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# Orthogonality Conditions for the Wavelet Scaling Function

Module by: Ivan Selesnick. E-mail the author

To generate orthogonal wavelet bases, we will need the set B= φtn nZ B φ t n n to be an orthogonal set. This set consists of the scaling function φt φ t and its integer-translations. The orthogonality condition is written as

φtφtndt=δn t φ t φ t n δ n
(1)
But not just any φt φ t satisfies this orthogonality condition. How can we be sure that a set of coefficients hn h n will generate a scaling function φt φ t that satisfies Equation 1?

## Projections

Define the set BB to be B= φtn nZ B φ t n n BB is called an orthonormal set if φ n t φ m tdt=δnm t φ n t φ m t δ n m where φ n tφtn φ n t φ t n . Define the set 𝒱𝒱 to be

𝒱=Span φtn nZ =The set of functions of the form  nanφtn 𝒱 Span φ t n n The set of functions of the form   n a n φ t n
(2)
Given a signal xt x t , how can we find the signal x ^ t x ^ t in 𝒱𝒱 that is closest to xt x t ?

If BB is an orthonormal set, then it is simple to find the signal x ^ t x ^ t in 𝒱𝒱 that minimizes the square error ε=xt x ^ t2dt ε t x t x ^ t 2 where x ^ t=nan φ n t x ^ t n a n φ n t We just need to find the coefficients an a n , which can be done by differentiating εε with respect to ak a k and setting it to zero.

dd an ε=dd an xtkak φ k t2dt=dxtkak φ k t2d an dt=2(xtkak φ k t) φ n tdt=2xt φ n tdt2kak φ k t φ n tdt=2xt φ n tdt2kakδkn=2xt φ n tdt2an a n ε a n t x t k a k φ k t 2 t a n x t k a k φ k t 2 t 2 x t k a k φ k t φ n t 2 t x t φ n t 2 k a k t φ k t φ n t 2 t x t φ n t 2 k a k δ k n 2 t x t φ n t 2 a n
(3)
Setting dd an ε a n ε to zero gives
an=xt φ n tdtxt, φ n t a n t x t φ n t x t φ n t
(4)
Therefore, if φt φ t is orthogonal to its integer-translates, then the coefficients an a n that give the best square-error signal in 𝒱𝒱 are given by integrating xt x t with φ n t φ n t . We get x ^ t=n(xt, φ n t) φ n t x ^ t n x t φ n t φ n t This is called the projection of xt x t onto 𝒱𝒱.

## The Normalization induced by Orthogonality

It turns out that if a scaling function φt φ t satisfies the orthogonality condition, φtφtndt=δn t φ t φ t n δ n then φtdt=±1 t φ t ± 1 To derive this normalization on φt φ t induced by Equation 1, begin with Equation 1. φtφtndt=δn t φ t φ t n δ n Sum each side over nn, nφtφtndt=nδn n t φ t φ t n n δ n Then we have φtnφtndt=1 t φ t n φ t n 1 Recall from this equation that nφtn=φtdt n φ t n t φ t (provided φt φ t is continuous), so that we can write φtφαd α dt=1 t φ t α φ α 1 or φtdtφαd α =1 t φ t α φ α 1 φtdt2=1 t φ t 2 1 or φtdt=±1 t φ t ± 1

(φtφtndt=δn)(φtdt=±1) t φ t φ t n δ n t φ t ± 1
(5)

## Orthogonality conditions in h(n)

If we assume that the scaling function φt φ t does satisfy Equation 1, what conditions can we derive for hn h n ?

Assume Equation 1 is true. Then use the dilation equation to expand Equation 1. Substitute φt=2khkφ2tk φ t 2 k h k φ 2 t k and φtn=2mhmφ2(tn)m φ t n 2 m h m φ 2 t n m into Equation 1. 2khkφ2tk(2mhmφ2t2nm)dt=δn t 2 k h k φ 2 t k 2 m h m φ 2 t 2 n m δ n Then we have 2khkmhmφ2tkφ2t2nmdt=δn 2 k h k m h m t φ 2 t k φ 2 t 2 n m δ n Use the change of variables α=2tk α 2 t k to get khkmhmφαφα+k2nmdt=δn k h k m h m t φ α φ α k 2 n m δ n Using Equation 1, khkmhmδk2nm=δn k h k m h m δ k 2 n m δ n Because δk2nm δ k 2 n m is zero except when m=k2n m k 2 n , we have

khkhk2n=δn k h k h k 2 n δ n
(6)
This is the condition on hn h n that corresponds to condition Equation 1.
(φtφtndt=δn)(khkhk2n=δn) t φ t φ t n δ n k h k h k 2 n δ n
(7)

## Constraint induced by the doubleshift orthogonality condition

Let's write out Equation 6 explicitly when hn h n is of length 4.



k          = 0        1        2          3          4       5
h(k)       = h(0)     h(1)     h(2)       h(3)       0       0
h(h-0)     = h(0)     h(1)     h(2)       h(3)       0       0
h(k)h(k-0) = h(0)^2 + h(1)^2 + h(2)^2  +  h(3)^2   + 0   +   0

h(k)       = h(0)     h(1)     h(2)       h(3)       0       0
h(k-2)     = 0        0        h(0)       h(1)       h(2)    h(3)
h(k)h(k-2) = 0        0        h(0)h(2)   h(1)h(3)   0       0

0   +    0    +   h(0)h(2) + h(1)h(3) + 0   +   0



We get

h20+h21+h22+h23=1 h 0 2 h 1 2 h 2 2 h 3 2 1
(8)
h0h2+h1h3=0 h 0 h 2 h 1 h 3 0
(9)
(khkhk2n=δn)hn  is of even length k h k h k 2 n δ n h n   is of even length
(10)

## Autocorrelation form of the doubleshift orthogonality condition

Let's define the autocorrelation of hn h n to be the sequence rnhn*hn=khkhkn r n h n h n k h k h k n Autocorrelation sequences have several properties.

1. rn=rn r n r n
2. Rz=𝒵rn=HzH1z R z 𝒵 r n H z H 1 z . Note that
Rz=𝒵rn=𝒵hn*hn=𝒵hn𝒵hn=HzH1z R z 𝒵 r n 𝒵 h n h n 𝒵 h n 𝒵 h n H z H 1 z
(11)
3. R f ω=DTFTrn=| H f ω|2 R f ω DTFT r n H f ω 2 . This comes from the following equations.
R f ω=Reiω=HeiωHe(iω)= H f ω H f ω-=| H f ω|2 R f ω R ω H ω H ω H f ω H f ω H f ω 2
(12)
We can write the orthogonality condition khkhk2n=δn k h k h k 2 n δ n in terms of the autocorrelation very compactly as r2n=δn r 2 n δ n .
khkhk2n=δnr2n=δn k h k h k 2 n δ n r 2 n δ n
(13)
In other words, if the scaling function φt φ t is orthogonal to its integer-translates, then the autocorrelation rn r n must be a halfband filter. That is, rn=0 r n 0 for even nn, except for n=0 n 0 , r0=1 r 0 1 .

## Fourier form of the doubleshift orthogonality condition

The Fourier form of the orthogonality condition is based on the following property. r2n=δnrn+1 nrn=2δn r 2 n δ n r n 1 n r n 2 δ n Let's take the 𝒵-transform of both sides of this equation. 𝒵rn+1 nrn=𝒵2δn 𝒵 r n 1 n r n 𝒵 2 δ n Note that 𝒵1nrn=Rz 𝒵 1 n r n R z Therefore, we have the equivalence r2n=δnRz+Rz=2 r 2 n δ n R z R z 2 Now let's write Equation 6 in terms of the Fourier transform. DTFTrn+1nrn=DTFT2δn DTFT r n 1 n r n DTFT 2 δ n Note that DTFT1 nrn=Rfωπ DTFT 1 n r n R ω f Therefore, we have the equivalence r2n=δnRfω+Rfωπ=2 r 2 n δ n R ω f R ω f 2 Equation 14 summarizes the different forms of the orthogonality condition.

khkhk2n=δn r2n=δn k h k h k 2 n δ n r 2 n δ n
(14)
Rz+Rz=2 R z R z 2 Rfω+Rfωπ=2 R ω f R ω f 2 Using the relation Rz=HzH1z R z H z H 1 z , or equivalently Rfω=HfωHfω- R ω f H ω f H ω f , we get Equation 15.
khkhk2n=δnHzH1z+HzH1z=2 k h k h k 2 n δ n H z H 1 z H z H 1 z 2
(15)
|Hfω|2+|Hfωπ|2=2 H ω f 2 H ω f 2 2 Equation 14 gives the Fourier transform and 𝒵𝒵-transform forms of the orthogonality condition using the autocorrelation sequence rn=hn*hn r n h n h n . Equation 15 gives the Fourier transform and 𝒵𝒵-transform forms of the orthogonality condition using the scaling filter hn h n itself.

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