Time-frequency ridges from wavelet synchrosqueezing
Obtain the wavelet synchrosqueezed transform of a quadratic chirp and extract the maximum time-frequency ridge, in
fridge, and the associated row indices, in
Load the chirp signal and obtain its synchrosqueezed transform.
load quadchirp; [sst,f] = wsst(quadchirp);
Extract the maximum time-frequency ridge.
[fridge,iridge] = wsstridge(sst);
Plot the synchrosqueezed transform.
pcolor(tquad,f,abs(sst)) shading interp title('Synchrosqueezed Transform')
Overlay the plot of the maximum energy frequency ridge.
hold on plot(tquad,fridge) title('Synchrosqueezed Transform with Overlaid Ridge')
Extract the two highest energy modes from a multicomponent signal.
Obtain and plot the wavelet synchrosqueezed transform.
load multicompsig; sig = sig1+sig2; [sst,F] = wsst(sig,sampfreq); contour(t,F,abs(sst)); xlabel('Time'); ylabel('Hz'); grid on; title('Synchrosqueezed Transform of Two-Component Signal');
Using a penalty of 10, extract the two highest energy modes and plot the result.
[fridge,iridge] = wsstridge(sst,10,F,'NumRidges',2); hold on; plot(t,fridge,'k','linewidth',2);
sst— Synchrosqueezed transform
Synchrosqueezed transform, specified as a matrix.
sst is a
time-frequency matrix and is the output of
penalty— Frequency bins scaling penalty
Frequency bins scaling penalty, specified as a nonnegative scalar. This input penalizes changes in frequency by multiplying the penalty value by the squared distance between frequency bins. Use a penalty term when you extract multiple ridges, or when you have a single modulated component in additive noise. The penalty term prevents jumps in frequency that occur when the region of highest energy in the time-frequency plane changes abruptly.
comma-separated pairs of
the argument name and
Value is the corresponding value.
Name must appear inside quotes. You can specify several name and value
pair arguments in any order as
NumRidges— Number of highest energy time-frequency ridges
Number of highest energy time-frequency ridges to extract, specified as the
comma-separated pair consisting of
'NumRidges' and a positive integer. If
this integer is greater than 1,
wsstridge iteratively determines the
maximum energy time-frequency ridge by removing the previously computed ridges and the
default or specified
'NumFrequencyBins' on either side of each ridge
NumFrequencyBins— Number of frequency bins to remove
Number of frequency bins to remove from synchrosqueezed transform
sst when extracting multiple ridges, specified as the comma-separated
pair consisting of
'NumFrequencyBins' and a positive integer. This integer
must be less than or equal to
round(size(sst,1)/4). You can specify the
number of frequency bins to remove only if you extract more than one ridge. After extracting
the highest energy time-frequency ridge,
wsstridge removes the
sst values corresponding to the
iridge indices at
each time step. The energy is removed along the time-frequency ridge extended on both sides
iridge index by the specified number of frequency bins. If the
index of the extended time-frequency ridge exceeds the number of frequency bins at any time
wsstridge truncates the removal region at the first or last
frequency bin. To specify
'NumFrequencyBins', you must specify
fridge— Time-frequency ridge frequencies
Time-frequency ridge frequencies, returned as a vector or matrix. The frequencies
correspond to the time-frequency ridge at each time step.
fridge is an
nr matrix where N is the number
of time samples (columns) in
nr is the number
of ridges. The first column of the matrix contains the frequencies for the maximum energy
time-frequency ridge in
sst. Subsequent columns contain the frequencies
for the time-frequency ridges in decreasing energy order. By default,
fridge contains frequencies in cycles per sample.
iridge— Time-frequency ridge indices
Time-frequency ridge row indices of
sst, returned as a vector or
matrix. The row indices in
iridge correspond to the row index of the
maximum time-frequency ridge for each
iridge is an N-by-
nr matrix where
N is the number of time samples (columns) in
nr is the number of ridges. The first column of the matrix contains the
indices for the maximum energy time-frequency ridge in
columns contain the indices for the time-frequency ridges in decreasing energy order.
The function uses a penalized forward-backward greedy algorithm to extract the maximum-energy ridges from a time-frequency matrix. The algorithm finds the maximum time-frequency ridge by minimizing –ln A at each time point, where A is the absolute value of the matrix. Minimizing –ln A is equivalent to maximizing the value of A. The algorithm optionally constrains jumps in frequency with a penalty that is proportional to the distance between frequency bins.
The following example illustrates the time-frequency ridge algorithm using a penalty that is
two times the distance between frequency bins. Specifically, the distance between the elements
(m,n) is defined as
(j-m)2. The time-frequency matrix has three
frequency bins and three time steps. The matrix columns correspond to time steps, and the matrix
rows correspond to frequency bins. The values in the second row represent a sine wave.
Suppose you have the matrix:
1 4 4 2 2 2 5 5 4
Update the value for the (1,2) element as follows.
Leave the values at the first time point unaltered. Begin the algorithm with the (1,2) element of the matrix, which presents the first frequency bin at the second time point. The bin value is 4. Penalize the values in the first column based on their distance from the (1,2) element. Applying the penalty to the first column produces
original value + penalty × distance 1 + 2 × 0 = 1 2 + 2 × 1 = 4 5 + 2 × 4 = 13
1 4 4 2 13 5
Add the minimum value in column 1 to the current bin value, 4. The updated value for (1,2) becomes 5, which came from bin 1.
Update the values for the remaining elements in column 2 as follows.
Recompute the original column 1 values with the penalty factor using the same process as in Step 2a. Obtain the remaining second column values using the same process as in Step 2b. For example, when updating the (2,2) element, which has bin value 2, applying the penalty to the column yields
original value + penalty × distance 1 + 2 × 1 = 3 2 + 2 × 0 = 2 5 + 2 × 1 = 7
1 5(1) 4 2 4(2) 2 5 9(2) 4
Repeat Step 2 for the third column. But now the penalty is applied to the updated second column. For example, when updating the (1,3) element, the penalty is
5 + 2 × 0 = 5 4 + 2 × 1 = 6 9 + 2 × 4 = 17
1 5(1) 9(1) 2 4(2) 6(2) 5 9(2) 10(2)
Starting at the last column of the matrix, find the minimum value. Walk back in time
through the matrix by going from the current bin to the origin of that bin at the previous time
point. Keep track of the bin indices, which form the path composing the ridge. The algorithm
smooths the transition by using the origin bin instead of the bin with the minimum value. For
this example, the ridge indices are
2, which matches the energy path of the sine wave in row 2 of the matrix
shown in Step 1.
If you are extracting multiple ridges, the algorithm removes the first ridge from the time-frequency matrix and repeats the process.
 Daubechies, I., J. Lu, and H.-T. Wu. "Synchrosqueezed wavelet transforms: an empirical mode decomposition-like tool." Applied and Computational Harmonic Analysis. Vol. 30, Number 2, 2011, pp. 243–261.
 Thakur, G., E. Brevdo, N. S. Fučkar, and H.-T. Wu. "The Synchrosqueezing algorithm for time-varying spectral analysis: Robustness properties and new paleoclimate applications." Signal Processing. Vol. 93, Number 4, 2013, pp. 1079–1094.