1
$\begingroup$

I have a set of RF signal samples of 2s length each recorded at 2MHz sampling rate, such as this IQ file: https://www.dropbox.com/s/dd6fr4va4alpazj/move_x_movedist_1_speed_25k_sample_6.wav?dl=0

I can effectively gather the zeroth, first and second order coefficients using kymatio and plot them using the following code:

import scipy.io.wavfile
import numpy as np
import matplotlib.pyplot as plt
from kymatio.numpy import Scattering1D

path = r"move_x_movedist_1_speed_25k_sample_6.wav"

# Read in the sample WAV file
fs, x = scipy.io.wavfile.read(path)
x = x.T
print(fs)
# Once the recording is in memory, we normalise it to +1/-1
x = x / np.max(np.abs(x))
print(x)
# Set up parameters for the scattering transform
## number of samples, T
N = x.shape[-1]
print(N)
## Averaging scale as power of 2, 2**J, for averaging
## scattering scale of 2**6 = 64 samples
J = 6
## No. of wavelets per octave (resolve frequencies at
## a resolution of 1/16 octaves)
Q = 16
# Create object to compute scattering transform
scattering = Scattering1D(J, N, Q)
# Compute scattering transform of our signal sample
Sx = scattering(x)
# Extract meta information to identify scattering coefficients
meta = scattering.meta()
# Zeroth-order
order0 = np.where(meta['order'] == 0)
# First-order
order1 = np.where(meta['order'] == 1)
# Second-order
order2 = np.where(meta['order'] == 2)

#%%
# Plot original signal
plt.figure(figsize=(8, 2))
plt.plot(x)
plt.title('Original Signal')
plt.show()

# Plot zeroth-order scattering coefficient (average of
# original signal at scale 2**J)
plt.figure(figsize=(8,8))
plt.subplot(3, 1, 1)
plt.plot(Sx[order0][0])
plt.title('Zeroth-Order Scattering')
# Plot first-order scattering coefficient (arrange
# along time and log-frequency)
plt.subplot(3, 2, 1)
plt.imshow(Sx[0][order1], aspect='auto')
plt.title('First-order scattering [1]')
plt.subplot(3, 2, 2)
plt.imshow(Sx[1][order1], aspect='auto')
plt.title('First-order scattering [2]')
# Plot second-order scattering coefficient (arranged
# along time but has two log-frequency indicies -- one
# first- and one second-order frequency. Both are mixed
# along the vertical axis)
plt.subplot(3, 3, 1)
plt.imshow(Sx[0][order2], aspect='auto')
plt.title('Second-order scattering [0]')
plt.subplot(3, 3, 2)
plt.imshow(Sx[1][order2], aspect='auto')
plt.title('Second-order scattering [1]')
plt.show()

However, my task is to classify each of the samples using a neural network architecture. The problem I am having is that the shape and size of these coefficients is quite large and simply storing them in memory is not feasible (with around ~2k samples overall).

Therefore, I am wanting to know if there is a good way of extracting features I can use to represent this (i.e. if I take the MFCC i can create feature columns of 1-13 MFCC coefficients such as via mfcc = librosa.feature.mfcc(y=x, sr=fs, n_mfcc=14, hop_length=hop_length, n_fft=M)[1:]). Or perhaps there are other ways of shortening this data so it is actually useable in a neural network for training (i.e. taking the spatial averages of each of the coefficients: ¯Sm,J = ∑x ˜Sm,J ((λ1,··· ,λm), x) but this spatial info whilst reducing the dimension).

Any help would be great!

EDIT: Here are the plots for power spectrum and time-frequency from matlab signal analyser. How could this be used to identify the spectral occupancy for downsampling to minimise data.

enter image description here

$\endgroup$
4
  • $\begingroup$ MFCC are used to represent speech... what possible use could they have in RF? $\endgroup$
    – MBaz
    Mar 22, 2022 at 13:44
  • $\begingroup$ I wasn't suggesting that MFCC would be good for RF, it was me taking an example of how (in audio classification) MFCC coefficients can be used as feature columns for a simple MLP network. One approach with the wavelets is of course I can just classify the images produced by plotting each of the order coefficients, however some suggest that classifying images of these things (i.e. spectrograms) are not that good for signals. $\endgroup$
    – rshah
    Mar 22, 2022 at 13:47
  • 1
    $\begingroup$ A first step that can be applied to minimize the amount of representation data needed is to convert the waveform to its complex analytic signal representation and shifted to baseband (remove the carrier and maintain a complex signal centered at DC) and stored at the minimum bandwidth possible. This will reduce the data from which further features can be extracted. Please comment if this would be of interest and I can provide specific details. A big question for its effectiveness is what the spectral occupancy actually is for your signal of interest, and once processed if 2 MHz fs is needed. $\endgroup$ Mar 24, 2022 at 20:50
  • $\begingroup$ As far as "classification" wouldn't the question and interest be in how this waveforms varies compared to other unique waveforms from the classification set? $\endgroup$ Mar 26, 2022 at 2:09

1 Answer 1

2
+50
$\begingroup$

Expanding on my related answer but for wavelet scattering:

  1. Higher T -> greater time-shift invariance.
  2. Higher Q -> greater frequency resolution, lower time resolution and time-warp stability.
  3. Higher J -> larger largest-scale feature (largest kernel convolution). Should be set to whatever we think our largest-scale relevant structure is - e.g. a sentence is longer than a word, a word is longer than a character.

Global averaging, i.e. T==len(x), can perform surprisingly well, while drastically slashing input size. Would also help to split up input along time and then join scatterings afterwards ("no-overlap-join convolution") as it's faster than one big convolution.

A pressing question is also whether such a high sampling rate is indeed required (as Dan pointed), and whether relevant features span the entire 0-2MHz bandwidth. If the effective bandwidth is only e.g. 1-1.5MHz, then data can be shrunk four-fold with downsampling techniques before feeding to the scattering network, greatly reducing processing time.

Re: downsampling From what I can tell per the graphs, downsampling by 2 shouldn't hurt, but 1) that's not a lot, and 2) it might indeed hurt if what's discarded is important, which depends on whether there's e.g. power law scaling as in e.g. EEG. The freqs seem rather evenly distributed, so if nothing else is known of thier "importance", maybe not much can be done. One approach:

  1. Downsample by 8, train, test
  2. Repeat for downsampling by 4, see if there's improvement
  3. Yes: repeat for downsampling by 2. No: repeat for downsampling by 16.

P.S, this should also help, but no meta.

$\endgroup$
6
  • $\begingroup$ Thanks for the answer! Related to both this and Dan's comment, are there any preferred downsampling techniques for signals? Scipy has its decimate function, however I'm not sure how to truly find the "effective bandwidth". $\endgroup$
    – rshah
    Mar 25, 2022 at 8:26
  • 1
    $\begingroup$ @rshah I'm afraid that's a separate question, but I added a partial response. Worth asking separately. $\endgroup$ Mar 25, 2022 at 22:47
  • 1
    $\begingroup$ I agree with Over that downsample by 2 won't hurt. Anything lower and you are cutting into that spectral content that is obviously there below 500KHz. So not saving much in this case (In contrast to seeing a very small bandwidth clustered somewhere with nothing anywhere else). $\endgroup$ Mar 26, 2022 at 2:08
  • 2
    $\begingroup$ @DanBoschen Right, if we're just discarding highs then scipy.decimate (or for GPU cusignal.decimate) is the way to go. Though again we should be careful on "low energy = unimportant" per things like power law scaling; an empirical survey could go a long way here. $\endgroup$ Mar 26, 2022 at 2:57
  • 2
    $\begingroup$ Very smart, wasn't thinking like that but totally makes sense. $\endgroup$ Mar 26, 2022 at 3:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.