In this example you will generate symlet scaling filter coefficients whose norm is equal to 1. You will also confirm the coefficients satisfy a necessary relation.

Compute the scaling filter coefficients of the order 10 symlet whose sum equals$\sqrt{2}$.

n = 10;
w = symaux(n,sqrt(2));

Confirm the sum of the coefficients is equal to $\sqrt{2}$ and the norm is equal to 1.

sqrt(2)-sum(w)

ans = 0

1-sum(w.^2)

ans = 1.1324e-14

Since integer translations of the scaling function form an orthogonal basis, the coefficients satisfy the relation ${\sum }_{\mathit{n}}\mathit{w}\left(\mathit{n}\right)\mathit{w}\left(\mathit{n}-2\mathit{k}\right)=\delta \left(\mathit{k}\right)$. Confirm this by taking the autocorrelation of the coefficients and plotting the result.

corrw = xcorr(w,w);
stem(corrw)
grid on
title('Autocorrelation of scaling coefficients')

stem(corrw(2:2:end))
grid on
title('Even-indexed autocorrelation values')

This example shows that symlet and Daubechies scaling filters of the same order are both solutions of the same polynomial equation.

Generate the order 4 Daubechies scaling filter and plot it.

wdb4 = dbaux(4)

wdb4 = 1×8

    0.1629    0.5055    0.4461   -0.0198   -0.1323    0.0218    0.0233   -0.0075

stem(wdb4)
title('Order 4 Daubechies Scaling Filter')

wdb4 is a solution of the equation: P = conv(wrev(w),w)*2, where P is the "Lagrange trous" filter for N = 4. Evaluate P and plot it. P is a symmetric filter and wdb4 is a minimum phase solution of the previous equation based on the roots of P.

P = conv(wrev(wdb4),wdb4)*2;
stem(P)
title('''Lagrange trous'' filter')

Generate wsym4, the order 4 symlet scaling filter and plot it. The Symlets are the "least asymmetric" Daubechies' wavelets obtained from another choice between the roots of P.

wsym4 = symaux(4)

wsym4 = 1×8

    0.0228   -0.0089   -0.0702    0.2106    0.5683    0.3519   -0.0210   -0.0536

stem(wsym4)
title('Order 4 Symlet Scaling Filter')

Compute conv(wrev(wsym4),wsym4)*2 and confirm that wsym4 is another solution of the equation P = conv(wrev(w),w)*2.

P_sym = conv(wrev(wsym4),wsym4)*2;
err = norm(P_sym-P)

err = 1.8677e-15

For a given support, the orthogonal wavelet with a phase response that most closely resembles a linear phase filter is called least asymmetric. Symlets are examples of least asymmetric wavelets. They are modified versions of the classic Daubechies db wavelets. In this example you will show that the order 4 symlet has a nearly linear phase response, while the order 4 Daubechies wavelet does not.

First plot the order 4 symlet and order 4 Daubechies scaling functions. While neither is perfectly symmetric, note how much more symmetric the symlet is.

[phi_sym,~,xval_sym]=wavefun('sym4',10);
[phi_db,~,xval_db]=wavefun('db4',10);
subplot(2,1,1)
plot(xval_sym,phi_sym)
title('sym4 - Scaling Function')
grid on
subplot(2,1,2)
plot(xval_db,phi_db)
title('db4 - Scaling Function')
grid on

Generate the filters associated with the order 4 symlet and Daubechies wavelets.

scal_sym = symaux(4,sqrt(2));
scal_db = dbaux(4,sqrt(2));

Compute the frequency response of the scaling synthesis filters.

[h_sym,w_sym] = freqz(scal_sym);
[h_db,w_db] = freqz(scal_db);

To avoid visual discontinuities, unwrap the phase angles of the frequency responses and plot them. Note how well the phase angle of the symlet filter approximates a straight line.

h_sym_u = unwrap(angle(h_sym));
h_db_u = unwrap(angle(h_db));
figure
plot(w_sym/pi,h_sym_u,'.')
hold on
plot(w_sym([1 end])/pi,h_sym_u([1 end]),'r')
grid on
xlabel('Normalized Frequency ( x \pi rad/sample)')
ylabel('Phase (radians)')
legend('Phase Angle of Frequency Response','Straight Line')
title('Symlet Order 4 - Phase Angle')

figure
plot(w_db/pi,h_db_u,'.')
hold on
plot(w_db([1 end])/pi,h_db_u([1 end]),'r')
grid on
xlabel('Normalized Frequency ( x \pi rad/sample)')
ylabel('Phase (radians)')
legend('Phase Angle of Frequency Response','Straight Line')
title('Daubechies Order 4 - Phase Angle')

The sym4 and db4 wavelets are not symmetric, but the biorthogonal wavelet is. Plot the scaling function associated with the bior3.5 wavelet. Compute the frequency response of the synthesis scaling filter for the wavelet and verify that it has linear phase.

[~,~,phi_bior_r,~,xval_bior]=wavefun('bior3.5',10);
figure
plot(xval_bior,phi_bior_r)
title('bior3.5 - Scaling Function')
grid on

[LoD_bior,HiD_bior,LoR_bior,HiR_bior] = wfilters('bior3.5');
[h_bior,w_bior] = freqz(LoR_bior);
h_bior_u = unwrap(angle(h_bior));
figure
plot(w_bior/pi,h_bior_u,'.')
hold on
plot(w_bior([1 end])/pi,h_bior_u([1 end]),'r')
grid on
xlabel('Normalized Frequency ( x \pi rad/sample)')
ylabel('Phase (radians)')
legend('Phase Angle of Frequency Response','Straight Line')
title('Biorthogonal 3.5 - Phase Angle')

This example demonstrates that for a given support, the cumulative sum of the squared coefficients of a scaling filter increase more rapidly for an extremal phase wavelet than other wavelets.

Generate the scaling filter coefficients for the db15 and sym15 wavelets. Both wavelets have support of width $2\times 15-1=29$.

[~,~,LoR_db,~] = wfilters('db15');
[~,~,LoR_sym,~] = wfilters('sym15');

Next, generate the scaling filter coefficients for the coif5 wavelet. This wavelet also has support of width $6\times 5-1=29$.

[~,~,LoR_coif,~] = wfilters('coif5');

Confirm the sum of the coefficients for all three wavelets equals $\sqrt{2}$.

sqrt(2)-sum(LoR_db)

ans = 2.2204e-16

sqrt(2)-sum(LoR_sym)

ans = 0

sqrt(2)-sum(LoR_coif)

ans = 2.2204e-16

Plot the cumulative sums of the squared coefficients. Note how rapidly the Daubechies sum increases. This is because its energy is concentrated at small abscissas. Since the Daubechies wavelet has extremal phase, the cumulative sum of its squared coefficients increases more rapidly than the other two wavelets.

plot(cumsum(LoR_db.^2),'rx-')
hold on
plot(cumsum(LoR_sym.^2),'mo-')
plot(cumsum(LoR_coif.^2),'b*-')
legend('Daubechies','Symlet','Coiflet')
title('Cumulative Sum')