How Can I Find Peak Values of Defined Areas from Spectrogram Data using numpy? - numpy

I have spectrogram data from an audio analysis which looks like this:
On one axis I have frequencies in Hz and in the other times in seconds. I added the grid over the map to show the actual data points. Due to the nature of the used frequency analysis, the best results never give evenly spaced time and frequency values.
To allow comparison data from multiple sources, I would like to normalize this data. For this reason, I would like to calculate the peak values (maximum and minimum values) for specified areas in the map.
The second visualization shows the areas where I would like to calculate the peak values. I marked an area with a green rectangle to visualize this.
While for the time values, I would like to use equally spaced ranges (e.g 0.0-10.0, 10.0-20.0, 20.0-30.0), The frequency ranges are unevenly distributed. In higher frequencies, they will be like 450-550, 550-1500, 1500-2500, ...
You can download an example data-set here: data.zip. You can unpack the datasets like this:
with np.load(DATA_PATH) as data:
frequency_labels = data['frequency_labels']
time_labels = data['time_labels']
spectrogram_data = data['data']
DATA_PATH has to point to the path of the .npz data file.
As input, I would provide an array of frequency and time ranges. The result should be another 2d NumPy ndarray with either the maximum or the minimum values. As the amount of data is huge, I would like to rely on NumPy as much as possible to speed up the calculations.
How do I calculate the maximum/minimum values of defined areas from a 2d data map?

Related

When should Data Binning be used in data processing?

In data pre-processing, Data Binning is a technique to convert continuous values of a feature to categorical ones. For example, sometimes, the values of age feature in datasets are replaced with one of intervals such as:
[10,20),
[20,30),
[30,40].
When is the best time to use Data Binning? Does it (always) lead to a better result in a predication system or it may work as a trial and error?
Trial and error mostly. When you apply binning to a continuous variable you automatically throw away some information. Many algorithms would prefer a continuous input to make a prediction and many would bin the continuous input themselves. Binning would be wise to apply if your continuous variable is noisy, meaning the values for your variable were not recorded very accurately. Then, binning could reduce this noise. There are binning strategies such as equal width binning or equal frequency binning. I would recommend avoiding equal width binning when your continuous variable is unevenly distributed.

finding transition point of data slope

I am wondering if there is a method to approach this problem.
The reason I need this is because for a certain trend of data I need to use a specific formula and for the next trend of the data I need to use a different formula.
Also, the data is not simple but there are two distinct slopes.
All data points are in excel cells.I haven't started the code yet. I am thinking about using (0,1,2,3,4) data points and finding slope and keep moving by 1 (1,2,3,4,5) then somehow calculate a difference in the 2 slopes and when they are significant. to call that the transition point
You may be able to reduce the problem to finding inflection points. This can be defined as point where the data flattens briefly to either resume a trend, change it (but in the same direction), or reverse it. You can do this by finding small time clusters with slope of zero. Or a better idea would be to divide your y data into horizontal bins. If a certain threshold of number of data points in a bin is reached, a change in trend is in progress. You can vary the inflection sensitivity by varying the bin size and/or varying the minimum number of points in a bin.

Frequency Range from FFT using vDSP

I have an array of values as input which can be plotted as follows…
Using vDSP_zvmagsD I get an array that I can plot as follows…
How do I get the frequency range that I need to label the x-axis?
The size of your frequency bins depend on the sampling rate of your input signal and the size of your FFT window:
input frequency / input samples = hz/frequency bin
Here you can find a more detailed answer by electronics stackexchange user Mark. There you will also find useful information about the tradeof between frequency resolution (bin size) and temporal resolution (when does which frequency occur)

Where to start with Fourier Analysis

I'm reading data from the microphone and want to perform some analysis on it. I'm attempting to generate a spectrum analyser something like this:
What I have at the moment is this:
My understanding is that I need to perform a Fourier analysis - a Fast Fourier Transform ? - to extract the component frequencies and their amplitudes.
Can someone confirm my understanding is correct and exactly what type of Fourier transform I need to apply?
At the moment, I'm getting frames containing 4k samples from the mic (using NAudio). The buffer I've got is 16bits/sample (Signed Short). For reference, the above plot shows approx half a frame
I'm coding in VB so any .Net libraries/examples (preferably on NuGet) would be of most use. I believe implementations vary considerably so the less I have to massage my data, the better.
The top plot is that of a spectrograph, where each vertical time line is colored based on the magnitudes of the result from an FFT (likely windowed) of a slice in time (possibly overlapped) of the input waveform. The number of vertical points to plot (the frequency resolution) is related to the length of the FFT. Almost any FFT will do. If you use the most common complex-to-complex FFT, just set the imaginary portion of each complex input sample to zero, copy a slice in time of samples of your input waveform to the "real" part, FFT, and take the magnitude or log magnitude of each complex result bin, then map these values to colors per your preference.

VB FFT - stuck understanding relationship of results to frequency

Trying to understand an fft (Fast Fourier Transform) routine I'm using (stealing)(recycling)
Input is an array of 512 data points which are a sample waveform.
Test data is generated into this array. fft transforms this array into frequency domain.
Trying to understand relationship between freq, period, sample rate and position in fft array. I'll illustrate with examples:
========================================
Sample rate is 1000 samples/s.
Generate a set of samples at 10Hz.
Input array has peak values at arr(28), arr(128), arr(228) ...
period = 100 sample points
peak value in fft array is at index 6 (excluding a huge value at 0)
========================================
Sample rate is 8000 samples/s
Generate set of samples at 440Hz
Input array peak values include arr(7), arr(25), arr(43), arr(61) ...
period = 18 sample points
peak value in fft array is at index 29 (excluding a huge value at 0)
========================================
How do I relate the index of the peak in the fft array to frequency ?
If you ignore the imaginary part, the frequency distribution is linear across bins:
Frequency#i = (Sampling rate/2)*(i/Nbins).
So for your first example, assumming you had 256 bins, the largest bin corresponds to a frequency of 1000/2 * 6/256 = 11.7 Hz.
Since your input was 10Hz, I'd guess that bin 5 (9.7Hz) also had a big component.
To get better accuracy, you need to take more samples, to get smaller bins.
Your second example gives 8000/2*29/256 = 453Hz. Again, close, but you need more bins.
Your resolution here is only 4000/256 = 15.6Hz.
It would be helpful if you were to provide your sample dataset.
My guess would be that you have what are called sampling artifacts. The strong signal at DC ( frequency 0 ) suggests that this is the case.
You should always ensure that the average value in your input data is zero - find the average and subtract it from each sample point before invoking the fft is good practice.
Along the same lines, you have to be careful about the sampling window artifact. It is important that the first and last data point are close to zero because otherwise the "step" from outside to inside the sampling window has the effect of injecting a whole lot of energy at different frequencies.
The bottom line is that doing an fft analysis requires more care than simply recycling a fft routine found somewhere.
Here are the first 100 sample points of a 10Hz signal as described in the question, massaged to avoid sampling artifacts
> sinx[1:100]
[1] 0.000000e+00 6.279052e-02 1.253332e-01 1.873813e-01 2.486899e-01 3.090170e-01 3.681246e-01 4.257793e-01 4.817537e-01 5.358268e-01
[11] 5.877853e-01 6.374240e-01 6.845471e-01 7.289686e-01 7.705132e-01 8.090170e-01 8.443279e-01 8.763067e-01 9.048271e-01 9.297765e-01
[21] 9.510565e-01 9.685832e-01 9.822873e-01 9.921147e-01 9.980267e-01 1.000000e+00 9.980267e-01 9.921147e-01 9.822873e-01 9.685832e-01
[31] 9.510565e-01 9.297765e-01 9.048271e-01 8.763067e-01 8.443279e-01 8.090170e-01 7.705132e-01 7.289686e-01 6.845471e-01 6.374240e-01
[41] 5.877853e-01 5.358268e-01 4.817537e-01 4.257793e-01 3.681246e-01 3.090170e-01 2.486899e-01 1.873813e-01 1.253332e-01 6.279052e-02
[51] -2.542075e-15 -6.279052e-02 -1.253332e-01 -1.873813e-01 -2.486899e-01 -3.090170e-01 -3.681246e-01 -4.257793e-01 -4.817537e-01 -5.358268e-01
[61] -5.877853e-01 -6.374240e-01 -6.845471e-01 -7.289686e-01 -7.705132e-01 -8.090170e-01 -8.443279e-01 -8.763067e-01 -9.048271e-01 -9.297765e-01
[71] -9.510565e-01 -9.685832e-01 -9.822873e-01 -9.921147e-01 -9.980267e-01 -1.000000e+00 -9.980267e-01 -9.921147e-01 -9.822873e-01 -9.685832e-01
[81] -9.510565e-01 -9.297765e-01 -9.048271e-01 -8.763067e-01 -8.443279e-01 -8.090170e-01 -7.705132e-01 -7.289686e-01 -6.845471e-01 -6.374240e-01
[91] -5.877853e-01 -5.358268e-01 -4.817537e-01 -4.257793e-01 -3.681246e-01 -3.090170e-01 -2.486899e-01 -1.873813e-01 -1.253332e-01 -6.279052e-02
And here is the resulting absolute values of the fft frequency domain
[1] 7.160038e-13 1.008741e-01 2.080408e-01 3.291725e-01 4.753899e-01 6.653660e-01 9.352601e-01 1.368212e+00 2.211653e+00 4.691243e+00 5.001674e+02
[12] 5.293086e+00 2.742218e+00 1.891330e+00 1.462830e+00 1.203175e+00 1.028079e+00 9.014559e-01 8.052577e-01 7.294489e-01
I'm a little rusty too on math and signal processing but with the additional info I can give it a shot.
If you want to know the signal energy per bin you need the magnitude of the complex output. So just looking at the real output is not enough. Even when the input is only real numbers. For every bin the magnitude of the output is sqrt(real^2 + imag^2), just like pythagoras :-)
bins 0 to 449 are positive frequencies from 0 Hz to 500 Hz. bins 500 to 1000 are negative frequencies and should be the same as the positive for a real signal. If you process one buffer every second frequencies and array indices line up nicely. So the peak at index 6 corresponds with 6Hz so that's a bit strange. This might be because you're only looking at the real output data and the real and imaginary data combine to give an expected peak at index 10. The frequencies should map linearly to the bins.
The peaks at 0 indicates a DC offset.
It's been some time since I've done FFT's but here's what I remember
FFT usually takes complex numbers as input and output. So I'm not really sure how the real and imaginary part of the input and output map to the arrays.
I don't really understand what you're doing. In the first example you say you process sample buffers at 10Hz for a sample rate of 1000 Hz? So you should have 10 buffers per second with 100 samples each. I don't get how your input array can be at least 228 samples long.
Usually the first half of the output buffer are frequency bins from 0 frequency (=dc offset) to 1/2 sample rate. and the 2nd half are negative frequencies. if your input is only real data with 0 for the imaginary signal positive and negative frequencies are the same. The relationship of real/imaginary signal on the output contains phase information from your input signal.
The frequency for bin i is i * (samplerate / n), where n is the number of samples in the FFT's input window.
If you're handling audio, since pitch is proportional to log of frequency, the pitch resolution of the bins increases as the frequency does -- it's hard to resolve low frequency signals accurately. To do so you need to use larger FFT windows, which reduces time resolution. There is a tradeoff of frequency against time resolution for a given sample rate.
You mention a bin with a large value at 0 -- this is the bin with frequency 0, i.e. the DC component. If this is large, then presumably your values are generally positive. Bin n/2 (in your case 256) is the Nyquist frequency, half the sample rate, which is the highest frequency that can be resolved in the sampled signal at this rate.
If the signal is real, then bins n/2+1 to n-1 will contain the complex conjugates of bins n/2-1 to 1, respectively. The DC value only appears once.
The samples are, as others have said, equally spaced in the frequency domain (not logarithmic).
For example 1, you should get this:
alt text http://home.comcast.net/~kootsoop/images/SINE1.jpg
For the other example you should get
alt text http://home.comcast.net/~kootsoop/images/SINE2.jpg
So your answers both appear to be correct regarding the peak location.
What I'm not getting is the large DC component. Are you sure you are generating a sine wave as the input? Does the input go negative? For a sinewave, the DC should be close to zero provided you get enough cycles.
Another avenue is to craft a Goertzel's Algorithm of each note center frequency you are looking for.
Once you get one implementation of the algorithm working you can make it such that it takes parameters to set it's center frequency. With that you could easily run 88 of them or what ever you need in a collection and scan for the peak value.
The Goertzel Algorithm is basically a single bin FFT. Using this method you can place your bins logarithmically as musical notes naturally go.
Some pseudo code from Wikipedia:
s_prev = 0
s_prev2 = 0
coeff = 2*cos(2*PI*normalized_frequency);
for each sample, x[n],
s = x[n] + coeff*s_prev - s_prev2;
s_prev2 = s_prev;
s_prev = s;
end
power = s_prev2*s_prev2 + s_prev*s_prev - coeff*s_prev2*s_prev;
The two variables representing the previous two samples are maintained for the next iteration. This can be then used in a streaming application. I thinks perhaps the power calculation should be inside the loop as well. (However it is not depicted as such in the Wiki article.)
In the tone detection case there would be 88 different coeficients, 88 pairs of previous samples and would result in 88 power output samples indicating the relative level in that frequency bin.
WaveyDavey says that he's capturing sound from a mic, thru the audio hardware of his computer, BUT that his results are not zero-centered. This sounds like a problem with the hardware. It SHOULD BE zero-centered.
When the room is quiet, the stream of values coming from the sound API should be very close to 0 amplitude, with slight +- variations for ambient noise. If a vibratory sound is present in the room (e.g. a piano, a flute, a voice) the data stream should show a fundamentally sinusoidal-based wave that goes both positive and negative, and averages near zero. If this is not the case, the system has some funk going on!
-Rick