Lecture 48 – Towards selecting
wavelets through vanishing moments
Dr Aditya Abhyankar
Dr.
Wavelet Transform
y
Decomposes signal into
two separate series
y
Single series to
represent most
coarse version
y
S l FFunction
Scaling
y
Double series to
represent refined
version
y
Wavelet Function
Wavelet Transform: Specialty
y
Scalingg and Translation are indeed
Hallmarks of Wavelet transform
y
They lead us to MultiResolutioAnalysis
(MRA) !!
Relationship
As the spaces
p
and spans
p
are clear now
y Intuitionally, we observe a relationship
between tthese
betwee
ese spaces!
y
.......V−2 ⊂ V−1 ⊂ V0 ⊂ V1 ⊂ V2 .......
y
Intuitively we can see that as we move
towards right, i.e. up the ladder, we are
moving towards L2 (ℜ)
2 Band MRA Filter Bank
H0(z)
2
2
H1(z)
p[n]
+
G0(z)
Analysis
2
2
G1( )
G1(z)
S th i
Synthesis
2 Band MRA Filter Bank
V0
p0 [n]
p[n]
H0(z)
2
V1
q0 [n]
G0( )
G0(z)
2
W0
Framework
Gave us p
power to move upp or down the
ladder
y We
e can
ca now
ow indeed
ee zoom-in
oo
or zoomo
oo
out of any part of the signal
y This makes the entire analysis ‘scalable’!!
scalable !!
y Scalability stems out of multi-resolution
framework !
y
Framework
Leads us to two questions
q
1) How do we go about selecting the
mother
ot e wavelet
wave et and
a scale
sca e of
o analysis?
a a ys s?
2) What is the procedure to calculate
scaling and wavelet coefficients?
y
How Fourier Works – Basis
Functions !!
How Fourier Works!!
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
cos(2π 1n)
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
cos(2π 1n) < y[n], cos(2π 1n) >
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
sin(2π 1n)
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
sin(2π 1n) < y[n],sin(2π 1n) >
5.4573e-013
5.4573e
013
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
cos(2π 3n)
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
cos(2π 3n) < y[n], cos(2π 3n) >
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
sin(2π 3n)
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
sin(2π 3n) < y[n],sin(2π 3n) >
1.4997e+003
1.4997e
003
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
How Fourier Works!!
cos(2π 8n)
y[n] = 0.6sin(2π 3n) + 0.8cos(2π 8n)
How Fourier Works!!
cos(2π 8n) < y[n], cos(2π 8n) >
2.0004e+003
y[n] = 0.6sin(2
0 6 i (2π 3n) + 0.8cos(2
0 8 (2π 8n)
Application
Detectingg hidden jump
j p discontinuityy
y Consider function
y
1
⎧
t
≤
t
<
,
0
⎪⎪
2
g (t ) = ⎨
⎪t − 1,, 1 ≤ t < 1
⎪⎩
2
y
Clear jjump
p at t=0.5
Application
Detectingg hidden jump
j p discontinuityy
y Let’s integrate
y
⎧ t2
1
,0 ≤ t <
⎪⎪
2
2
h(t ) = ∫ g (t )dt = ⎨ 2
⎪t − t + 1 , 1 ≤ t < 1
⎪⎩ 2
2 2
y
Cusp
p jjump
p at t=0.5
Application
Detectingg hidden jump
j p discontinuityy
y Let’s integrate again
y
3
⎧
t
1
,0 ≤ t <
⎪⎪
6
2
f (t ) = ∫ h(t )dt = ⎨ 3 2
⎪t − t + t − 1 , 1 ≤ t < 1
⎩⎪ 6 2 2 8 2
y
Appears
pp
smooth to eye
y
Coefficients
Who ggives us coefficients of scalingg
equation?
y Haar
aa
y
In search of coefficients
y
We can thinks of usingg three guiding
g
g
theorems !
In search of coefficients
We can thinks of usingg three guiding
g
g
theorems !
y Theorem
eo e 1::
y
For the scaling equation φ ( x) = ∑ hk 2φ (2 x − k ), with
k
M
k k =N
non-vanishingg coefficients {h }
onlyy for N ≤ k ≤ M ,
its φ ( x) is with a compact support contained in interval [ N , M ]
In search of coefficients
We can thinks of usingg three guiding
g
g
theorems !
y Theorem
eo e 2::
y
If the scaling function φ ( x) has compact support on
0 ≤ x ≤ N -1 and if, {φ ( x - k )} are linearly independent,
then hn = h( n) = 0, for n<0 and n > N -1.
Hence N is the length of the sequence.
In search of coefficients
We can thinks of using three guiding
theorems !
y Theorem 3:
y
If the
h scaling
li coefficieents
ffi i
{hk }
satisfy the condition for existence and
orthogonality of φ ( x), then
ϕ ( x) = ∑ g k 2φ (2 x − k )
k
where, g k = ± (−1) k hN − k
∞
and,
∫ ϕ ( x - l )φ ( x - k )dx = δ
-∞
l ,k
= 0, l ≠ k
Properties of scaling coefficients
Properties of scaling coefficients
Thank You!
Questions ??