EE 5632 小波轉換與應用 Chapter 1 Introduction

(1) Filter h(0), h(1), h(2), … (2) Filter bank: a set of filters. analysis bank synthesis bank downsampling : to maintain critical sampling PR FIR filter banks: the reconstructed output from the synthesis bank is identical to the original input x to the analysis bank (with time delay). In matrix language, a banded matrix (for the analysis bank) has a banded inverse (for the synthesis bank).

Downsampling is not a time-invariant operation Downsampling is not a time-invariant operation. If we delay all components of y by one time unit the output from downsampling is totally different. The new samples y(-1), y(1), y(3) are entirely separate and independent from the original samples y(0), y(2), y(4). Those two subsamplet signals are two “phases” of y, not connected. Each phase of y comes from filtering the phases of x. The whole operation together, filtering followed by downdsampling, becomes a matrix multiplication – by the polyphase matrix.

Wavelets Mother wavelet : w(t) Wavelets at scale j with shift k : Orthogonality: Any admissible function f(t) can be expressed as With   In connection with filter banks , it is the “highpass filter” that leads to w(t). The “lowpass filter“ leads to scaling function .

Multiresolution Signal at level j (local averages) details at level j (local differences) V= scaling space at level j ⊕ V= scaling space at level j+1 W= wavelet space at level j The signal is divided into different scales of resolution, rather than different frequencies. The “time-scale plane” takes the place for wavelets that the “time-frequency plane” takes for filters.

Fig. 1.1 The matching of long time with low frequency and short time with high frequency occurs in a natural way for wavelets.

Frequency Domain Representation or A direct conversion to Z-transform Convolution by h in time becomes multiplication by H in frequency : In Z-domain Y(z)=H(z) X(z) The frequency response is the transform of the impulse response :

Convolution by Hand x(2) x(1) x(0) 3 2 4 = x h(2) h(1) h(0) 1 5 2 = h (2) (1) (0) 6 4 8 (3) (2) (1) 15 10 20 (4) (3) (2) 3 2 4 y(4) y(3) y(2) y(1) y(0) 3 17 20 24 8 = y

Check : at x(n)=4,2,3 add to X(0)= 9 h(n)=2,5,1 add to H(0)= 8 The sum for y= h*x is H(0)X(0)= 72 Another check : at The alternating sum 4-2+3 gives X(π)=5 Similary, H(π)=-2 Then Y(π)=(5)(-2) This agrees with the alternating sum 3-17+20-24+8=-10

1.2 Lowpass Filter = Moving Average y(n)= x(n)+ x(n-1) filter coeff. h(0)= h(1)= It is a moving average, because the output averages the current component x(n) with the previous one. Old components drop away as the average moves forward with the time. If x= (…, 0, 0, 1, 0, 0,…), which is unit impulse, then y= (…, 0, 0, , , 0,…), which is the impulse response. The moving average, when expressed in matrix form, is

A causal FIR filter has h(n)=0 for all negative n and for large positive n. The matrix is banded and lower triangular. Only a finite number of coefficients h(0), h(1), … , h(N) can be nonzero. The filter has N+1 “taps”.

Frequency Response If x(n)= , then y(n)= x(n)+ x(n-1) = + =( + ) = + =( + ) The frequency response function At H= + =1 i.e. x=(…, 1, 1, 1,…) y=(…, 1, 1, 1,…) “lowpass filter”

and   -π 0 π ω -π 0 π ω Fig. 1.2 The magnitude and the phase of

Fig. 1.3: Nyquist diagram of corresponds to x(n)=  y(n)=(…, 0, 0, 0, …) This confirms H(π)=0.   -1 0 )-ω/2 ) -ω 1 Fig. 1.3: Nyquist diagram of

h(k)= h(N-k) (symmetry) Linear phase h(k)= -h(N-k) (antisymmetry) The center of symmetry at N/2 produces a factor in . This means a linear term –N /2 in the phase. A symmetric filter is lowpass, while an antisymmetric filter is highpass. (why?)

1.3 Highpass Filter =Moving Difference highpass : y(n)= x(n) - x(n-1) filter coeff. : h(0)= h(1)= - impulse response : h=(…, 0, 0, , - , 0,…) I/O relation in matrix form:

Frequency Response With Fig. 1.4 The magnitude and phase of -π 0 -π 0 -π 0 π ω π ω Fig. 1.4 The magnitude and phase of

Linear phase with a discontinuity at ω=0

Invertibility and Noninvertibility The averaging and differencing filters are not invertible because H0(π)=0 and H1(0)=0. The frequency response of an invertible filter must have H(ω)≠0 at all frequencies. If we still attempt to invert the moving average, we can compute (1.17)

We seem to have found an inverse , but it is not a legal filter . This H-1 is not stable, because the bounded input y=(…,1,-1, 1,-1,1…) will produce an unbounded output H-1y. The series expansion in (1.17) breaks down at ω= π, where 1+ = 0 =2(1+1+1+1+…) To invert a FIR filter with a FIR filter, we need to go to a filter bank.

1.4 Filter Bank = Lowpass and Highpass The lowpass and highpass filter separate the signal into frequency bands. But the signal length has doubled. Downsampling (↓2) y = (…, y(-4), y(-2), y(0), y(2), y(4) ,…) Normalization : to compensate for losing half the components in (↓2), we multiply the surviving y(2n) by . This normalizing factor is usually included with the filter bank, so that lowpass : H0(ω) changes to C(ω)= H0 (ω) highpass : H1(ω) changes to D(ω)= H1(ω). with filter coeff. c(0)=c(1)= d(0)= d(1) = -

Decimation filters in the Time Domain L=(↓2)C= B=(↓2)D=  

The row vectors are orthonormal The row vectors are orthonormal. The column vectors are also orthonormal. which represents the synthesis bank.  The channels L=(↓2) C and B=(↓ 2) D of an orthogonal filter bank are represented in the time domain.

by a combined orthogonal matrix The synthesis bank is the transpose of the analysis bank. When one follows the other we have perfect reconstruction. For causality we add a delay. The whole filter bank is called orthogonal. When analysis bank and synthesis bank are inverses but not necessarily transposes, the filter bank is biorthogonal.

Block Form of a Filter bank D 2 x y0 y1 V0 =( 2) Cx = Lx V1 =( 2) Dx = Bx

To recover the original x, we form W0= and W1= and w0+ w1= x(n-1) The total effect of the whole filter bank is a delay.

The synthesis bank The analysis bank has two steps, filtering and downsampling. Its inverse, the synthetic filter bank, also has two steps, upsampling and filtering. Upsampling (↑ 2)

(↓2) ( ↑ 2) (↓2) ≠ y But (↓2) (↑ 2) y = y (↑ 2) is a right-inverse of (↓2) but not a left-inverse of (↓2). (↑ 2) is the “pseudoinverse” of (↓2). The second step in the synthesis bank is filtering.

F G u0 u1 W0 W1 ( 2) Cx ( 2) Dx 2 ⊕ x(n-1) u0 = and u1 =

F filters to give = w0

This can be achieved by F u(n)= [u(n)+u(n-1)] with filter coeff. , And the second synthesis filter is G filters to give = w1

which can be achieved by G u(n)= [-u(n)+u(n-1)] with filter coff. , The Haar filter bank in the whole performs the 2-point DFT. The whole filter bank: F G v y 2 ⊕ x(n-1) C D x(n) u

Scaling Function and Wavelets The two-scale equation (dilation equation) for the scaling function is or in terms of the lowpass filter coefficient h(k), Some issues: 1.   There may or may not be a solution . 2.   The solution is zero outside the interval 0 t<N. 3.   The solution seldom has an elementary formula. 4.   The solution is not likely to be a smooth function.

For the Haar example, 2h(0)=1, 2h(1)=1, the dilation eq. is Its solution is the box function 1 t 1 t The dilation equation is linear, so any multiple of the box function is also a solution. It’s common to normalize it.

Thm 1.1 If , then h(0)+h(1)+… + h(N)=1 Proof: If , then . Since , This is consistent with lowpass filter convention. The box function is like the averaging filter – it smoothes the input. average over moving interval.

The Wavelet Equation or in terms of the original highpass filter coefficients In our Haar example With being a box function, the wavelet is a difference of half-boxes. Explicitly,

w(2t) w(2t-1) 1 t w(t) 1 t Haar wavelet

The Haar wavelet has compact support, which comes from FIR The Haar wavelet has compact support, which comes from FIR. Generally, wavelets need not have compact support! They can come from IIR filters instead of FIR filters, oscillate above and below zero along the whole line ,decaying as |t| . This still qualifies as a wavelet. Orthogonality :

Thm 1.2 The rescaled Haar wavelets form an orthonormal basis:

Wavelet Transforms by Multiresolution For an orthonormal basis Synthesis of a function : Analysis of a function: In discrete-time case, the input is x(n), Synthesis in discrete time : x=S b Analysis in discrete time : b=A x Where the columns of the L× L matrix S are the discrete wavelets. For orthonormal wavelets, A= ST . In general, A= S-1 The rows of A= are biorthogonal to the columns of S: (row i of A)‧(column j of S)=

Tree-Structured Filter Bank 2 C D x(n) bj-1,k bjk The highpass filter D computes differences of the input. The downsampling step symbolized by (↓2) keeps the even-numbered differences (x(2k)-x(2k-1))/ . The lowpass filter C computes averages.

The averages and differences of all levels follow the fundamental recursion : Averages(lowpass filter) Differences(highpass filter) We can see this logarithmic tree as a pyramid of averages and differences. The averages are sent up the pyramid, to be averaged again. Whenever a difference is computed, it is final.

Starting at level J=3, where the input x has L= 2J = 8 components, the count of wavelet coefficients is 4 differences + 2 differences +1 difference +overall average= 8

In terms of matrices, for L=4 with

Fast Wavelet Transform

Thm 1.3 The fast wavelet transform computes the L coefficients b=Ax in less than 2TL multiplications. The synthesis tree has the inverse matrices in opposite order: L B x LT BT

Haar Wavelets and Recursion From the two-scale eq. for Haar wavelets and , We have Multiply the two equations by f(t) and integrate:   Scaling coefficient: Wavelet coefficient:

Ajk aj-1,k a1k a0k bj-1,k b1k b0k

Multiresolution in Continuous Time where Vj = all combinations of scaling function at level j Wj = all combinations of wavelets at level j e.g. V3 = V2 ⊕ W2 = V1 ⊕ W1 ⊕ W2 = V0 ⊕ W0 ⊕ W1 ⊕ W2 A function in V3 can be expressed as

Discrete time Continuous time filter bank tree multiresolution downsampling rescaling t 2t lowpass filter averaging with highpass filter detailing with w(t) orthogonal matrices orthogonal bases analysis bank output wavelet coefficients synthesis bank output sum of wavelet series product of filter matrices fast wavelet transform

For biorthogonal case, it has two multiresolutions, in parallel with Vj+1 = Vj ⊕ Wj .