Problem:
I have a long vector which I want to downsample.
By downsampling I mean take the long vector and shorten it by averaging parts of it.
I'm using Python (2.7)
For example I have a mono audio file which was recorded at 44,100 Hz.
The recording is 10 seconds long, providing a vector of length 441,000.
Say I want to downsample it to 8KHz. The output vector should be 80,000 long.
Of course, much of information will be lost by downsampling.
Suggested solution:
The Sample rates ratio is:
R = 44100 / 8000 = 5.5125
So, I would like to take every 5.5125 samples of the original vector and average it into a single number.
Is there a library function which does this (i.e. from Numpy, statistics, etc. Python libraries)?
If not, can you recommend a neat way to run this? How can I average a non-integer index (5.5125)?
How can I "break" the original vector into smaller parts of 5.5125 size in a neat way?
Related
I have short audio files which I'm trying to analyze using Librosa, in particular the spectral centroid function. However, this function outputs an array of different values representing the spectral centroid at different frames within the audio file. The documentation says that the frame size can be changed by specifying the parameter n_fft when calling the function. It would be more beneficial to me if this function analyzed the entire audio file at once rather than outputting the result at multiple points in time. Is there a way for me to specify that I want the function to be called with, say, a frame size of the entire audio file instead of the default time which is 2048 samples? Is there another better way?
Cheers and thank you!
The length of the FFT window (n_fft) specifies not only how many samples you need, but also the frequency resolution of the result (longer n_fft, better resolution). To ensure comparable results for many files you probably want to use the same n_fft value for all of them.
With that out of the way, say your files all have no more than 16k samples. Then you may still achieve a reasonable runtime (FFT runs in O(N log N)). Obviously, this will get worse as your file size increases. So you could call spectral_centroid(y=y, n_fft=16384, hop_length=16384, center=False) and because hop_length is set to the same value as n_fft you would compute the FFT for non-overlapping windows. And because n_fft is greater than the max number of samples in all your files (in this example), you should get only one value. Note that I set center to False to avoid an adjustment that is not necessary for your scenario.
Alternatively to choosing a long transform window, you could also compute many values for overlapping windows (or frames) using the STFT (which is what librosa does anyway) and simply average the resulting values like this:
import numpy as np
import librosa
y, sr = librosa.load(librosa.ex('trumpet'))
cent = librosa.feature.spectral_centroid(y=y, sr=sr, center=False)
avg_cent = np.mean(cent)
print(avg_cent)
2618.004809523263
The latter solution is in line with what is usually done in MIR and my recommendation. Note that this also allows you to use other statistics functions like the median, which may or may not be something you are interested in. In other words, you can determine the distribution of the centroids, which arguably carries more meaning.
I want to know how much energy is at a specific frequency in a signal. I am using FFT to get the spectrum, and the frequency step is determined by the length of my signal.
My spectrum looks, for example, like this :
I want to get the spectrum peak at a specific frequency, -0.08. However, the discretization of the spectrum only give me a peaks at -0.0729 and -0.0833.
Is there a way to shift the spectrum to make sure there is a data point at the frequency I want? Or a way to get the value without necessarily using fft?
Thank you very much!
What you're actually doing when you take a DFT (or any Fourier Transform) is measuring how much of your signal "intersects" with sines of certain frequencies. This is done by summing product of your signal with the complex conjugate of the wave of whatever frequency. Technically, this is called an inner product, which is a generalization of the dot product, and measures how "close" a signal is to another. So if you're only interested in one frequency, don't take the whole DFT, just look at one you want.
I'm not sure what your units are, so I'll assume you want the peak at f0 = -0.08 Hz (If your units are something else, like normalized to the sampling frequency, then you'll need to account for that). This corresponds to the complex exponential exp(2*pi*j*f0*t). Because you're sampling, your t is discrete, so t = n/fs, where fs is the sampling frequency (in Hz).
# assuming you're using numpy arrays
w = exp(-2*pi*1j*f0*arange(len(signal))/fs)
peak = abs(sum(signal*w))
There are different definitions of the DFT; I'm pretty sure numpy's corresponds to what I have above. The extra minus in the exponential is because it's the complex conjugate.
Notice that it's unlikely that w is actually periodic. If the number of samples is large enough this doesn't really matter. A good heuristic is at least 10 periods.
If you have discrete data but need an output for a continuous variable you'll necessarily need some kind of interpolation function. For a value by request style I would advise Scipy interp1d (example of the use of a interp1d function). I believe it's the fastest way to achieve your intended results.
I'm using a slightly modified version of this python code to do frequency analysis:
FFT wrong value?
Lets say I have a pack of sine waves in the time domain that are very close together in frequency, while sharing the same amplitude. This is how they look like in the frequency domain, using FFT on 1024 samples from which I strip out the second half, giving 512 bins of resolution:
This is when I apply a FFT over the same group of waves but this time with 128 samples (64 bins):
I expected a plateau-ish frequency response but it looks like the waves in the center are being cancelled. What are those "horns" I see? Is this normal?
I believe your result is correct. The peaks are at ±f1 and ±f2), corresponding to the respective frequency components of the two signals shown in your first plot.
I assume that you are shifting the DC component back to the center? What "waves in the center" are you referring to?
There are a couple of other potential issues that you should be aware of:
Aliasing: by inspection it appears that you have enough samples across your signal but keep in mind that artificial (or aliased) frequencies can be created by the FFT, if there are not enough sample points to capture the underlying frequency. Specifically, if your frequency is f, then you need your data sample spacing to be at least, Δx = 1/(2*f), or smaller.
Windowing: your signal is windowed (has a finite extent) so there will also be some broadening, ringing, or redistribution of power about each spatial frequency due to edge affects.
Since I don't know the details of your data, I went ahead and created a sinusoid and then sampled the data close to what appears to be your sampling rate. For example, below is a sinusoid with 64 points and with a signal frequency at 10 cycles (count the peaks):
The FFT result is then:
which shows the same quantitative features as yours, but without having your data, its difficult for me to match your exact situation (spacing and taper).
Next I applied a super-Gauss window function (shown below) to simulate the finite extent of your data:
After applying the window to the input signal we have:
The corresponding FFT result shows some additional power redistribution, due to the finite extent of the data:
Although I can't match your exact situation, I believe your results appear as expected and some qualitative features of your data have been identified. Hope this helps.
Sine waves closely spaced in the frequency domain will occasionally nearly cancel out in the time domain. Since your second FFT is 8 times shorter than your first FFT, you may have windowed just such an short area of cancellation. Try a different time location of shorter time window to see something different (or different phases of sinusoids).
I want to amplify audio input at specific frequencies, and I use numpy.fft.
So my question is: When changing the amplitudes of the signal, what happens with phase?
For example, if I multiply amplitudes in some frequency range, by some factor, let's say 2, do I need to change the phases, and if so, what should I do with them?
I've done the amplification without changing phases, and the result was not what I wanted. It's pretty much the same signal, with some unwanted noise.
You shouldn't need the change the phase for something like this. More likely the problem is that you need to be a bit more gentle about applying the boost. It sounds like you are taking some frequency window and multiplying by a constant while leaving everything else unchanged. This will cause ringing in the time domain with a very long tail. You need to smooth the transition from the gain=1 region to the gain=2 region, for instance by using a gaussian waveform with code that looks something like this:
x, t = get_waveform()
f0, df = get_parameters() # Center frequency and bandwidth of gain region
f = np.fft.rfft(x)
freqs = np.fft.fftfreq(len(x), t[1]-t[0])
freqs = freqs[0:len(f)] # rfft has only non-negative frequency components
gain_window = 1 + np.exp(-(freqs-f0)**2/(df)**2)
f = f * gain_window
x = np.fft.irfft(f)
return x
If that works, you can experiment with more aggressive functions that have sharper turn-on and a flatter top.
The FFT may not actually be what you want. FFTs are not normally used for real-time / streaming applications. This is because in the naive approach you have to collect the whole sample buffer before you start processing. For simple filtering applications it is often easier to do filtering directly in the time domain. This is what FIR and IIR filters do.
In order to filter with the fourier transform in real time what you have to do is break your data stream into overlapping blocks of a fixed length, FFT, filter, reverse FFT, and stich them back together without introducing glitches. This is possible, but it is tricky to get right. For a full-blown multi-channel EQ it might still be the best option, but you should at least consider time domain filtering.
If this is not a real-time application, then FFT is the way to go. For medium sized data sets (up to a few hundred megabytes) you can just FFT the whole data set. For much larger data sets you still have to break the data up into blocks, but they can be much larger blocks and you don't have to worry about the latency introduced.
Also, remember the FFT treats the signal as periodic, so if your signal doesn't go to zero at the beginning and end you will need to do some sort of windowing.
i don't know what to do after obtaining a set of complex numbers from FFT on a wav file.How could i obtain the corresponding frequencies.This is output i got after performing FFT which is shown below
[ 12535945.00000000 +0.j -30797.74496367 +6531.22295858j
-26330.14948055-11865.08322966j ..., 34265.08792783+31937.15794965j
-26330.14948055+11865.08322966j -30797.74496367 -6531.22295858j]
Actually the abs(x) operation only converts a real/imaginary pair from your result list into a magnitude. Do that unless you want to keep the imaginary portion for future use. So after conversion, each number in the result list represents a magnitude of signal at a certain frequency in your frequency spectrum. So frequency is represented by list index. When you plot the data on an XY graph, what you see is the magnitude of frequencies that your source signal contains. Don't forget that only your first half of data is valid. The other half is usually a mirror image of the first half due to aliasing.
For example say you run a 1024 point FFT on a wav file that contains data sampled at 10Khz. The FFT will take that 10Khz spectrum and divide that into 1024 'bins'. Then the FFT will decide how much each chunk of spectrum is present in the source wav file. Your output should be those bins. Generally when I do a frequency analysis, the actual numbers I get back aren't what's important. Its the magnitudes relative to surrounding bins that I'm interested in.
For a little more detail, we're relying on the principle of superposition which states that any time-varying signal containing many frequencies can be split up into many signals containing one component frequency each and vice versa. So the FFT output reflects this property. Each value of your output list represents a magnitude for a signal at a single frequency (usually called a 'bin') that is present in your source signal. Combine all those signals together and you should get your source signal back.
Oh and in case you didn't know, only the first half of your result list is valid due to Nyquist's Rule (or law, not sure) which says that all sampling systems can only reproduce frequencies in a signal that is at most half the sampling frequency. So if you sample a signal at 10Khz, you can only reproduce frequencies up to 5Khz from the data taken during your sampling. The same principle is the reason why only the first half of your FFT data is valid. The second half is an alias of the first half.
Sorry for the long-winded explanation, your question doesn't indicate what experience you have so I thought an explanation of the general gist of an FFT is needed.
As #KennyTM already explained on the duplicate question:
The frequency is determined by the index of the array. Each element corresponds to a frequency.
To determine the frequency of that each element represents, you need to know the sampling frequency of your data and the length of the array.
Basically, it would be something like:
sampling_freq = 1000.0 # in Hz
freq = np.linspace(0, (1.0 / sampling_freq / 2.0), (x.size / 2) + 1)
For one half of the fft array (which is symmetric about the center). My memory is rusty, though, so this may be a bit off...
Either way, numpy has a helper function to do it for you: numpy.fft.fftfreq
If I'm not mistaken, the frequency can be obtained by calculating the magnitude of the complex number. So a simple abs(x) on each of those complex numbers should return the frequency.