Projections

Define the set BB to be B={φt-n|n∈ℤ} B φ t n n BB is called an orthonormal set if ∫ φ n t φ m tdt=δn-m t φ n t φ m t δ n m where φ n t≔φt-n ≔ φ n t φ t n . Define the set 𝒱𝒱 to be

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.

Setting ddanε a n ε to zero gives 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, ∫φtφt-ndt=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. ∫φtφt-ndt=δn t φ t φ t n δ n Sum each side over nn, ∑n∫φtφt-ndt=∑nδn n t φ t φ t n n δ n Then we have ∫φt∑nφt-ndt=1 t φ t n φ t n 1 Recall from this equation that ∑nφt-n=∫φtdt n φ t n t φ t (provided φt φ t is continuous), so that we can write ∫φt∫φαdαdt=1 t φ t α φ α 1 or ∫φtdt∫φαdα=1 t φ t α φ α 1 ∫φtdt2=1 t φ t 2 1 or ∫φtdt=±1 t φ t ± 1

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=2∑khkφ2t-k φ t 2 k h k φ 2 t k and φt-n=2∑mhmφ2t-n-m φ t n 2 m h m φ 2 t n m into Equation 1. ∫2∑khkφ2t-k2∑mhmφ2t-2n-mdt=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 2∑khk∑mhm∫φ2t-kφ2t-2n-mdt=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=2t-k α 2 t k to get ∑khk∑mhm∫φαφα+k-2n-mdt=δn k h k m h m t φ α φ α k 2 n m δ n Using Equation 1, ∑khk∑mhmδk-2n-m=δn k h k m h m δ k 2 n m δ n Because δk-2n-m δ k 2 n m is zero except when m=k-2n m k 2 n , we have

This is the condition on hn h n that corresponds to condition Equation 1.

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

Autocorrelation form of the doubleshift orthogonality condition

Let's define the autocorrelation of hn h n to be the sequence rn≔hn*h-n=∑khkhk-n ≔ r n h n h n k h k h k n Autocorrelation sequences have several properties.

We can write the orthogonality condition ∑khkhk-2n=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 . 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=δn⇔rn+-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 𝒵-1nrn=R-z 𝒵 1 n r n R z Therefore, we have the equivalence r2n=δn⇔Rz+R-z=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 DTFT-1 nrn=Rfω-π DTFT 1 n r n R ω f Therefore, we have the equivalence r2n=δn⇔Rfω+Rfω-π=2 ⇔ r 2 n δ n R ω f R ω f 2 Equation 14 summarizes the different forms of the orthogonality condition.

⇔Rz+R-z=2 ⇔ R z R z 2 ⇔Rfω+Rfω-π=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. ⇔|Hfω|2+|Hfω-π|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*h-n 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.