I'm able to read a wav files and its values. I need to find peaks and pits positions and their values. First time, i tried to smooth it by (i-1 + i + i +1) / 3 formula then searching on array as array[i-1] > array[i] & direction == 'up' --> pits style solution but because of noise and other reasons of future calculations of project, I'm tring to find better working area. Since couple days, I'm researching FFT. As my understanding, fft translates the audio files to series of sines and cosines. After fft operation the given values is a0's and a1's for a0 + ak * cos(k*x) + bk * sin(k*x) which k++ and x++ as this picture
http://zone.ni.com/images/reference/en-XX/help/371361E-01/loc_eps_sigadd3freqcomp.gif
My question is, does fft helps to me find peaks and pits on audio? Does anybody has a experience for this kind of problems?
It depends on exactly what you are trying to do, which you haven't really made clear. "finding the peaks and pits" is one thing, but since there might be various reasons for doing this there might be various methods. You already tried the straightforward thing of actually looking for the local maximum and minima, it sounds like. Here are some tips:
you do not need the FFT.
audio data usually swings above and below zero (there are exceptions, including 8-bit wavs, which are unsigned, but these are exceptions), so you must be aware of positive and negative values. Generally, large positive and large negative values carry large amounts of energy, though, so you want to count those as the same.
due to #2, if you want to average, you might want to take the average of the absolute value, or more commonly, the average of the square. Once you find the average of the squares, take the square root of that value and this gives the RMS, which is related to the power of the signal, so you might do something like this is you are trying to indicate signal loudness, intensity or approximate an analog meter. The average of absolutes may be more robust against extreme values, but is less commonly used.
another approach is to simply look for the peak of the absolute value over some number of samples, this is commonly done when drawing waveforms, and for digital "peak" meters. It makes less sense to look at the minimum absolute.
Once you've done something like the above, yes you may want to compute the log of the value you've found in order to display the signal in dB, but make sure you use the right formula. 10 * log_10( amplitude ) is not it. Rule of thumb: usually when computing logs from amplitude you will see a 20, not a 10. If you want to compute dBFS (the amount of "headroom" before clipping, which is the standard measurement for digital meters), the formula is -20 * log_10( |amplitude| ), where amplitude is normalize to +/- 1. Watch out for amplitude = 0, which gives an infinite headroom in dB.
If I understand you correctly, you just want to estimate the relative loudness/quietness of an audio digital sample at a given point.
For this estimation, you don't need to use FFT. However your method of averaging the signal does not produce the appropiate picture neither.
The digital signal is the value of the audio wave at a given moment. You need to find the overall amplitude of the signal at that given moment. You can somewhat see it as the local maximum value for a given interval around the moment you want to calculate. You may have a moving max for the signal and get your amplitude estimation.
At a 16 bit sound sample, the sound signal value can go from 0 up to 32767. At a 44.1 kHz sample rate, you can find peaks and pits of around 0.01 secs by finding the max value of 441 samples around a given t moment.
max=1;
for (i=0; i<441; i++) if (array[t*44100+i]>max) max=array[t*44100+i];
then for representing it on a 0 to 1 scale you (not really 0, because we used a minimum of 1)
amplitude = max / 32767;
or you might represent it in relative dB logarithmic scale (here you see why we used 1 for the minimum value)
dB = 20 * log10(amplitude);
all you need to do is take dy/dx, which can getapproximately by just scanning through the wave and and subtracting the previous value from the current one and look at where it goes to zero or changes from positive to negative
in this code I made it really brief and unintelligent for sake of brevity, of course you could handle cases of dy being zero better, find the 'centre' of a long section of a flat peak, that kind of thing. But if all you need is basic peaks and troughs, this will find them.
lastY=0;
bool goingup=true;
for( i=0; i < wave.length; i++ ) {
y = wave[i];
dy = y - lastY;
bool stillgoingup = (dy>0);
if( goingup != direction ) {
// changed direction - note value of i(place) and 'y'(height)
stillgoingup = goingup;
}
}
Related
I was wondering if it would be possible to poll the AnalyzerNode from the WebAudio API and use it to construct a PeriodicWave that is synthesized via an OscillatorNode?
My intuition is that something about the difference in amplitudes between analyzer frames can help calculate the right phase for a PeriodicWave, but I'm not sure how to go about implementing it. Any help on the right algorithm to use would be appreciated!
As luck would have it, I was working on a similar project just a few weeks ago. I put together a JSFiddle to explore the idea of reconstructing a phase-randomized version of a waveform using frequency data from an AnalyserNode. You can find that experiment here:
https://jsfiddle.net/mattdiamond/w4u7x8zk/
Here's the code that takes in the frequency data output from an AnalyserNode and generates a PeriodicWave:
function generatePeriodicWave(freqData) {
const real = [];
const imag = [];
freqData.forEach((x, i) => {
const amp = fromDecibels(x);
const phase = getRandomPhase();
real.push(amp * Math.cos(phase));
imag.push(amp * Math.sin(phase));
});
return context.createPeriodicWave(real, imag);
}
function fromDecibels(x) {
return 10 ** (x / 20);
}
function getRandomPhase() {
return Math.random() * 2 * Math.PI - Math.PI;
}
Since the AnalyserNode converts the FFT amplitude values to decibels, we need to recover those original values first (which we do by simply using the inverse of the formula that was used to convert them to decibels). We also need to provide a phase for each frequency, which we select at random from the range -π to π.
Now that we have an amplitude and phase, we construct a complex number by multiplying the amplitude by the cosine and sine of the phase. This is because the amplitude and phase correspond to a polar coordinate, and createPeriodicWave expects a list of real and imaginary numbers corresponding to Cartesian coordinates in the complex plane. (See here for more information on the mathematics behind this conversion.)
Once we've generated the PeriodicWave, all that's left to do is load it into an OscillatorNode, set the desired frequency, and start the oscillator. You'll notice that the default frequency is set to context.sampleRate / FFT_SIZE (you can ignore the toFixed, that was just for the sake of the UI). This causes the oscillator to play the wave at the same rate as the original samples. Increasing or decreasing the frequency from this value will pitch-shift the audio up or down, respectively.
You'll also notice that I chose 2^15 as the FFT size, which is the maximum size that the AnalyserNode allows. For my purposes -- creating interesting looped drones -- a larger FFT results in a more interesting and less "loopy" drone. (A while back I created a webpage that allowed users to generate drones from much larger FFTs... that experiment utilized a third-party FFT library instead of the AnalyserNode.) I'm not sure if this is the right FFT size for your purposes, but it's something to consider.
Anyway, I think that covers the core of the algorithm. Hope this helps! (And feel free to ask more questions in the comments if anything's unclear.)
First time posting, thanks for the great community!
I am using AudioKit and trying to add frequency weighting filters to the microphone input and so I am trying to understand the values that are coming out of the AudioKit AKFFTTap.
Currently I am trying to just print the FFT buffer converted into dB values
for i in 0..<self.bufferSize {
let db = 20 * log10((self.fft?.fftData[Int(i)])!)
print(db)
}
I was expecting values ranging in the range of about -128 to 0, but I am getting strange values of nearly -200dB and when I blow on the microphone to peg out the readings it only reaches about -60. Am I not approaching this correctly? I was assuming that the values being output from the EZAudioFFT engine would be plain amplitude values and that the normal dB conversion math would work. Anyone have any ideas?
Thanks in advance for any discussion about this issue!
You need to add all of the values from self.fft?.fftData (consider changing negative values to positive before adding) and then change that to decibels
The values in the array correspond to the values of the bins in the FFT. Having a single bin contain a magnitude value close to 1 would mean that a great amount of energy is in that narrow frequency band e.g. a very loud sinusoid (a signal with a single frequency).
Normal sounds, such as the one caused by you blowing on the mic, spread their energy across the entire spectrum, that is, in many bins instead of just one. For this reason, usually the magnitudes get lower as the FFT size increases.
Magnitude of -40dB on a single bin is quite loud. If you try to play a tone, you should see a clear peak in one of the bins.
I am designing a fractional delay filter, and my lagrange coefficient of order 5 h(n) have 6 taps in time domain. I have tested to convolute the h(n) with x(n) which is 5000 sampled signal using matlab, and the result seems ok. When I tried to use FFT and IFFT method, the output is totally wrong. Actually my FFT is computed with 8192 data in frequency domain, which is the nearest power of 2 for 5000 signal sample. For the IFFT portion, I convert back the 8192 frequency domain data back to 5000 length data in time domain. So, the problem is, why this thing works in convolution, but not in FFT multiplication. Does converting my 6 taps h(n) to 8192 taps in frequency domain causes this problem?
Actually I have tried using overlap-save method, which perform the FFT and multiplication with smaller chunks of x(n) and doing it 5 times separately. The result seems slight better than the previous, and at least I can see the waveform pattern, but still slightly distorted. So, any idea where goes wrong, and what is the solution. Thank you.
The reason I am implementing the circular convolution in frequency domain instead of time domain is, I am try to merge the Lagrange filter with other low pass filter in frequency domain, so that the implementation can be more efficient. Of course I do believe implement filtering in frequency domain will be much faster than convolution in time domain. The LP filter has 120 taps in time domain. Due to the memory constraints, the raw data including the padding will be limited to 1024 in length, and so with the fft bins.
Because my Lagrange coefficient has only 6 taps, which is huge different with 1024 taps. I doubt that the fft of the 6 taps to 1024 bins in frequency domain will cause error. Here is my matlab code on Lagrange filter only. This is just a test code only, not implementation code. It's a bit messy, sorry about that. Really appreciate if you can give me more advice on this problem. Thank you.
t=1:5000;
fs=2.5*(10^12);
A=70000;
x=A*sin(2*pi*10.*t.*(10^6).*t./fs);
delay=0.4;
N=5;
n = 0:N;
h = ones(1,N+1);
for k = 0:N
index = find(n ~= k);
h(index) = h(index) * (delay-k)./ (n(index)-k);
end
pad=zeros(1,length(h)-1);
out=[];
H=fft(hh,1024);
H=fft([h zeros(1,1024-length(h))]);
for i=0:1:ceil(length(x)/(1024-length(h)+1))-1
if (i ~= ceil(length(x)/(1024-length(h)+1))-1)
a=x(1,i*(1024-length(h)+1)+1:(i+1)*(1024-length(h)+1));
else
temp=x(1,i*(1024-length(h)+1)+1:length(x));
a=[temp zeros(1,1024-length(h)+1-length(temp))];
end
xx=[pad a];
X=fft(xx,1024);
Y=H.*X;
y=abs(ifft(Y,1024));
out=[out y(1,length(h):length(y))];
pad=y(1,length(a)+1:length(y));
end
Some comments:
The nearest power of two is actually 4096. Do you expect the remaining 904 samples to contribute much? I would guess that they are significant only if you are looking for relatively low-frequency features.
How did you pad your signal out to 8192 samples? Padding your sample out to 8192 implies that approximately 40% of your data is "fictional". If you used zeros to lengthen your dataset, you likely injected a step change at the pad point - which implies a lot of high-frequency content.
A short code snippet demonstrating your methods couldn't hurt.
I'm not a DSP expert, but I understand that there are two ways that I can apply a discrete time-domain filter to a discrete time-domain waveform. The first is to convolve them in the time domain, and the second is to take the FFT of both, multiply both complex spectrums, and take IFFT of the result. One key difference in these methods is the second approach is subject to circular convolution.
As an example, if the filter and waveforms are both N points long, the first approach (i.e. convolution) produces a result that is N+N-1 points long, where the first half of this response is the filter filling up and the 2nd half is the filter emptying. To get a steady-state response, the filter needs to have fewer points than the waveform to be filtered.
Continuing this example with the second approach, and assuming the discrete time-domain waveform data is all real (not complex), the FFT of the filter and the waveform both produce FFTs of N points long. Multiplying both spectrums IFFT'ing the result produces a time-domain result also N points long. Here the response where the filter fills up and empties overlap each other in the time domain, and there's no steady state response. This is the effect of circular convolution. To avoid this, typically the filter size would be smaller than the waveform size and both would be zero-padded to allow space for the frequency convolution to expand in time after IFFT of the product of the two spectrums.
My question is, I often see work in the literature from well-established experts/companies where they have a discrete (real) time-domain waveform (N points), they FFT it, multiply it by some filter (also N points), and IFFT the result for subsequent processing. My naive thinking is this result should contain no steady-state response and thus should contain artifacts from the filter filling/emptying that would lead to errors in interpreting the resulting data, but I must be missing something. Under what circumstances can this be a valid approach?
Any insight would be greatly appreciated
The basic problem is not about zero padding vs the assumed periodicity, but that Fourier analysis decomposes the signal into sine waves which, at the most basic level, are assumed to be infinite in extent. Both approaches are correct in that the IFFT using the full FFT will return the exact input waveform, and both approaches are incorrect in that using less than the full spectrum can lead to effects at the edges (that usually extend a few wavelengths). The only difference is in the details of what you assume fills in the rest of infinity, not in whether you are making an assumption.
Back to your first paragraph: Usually, in DSP, the biggest problem I run into with FFTs is that they are non-causal, and for this reason I often prefer to stay in the time domain, using, for example, FIR and IIR filters.
Update:
In the question statement, the OP correctly points out some of the problems that can arise when using FFTs to filter signals, for example, edge effects, that can be particularly problematic when doing a convolution that is comparable in the length (in the time domain) to the sampled waveform. It's important to note though that not all filtering is done using FFTs, and in the paper cited by the OP, they are not using FFT filters, and the problems that would arise with an FFT filter implementation do not arise using their approach.
Consider, for example, a filter that implements a simple average over 128 sample points, using two different implementations.
FFT: In the FFT/convolution approach one would have a sample of, say, 256, points and convolve this with a wfm that is constant for the first half and goes to zero in the second half. The question here is (even after this system has run a few cycles), what determines the value of the first point of the result? The FFT assumes that the wfm is circular (i.e. infinitely periodic) so either: the first point of the result is determined by the last 127 (i.e. future) samples of the wfm (skipping over the middle of the wfm), or by 127 zeros if you zero-pad. Neither is correct.
FIR: Another approach is to implement the average with an FIR filter. For example, here one could use the average of the values in a 128 register FIFO queue. That is, as each sample point comes in, 1) put it in the queue, 2) dequeue the oldest item, 3) average all of the 128 items remaining in the queue; and this is your result for this sample point. This approach runs continuously, handling one point at a time, and returning the filtered result after each sample, and has none of the problems that occur from the FFT as it's applied to finite sample chunks. Each result is just the average of the current sample and the 127 samples that came before it.
The paper that OP cites takes an approach much more similar to the FIR filter than to the FFT filter (note though that the filter in the paper is more complicated, and the whole paper is basically an analysis of this filter.) See, for example, this free book which describes how to analyze and apply different filters, and note also that the Laplace approach to analysis of the FIR and IIR filters is quite similar what what's found in the cited paper.
Here's an example of convolution without zero padding for the DFT (circular convolution) vs linear convolution. This is the convolution of a length M=32 sequence with a length L=128 sequence (using Numpy/Matplotlib):
f = rand(32); g = rand(128)
h1 = convolve(f, g)
h2 = real(ifft(fft(f, 128)*fft(g)))
plot(h1); plot(h2,'r')
grid()
The first M-1 points are different, and it's short by M-1 points since it wasn't zero padded. These differences are a problem if you're doing block convolution, but techniques such as overlap and save or overlap and add are used to overcome this problem. Otherwise if you're just computing a one-off filtering operation, the valid result will start at index M-1 and end at index L-1, with a length of L-M+1.
As to the paper cited, I looked at their MATLAB code in appendix A. I think they made a mistake in applying the Hfinal transfer function to the negative frequencies without first conjugating it. Otherwise, you can see in their graphs that the clock jitter is a periodic signal, so using circular convolution is fine for a steady-state analysis.
Edit: Regarding conjugating the transfer function, the PLLs have a real-valued impulse response, and every real-valued signal has a conjugate symmetric spectrum. In the code you can see that they're just using Hfinal[N-i] to get the negative frequencies without taking the conjugate. I've plotted their transfer function from -50 MHz to 50 MHz:
N = 150000 # number of samples. Need >50k to get a good spectrum.
res = 100e6/N # resolution of single freq point
f = res * arange(-N/2, N/2) # set the frequency sweep [-50MHz,50MHz), N points
s = 2j*pi*f # set the xfer function to complex radians
f1 = 22e6 # define 3dB corner frequency for H1
zeta1 = 0.54 # define peaking for H1
f2 = 7e6 # define 3dB corner frequency for H2
zeta2 = 0.54 # define peaking for H2
f3 = 1.0e6 # define 3dB corner frequency for H3
# w1 = natural frequency
w1 = 2*pi*f1/((1 + 2*zeta1**2 + ((1 + 2*zeta1**2)**2 + 1)**0.5)**0.5)
# H1 transfer function
H1 = ((2*zeta1*w1*s + w1**2)/(s**2 + 2*zeta1*w1*s + w1**2))
# w2 = natural frequency
w2 = 2*pi*f2/((1 + 2*zeta2**2 + ((1 + 2*zeta2**2)**2 + 1)**0.5)**0.5)
# H2 transfer function
H2 = ((2*zeta2*w2*s + w2**2)/(s**2 + 2*zeta2*w2*s + w2**2))
w3 = 2*pi*f3 # w3 = 3dB point for a single pole high pass function.
H3 = s/(s+w3) # the H3 xfer function is a high pass
Ht = 2*(H1-H2)*H3 # Final transfer based on the difference functions
subplot(311); plot(f, abs(Ht)); ylabel("abs")
subplot(312); plot(f, real(Ht)); ylabel("real")
subplot(313); plot(f, imag(Ht)); ylabel("imag")
As you can see, the real component has even symmetry and the imaginary component has odd symmetry. In their code they only calculated the positive frequencies for a loglog plot (reasonable enough). However, for calculating the inverse transform they used the values for the positive frequencies for the negative frequencies by indexing Hfinal[N-i] but forgot to conjugate it.
I can shed some light to the reason why "windowing" is applied before FFT is applied.
As already pointed out the FFT assumes that we have a infinite signal. When we take a sample over a finite time T this is mathematically the equivalent of multiplying the signal with a rectangular function.
Multiplying in the time domain becomes convolution in the frequency domain. The frequency response of a rectangle is the sync function i.e. sin(x)/x. The x in the numerator is the kicker, because it dies down O(1/N).
If you have frequency components which are exactly multiples of 1/T this does not matter as the sync function is zero in all points except that frequency where it is 1.
However if you have a sine which fall between 2 points you will see the sync function sampled on the frequency point. It lloks like a magnified version of the sync function and the 'ghost' signals caused by the convolution die down with 1/N or 6dB/octave. If you have a signal 60db above the noise floor, you will not see the noise for 1000 frequencies left and right from your main signal, it will be swamped by the "skirts" of the sync function.
If you use a different time window you get a different frequency response, a cosine for example dies down with 1/x^2, there are specialized windows for different measurements. The Hanning window is often used as a general purpose window.
The point is that the rectangular window used when not applying any "windowing function" creates far worse artefacts than a well chosen window. i.e by "distorting" the time samples we get a much better picture in the frequency domain which closer resembles "reality", or rather the "reality" we expect and want to see.
Although there will be artifacts from assuming that a rectangular window of data is periodic at the FFT aperture width, which is one interpretation of what circular convolution does without sufficient zero padding, the differences may or may not be large enough to swamp the data analysis in question.
According to "Introduction to Neural Networks with Java By Jeff Heaton", the input to the Kohonen neural network must be the values between -1 and 1.
It is possible to normalize inputs where the range is known beforehand:
For instance RGB (125, 125, 125) where the range is know as values between 0 and 255:
1. Divide by 255: (125/255) = 0.5 >> (0.5,0.5,0.5)
2. Multiply by two and subtract one: ((0.5*2)-1)=0 >> (0,0,0)
The question is how can we normalize the input where the range is unknown like our height or weight.
Also, some other papers mention that the input must be normalized to the values between 0 and 1. Which is the proper way, "-1 and 1" or "0 and 1"?
You can always use a squashing function to map an infinite interval to a finite interval. E.g. you can use tanh.
You might want to use tanh(x * l) with a manually chosen l though, in order not to put too many objects in the same region. So if you have a good guess that the maximal values of your data are +/- 500, you might want to use tanh(x / 1000) as a mapping where x is the value of your object It might even make sense to subtract your guess of the mean from x, yielding tanh((x - mean) / max).
From what I know about Kohonen SOM, they specific normalization does not really matter.
Well, it might through specific choices for the value of parameters of the learning algorithm, but the most important thing is that the different dimensions of your input points have to be of the same magnitude.
Imagine that each data point is not a pixel with the three RGB components but a vector with statistical data for a country, e.g. area, population, ....
It is important for the convergence of the learning part that all these numbers are of the same magnitude.
Therefore, it does not really matter if you don't know the exact range, you just have to know approximately the characteristic amplitude of your data.
For weight and size, I'm sure that if you divide them respectively by 200kg and 3 meters all your data points will fall in the ]0 1] interval. You could even use 50kg and 1 meter the important thing is that all coordinates would be of order 1.
Finally, you could a consider running some linear analysis tools like POD on the data that would give you automatically a way to normalize your data and a subspace for the initialization of your map.
Hope this helps.