I want to take audio PCM data and find peaks in it. Specifically, I want to return the frequency and time at which a peak occurs.
My understanding of this is that I have to take the PCM data and dump it into an array, setting it as the real values with the complex parts set to 0. I then take the FFT, and I get an array back. If each number in the array is a magnitude value, how do I get the frequency associated with each one? Also, do I take the magnitude of the real & complex part or just discard the complex values?
Finally, if I wanted to find the peaks in a single song, do I just set a small window to FFT and slide it across all of the audio? Any suggestions on how large that window should be?
If the samplerate of your PCM data is F, then the highest frequency component in the FFT is F/2. Suppose your PCM data was sampled at 44100Hz, then your FFT values will run from 0Hz (DC) to 22050Hz. If you start with N samples, (N being a power of 2), then the FFT may return N/2 values representing all positive frequencies from 0 to F/2, or it may return N values that also include the negative frequencies from -F/2 to 0. You should check the specification of your FFT algorithm to find out to which frequency each array item is mapped.
To find the peaks, you need to look at the magnitude of the FFT values. So you need to add the squared real and imaginary parts of each complex value.
Suppose your FFT of N PCM samples returns N/2 complex values representing positive frequencies. Then the distance between 2 complex samples is F/2N Hz. With F=44100Hz and N=1024 samples, this would be 21.5Hz. This is your frequency resolution. If you need to find lower frequency beats, the FFT window will need to be extended.
well,
A raw array of size 512 of complex numbers expressing the input wave, when processed with FFT we will replace the imaginary parts with zero (according to intended use), leaving the real parts, then pass the array to the FFT with Sample rate : 8192 Hz.
Now we have a 512 array of FFTed real values, each value is an irrational number, every irrational number express several useful values.
To get the fundamental frequency we have to divide the sample rate by the buffer size:
8192/512 = 32;
32 is the resolution of the FFT values means that we're getting to know the high amplitude frequencies near the numbers that are multiples of 32.
Like if we have a wave of
frequency : 3 48 23 128
Amplitude : 10 5 12 8 dB (ref = 1)
after FFT we get:
frequency : 0 32 64 128
Amplitude : 9 8 2 8
FFT is frequency domain means it arranges according to frequency
Time-domain on the other side means arranging by time we listen to music from second zero to second N.
FFT can only listen when it arranged by Frequency from frequency 0 to frequency N.
So it arranges frequencies in ascending order, since it didn't take all the actual samples from the audio (which are approaching infinite) like taking every nanosecond & less to the FFT, luckily this doesn't happen FFT takes samples from the audio, takes a sample every (1/sample rate) second. this samples get buffered (in our case : 512), each 512 samples buffered into FFT, the output is 512 FFT values.
Since FFT arranges frequencies, it messes with the time samples, samples now arranged according to their frequencies.
The frequencies shown on regular base which is the fundamental frequency which is sample rate divided by buffer size, which is in our case 8192/512 = 32.
So, frequencies power shown every 32 frequencies, the power of the nearest frequency is shown according to how much the power frequency is near to the index.
High resolution can be achieved by using higher sample rate.
To show frequencies we print the index in ascending corresponding to the Amplitude.
Amplitude = 20log10(output/ref)
Amplitudes printed next to each Index show the power of the frequency & get more accurate according to the precision of the resolution.
Conclusion, FFT produces an index of amplitudes, each amplitude expresses the power of its corresponding index (frequency).
You may actually be looking for a spectrogram, which is basically an FFT of the data in a small window that's slid along the time axis. If you have software that implements this, it might save you some effort. It's what's commonly used for analysing time varying acoustic signals, and is a very useful way to look at sounds. Also, there are some tricks, for example, with windowing data for FFTs, that the spectrogram will probably get right, but will be harder (though not very hard) for you to do correctly.
Related
I found a similar question posted but not with a suitable answer.
I want to have sine wave buffers, doing audio synthesis, and wanting to figure an optimal size for buffers to avoid artifacts where the sine wave's period measured in samples is a non-integer.
For instance, a sine at A4, 440.0 Hz, with sample rate of 44100, repeats every 1/440 * 44100 samples, or every 100.22727272727272 samples. If you simply create a circular buffer of 100 (or 101) samples to contain a single cycle of the sine, playback has noticeable artifacts do to the disjoint.
Based on the response to the previous poster's version, the proposed solution was to express the period as a whole fraction and find the greatest common divisor to find the number of cycles and samples needed. However in practice I keep finding ratios where the greatest common divisor calculates as 1, which intuitively seems unlikely to me.
Using the above example, in Python:
freq = 440.0
rate = 44100
period = 1.0 / freq * rate
ratio = period.as_integer_ratio()
from fractions import gcd
divisor = gcd(*ratio)
Divisor equals 1, apparently meaning that there is no common divisor, meaning that it requires infinite cycles of the frequency to ever find an integer number of samples to perfectly store the sine.
Am I doing this wrong? Is there another solution?
NOTE in practice, I've found that setting the buffer size somewhere around 1000x the actual period of the sine eliminates artifacts to the naked ear, but I would like to lower my memory usage and so would like to mathematically determine smaller allowable buffer sizes based on arbitrary frequencies.
What gcd(*ratio) returning 1 tells you is that the fraction you obtained with period.as_integer_ratio() has already been simplified, not that "it requires infinite cycles of the frequency to ever find an integer number of samples to perfectly store the sine".
However that simplification might not be so great given round-off errors when computing period = 1.0 / freq * rate. You would be better off using:
divisor = fractions.gcd(rate, freq)
ratio = (rate/divisor, freq/divisor)
At that point since ratio is a simplified fraction (in your specific case (2205, 22)), ratio[1] cycles would fit in an integer number of samples (more specifically ratio[0]*ratio[1] or in your specific case 48510 samples).
I am currently running Python's Numpy fft on 44100Hz audio samples which gives me a working frequency range of 0Hz - 22050Hz (thanks Nyquist). Once I use fft on those time domain values, I have 128 points in my fft spectrum giving me 172Hz for each frequency bin size.
I would like to tighten the frequency bin to 86Hz and still keep to only 128 fft points, instead of increasing my fft count to 256 through an adjustment on how I'm creating my samples.
The question I have is whether or not this is theoretically possible. My thought would be to run fft on any Hz values between 0Hz to 11025Hz only. I don't care about anything above that anyway. This would cut my working spectrum in half and put my frequency bins at 86Hz and while keeping to my 128 spectrum bins. Perhaps this can be accomplished via a window function in the time domain?
Currently the code I'm using to create my samples and then convert to fft is:
import numpy as np
sample_rate = 44100
chunk = 128
record_seconds = 2
stream = self.audio.open(format=pyaudio.paInt16, channels=1,
rate=sample_rate, input=True, frames_per_buffer=6300)
sample_list = []
for i in range(0, int(sample_rate / chunk * record_seconds)):
data = stream.read(chunk)
sample_list.append(np.fromstring(data, dtype=np.int16))
### then later ###:
for samp in sample_list:
samp_fft = np.fft.fft(samp) ...
I hope I worded this clearly enough. Let me know if I need to adjust my explanation or terminology.
What you are asking for is not possible. As you mentioned in a comment you require a short time window. I assume this is because you're trying to detect when a signal arrives at a certain frequency (as I've answered your earlier question on the subject) and you want the detection to be time sensitive. However, it seems your bin size is too large for your requirements.
There are only two ways to decrease the bin size. 1) Increase the length of the FFT. Unfortunately this also means that it will take longer to acquire the data. 2) lower the sample rate (either by sample rate conversion or at the hardware level) but since the samples arrive slower it will also take longer to acquire the data.
I'm going to suggest to you a 3rd option (from what I've gleaned from this and your other questions is possibly a better solution) which is: Perform the frequency detection in the time domain. What this would require is a time-domain bandpass filter followed by an RMS meter. Implementation wise this would be one or more biquad filters that you could implement in python for the filter - there are probably implementations already available. The tricky part would be designing the filter but I'd be happy to help you in chat. The RMS meter is basically taking the square root of the sum of the squares of the output samples from the filter.
Doubling the size of the FFT would be the obvious thing to do, but if there is a good reason why you can't do this then consider 2x downsampling prior to the FFT to get the effective sample rate down to 22050 Hz:
- Apply low pass filter with cut off at 11 kHz
- Discard every other sample from filtered output
- Apply FFT to down-sampled data
If you are not trying to resolve between close adjacent frequency peaks or noise, then, to half the frequency bin spacing, you can zero-pad your data to double the FFT length, without having to wait for more data. Then, if you only want the lower half of the frequency range 0..Fs/2, just throw away the middle half of the FFT result vector (which is usually far more efficient than trying to compute the lower half of the frequency range via non-FFT means).
Note that zero-padding gives the same result as high-quality interpolation (as in smoothing a plot of the original FFT result points). It does not increase peak separation resolution, but might make it easier to pick out more precise peak locations in the plot if the noise level is low enough.
I've seen the various FFT questions on here but I'm confused on part of the implementation. Instead of performing the FFT in real time, I want to do it offline. Lets say I have the raw data in float[] audio. The sampling rate is 44100 and so audio[0] to audio[44099] will contain 1 seconds worth of audio. If my FFT function handles the windowing (e.g. Hanning), do I simply put the entire audio buffer into the function in one go? Or, do I have to cut the audio into chunks of 4096 (my window size) and then input that into the FFT which will then perform the windowing function on top?
You may need to copy your input data to a separate buffer and get it in the correct format, e.g. if your FFT is in-place, or if it requires interleaved complex data (real/imaginary). However if your FFT routine can take a purely real input and is not in-place (i.e. non-destructive) then you may just be able to pass a pointer to the original sample data, along with an appropriate size parameter.
Typically for 1s of audio, e.g. speech or music, you would pick an FFT size which corresponds to a reasonably stationary chunk of audio, e.g. 10 ms or 20 ms. So at 44.1 kHz your FFT size might be say 512 or 1024. You would then generate successive spectra by advancing through your buffer and doing a new FFT at each starting point. Note that it's common practice to overlap these successive buffers, typically by 50%. So if N = 1024 your first FFT would be for samples 0..1023, your second would be for samples 512..1535, then 1024..2047, etc.
The choice of whether to calculate one FFT over the entire data set (in the OP's case, 44100 samples representing 1-second of data), or whether to do a series of FFT's over smaller subsets of the full data set, depends on the data, and on the intended purpose of the FFT.
If the data is relatively static spectrally over the full data set, then one FFT over the entire data set is probably all that's needed.
However, if the data is spectrally dynamic over the data set, then multiple sliding FFT's over small subsets of the data would create a more accurate time-frequency representation of the data.
The plot below shows the power spectrum of an acoustic guitar playing an A4 note. The audio signal was sampled at 44.1 KHz and the data set contains 131072 samples, almost 3 seconds of data. This data set was pre-multiplied with a Hann window function.
The plot below shows the power spectrum of a subset of 16384 samples (0 to 16383) taken from the full data set of the acoustic guitar A4 note. This subset was also pre-multiplied with a Hann window function.
Notice how the spectral energy distribution of the subset is significantly different from the spectral energy distribution of the full data set.
If we were to extract subsets from the full data set, using a sliding 16384 sample frame, and calculate the power spectrum of each frame, we would create an accurate time-frequency picture of the full data set.
References:
Real audio signal data, Hann window function, plots, FFT, and spectral analysis were done here:
Fast Fourier Transform, spectral analysis, Hann window function, audio data
The chunk size or window length you pick controls the frequency resolution and the time resolution of the FFT result. You have to determine which you want or what trade-off to make.
Longer windows give you better frequency resolution, but worse time resolution. Shorter windows, vice versa. Each FFT result bin will contain a frequency bandwidth of roughly 1 to 2 times the sample rate divided by the FFT length, depending on the window shape (rectangular, von Hann, etc.), not just one single frequency. If your entire data chunk is stationary (frequency content doesn't change), then you may not need any time resolution, and can go for 1 to 2 Hz frequency "resolution" in your 1 second of data. Averaging multiple short FFT windows might also help reduce the variance of your spectral estimations.
I have a program that plots the spectrum analysis (Amp/Freq) of a signal, which is preety much the DFT converted to polar. However, this is not exactly the sort of graph that, say, winamp (right at the top-left corner), or effectively any other audio software plots. I am not really sure what is this sort of graph called (if it has a distinct name at all), so I am not sure what to look for.
I am preety positive about the frequency axis being base two exponential, the amplitude axis puzzles me though.
Any pointers?
Actually an interesting question. I know what you are saying; the frequency axis is certainly logarithmic. But what about the amplitude? In response to another poster, the amplitude can't simply be in units of dB alone, because dB has no concept of zero. This introduces the idea of quantization error, SNR, and dynamic range.
Assume that the received digitized (i.e., discrete time and discrete amplitude) time-domain signal, x[n], is equal to s[n] + e[n], where s[n] is the transmitted discrete-time signal (i.e., continuous amplitude) and e[n] is the quantization error. Suppose x[n] is represented with b bits, and for simplicity, takes values in [0,1). Then the maximum peak-to-peak amplitude of e[n] is one quantization level, i.e., 2^{-b}.
The dynamic range is the defined to be, in decibels, 20 log10 (max peak-to-peak |s[n]|)/(max peak-to-peak |e[n]|) = 20 log10 1/(2^{-b}) = 20b log10 2 = 6.02b dB. For 16-bit audio, the dynamic range is 96 dB. For 8-bit audio, the dynamic range is 48 dB.
So how might Winamp plot amplitude? My guesses:
The minimum amplitude is assumed to be -6.02b dB, and the maximum amplitude is 0 dB. Visually, Winamp draws the window with these thresholds in mind.
Another nonlinear map, such as log(1+X), is used. This function is always nonnegative, and when X is large, it approximates log(X).
Any other experts out there who know? Let me know what you think. I'm interested, too, exactly how this is implemented.
To generate a power spectrum you need to do the following steps:
apply window function to time domain data (e.g. Hanning window)
compute FFT
calculate log of FFT bin magnitudes for N/2 points of FFT (typically 10 * log10(re * re + im * im))
This gives log magnitude (i.e. dB) versus linear frequency.
If you also want a log frequency scale then you will need to accumulate the magnitude from appropriate ranges of bins (and you will need a fairly large FFT to start with).
Well I'm not 100% sure what you mean but surely its just bucketing the data from an FFT?
If you want to get the data such that you have (for a 44Khz file) frequency points at 22Khz, 11Khz 5.5Khz etc then you could use a wavelet decomposition, i guess ...
This thread may help ya a bit ...
Converting an FFT to a spectogram
Same sort of information as a spectrogram I'd guess ...
What you need is power spectrum graph. You have to compute DFT of your signal's current window. Then square each value.
The digital sound is playing using DirectSound device. It is necessary to display sound activity in decibels - like analog devices do.
What is the right way to calculate sound pressure from the WAVE PCM data (44100 Hz, 16-bit)?
if you just need an "idea" of the sound pressure, you can simply compute the log-energy on some time franmes of the signal: split the signal every N samples, compute 10*log(sum(xn**2)) where x are the N samples, and you get a value in the dB domain. If you need to precisely display a measure (that is your 0 dB matches say a mixtable 0dB), it is a bit more complicated.
See here for more details:
http://music.columbia.edu/pipermail/music-dsp/2002-April/048341.html
Sound pressure is a measure of force per unit area. To determine this you would have to have information about the speaker(s) on which the audio is played. You can obtain a decibel level with respect to an arbitrary reference (as opposed to the threshold of hearing) with the algorithm proposed by cournape.
Calculate the average signal power over a time interval, compute the base-10 logarithm and multiply by 19. The average power is calculated by averaging the the square of each sample over the interval. Note that positive and negative values are necessary (i.e. it must be an AC signal). So, make sure the PCM values are either floating-point, 2's complement or offset unsigned values accordingly.
Also, by applying Parseval's theorum and the Fourier transform you can also generate signal levels for different frequency bands.