Slides7 - Course 18.327 and 1.130 Course Wavelets and...

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Unformatted text preview: Course 18.327 and 1.130 Course Wavelets and Filter Banks Orthogonal Filter Banks; Orthogonal Paraunitary Matrices; Orthogonality Condition (Condition O) in the Time Domain, Modulation Domain and Polyphase Domain Unitary Matrices Unitary The constant complex matrix A is said to be unitary if The A† A = I example: example: 1 A= �2 A-1 1 -i i -1 -1 -1 i = � 2 -i 1 11i A* = � 2 -i -1 A† = A*T A* 1 1 -i = � 2 i -1 � A† = A-1 2 Paraunitary Matrices Paraunitary The matrix function H(z) is said to be paraunitary if The it is unitary for all values of the parameter z HT(z-1) H(z) = I Frequency Domain: HT(-w) H(w) = I or H*T(w) H(w) = I or for all z „ 0 -----------------(1) for ----------------- for all w Note: we are assuming that h[n] are real. 3 Orthogonal Filter Banks Orthogonal Centered form (PR with no delay): Centered y0[n] > h0[n] > fl2 > ›2 > h0[-n] > x[n] x[n] > +> y1[n] > ›2 > h1[-n] > > h1[n] > fl2 Synthesis bank = transpose of analysis bank h0[n] causal � f0[n] ” h0[-n] anticausal causal 4 What are the conditions on h0[n], h1[n], in the What (i) time domain? (ii) polyphase domain? (iii) modulation domain? 5 Time Domain Time Analysis: N = 3 (filter length = 4) Analysis: M y0[0] y0[1] y0[2] y0[3] M K h0[3] h0[2] h0[1] h0[0] h0[3] h0[2] h0[1] h0[0] h0[3] h0[2] h0[1] h0[0] h0[3] h0[2] h0[1] h0[0] L M x [-3] x [-2] x [-1] x [-0] x [1] = x [2] M K x [3] h1[3] h1[2] h1[1] h1[0] y1[0] x [4] h1[3] h1[2] h1[1] h1[0] y1[1] x [5] [5] h1[3] h1[2] h1[1] h1[0] y1[2] h1[3] h1[2] h1[1] h1[0] x [6] y1[3] M 123 L W M -----(2) 6 Synthesis: Synthesis: M x[-3] x[ x[-2] x[-1] x[0] x[1] x[2] x[3] = x[4] x[5] x[6] M M M h0[3] h0[2] h0[1] h0[3] h0[0] h0[2] h0[1] h0[3] h0[0] h0[2] h0[1] h0[3] h0[0] h0[2] h0[1] h0[0] M . h1[3] h1[2] h1[1] h1[3] h1[0] h1[2] h1[1] h1[3] h1[0] h1[2] h1[1] h1[3] h1[0] h1[2] h1[1] h1[0] M . 123 WT M y0 [0] [0] y0 [1] [1] y0 [2] [2] y0 [3] [3] M M y1 [0] [0] y1 [1] [1 y1 [2] [2] y1 [3] [3] M -----(3) ----7 Orthogonality condition (Condition O) is Orthogonality WTW = I = WWT � W orthogonal matrix orthogonal Block Form: L W= B LTL + BTB = I LLT LBT LL LB BLT BBT BL BB = I 0 0 I LLT = I � � h0[n] h0[n – 2k] = d[k] --------------(4) n LBT = 0 � � h0[n] h1[n – 2k] = 0 LB n --------------(5) BBT = I � � h1[n] h1[n – 2k] = d[k] --------------(6) BB n 8 Good choice for h1[n]: Good h1[n] = (-1)n h0[N-n] ; N odd --------------(7) Alternating flip Alternating Example: N = 3 = h0[3] h1[0] = -h0[2] h1[1] = h0[1] h1[2] = -h0[0] h1[3] With this choice, Equation (5) is automatically satisfied: =0 k = -1: h0[0]h0[1] - h0[1]h0[0] k = 0: h0[0]h0[3] - h0[1]h0[2] + h0[2]h0[1] – h0[3]h0[0] = 0 =0 k = 1: h0[2]h0[3] – h0[3]h0[2] k = –2: no overlap 9 Also, Equation (6) reduces to Equation (4) Also, d[k] = � h1[n] h1[n-2k] = � (-1)n h0[N-n] (-1)n-2k h0[N-n+2k] n n = � h0[l] h0[l + 2k] 2k] l So, Condition O on the lowpass filter + alternating flip for highpass filter lead to orthogonality 10 10 Polyphase Domain Polyphase x[n] > fl2 > z-1 > fl2 Hp(z) = y0[n] > xodd[n-1] Hp(z) > H0,even(z) H1,even(z) xeven[n] > xeven[n] > y1[n] HpT(z-1) > H0,odd(z) 0,odd H1,odd(z) xodd[n-1] > ›2 > ›2 x[n-1] +> > z Polyphase Polyphase Matrix 11 11 Condition O: Condition HpT(z-1) Hp(z) = I � Hp(z) is paraunitary H0,even(z-1) H1,even(z-1) 1,even H0,even(z) H0,odd(z) 0,odd 10 = H0,odd(z-1) H1,odd(z-1) 1,odd H1,even(z) H1,odd(z) 1,odd 01 Reverse the order of multiplication: H0,even(z) H0,odd(z) 0,odd H0,even(z-1) H1,even(z-1) 1,even 10 = H1,even(z) H1,odd(z) 1,odd H0,odd(z-1) H1,odd(z-1) 1,odd 01 12 12 Express Condition O as a condition on H0,even(z), Express H0,odd(z): H0,even(z) H0,even(z-1) + H 0,odd(z) H0,odd(z-1) = 1 0,even 0,odd ------(8) Frequency domain: ‰H0,even(w)‰2 + ‰H0,odd(w)‰2 = 1 -------(9) ------- 13 13 The alternating flip construction for H1(z) ensures The that the remaining conditions are satisfied. H0(z) = H0,even(z2) + z -1H0,odd(z2) alternating flip H1(z) = -z-N H0(-z-1) = -z-N {H0,even(z-2) - z H0,odd(z-2)} {H = -z-N H0,even(z-2) + z -N+1 H0,odd(z-2) 123 123 z-1 H1,odd(z2) H1,even(z2) So H1,even(z) = z(-N+1)/2 H0,odd(z-1) H1,odd(z) = -z(-N+1)/2 H0,even(z-1) � H0,even(z) H1,even(z-1) + H0,odd(z) H1,odd(z-1) = 0 and H1,even(z) H1,even(z-1) + H1,odd(z) H1,odd(z-1) = 1 1,even 14 14 Modulation Domain Modulation x[n] > H0(z) > > fl2 > H1(z) > y0[n] fl2 > ›2 > H0(z-1) > ›2 > H1(z-1) > y1[n] > PR conditions: H0(z) H0(z-1) + H1(z) H1(z-1) = 2 -------(10) H0(-z) H0(z-1) + H1(-z) H1(z-1) = 0 --------(11) [H0(z-1) H1(z-1)] H0(z) H1(z) x[n] +> No distortion Alias cancellation H0(-z) = 20 H1(-z) 123 123 Hm(z) modulation matrix 15 15 Replace z with –z in Equations (10) and (11) Replace H0(-z) H0(-z-1) + H1(-z) H1(-z-1) = 2 H0(z) H0(-z-1) + H1(z) H1(-z-1) = 0 H0(z-1) H1(z-1) H0(-z-1) H1(-z-1) H0(z) H0(-z) H1(z) H1(-z) 123 123 123 123 HmT(z-1) Condition O: HmT(z-1) Hm(z) Hm(z) = 2 0 0 2 123 123 2I = 2I � Hm(z) is paraunitary 16 16 Reverse the order of multiplication: Reverse H0(z) H0(-z) H0(z-1) H1(z-1) 2 0 0 2 = H1(z) H1(-z) H0(-z-1) H1(-z-1) Express Condition O as a condition on H0(z): H0(z) H0(z-1) + H0(-z) H0(-z-1) = 2 ------------(12) Frequency Domain: ‰H0(w)‰2 + ‰H0(w + p)‰2 2 2 =2 -------------(13) ------------- Again, the remaining conditions are automatically Again, satisfied by the alternating flip choice, H1(z) = -z-N H0(-z-1) 17 17 Summary Summary Condition O as a constraint on the lowpass filter: Condition • Matrix form: LLT = I Matrix LL • Coefficient form: � h[n]h[n-2k] = d[k] Coefficient h[n]h[n • Polyphase form: Polyphase H0,even(z) H0,even(z-1) = H0,odd(z) H0,odd(z-1) = 1 0,odd • Modulation form: H0(z) H0(z-1) + H0(-z) H0(-z-1) = 2 Modulation n Then choose H1(z) = -z-N H0(-z-1) i.e., h1[n] = (-1)n h0[N-n] ; N odd 18 18 ...
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This note was uploaded on 12/04/2011 for the course ESD 18.327 taught by Professor Gilbertstrang during the Spring '03 term at MIT.

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