % time span
t = 0:dt:Tend; % time vector
N = length(t); % number samples
Sig=sin(2*pi*fa*t);
for n = 1:kl
ids = (1:100) + (n-1)*100;
Sig(ids) = Sig(ids).*Lvl(n);
end
% FFT
Y = fft(Sig);
fv = (0:N-1)/(N*dt); % frequency range
% FFT shift:
Y_shift = fftshift(Y);
fv_shift = (-N/2:N/2-1)/(N*dt); % zero centered frequency vector
% Plot
figure(1); clf
subplot(311)
plot(t,Sig)
title('Signal')
subplot(312)
plot(fv,abs(Y))
title('FFT Magnitude')
subplot(313)
plot(fv_shift,abs(Y_shift))
title('FFT Magnitude zero shift')
I am doing a fft (stft) on a audio-file, which contains voice and music. I took out a range of 1-200 from signal, which i saved in a variable called, for example, frequency_band1. How can i calculate the frequency bins from variable, which stores my specificially chosen signal, on which the fft was applied before.? I think, read somewhere, that i needed to calculate fs/fftpoints. In my case, 44100/4096. Do i multiply it with my variable frequency_band1*(44100/4096)? This is done on matlab, and i need to do it, with a stft.
The output values of a DFT (FFT) are at frequencies (0:n-1)/n*fs where n is DFT size and fs is sample frequency.
Here's an example. This creates a sinusoid at a known frequency, performs the DFT of the whole signal (so n is signal size), and plots the spectrum using the above as frequency values.
fs = 44100; % sample frequency
t = 0:1/fs:.1; % 0.1-second time axis
x = sin(2*pi*300*t); % sinusoid of frequency 220 Hz
X = fft(x); % DFT of x
f = (0:numel(X)-1)/numel(X)*fs; % frequency axis of DFT
plot(f, abs(X)) % plot the spectrum in absolute value
grid % add grid
axis([0 1000 -500 2500])
You can check that the spectral line of this signal is where it should, that is, at 300 (Hz).
Here in this code i am doing a stft on my wav-file. There is no problem with that. At the beginning, i am defining my parameter, afterwards using my wav file and then applying the stft. Basically what i am doing is a real-time spectral analysis. Anyway my question is, how do i a frequency band? I want my signal to be separated in LOW/MEDIUM/HIGH. I want my vector to be saved, from 0-250 Hz in the LOW-Band, 250-5000 Hz in the MEDIUM-Band, 5-22.05k Hz in the HIGH-Band. I advise you, to try my code in Matlab, if you don't understand it. Just take any wav-file. Btw my signal is plotted in the variable "Yres". Any solution is appreciated!
NFA=2; % Number is used for plotting every 2nd picture
t_seg=0.05; % Length of segment in ms
fftlen = 4096;
% Lenght of "fft",because our segment contains 2205 points
[y,fs]=audioread('UnchainMyHeart.wav');
% audioread = functions reads WAV-file
% y = A vector which contains my audio signal
% fs = sample frequency (44100)
% 'UnchainMyHeart' = WAV-file
t=linspace(0,length(y)/fs,length(y));
% linspace = Creating time vector
% 0 = Start time
% length(y)/fs = End time
% length(y) = Number of samples in y
plot(t,y)
% plotting signal in the time domain
segl =floor(t_seg*fs);
% Applying fft function on the variable "segl"
windowshift=segl/2;
% Defining the size of the window, which goes to the next "segl"
window=hann(segl);
% hann function
window=window.';
si=1;
%Start index
ei=segl;
%End index
AOS= length(y)/windowshift - 1;
% AOS is the number of "segl" we use (About 433)
f1=figure;
% Opening new window
f=0:1:fftlen-1;
f=f/(fftlen-1)*fs;
% Defining frequency vector
Ya=zeros(1,fftlen);
plot(f,Ya),axis([0 fs -90 50])
grid on
n=0;
%start variable
for m= 1:1:AOS
y_a = y(si:ei);
y_a= y_a.*window;
Ya=fft(y_a, fftlen);
n=n+1;
if n==1
Yres=abs(Ya);
else
Yres=Yres+abs(Ya);
end
if n==NFA
Yres=Yres/NFA;
n=0;
drawnow;
%Tut die Grafikobjekte immer auf den neuesten Stand updaten
figure(f1);
plot(f(1:end/2), 20*log10(abs(Yres(1:end/2))));
ylim([-90 50]);
title('Spektrum eines Audiosignal');
xlabel('f(Hz)');
ylabel('dB');
grid on;
end
si=si+windowshift;
% Updating start index
ei=ei+windowshift;
% Updating end index
end
This may not be the best answer! But this may help you get started on something. You can use spectrogram() function from MATLAB's Signal Processing Toolbox.
Let's suppose you have an audio file named ''UnchainMyHeart.wav'(in your case) with one channel. The code goes as follows:
% Reading the audio file
[y1,fs] = audioread('UnchainMyHeart.wav');
% Parameters for STFT (or spectrogram)
windowDuration = 30e-3; overlapDuration = 15e-3;
windowLength = round(windowDuration*fs); % window length
overlapLength = round(overlapDuration*fs); % overlapping of windows
nfft = 1024;
% Executing STFT for the signal
[S1,F1,T1,P1] = spectrogram(x1,hanning(windowLength), ...
overlapLength, nfft, fs, 'yaxis');
S1 and P1 contain STFT and Power Spectrum Density(PSD) of the signal for a time interval of each section with a time interval whose estimations are contained in T1.
For your question, you are looking for F1 which is a vector of cyclical frequencies expressed in terms of sampling frequency, fs. For example: if you have a sampling frequency of 48 kHz (fs) and nfft of 1024, then you will have 513 [(1024/2) +1)] frequency values spaced by (fs/nfft). i.e. 46.875. So your frequency components will be 0, 46.875, 46.875*2, ..., 46.875*512. The maximum you will have is 24 kHz due to Nyquist criterion.
Now, you can easily write a simple routine specifying the ranges as you said. The same technique can be used in your code which is an implementation of stft. I would suggest using MATLAB's built-in function unless your problem requires an implementation. Hope this helps!
If needed, I can answer why the parameters for STFT are chosen as included in the code.
I'm working on sound signals of a walking pattern, which has obvious regular patterns:
Then I thought I can get the frequency of walking (approximately 1.7Hz from the image) using FFT function:
x = walk_5; % Walking sound with a size of 711680x2 double
Fs = 48000; % sound frquency
L=length(x);
t=(1:L)/Fs; %time base
plot(t,x);
figure;
NFFT=2^nextpow2(L);
X=fft(x,NFFT);
Px=X.*conj(X)/(NFFT*L); %Power of each freq components
fVals=Fs*(0:NFFT/2-1)/NFFT;
plot(fVals,Px(1:NFFT/2),'b','LineSmoothing','on','LineWidth',1);
title('One Sided Power Spectral Density');
xlabel('Frequency (Hz)')
ylabel('PSD');
But then it doesn't give me what I expected:
FFT result:
zoom image has lots of noises:
and there is no information near 1.7Hz
Here is the graph from log domain using
semilogy(fVals,Px(1:NFFT));
It's pretty symmetric though:
I couldn't find anything wrong with my code. Do you have any solutions to easily extract the 1.7Hz from the walking pattern?
here is the link for the audio file in mat
https://www.dropbox.com/s/craof8qkz9n5dr1/walk_sound.mat?dl=0
Thank you very much!
Kai
I suggest you to forget about DFT approach since your signal is not appropriate for this type of analysis due to many reasons. Even by looking on the spectrum in range of frequencies that you are interested in, there is no easy way to estimate the peak:
Of course you could try with PSD/STFT and other funky methods, but this is an overkill. I can think of two, rather simple methods, for this task.
First one is based simply on the Auto Correlation Function.
Calculate the ACF
Define the minimum distance between them. Since you know that expected frequency is around 1.7Hz, then it corresponds to 0.58s. Let's make it 0.5s as the minimum distance.
Calculate the average distance between peaks found.
This gave me an approximate frequency of 1.72 Hz .
Second approach is based on the observation to your signal already has some peaks which are periodic. Therefore we can simply search for them using findpeaks function.
Define the minimum peak distance in a same way as before.
Define the minimum peak height. For example 10% of maximum peak.
Get the average difference.
This gave me an average frequency of 1.7 Hz.
Easy and fast method. There are obviously some things that can be improved, such as:
Refining thresholds
Finding both positive and negative peaks
Taking care of some missing peaks, i.e. due to low amplitude
Anyway that should get you started, instead of being stuck with crappy FFT and lazy semilogx.
Code snippet:
load walk_sound
fs = 48000;
dt = 1/fs;
x = walk_5(:,1);
x = x - mean(x);
N = length(x);
t = 0:dt:(N-1)*dt;
% FFT based
win = hamming(N);
X = abs(fft(x.*win));
X = 2*X(1:N/2+1)/sum(win);
X = 20*log10(X/max(abs(X)));
f = 0:fs/N:fs/2;
subplot(2,1,1)
plot(t, x)
grid on
xlabel('t [s]')
ylabel('A')
title('Time domain signal')
subplot(2,1,2)
plot(f, X)
grid on
xlabel('f [Hz]')
ylabel('A [dB]')
title('Signal Spectrum')
% Autocorrelation
[ac, lag] = xcorr(x);
min_dist = ceil(0.5*fs);
[pks, loc] = findpeaks(ac, 'MinPeakDistance', min_dist);
% Average distance/frequency
avg_dt = mean(gradient(loc))*dt;
avg_f = 1/avg_dt;
figure
plot(lag*dt, ac);
hold on
grid on
plot(lag(loc)*dt, pks, 'xr')
title(sprintf('ACF - Average frequency: %.2f Hz', avg_f))
% Simple peak finding in time domain
[pkst, loct] = findpeaks(x, 'MinPeakDistance', min_dist, ...
'MinPeakHeight', 0.1*max(x));
avg_dt2 = mean(gradient(loct))*dt;
avg_f2 = 1/avg_dt2;
figure
plot(t, x)
grid on
hold on
plot(loct*dt, pkst, 'xr')
xlabel('t [s]')
ylabel('A')
title(sprintf('Peak search in time domain - Average frequency: %.2f Hz', avg_f2))
Here's a nifty solution:
Take the absolute value of your raw data before taking the FFT. The data has a ton of high frequency noise that is drowning out whatever low frequency periodicity is present in the signal. The amplitude of the high frequency noise gets bigger every 1.7 seconds, and the increase in amplitude is visible to the eye, and periodic, but when you multiply the signal by a low frequency sine wave and sum everything you still end up with something close to zero. Taking the absolute value changes this, making those amplitude modulations periodic at low frequencies.
Try the following code comparing the FFT of the regular data with the FFT of abs(data). Note that I took a few liberties with your code, such as combining what I assume were the two stereo channels into a single mono channel.
x = (walk_5(:,1)+walk_5(:,2))/2; % Convert from sterio to mono
Fs = 48000; % sampling frquency
L=length(x); % length of sample
fVals=(0:L-1)*(Fs/L); % frequency range for FFT
walk5abs=abs(x); % Take the absolute value of the raw data
Xold=abs(fft(x)); % FFT of the data (abs in Matlab takes complex magnitude)
Xnew=abs(fft(walk5abs-mean(walk5abs))); % FFT of the absolute value of the data, with average value subtracted
figure;
plot(fVals,Xold/max(Xold),'r',fVals,Xnew/max(Xnew),'b')
axis([0 10 0 1])
legend('old method','new method')
[~,maxInd]=max(Xnew); % Index of maximum value of FFT
walkingFrequency=fVals(maxInd) % print max value
And plotting the FFT for both the old method and the new, from 0 to 10 Hz gives:
As you can see it detects a peak at about 1.686 Hz, and for this data, that's the highest peak in the FFT spectrum.
I have a question while computing the spectrum of a time series in Matlab. I have read the documentations concerning 'fft' function. However I have seen two ways of implementation and both wgive me different results. I would appreciate to have some answer about this difference:
1st Method:
nPoints=length(timeSeries);
Time specifications:
Fs = 1; % samples per second
Fs = 50;
freq = 0:nPoints-1; %Numerators of frequency series
freq = freq.*Fs./nPoints;
% Fourier Transform:
X = fft(timeSeries)/nPoints; % normalize the data
% find find nuquist frequency
cutOff = ceil(nPoints./2);
% take only the first half of the spectrum
X = abs(X(1:cutOff));
% Frequency specifications:
freq = freq(1:cutOff);
%Plot spectrum
semilogy(handles.plotLoadSeries,freq,X);
2nd Method:
NFFT = 2^nextpow2(nPoints); % Next power of 2 from length of y
Y = fft(timeSeries,NFFT)/nPoints;
f = 1/2*linspace(0,1,NFFT/2+1);
% % Plot single-sided amplitude spectrum.
% plot(handles.plotLoadSeries, f,2*abs(Y(1:NFFT/2+1)))
semilogy(handles.plotLoadSeries,f,2*abs(Y(1:NFFT/2+1)));
I thought that it is not necessary to use 'nextpow' function in 'fft' function in Matlab. Finally, which is the good one?
THanks
The short answer: you need windowing for spectrum analysis.
Now for the long answer... In the second approach, you are using an optimised FFT algorithm useful when the length of the input vector is a power of two. Let's assume that your original signal has 401 samples (as in my example below) from an infinitely long signal; nextpow2() will give you NFFT=512 samples. When you feed the shorter, 401-sample signal into the fft() function, it is implicitly zero-padded to match the requested length of 512 (NFFT). But (here comes the tricky part): zero-padding your signal is equivalent to multiplying an infinitely long signal by a rectangular function, an operation that in the frequency domain translates to a convolution with a sinc function. This would be the reason behind the increased noise floor at the bottom of your semilogarithmic plot.
A way to avoid this noise increase is to create manually the 512-sample signal you want to feed into fft(), using a smoother window function instead of the default rectangular one. Windowing means just multiplying your signal by a tapered, symmetric one. There are tons of literature on choosing a good windowing function, but a typically accurate one with low sidelobes (low noise increase) is the Hamming function, implemented in MATLAB as hamming().
Here is a figure illustrating the issue (in the frequency domain and time domain):
...and the code to generate this figure:
clear
% Create signal
fs = 40; % sampling freq.
Ts = 1/fs; % sampling period
t = 0:Ts:10; % time vector
s = sin(2*pi*3*t); % original signal
N = length(s);
% FFT (length not power of 2)
S = abs(fft(s)/N);
freq = fs*(0:N-1)/N;
% FFT (length power of 2)
N2 = 2^nextpow2(N);
S2 = abs(fft(s, N2)/N2);
freq2 = fs*(0:N2-1)/N2;
t2 = (0:N2-1)*Ts; % longer time vector
s2 = [s,zeros(1,N2-N)]; % signal that was implicitly created for this FFT
% FFT (windowing before FFT)
s3 = [s.*hamming(N).',zeros(1,N2-N)];
S3 = abs(fft(s3, N2)/N2);
% Frequency-domain plot
figure(1)
subplot(211)
cla
semilogy(freq,S);
hold on
semilogy(freq2,S2,'r');
semilogy(freq2,S3,'g');
xlabel('Frequency [Hz]')
ylabel('FFT')
grid on
legend( 'FFT[401]', 'FFT[512]', 'FFT[512] with windowing' )
% Time-domain plot
subplot(212)
cla
plot(s)
hold on
plot(s3,'g')
xlabel('Index')
ylabel('Amplitude')
grid on
legend( 'Original samples', 'Windowed samples' )