Obtain the MODWTMRA of a simple time-series signal and demonstrate perfect reconstruction.

Create a time-series signal

t = 1:10;
x = sin(2*pi*200*t);

Obtain the MODWT and the MODWTMRA and sum the MODWTMRA rows.

m = modwt(x);
mra = modwtmra(m);
xrec = sum(mra);

Use the maximum of the absolute values to show that the difference between the original signal and the reconstruction is extremely small. The largest absolute value is on the order of $$10^{-25}$$, which demonstrates perfect reconstruction.

max(abs(x-xrec))

ans = 5.5738e-25

Construct an MRA of an ECG signal down to level four using the db2 wavelet. The data are taken from Percival & Walden (2000), p.125 (data originally provided by William Constantine and Per Reinhall, University of Washington). The sampling frequency for the ECG signal is 180 hertz.

load wecg;
lev = 4;
wtecg = modwt(wecg,'db2',lev);
mra = modwtmra(wtecg,'db2');

Plot the ECG waveform and the MRA.

t = (0:numel(wecg)-1)/180;
subplot(6,1,1)
plot(t,wecg)
for kk = 2:lev+2
    subplot(6,1,kk)
    plot(t,mra(kk-1,:))
end
xlabel('Time (s)')
set(gcf,'Position',[0 0 500 700])

Construct a multiresolution analysis for the Southern Oscillation Index data. The sampling period is one day. Plot the level eight details corresponding to a scale of $$2^8$$ days. The details at this scale capture oscillations on a scale of approximately one year.

load soi
wtsoi = modwt(soi);
mrasoi = modwtmra(wtsoi);
plot(mrasoi(8,:))
title('Level 8 Details')

Obtain the MRA for the Deutsch Mark - U.S. Dollar exchange rate data using the minimum bandwidth scaling and wavelet filters with four coefficients.

load DM_USD;
Lo = [0.4801755, 0.8372545, 0.2269312, -0.1301477];
Hi = qmf(Lo);
wdm = modwt(DM_USD,Lo,Hi);
mra = modwtmra(wdm,Lo,Hi);

Obtain the MRA for an ECG signal using 'reflection' boundary handling. The data are taken from Percival & Walden (2000), p.125 (data originally provided by William Constantine and Per Reinhall, University of Washington).

load wecg;
wtecg = modwt(wecg,'reflection');
mra = modwtmra(wtecg,'reflection');

Show that the number of columns in the MRA is equal to the number of elements in the original signal.

isequal(size(mra,2),numel(wecg))

ans = logical
   1

This example demonstrates the differences between the functions MODWT and MODWTMRA. The MODWT partitions a signal's energy across detail coefficients and scaling coefficients. The MODWTMRA projects a signal onto wavelet subspaces and a scaling subspace.

Choose the sym6 wavelet. Load and plot an electrocardiogram (ECG) signal. The sampling frequency for the ECG signal is 180 hertz. The data are taken from Percival and Walden (2000), p.125 (data originally provided by William Constantine and Per Reinhall, University of Washington).

load wecg
t = (0:numel(wecg)-1)/180;
wv = 'sym6';
plot(t,wecg)
grid on
title(['Signal Length = ',num2str(numel(wecg))])
xlabel('Time (s)')
ylabel('Amplitude')

Take the MODWT of the signal.

wtecg = modwt(wecg,wv);

The input data are samples of a function $$f(x)$$ evaluated at $$N$$-many time points. The function can be expressed as a linear combination of the scaling function $$\varphi (x)$$ and wavelet $$\psi (x)$$at varying scales and translations: $$f(x)=\sum _{k=0}^{N-1}{c}_k{2}^{-J_0/2}\varphi (2^{-J_0}x-k)+\sum _{j=1}^{J_0}{f}_j(x)$$ where $$f_j(x)=\sum _{k=0}^{N-1}{d}_{j,k}{2}^{-j/2}\psi (2^{-j}x-k)$$ and $$J_0$$ is the number of levels of wavelet decomposition. The first sum is the coarse scale approximation of the signal, and the $$f_j(x)$$ are the details at successive scales. MODWT returns the $$N$$-many coefficients $$\{c_k\}$$and the $$(J_0\times N)$$-many detail coefficients $$\{d_{j,k}\}$$ of the expansion. Each row in wtecg contains the coefficients at a different scale.

When taking the MODWT of a signal of length $$N$$, there are $$\text{floor}({\log }_2(N))$$-many levels of decomposition (by default). Detail coefficients are produced at each level. Scaling coefficients are returned only for the final level. In this example, since $$N=2048$$, $$J_0=\text{floor}(\log 2(2048))=11$$ and the number of rows in wtecg is $$J_0+1=11+1=12$$.

The MODWT partitions the energy across the various scales and scaling coefficients: $${\left|\right|X\left|\right|}^2=\sum _{j=1}^{J_0}{\left|\right|W_j\left|\right|}^2+{\left|\right|V_{J_0}\left|\right|}^2$$ where $$X$$ is the input data, $$W_j$$ are the detail coefficients at scale $$j$$, and $$V_{J_0}$$ are the final-level scaling coefficients.

Compute the energy at each scale, and evaluate their sum.

energy_by_scales = sum(wtecg.^2,2);
Levels = {'D1';'D2';'D3';'D4';'D5';'D6';'D7';'D8';'D9';'D10';'D11';'A11'};
energy_table = table(Levels,energy_by_scales);
disp(energy_table)

    Levels     energy_by_scales
    _______    ________________

    {'D1' }         14.063     
    {'D2' }         20.612     
    {'D3' }         37.716     
    {'D4' }         25.123     
    {'D5' }         17.437     
    {'D6' }         8.9852     
    {'D7' }         1.2906     
    {'D8' }         4.7278     
    {'D9' }         12.205     
    {'D10'}         76.428     
    {'D11'}         76.268     
    {'A11'}         3.4192     

energy_total = varfun(@sum,energy_table(:,2))

energy_total=table
    sum_energy_by_scales
    ____________________

           298.28       

Confirm the MODWT is energy-preserving by computing the energy of the signal and comparing it with the sum of the energies over all scales.

energy_ecg = sum(wecg.^2);
max(abs(energy_total.sum_energy_by_scales-energy_ecg))

ans = 7.4402e-10

Take the MODWTMRA of the signal.

mraecg = modwtmra(wtecg,wv);

MODWTMRA returns the projections of the function $$f(x)$$ onto the various wavelet subspaces and final scaling space. That is, MODWTMRA returns $$\sum _{k=0}^{N-1}{c}_k{2}^{-J_0/2}\varphi (2^{-J_0}x-k)$$and the $$J_0$$-many $$\{f_j(x)\}$$evaluated at $$N$$-many time points. Each row in mraecg is a projection of $$f(x)$$ onto a different subspace. This means the original signal can be recovered by adding all the projections. This is not true in the case of the MODWT. Adding the coefficients in wtecg will not recover the original signal.

Choose a time point, add the projections of $$f(x)$$ evaluated at that time point and compare with the original signal.

time_point = 1000;
abs(sum(mraecg(:,time_point))-wecg(time_point))

ans = 3.0846e-13

Confirm that, unlike MODWT, MODWTMRA is not an energy-preserving transform.

energy_ecg = sum(wecg.^2);
energy_mra_scales = sum(mraecg.^2,2);
energy_mra = sum(energy_mra_scales);
max(abs(energy_mra-energy_ecg))

ans = 115.7053

The MODWTMRA is a zero-phase filtering of the signal. Features will be time-aligned. Demonstrate this by plotting the original signal and one of its projections. To better illustrate the alignment, zoom in.

plot(t,wecg,'b')
hold on
plot(t,mraecg(4,:),'-')
hold off
grid on
xlim([4 8])
legend('Signal','Projection','Location','northwest')
xlabel('Time (s)')
ylabel('Amplitude')

Make a similar plot using the MODWT coefficients at the same scale. Note that features will not be time-aligned. The MODWT is not a zero-phase filtering of the input.

plot(t,wecg,'b')
hold on
plot(t,wtecg(4,:),'-')
hold off
grid on
xlim([4 8])
legend('Signal','Coefficients','Location','northwest')
xlabel('Time (s)')
ylabel('Amplitude')