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Handout11 - Course 18.327 and 1.130 Wavelets and Filter...

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Unformatted text preview: Course 18.327 and 1.130 Wavelets and Filter Banks Orthogonal wavelet bases: connection to orthogonal filters; orthogonality in the frequency domain. Biorthogonal wavelet bases. Orthogonal Wavelets 2D Vector Space: Basis vectors are i, j y vy � orthonormal basis v vx x vx and vy are the projections of v onto the x and y axes: vx = � v , i � i vy = � v , j � j 2 1 � v , i � = [vx vy] 0 1 = vx Inner Product Orthogonal multiresolution spaces: Vj has an orthonormal basis {2 j/2 �(2j t œ k): - � � k � �} �j,k(t) ��� Wj has an orthonormal basis {2j/2 w(2j t œ k): -� � k � �} ��� ��� wj,k (t) 3 Orthonormal means ��j,k(t) , �j,l(t)� = �2j/2�(2jt œ k) 2j/2�(2jt - l)dt = �[k - l] �� � � -� � � �wj,k(t) , wj,l(t)� = � 2j/2w(2jt œ k) 2j/2w(2jt - l)dt = �[k - l] -� For orthogonal multiresolution spaces, we have � � � So � Vj Wj ��j,k(t) , wj,l(t)� = 0 4 2 Projection of an L2 function, f(t), onto Vj: with fj(t) = � aj[k] �j,k(t) k aj[k] = �f(t), �j,k(t) � Haar example f(t) 0 2j/2aj[1] 2j/2aj[0] fj(t) 2/2j 1/2j 3/2j 4/2j t 5 Projection of f(t) onto Wj: gj(t) = � bj[k]wj,k(t) k with bj[k] = �f(t), wj,k(t)� 6 3 Biorthogonal Wavelet Bases Two scaling functions and two wavelets: Synthesis: �(t) = 2 �f0[k] �(2t œ k) w(t) = 2 �f1[k] �(2t œ k) k k Analysis: ~ ~ �(t) = 2 � h0[-k] �(2t œ k) k ~ ~ w(t) = 2 � h1[-k] �(2t œ k) k 7 Two sets of multiresolution spaces: {0} � … V0 � V1 � …� Vj � … � L2(�) ~ ~ ~ {0} � … V0 � V1 � … � Vj � … � L2(�) Vj + Wj = Vj+1 ~ ~ ~ Vj + Wj = Vj+1 Spaces are orthogonal w.r.t. each other i.e. Vj h ~ Wj ~ Vj h V0 has a basis {�(t œ k) : - � � k � �} Wj V0 has a basis {�(t œ k) : - � � k � �} ~ ~ W0 has a basis {w(t œ k): - � � k � �} ~ W0 has a basis {w(t œ k): - � � k � �} ~ 8 4 Bases are orthogonal w.r.t. each other i.e. ~ ~ �� �� �� (t) � (t œ k) dt = �[k] �� (t) w(t œ k)dt = 0 ~ ~ �w(t) � (t œ k) dt = 0 �w(t) w(t œ k)dt = �[k] Equivalent to perfect reconstruction conditions on filters Representation of functions in a biorthogonal basis: f(t) = � ck�(t œ k) + � � dj,k 2j/2 w(2jt œ k) � k j=0 k ~ ck = � f(t) �(t œ k) dt ~ d = 2j/2 � f(t) w(2jt œ k) dt j,k 9 Similarly, we can represent f(t) in the dual basis � ~ ~ ~~ f(t) = � c �(t œ k) + � � d 2j/2 w(2j t œ k) k k j=0 k j,k ~ ck = � f(t) �(t œ k) dt ~ dj,k = 2j/2 � f(t) w(2jt œ k) dt Note: When f0[k] = h0[-k] and f1[k] = h1[-k], we have ~ ~ �(t) = �(t) � Vj = Vj ~ ~ w(t) = w(t) � Wj = Wj i.e. we have orthogonal wavelets! 10 5 Connection between orthogonal wavelets and orthogonal filters Start with the orthonormality requirement on scaling functions � �[n] = -� �(t) �(t œ n) dt � � And then change scale using the refinement equation: �[n] = � 2 �h0[k]�(2t œ k) 2 �h0[l]�(2(t œ n)-l) dt � -� = 4 � h0[k] � h0[l] � �(�) �(� + k œ 2n - l)d�/2 �l k -� = 2 � h0[k] � h0[l] �[-k + 2n + l] k l �[n] = 2 � h0[k] h0[k œ2n] k l k This is the —double shift“ orthogonality condition that we previously encountered for orthogonal filters. 11 So orthogonal filters are necessary for orthogonal wavelets. Are they sufficient? Consider the cascade algorithm �(i + 1) (t) = 2� h0[k]�(i) (2t œ k) N k=0 If the filters are orthogonal and � �(i) (t) �(i)(t œ n)dt = �[n] � -� then � �(i + 1)(t) �(i + 1)(t œ n)dt = �[n] � Our initial guess for the iteration was �(0)(t) = box on [0,1] -� which is orthonormal w.r.t. its shifts. By induction, we have � �(t) �(t œ n)dt = �[n] as the limit, provided that � -� the cascade algorithm converges. 12 6 Note: with the alternating flip requirement, which was necessary for the highpass filter in the case of orthogonal filters, we can show that � w(t) w(t œ n)dt = �[n] � -� and � �(t) w(t œ n)dt = 0 � -� 13 Orthogonality in the Frequency Domain Let a[n] = � �(t) �(t œ n)dt � -� Use Parseval‘s theorem 1� ^ a[n] = 2� � ^(�) �(�) e-i�n d� = � -� 1 �^ 2 i�n ��� 2� ���(�)� e -� d� Trick: split integral over entire � axis into a sum of integrals over 2� intervals � �^ 1 � ���( � + 2�k)�2 ei(� + 2�k)n d� ��� a[n] = 2� k=-� -� = 1 2� 2 i�n � � � ^ (� + 2�k)� e d� � �� -� k=-� 14 7 Take the Discrete Time Fourier Transform of both sides � ^ A(�) � � a[n]eœi�n = � � �(� + 2�k)�2 k=-� n If �(t œ n) are orthonormal, then a[n] = �[n] � A(�) = 1 So the scaling function and it shifts are orthogonal if � ^ 2 �� � ��(� + 2�k)� = 1 k=-� Note: if we set � = 0, then ^ ���(2�k)�2 = 1 k ^ and since �(0) = 1, we see that ^ �(2�k) = 0 for k � 0 15 Connection between orthogonal wavelets and orthogonal filters (in frequency domain): Start with an orthogonal scaling function: � ^ 1 = � ��(� + 2�k)�2 �� k=-� and then change scale using the refinement equation in the frequency domain: ^ ^ �(�) = H0(�/2 ) �( �/2) � ^ 1 = � �H0( �/2 + �k)�2�� ( �/2 + �k)�2 k=-� 16 8 Trick: split sum into even and odd � 2^ 2 1 = � �H0( �/2 + 2�k)� �� ( �/2 + 2�k)� + k=-� � 2^ 2 ��H0( �/2 + �(2k + 1))� ��( �/2 + � (2k + 1)� k=-� � = �H0(�/2 )�2 � �� ( �/2 + 2�k)�2 + �^ � k=-� 2� ^ 2 �� �H0( �/2 + �)� � �� ( �/2 + � + 2�k)� k=-� But each of the two infinite sums is equal to 1 So the discrete time filter H0(�) must be orthogonal: 1 = �H0(�)�2 + � H0 (� + �)�2 17 9 ...
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