Wavelet Transform

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Lecture 48

Lecture 48 –– Towards selecting Towards selecting

wavelets through vanishing moments wavelets through vanishing moments wavelets through vanishing moments wavelets through vanishing moments

Dr Aditya Abhyankar Dr. Aditya Abhyankar

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Wavelet Transform Wavelet Transform Wavelet Transform Wavelet Transform

y Decomposes signal into two separate series

y Single series to Single series to _y Double series to represent most

coarse version

y Double series to represent refined version

coarse version version S l F

y Scaling Function _y Wavelet Function

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Wavelet Transform: Specialty Wavelet Transform: Specialty Wavelet Transform: Specialty Wavelet Transform: Specialty

y Scaling and Translation are indeed g Hallmarks of Wavelet transform

y They lead us to MultiResolutioAnalysis (MRA) !!

(MRA) !!

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Relationship Relationship Relationship Relationship

y As the spaces and spans are clear nowp p

y Intuitionally, we observe a relationship between these spaces!

betwee t ese spaces!

V ₂ ⊂ V ₁ ⊂ V₀ ⊂ V₁ ⊂ V₂

...V₋ ⊂ V₋ ⊂ V ⊂ V ⊂ V ...

y Intuitively we can see that as we move towards right, i.e. up the ladder, we are moving towards

2( )

L ℜ

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2 Band MRA Filter Bank 2 Band MRA Filter Bank 2 Band MRA Filter Bank 2 Band MRA Filter Bank

H0(z) 2 2 ^H1(z)

[ ]

p n +

2 2 _{G1( )}

G0(z) 2 2 _G1(z)

S th i

Analysis Synthesis

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2 Band MRA Filter Bank

2 Band MRA Filter Bank V⁰ 2 Band MRA Filter Bank

2 Band MRA Filter Bank [ ]

p n p n⁰[ ]

0

H0(z) [ ]

p n

2

[ ] q n V1

0[ ] q n

G0( ) ²

W0

G0(z) ²

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Framework Framework Framework Framework

y Gave us power to move up or down the p p ladder

y We can now indeed zoom-in or zoom-e ca ow ee oo o oo out of any part of the signal

y This makes the entire analysis ‘scalable’!!

y This makes the entire analysis scalable !!

y Scalability stems out of multi-resolution framework !

framework !

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Framework Framework Framework Framework

y Leads us to two questionsq

1) How do we go about selecting the mother wavelet and scale of analysis?ot e wave et a sca e o a a ys s?

2) What is the procedure to calculate scaling and wavelet coefficients?

scaling and wavelet coefficients?

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How Fourier Works

How Fourier Works –– Basis Basis Functions !!

Functions !!

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How Fourier Works!!

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

cos(2 1 )π n

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

cos(2 1 )π n < y n[ ], cos(2 1 )π n >

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

sin(2 1 )π n

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

sin(2 1 )π n < y n[ ], sin(2 1 )π n >

5.4573e-013

[ ] 0 6 i (2 3 ) 0 8 (2 8 )

5.4573e 013

[ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

cos(2 3 )π n

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

[ ], cos(2 3 )

y n π n

< >

cos(2 3 )π n

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

sin(2 3 )π n

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

sin(2 3 )π n < y n[ ], sin(2 3 )π n >

1.4997e+003

[ ] 0 6 i (2 3 ) 0 8 (2 8 )

1.4997e 003

[ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

[ ] 0 6 i (2 3 ) 0 8 (2 8 ) [ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

cos(2 8 )π n

[ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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How Fourier Works!!

[ ], cos(2 8 )

y n π n

< >

cos(2 8 )π n

[ ] 0 6 i (2 3 ) 0 8 (2 8 )

2.0004e+003

[ ] 0.6 sin(2 3 ) 0.8 cos(2 8 )

y n = π n + π n

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Application Application Application Application

y Detecting hidden jump discontinuityg j p y

y Consider function

, 0 1

t t 2

⎧ ≤ <

⎪⎪ 2

( ) 1

1, 1

2 g t

t t

= ⎨⎪

⎪ − ≤ <

⎪⎩

y Clear jump at t=0.5

, 2

⎪⎩

j p

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y Let’s integrate

⎧ ² 1

2 , 0 2

( ) ( )

t t

h t g t dt

⎧ ≤ <

=

∫

= ⎨⎪⎪ ² ( ) ( )

1 1, 1

2 2 2

h t g t dt

t t t

= = ⎨

⎪ − + ≤ <

⎪⎩

∫

y Cusp jump at t=0.5

⎩

p j p

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y Let’s integrate again

⎧ ³ 1

6 , 0 2

( ) ( )

t t

f t h t dt

⎧ ≤ <

=

∫

= ⎨⎪⎪ ³ ² ( ) ( )

1 1, 1

6 2 2 8 2

f t h t dt

t t t

t

= = ⎨

⎪ − + − ≤ <

⎪⎩

∫

y Appears smooth to eye

⎩

pp y

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Coefficients Coefficients Coefficients Coefficients

y Who gives us coefficients of scaling g g equation?

y Haaraa

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In search of coefficients In search of coefficients In search of coefficients In search of coefficients

y We can thinks of using three guiding g g g theorems !

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y Theorem 1:eo e :

∑

For the scaling equation ( ) 2 (2 ), with non-vanishing coefficients { } only for ,

k k M k k N

x h x k

h N k M

φ = φ −

≤ ≤

∑

g { } y ,

its ( ) is with a compact support contained in interval [ , ]

k k N

x N M

φ

=

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y Theorem 2:eo e :

If the scaling function ( ) has compact support on φ x 0 -1 and if, { ( - )} are linearly independent, then _n ( ) 0, for n<0 and -1.

x N x k

h h n n N

φ

≤ ≤

= = >

Hence N is the length of the sequence.

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y We can thinks of using three guiding theorems !

theorems !

y Theorem 3:

If h li ffi i { }h If the scaling coefficieents { }

satisfy the condition for existence and hk

orthogonality of ( ), then

( ) _k 2 (2 )

x

x g x k

φ

ϕ = ∑ φ −

where, ( 1)

k

k N k

g h ₋

∞

= ± −

, -

and, ( - ) ( - )^∞ ϕ x l φ x k dx δ_{l k} 0,l k

∞

= = ≠

∫

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Properties of scaling coefficients Properties of scaling coefficients Properties of scaling coefficients Properties of scaling coefficients

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Properties of scaling coefficients Properties of scaling coefficients Properties of scaling coefficients Properties of scaling coefficients

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Thank You!

Questions ??