fridge = tfridge(tfm,f)
[fridge,iridge] = tfridge(tfm,f)
[fridge,iridge,lridge] = tfridge(tfm,f)
[___] = tfridge(tfm,f,penalty)
[___] = tfridge(___,'NumRidges',nr)
[___] = tfridge(___,'NumRidges',nr,'NumFrequencyBins',nbins)
[___] = tfridge(___,'NumRidges', extracts
nr time-frequency ridges with the highest
energy. This syntax accepts any combination of input arguments from
Create a matrix that resembles a time-frequency matrix with a sharp ridge. Visualize the matrix in three dimensions.
t = 0:0.05:10; f = 0:0.2:8; rv = 1; [F,T] = ndgrid(f,t); S = zeros(size(T)); S(abs((F-4)-cos((T-6).^2))<0.1) = rv; mesh(T,F,S) view(-30,60)
Add noise to the matrix and redisplay the plot.
S = S+rand(size(S))/10; mesh(T,F,S) view(-30,60) xlabel('Time') ylabel('Frequency')
Extract the ridge and plot the result.
[fridge,~,lridge] = tfridge(S,f); rvals = S(lridge); hold on plot3(t,fridge,rvals,'k','linewidth',4) hold off
Generate a signal that is sampled at 3 kHz for one second. The signal consists of two tones and a quadratic chirp.
The first tone has a frequency of 1000 Hz and unit amplitude.
The second tone has a frequency of 1200 Hz and unit amplitude.
The chirp has an initial frequency of 500 Hz and reaches 750 Hz at the end of the sampling. It has an amplitude of six.
fs = 3000; t = 0:1/fs:1-1/fs; x1 = 6*chirp(t,fs/6,t(end),fs/4,'quadratic'); x2 = sin(2*pi*fs/3*t); x3 = sin(2*pi*fs/2.5*t); x = x1+x2+x3;
Compute and display the Fourier synchrosqueezed transform of the signal.
[sst,f] = fsst(x,fs); mx = max(abs(sst(:)))*ones(size(t)); mesh(t,f,abs(sst)) view(2)
Extract and plot the two highest-energy signal components. Set no penalty for changing frequency.
penval = 0; fridge = tfridge(sst,f,penval,'NumRidges',2); hold on plot3(t,fridge,mx,'w','linewidth',5) hold off
The two tones have the same amplitude, and the algorithm jumps between them. Set the penalty for changing frequency to 1.
penval = 1; fridge = tfridge(sst,f,penval,'NumRidges',2); mesh(t,f,abs(sst)) view(2) xlabel('Time (s)') ylabel('Frequency (Hz)') hold on plot3(t,fridge,mx,'w','linewidth',5) hold off
Set the penalty to a high value for comparison. The chirp is penalized because its frequency is not constant.
penval = 1000; fridge = tfridge(sst,f,penval,'NumRidges',2); mesh(t,f,abs(sst)) view(2) xlabel('Time (s)') ylabel('Frequency (Hz)') hold on plot3(t,fridge,mx,'w','linewidth',5) hold off
Generate a signal composed of two quadratic chirps. The signal is sampled at 1 kHz for 3 seconds. The chirps are such that the instantaneous frequency is symmetric about the halfway point of the sampling interval. One chirp is concave and the other chirp is convex. The concave chirp has twice the amplitude of the convex chirp.
fs = 1e3; t = 0:1/fs:3; x = chirp(t-1.5,100,1.1,200,'quadratic',,'convex'); y = 2*chirp(t-1.5,300,1.1,400,'quadratic',,'concave'); % To hear, type soundsc(x+y,fs)
Compute and display the Fourier synchrosqueezed transform of the signal.
sig = x+y; [sst,f,t] = fsst(sig,fs); fsst(sig,fs,'yaxis')
Extract the two time-frequency ridges that have the highest energy. Specify a penalty of 1 for changes in frequency. Remove 1 frequency bin around the ridge with the highest energy before extracting the second ridge. Plot the ridges.
nfb = 1; [fr,ir] = tfridge(sst,f,1,'NumRidges',2,'NumFrequencyBins',nfb); plot(t,fr)
One bin is insufficient: The function finds a second ridge that is partly on the slope of the first. Increase to 50 the number of bins to remove and repeat the calculation.
nfb = 50; [fr,ir] = tfridge(sst,f,1,'NumRidges',2,'NumFrequencyBins',nfb); plot(t,fr)
Removing too many bins distorts lower-energy ridges. Decrease the number to 15 and repeat the calculation.
nfb = 15; [fr,ir] = tfridge(sst,f,1,'NumRidges',2,'NumFrequencyBins',nfb); plot(t,fr)
Invert the transform corresponding to the two ridges. Add the ridges to reconstruct the signal. Plot the difference between the reconstructed signal and the chirps.
itr = ifsst(sst,,ir,'NumFrequencyBins',nfb); xrec = sum(itr'); plot(t,xrec-(x+y)) ylim([-.1 .1])
% To hear, type soundsc(xrec,fs)
The agreement is good most of the time but deteriorates at the ends, where frequencies change most rapidly.
tfm— Time-frequency matrix
Time-frequency matrix, specified as a matrix.
the synchrosqueezed transform of a sinusoid.
Complex Number Support: Yes
f— Sampling frequencies
Sampling frequencies, specified as a vector. The length of
equal the number of rows in
penalty— Penalty for changing frequency
Penalty for changing frequency, specified as a nonnegative real scalar.
nr— Number of time-frequency ridges to extract
Number of time-frequency ridges to extract, specified as the
comma-separated pair consisting of
a positive integer scalar. You can specify this name-value pair anywhere
tfm in the input argument list.
nr is greater than 1,
Extracts the time-frequency ridge with the highest energy
tfm the energy contained
in the ridge it extracted and in the
frequency bins on either side of the ridge
Extracts the highest-energy ridge in the modified
Iterates until it has extracted
nbins— Number of bins to remove
Number of bins to remove when extracting multiple ridges, specified
as the comma-separated pair consisting of
a positive integer scalar.
nbins must be smaller
than 1/4 of the sampling frequencies. Indices close to the frequency
edges that have fewer than
nbins bins on one
side are reconstructed using a smaller number of bins.
fridge— Time-frequency ridges
Time-frequency ridges, returned as a matrix with
The number of rows in
fridge equals the number
of columns in
tfm. The first column contains
the frequencies that correspond to the highest-energy ridge. Subsequent
columns contain the frequencies of the other ridges in decreasing
order of energy.
iridge— Ridge row indices
lridge— Ridge linear indices
Ridge linear indices, returned as a matrix with
defined so that
tfm(lridge) is the amplitude of
the ridges. The number of rows in
the number of columns in
tfm. The first column
contains the indices that correspond to the highest-energy ridge.
Subsequent columns contain the indices of the other ridges in decreasing
order of energy.
lridge is equivalent to
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.