1. EE 小波轉換與應用
Chapter 1
Introduction

2. (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).

3. 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.

4. 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

5. 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.

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

7. Frequency Domain Representationor
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 :

8. 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)
(3) (2) (1)
(4) (3) (2)
y(4) y(3) y(2) y(1) y(0) = y

9. 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 =-10

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

11. 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”.

12. 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”

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

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

15. 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?)

16. 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:

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

18. Linear phase with a discontinuity at ω=0

19. 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)

20. 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
= 0
=2( …)
To invert a FIR filter with a FIR filter, we need to go to a filter bank.

21. 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) = -

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

23. 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.

24. 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.

25. Block Form of a Filter bankD 2 x y0 y1
V0 =( 2) Cx = Lx
V1 =( 2) Dx = Bx

26. 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.

27. 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)

28. (↓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.

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

30. F filters to give = w0

31. 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

32. 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

33. 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.

34. 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.

35. 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.

36. 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,

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

38. 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 :

39. Thm 1.2 The rescaled Haar wavelets
form an orthonormal basis:

40. 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)=

41. Tree-Structured Filter Bank2 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.

42. 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.

43. 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

44. In terms of matrices, for L=4
with

45. Fast Wavelet Transform

46. 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

47. 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:

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

49. 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

50. 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

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