I have an accelerometer 3 axis.
As far as we know that acceleration is the sum of static acceleration (gravity) and dynamic acceleration.
my goal's to extract a gravity acceleration which will show me the direction of the device.
i will apply a butterworth filter to extract a gravity acceleration. but i have a problem in choosing cut-off frequency and a filter order.
T = 0.16 s ; %Time of sample rate so Fs = 1/0.16 ; % sampling rate? is this correct?
After reading a few articles, i found that the cut-off varie between 0.1 to 0.5 , here i will choose a 0.5 (because as i don't know they are based on their choice.
this is the program which i will execute in Matlab to extract gravity acceleration from the 3 axis.
Fc = 0.5 ; %cut-off frequency
Fs = 6.26 hz (1/0.16) ; % sampling rate order = 4;
[b,a] = butter(order,fc(fs/2),'low');
x = filter (b,a,x0);
y = filter(b,a,y0);
z = filter(b,a,z0);
What you are doing is, you "slow" the measurements down with a butterworth filter. Thus, in practice you try to get rid of the "fast" part. What this means in frequency domain is: you want a lowpass (low frequency = slow signal goes through, while high = fast gets filtered out). So I guess with what you know already you should be able to find a reasonable value.
Generally for estimating the angles you have gyroscopes at hand. Then you would rather go for kalman filtering, since this doesn't add as much of a delay to your measurements (if it is time dependent).
As already commented, keep in mind, that you are dealing with sampled data (nyquist freq. is a natural restriction).
Cutoff frequency depends upon the interested range of signal you want to extract from a noisy signal. Assume: You use your mobile phone inside the trouser pocket and collect raw data during activities. Then your cutoff frequency is directly related to the rate of human activity( Eg: Running, walking ,jogging etc).
Note: As the filter order N increases, the actual frequency responses approach the ideal.
Filter order will also be a factor of the signal you want to extract to analyse.
Refer this publication for more info:
A gyroscopic data based pedometer algorithm by Jayalath.S
Link:
http://ieeexplore.ieee.org/xpl/articleDetails.jsp?tp=&arnumber=6553971&queryText%3DA+gyroscopic+data+based+pedometer+algorithm
Related
I'm trying to remove the high frequency noise from the following file.
It's a file of a woman reading the news, with a high pitched noise playing loudly over it. Towards the end of the file, someone else begins to speak, but in a different language.
I want to filter out this high pitched noise, and be able to clearly hear the woman reading the news. Looking at the frequency domain:
I have tried using a low pass filter, and band stop filter. The bandstop filter produces a signal that no longer has the high pitch ringing, but the audio isn't very clear and it's hard to make out what is being said - the same goes for the low pass filter. I surmise that this is due to me filtering out not only the noise, but the harmonics of the speech as well. It was also necessary that I amplify the audio signal after I filtered it, because it was quieter than before.
Is there some clever way for me to reconstruct the harmonics of the speech in order to hear what is being said more clearly? Or is there a clever way for me to filter the signal without losing too much audio clarity?
I can include any code I used in matlab if needed.
Note:
I shifted the signal to 0 in the image I linked
I did use filtfilt() instead of filter()
I used butter() for the filters
Given the fairly dynamic nature of the interference in your sample, stationary filters are not going to yield very satisfying results. To improve performance, you would need to dynamically adjust the filtering parameters based on estimates of the interference.
Fortunately in this case the interference is pretty strong and exhibits a fairly regular pattern which makes it easier to estimate. This can be seen from the signal's spectrogram.
For the following derivations we will be assuming the samples of the wavfile has been stored in the array x and that the sampling rate is fs (which is 8000Hz in the provided sample).
[Sx,f,t] = spectrogram(x, triang(1024), 1023, [], fs, 'onesided');
Given that the interference is strong, obtaining the frequency of the interference can be done by locating the peak frequency in each time slice:
frequency = zeros(size(Sx,2),1);
for k = 1:size(Sx,2)
[pks,loc] = findpeaks(Sx(:,k));
frequency(k) = fs * (loc(1)-1);
end
Seeing that the interference is periodic we can use the Discrete Fourier Transform to decompose this signal:
M = 32*fs;
Ff = fft(frequency, M);
plot(fs*[0:M-1]/M, 20*log10(abs(Ff));
axis(0, 2);
xlabel('frequency (Hz)');
ylabel('amplitude (dB)');
Using the first two harmonics as an approximation, we can model the frequency of the interference signal as:
T = 1.0/fs
t = [0:length(x)-1]*T
freq = 750.0127340203496
+ 249.99913423501602*cos(2*pi*0.25*t - 1.5702946346796276)
+ 250.23974282864816*cos(2*pi*0.5 *t - 1.5701043282285363);
At this point we would have enough to create a narrowband filter with a center frequency (which would change dynamically as we keep updating the filter coefficients) given by that frequency model. Note however that constantly recomputing and updating the filter coefficient is a fairly expensive process and given that the interference is strong, it is possible to do better by locking on to the interference phase. This can be done by correlating small blocks of the original signal with sine and cosine at desired frequency. We can then slightly tweak the phase to align the sine/cosine with the original signal.
% Compute the phase of the sine/cosine to correlate the signal with
delta_phi = 2*pi*freq/fs;
phi = cumsum(delta_phi);
% We scale the phase adjustments with a triangular window to try to reduce
% phase discontinuities. I've chosen a window of ~200 samples somewhat arbitrarily,
% but it is large enough to cover 8 cycles of the interference around its lowest
% frequency sections (so we can get a better estimate by averaging out other signal
% contributions over multiple interference cycles), and is small enough to capture
% local phase variations.
step = 50;
L = 200;
win = triang(L);
win = win/sum(win);
for i = 0:floor((length(x)-(L-step))/step)
% The phase tweak to align the sine/cosine isn't linear, so we run a few
% iterations to let it converge to a phase locked to the original signal
for iter = 0:1
xseg = x[(i*step+1):(i*step+L+1)];
phiseg = phi[(i*step+1):(i*step+L+1)];
r1 = sum(xseg .* cos(phiseg));
r2 = sum(xseg .* sin(phiseg));
theta = atan2(r2, r1);
delta_phi[(i*step+1):(i*step+L+1)] = delta_phi[(i*step+1):(i*step+L+1)] - theta*win;
phi = cumsum(delta_phi);
end
end
Finally, we need to estimate the amplitude of the interference. Here we choose to perform the estimation over the initial 0.15 seconds where there is a little pause before the speech starts so that the estimation is not biased by the speech's amplitude:
tmax = 0.15;
nmax = floor(tmax * fs);
amp = sqrt(2*mean(x[1:nmax].^2));
% this should give us amp ~ 0.250996990794946
These parameters then allow us to fairly precisely reconstruct the interference, and correspondingly remove the interference from the original signal by direct subtraction:
y = amp * cos(phi)
x = x-y
Listening to the resulting output, you may notice a remaining faint whooshing noise, but nothing compared to the original interference. Obviously this is a fairly ideal case where the parameters of the interference are so easy to estimate that the results almost look too good to be true. You may not get the same performance with more random interference patterns.
Note: the python script used for this processing (and the corresponding .wav file output) can be found here.
I have implemented a pulse oximeter on FPGA and I am collecting the data at sampling rate = 115200. So when I plot the data in Matlab I get the following waveform:
I would like to set up a simple low pass filter to filter out the noise.So the filtered data looks something like this:
How do I design this filter? How do I determine the cutoff frequency for butterworth filter ([b,a] = butter(n,Wn))?
W1=0.5/60; % lower cutoff
W2=4/60; %higher cutoff
[b,a]=butter(2,[W1,W2]); % Bandpass digital filter design
h = fvtool(b,a); % Visualize filter
low_pass_data1 = filtfilt(b,a,data); % applying filter to our data
The above mentioned algorithm works and I got it from the internet but I don't know how it works? There is no mention of the sampling rate in the above algorithm and yet it works! I don't even know if it's right.
Since you are designing a digital filter, the sampling rate is irrelevant. A digital filter takes a stream of values and outputs a stream of values, it has no chance of knowing at what rate they were taken, nor would it matter.
Therefore, frequencies in digital filters are always relative. Actually, the help of butter explains it:
The cutoff frequency Wn must be 0.0 < Wn < 1.0, with 1.0 corresponding to half the sample rate.
Therefore, if your natural sampling rate is 115200 Hz and you want to cut off at, I don't know, 10 kHz, then the frequency to pass to butter is 10000/(115200/2) = 0.1736.
In case you wonder why it is half the sampling rate: with a sampling rate of Fs, we can see the frequency content in [-Fs/2,Fs/2] (due to Shannon-Nyquist), therefore half the sampling rate is the highest frequency that makes sense.
I was studying for a signals & systems project and I have come across this code on high and low pass filters for an audio signal on the internet. Now I have tested this code and it works but I really don't understand how it is doing the low/high pass action.
The logic is that a sound is read into MATLAB by using the audioread or wavread function and the audio is stored as an nx2 matrix. The n depends on the sampling rate and the 2 columns are due to the 2 sterio channels.
Now here is the code for the low pass;
[hootie,fs]=wavread('hootie.wav'); % loads Hootie
out=hootie;
for n=2:length(hootie)
out(n,1)=.9*out(n-1,1)+hootie(n,1); % left
out(n,2)=.9*out(n-1,2)+hootie(n,2); % right
end
And this is for the high pass;
out=hootie;
for n=2:length(hootie)
out(n,1)=hootie(n,1)-hootie(n-1,1); % left
out(n,2)=hootie(n,2)-hootie(n-1,2); % right
end
I would really like to know how this produces the filtering effect since this is making no sense to me yet it works. Also shouldn't there be any cutoff points in these filters ?
The frequency response for a filter can be roughly estimated using a pole-zero plot. How this works can be found on the internet, for example in this link. The filter can be for example be a so called Finite Impulse Response (FIR) filter, or an Infinite Impulse Response (IIR) filter. The FIR-filters properties is determined only from the input signal (no feedback, open loop), while the IIR-filter uses the previous signal output to control the current signal output (feedback loop or closed loop). The general equation can be written like,
a_0*y(n)+a_1*y(n-1)+... = b_0*x(n)+ b_1*x(n-1)+...
Applying the discrete fourier transform you may define a filter H(z) = X(z)/Y(Z) using the fact that it is possible to define a filter H(z) so that Y(Z)=H(Z)*X(Z). Note that I skip a lot of steps here to cut down this text to proper length.
The point of the discussion is that these discrete poles can be mapped in a pole-zero plot. The pole-zero plot for digital filters plots the poles and zeros in a diagram where the normalized frequencies, relative to the sampling frequencies are illustrated by the unit circle, where fs/2 is located at 180 degrees( eg. a frequency fs/8 will be defined as the polar coordinate (r, phi)=(1,pi/4) ). The "zeros" are then the nominator polynom A(z) and the poles are defined by the denominator polynom B(z). A frequency close to a zero will have an attenuation at that frequency. A frequency close to a pole will instead have a high amplifictation at that frequency instead. Further, frequencies far from a pole is attenuated and frequencies far from a zero is amplified.
For your highpass filter you have a polynom,
y(n)=x(n)-x(n-1),
for each channel. This is transformed and it is possble to create a filter,
H(z) = 1 - z^(-1)
For your lowpass filter the equation instead looks like this,
y(n) - y(n-1) = x(n),
which becomes the filter
H(z) = 1/( 1-0.9*z^(-1) ).
Placing these filters in the pole-zero plot you will have the zero in the highpass filter on the positive x-axis. This means that you will have high attenuation for low frequencies and high amplification for high frequencies. The pole in the lowpass filter will also be loccated on the positive x-axis and will thus amplify low frequencies and attenuate high frequencies.
This description is best illustrated with images, which is why I recommend you to follow my links. Good luck and please comment ask if anything is unclear.
I am working on some experimental data which, at some point, need to be time-integrated and then high-pass filtered (to remove low frequency disturbancies introduced by integration and unwanted DC component).
The aim of my work is not related to filtering, but still I would like to analyze more in detail the filters I am using to give some justification (for example to motivate why I chosed to use a 4th order filter instead of a higher/lower one).
This is the filter I am using:
delta_t = 1.53846e-04;
Fs = 1/delta_t;
cut_F = 8;
Wn = cut_F/(Fs/2);
ftype = 'high';
[b,a] = butter(4,Wn,ftype);
filtered_signal = filtfilt(b,a,signal);
I already had a look here: High-pass filtering in MATLAB to learn something about filters (I never had a course on signal processing) and I used
fvtool(b,a)
to see the impulse response, step response ecc. of the filter I have used.
The problem is that I do not know how to "read" these plots.
What do I have to look for?
How can I understand if a filter is good or not? (I do not have any specification about filter performances, I just know that the lowest frequency I can admit is 5 Hz)
What features of different filters are useful to be compared to motivate the choice?
I see you are starting your Uni DSP class on filters :)
First thing you need to remember is that Matlab can only simulate using finite values, so the results you see are technically all discrete. There are 4 things that will influence your filtering results(or tell you if your filter is good or bad) which you will learn about/have to consider while designing a Finite response filter:
1, the Type of the filter (i.e. Hamming, Butterworth (the one you are using), Blackman, Hanning .etc)
2, the number of filter Coefficients (which determines your filter resolution)
3, the sampling frequency of the original signal (ideally, if you have infinite sampling frequency, you can have perfect filters; not possible in Matlab due to reason above, but you can simulate its effect by setting it really high)
4, the cut-off frequency
You can play around with the 4 parameters so that your filter does what you want it to.
So here comes the theory:
There is a trade-off in terms of the width of your main lobe vs the spectrum leakage of your filter. The idea is that you have some signal with some frequencies, you want to filter out the unwanted (i.e. your DC noise) and keep the ones you want, but what if your desired signal frequency is so low that it is very close to the DC noise. If you have a badly designed filter, you will not be able to filter out the DC component. In order to design a good filter, you will need to find the optimal number for your filter coefficients, type of filter, even cut-off frequency to make sure your filter works as you wanted.
Here is a low-pass filter that I wrote back in the days, you can play around with filters a lot by filtering different kinds of signals and plotting the response.
N = 21; %number of filter coefficients
fc = 4000; %cut-off frequency
f_sampling = fs; %sampling freq
Fc = fc/f_sampling;
n = -(N-1)/2:(N-1)/2;
delta = [zeros(1,(N-1)/2) 1 zeros(1,(N-1)/2)];
h = delta - 2*Fc*sinc(2*n*Fc);
output = filter(h,1,yoursignal);
to plot the response, you want to plot your output in the frequency domain using DFT or FFT(in Matlab) and see how the signal has been distorted due to the leakage and etc.
NFFT=256; % FFT length
output=1/N*abs(fft(output,NFFT)).^2; % PSD estimate using FFT
this gives you what is known as a periodigram, when you plot, you might want to do the 10*log10 to it, so it looks nicer
Hope you do well in class.
I have two sensors seperated by some distance which receive a signal from a source. The signal in its pure form is a sine wave at a frequency of 17kHz. I want to estimate the TDOA between the two sensors. I am using crosscorrelation and below is my code
x1; % signal as recieved by sensor1
x2; % signal as recieved by sensor2
len = length(x1);
nfft = 2^nextpow2(2*len-1);
X1 = fft(x1);
X2 = fft(x2);
X = X1.*conj(X2);
m = ifft(X);
r = [m(end-len+1) m(1:len)];
[a,i] = max(r);
td = i - length(r)/2;
I am filtering my signals x1 and x2 by removing all frequencies below 17kHz.
I am having two problems with the above code:
1. With the sensors and source at the same place, I am getting different values of 'td' at each time. I am not sure what is wrong. Is it because of the noise? If so can anyone please provide a solution? I have read many papers and went through other questions on stackoverflow so please answer with code along with theory instead of just stating the theory.
2. The value of 'td' is sometimes not matching with the delay as calculated using xcorr. What am i doing wrong? Below is my code for td using xcorr
[xc,lags] = xcorr(x1,x2);
[m,i] = max(xc);
td = lags(i);
One problem you might have is the fact that you only use a single frequency. At f = 17 kHz, and an estimated speed-of-sound v = 340 m/s (I assume you use ultra-sound), the wavelength is lambda = v / f = 2 cm. This means that your length measurement has an unambiguity range of 2 cm (sorry, cannot find a good link, google yourself). This means that you already need to know your distance to better than 2 cm, before you can use the result of your measurement to refine the distance.
Think of it in another way: when taking the cross-correlation between two perfect sines, the result should be a 'comb' of peaks with spacing equal to the wavelength. If they overlap perfectly, and you displace one signal by one wavelength, they still overlap perfectly. This means that you first have to know which of these peaks is the right one, otherwise a different peak can be the highest every time purely by random noise. Did you make a plot of the calculated cross-correlation before trying to blindly find the maximum?
This problem is the same as in interferometry, where it is easy to measure small distance variations with a resolution smaller than a wavelength by measuring phase differences, but you have no idea about the absolute distance, since you do not know the absolute phase.
The solution to this is actually easy: let your source generate more frequencies. Even using (band-limited) white-noise should work without problems when calculating cross-correlations, and it removes the ambiguity problem. You should see the white noise as a collection of sines. The cross-correlation of each of them will generate a comb, but with different spacing. When adding all those combs together, they will add up significantly only in a single point, at the delay you are looking for!
White Noise, Maximum Length Sequency or other non-periodic signals should be used as the test signal for time delay measurement using cross correleation. This is because non-periodic signals have only one cross correlation peak and there will be no ambiguity to determine the time delay. It is possible to use the burst type of periodic signals to do the job, but with degraded SNR. If you have to use a continuous periodic signal as the test signal, then you can only measure a time delay within one period of the periodic test signal. This should explain why, in your case, using lower frequency sine wave as the test signal works while using higher frequency sine wave does not. This is demonstrated in these videos: https://youtu.be/L6YJqhbsuFY, https://youtu.be/7u1nSD0RlwY .