scal2frq

This example shows how the pseudo-frequency changes as you double the scale.

Construct a vector of scales with 10 voices per octave over five octaves.

vpo = 10;
no = 5;
a0 = 2^(1/vpo);
ind = 0:vpo*no;
sc = a0.^ind;

Verify that the range of scales covers five octaves.

log2(max(sc)/min(sc))

ans = 
5.0000

If you plot the scales, you can use a data cursor to confirm that the scale at index $$n+10$$ is twice the scale at index $$n$$. Set the y-ticks to mark each octave.

plot(ind,sc)
title('Scales')
xlabel('Index')
ylabel('Scale')
grid on
set(gca,'YTick',2.^(0:5))

Convert the scales to pseudo-frequencies for the real-valued Morlet wavelet. First, assume the sampling period is 1. Since there are 10 voices per octave, display every tenth row in the table. Observe that for each doubling of the scale, the pseudo-frequency is cut in half.

pf = scal2frq(sc,"morl");
T = [sc(:) pf(:)];
T = array2table(T, ...
    'VariableNames',{'Scale','Pseudo-Frequency'});
disp(T(1:10:end,:))

    Scale    Pseudo-Frequency
    _____    ________________

      1            0.8125    
      2           0.40625    
      4           0.20313    
      8           0.10156    
     16          0.050781    
     32          0.025391    

Assume that data is sampled at 100 Hz. Construct a table with the scales, the corresponding pseudo-frequencies, and periods. Display every tenth row in the table.

Fs = 100;
DT = 1/Fs;
pf = scal2frq(sc,"morl",DT);
T = [sc(:)/Fs pf(:) 1./pf(:)];
T = array2table(T, ...
    'VariableNames',{'Scale','Pseudo-Frequency','Period'});
T(1:vpo:end,:)

ans=6×3 table
    Scale    Pseudo-Frequency     Period 
    _____    ________________    ________

    0.01           81.25         0.012308
    0.02          40.625         0.024615
    0.04          20.313         0.049231
    0.08          10.156         0.098462
    0.16          5.0781          0.19692
    0.32          2.5391          0.39385

Note the presence of the $$\Delta t=\frac{1}{Fs}$$ factor in scal2frq. This is necessary in order to achieve the proper scale-to-frequency conversion. The $$\Delta t$$ is needed to adjust the raw scales properly. For example, with:

f = scal2frq(1,'morl',0.01);

You are really asking what happens to the center frequency of the mother Morlet wavelet, if you dilate the wavelet by 0.01. In other words, what is the effect on the center frequency if instead of $$\psi (t)$$, you look at $$\psi (t/0.01)$$. The $$\Delta t$$ provides the correct adjustment factor on the scales.

You could have obtained the same results by first converting the scales to their adjusted sizes and then using scal2frq without specifying $$\Delta t$$.

scadjusted = sc.*0.01;
pf2 = scal2frq(scadjusted,'morl');
max(pf-pf2)

ans = 
0

The example shows how to create a contour plot of the CWT using approximate frequencies in Hz.

Create a signal consisting of two sine waves with disjoint support in additive noise. Assume the signal is sampled at 1 kHz.

Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = 1.5*cos(2*pi*100*t).*(t<0.25)+1.5*cos(2*pi*50*t).*(t>0.5 & t<=0.75);
x = x+0.05*randn(size(t));

Obtain the CWT of the input signal and plot the result.

[cfs,f] = cwt(x,Fs);
contour(t,f,abs(cfs).^2); 
axis tight;
grid on;
xlabel('Time');
ylabel('Approximate Frequency (Hz)');
title('CWT with Time vs Frequency');