Obtain the MODWPT of an electrocardiogram (ECG) signal using the default length 18 Fejer-Korovkin ('fk18') wavelet.

load wecg;
wpt = modwpt(wecg);

wpt is a 16-by-2048 matrix containing the sequency-ordered wavelet packet coefficients for the wavelet packet transform nodes. In this case, the nodes are at level 4. Each node corresponds to an approximate passband filtering of $$[n{f}_s/2^5,(n+1)f_s/2^5)$$, where n = 0,...,15, and $$f_s$$ is the sampling frequency. Plot the wavelet packet coefficients at node (4,2), which is level 4, node 2.

plot(wpt(3,:))
title('Node 4 Wavelet Packet Coefficients')

Obtain the MODWPT of Southern Oscillation Index data with the Daubechies extremal phase wavelet with two vanishing moments ('db2').

load soi;
wsoi = modwpt(soi,'db2');

Verify that the size of the resulting transform contains 16 nodes. Each node is in a separate row.

size(wsoi)

ans = 1×2

          16       12998

Obtain the MODWPT of an ECG waveform using the Fejer-Korovkin length 18 scaling and wavelet filters.

load wecg; 
[lo,hi] = wfilters('fk18');
wpt = modwpt(wecg,lo,hi);

Obtain the MODWPT and full wavelet packet tree of an ECG waveform using the default length 18 Fejer-Korovkin ('fk18') wavelet. Extract and plot the node coefficients at level 3, node 2.

load wecg;     
[wpt,packetlevels,cfreq] = modwpt(wecg,'FullTree',true);
p3 = wpt(packetlevels==3,:);
plot(p3(3,:))
title('Level 3, Node 2 Wavelet Coefficients')

Display the center frequencies at level 3.

cfreq(packetlevels==3,:)

ans = 8×1

    0.0312
    0.0938
    0.1562
    0.2188
    0.2812
    0.3438
    0.4062
    0.4688

Obtain and plot the MODWPT energy and relative energy of an ECG waveform.

load wecg
[wpt,~,cfreq,energy,relenergy] = modwpt(wecg);

Show that the sum of the MODWPT energies is equal to the sum of the energy in the original signal. The difference between the total MODWPT energy and the signal energy is small enough to be considered insignificant.

disp(['Difference between MODWPT energy and signal energy: ',num2str(sum(energy)-sum(wecg.^2))])

Difference between MODWPT energy and signal energy: 3.6122e-09

Plot the MODWPT energy by node.

figure
bar(1:16,energy)
xlabel('Node')
ylabel('Energy')
title('Energy by Node')

disp(['Total power in passband: ',num2str(energy(1))])

Total power in passband: 200.8446

Plot the relative energy and show the percentage of signal energy in the first passband [0,5.6250].

figure
bar(1:16,relenergy*100)
xlabel('Node')
ylabel('Percent Energy')
title('Energy Relative to Signal Energy by Node')

disp(['Percentage of signal power in passband: ',num2str(relenergy(1)*100)])

Percentage of signal power in passband: 67.3352

Obtain the time-aligned MODWPT of two intermittent sine waves in noise. The sine wave frequencies are 150 Hz and 200 Hz. The data is sampled at 1000 Hz.

dt = 0.001;     
t = 0:dt:1-dt;     
x = cos(2*pi*150*t).*(t>=0.2 & t<0.4)+ sin(2*pi*200*t).*(t>0.6 & t<0.9);     
y = x+0.05*randn(size(t));
[wpta,~,Falign] = modwpt(x,'TimeAlign',true);
[wptn,~,Fnon] = modwpt(x);

Compare the nonaligned and time-aligned time-frequency plots.

subplot(2,1,1);
contour(t,Fnon.*(1/dt),abs(wptn).^2); 
grid on;     
ylabel('Hz');     
title('Time-Frequency Plot (Nonaligned)');
subplot(2,1,2)     
contour(t,Falign.*(1/dt),abs(wpta).^2); 
grid on;     
xlabel('Time'); 
ylabel('Hz');     
title('Time-Frequency Plot (Aligned)');