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.
Related
I've been given some digitized sound recordings and asked to plot the sound pressure level per Hz.
The signal is sampled at 40KHz and the units for the y axis are simply volts.
I've been asked to produce a graph of the SPL as dB/Hz vs Hz.
EDIT: The input units are voltage vs time.
Does this make sense? I though SPL was a time domain measure?
If it does make sense how would I go about producing this graph? Apply the dB formula (20 * log10(x) IIRC) and do an FFT on that or...?
What you're describing is a Power Spectral Density. Matlab, for example, has a pwelch function that does literally what you're asking for. To scale to dBSPL/Hz, simply apply 10*log10([psd]) where psd is the output of pwelch. Let me know if you need help with the function inputs.
If you're working with a different framework, let me know which, 100% sure they'll have a version of this function, possibly with a different output format in which case the scaling might be different.
According to what I have read on the internet, the normal range of fundamental frequency of female voice is 165 to 255 Hz .
I am using Praat and also python library called Parselmouth to get the fundamental frequency values of female voice in an audio file(.wav). however, I got some values that are over 255Hz(eg: 400+Hz, 500Hz).
Is it normal to get big values like this?
It is possible, but unlikely, if you are trying to capture the fundamental frequency (F0) of a speaking voice. It sounds likely that you are capturing a more easily resonating overtone (e.g. F1 or F2) instead.
My experiments with Praat give me the impression that the with good parameters it will reliably extract F0.
What you'll want to do is to verify that by comparing the pitch curve with a spectrogram. Here's an example of a fitting made by Praat (female speaker):
You can see from the image that
Most prominent frequency seems to be F2
Around 200 Hz seems likely to be F0, since there's only noise below that (compared to before/after the segment)
Praat has calculated a good estimate of F0 for the voiced speech segments
If, after a visual inspection, it seems that you are getting wrong results, you can try to tweak the parameters. Window length greatly affects the frequency resolution.
If you can't capture frequencies this low, you should try increasing the window length - the intuition is that it gives the algorithm a better chance at finding slowly changing periodic features in the data.
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;
}
}
From the question: How do I obtain the frequencies of each value in an FFT?
I have a similar question. I understand the answers to the previous question, but I would like further clarification on frequency. Is frequency the same as the index?
Let's go for an example: let's assume we have an array (1X200) of data in MATLAB. When you apply 'abs(fft)' for that array it gives the same size array as the result (1X200). So, does this mean this array contains magnitude? Does this mean the indices of these magnitudes are the frequencies? Like 1, 2, 3, 4...200? Or, if this assumption is wrong, please tell me how to find the frequency from the magnitude.
Instead of using the FFT directly you can use MATLAB's periodogram function, which takes care of a lot of the housekeeping for you, and which will plot the X (frequency axis) correctly if you supply the sample rate. See e.g. this answer.
For clarification though, the index of the FFT corresponds to frequency, and the magnitude of the complex value at each frequency (index) tells you the amplitude of the signal at that frequency.
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.