Suppose h⁢n=0 h n 0 for n<0 n 0 and n>N−1 n N 1 . Then the support of φ⁢t φ t is 0 N−1 0 N 1 as we saw before . How can we find φ⁢t φ t from h⁢n h n ? φ⁢t=2⁢∑n=0N−1h⁢n⁢φ⁢2⁢t−n φ t 2 n 0 N 1 h n φ 2 t n It turns out that we can find samples of φ⁢t φ t using the following procedure.

For example, if h⁢n h n is of length N=4 N 4 , then from the dilation equation we have φ⁢0=2⁢(h⁢0⁢φ⁢0+h⁢1⁢φ⁢-1+h⁢2⁢φ⁢-2+h⁢3⁢φ⁢-3) φ 0 2 h 0 φ 0 h 1 φ -1 h 2 φ -2 h 3 φ -3 φ⁢1=2⁢(h⁢0⁢φ⁢2+h⁢1⁢φ⁢1+h⁢2⁢φ⁢0+h⁢3⁢φ⁢-1) φ 1 2 h 0 φ 2 h 1 φ 1 h 2 φ 0 h 3 φ -1 φ⁢2=2⁢(h⁢0⁢φ⁢4+h⁢1⁢φ⁢3+h⁢2⁢φ⁢2+h⁢3⁢φ⁢1) φ 2 2 h 0 φ 4 h 1 φ 3 h 2 φ 2 h 3 φ 1 φ⁢3=2⁢(h⁢0⁢φ⁢6+h⁢1⁢φ⁢5+h⁢2⁢φ⁢4+h⁢3⁢φ⁢3) φ 3 2 h 0 φ 6 h 1 φ 5 h 2 φ 4 h 3 φ 3 As we know that φ⁢t φ t is supported on 0 N−1 0 N 1 ( 0 3 0 3 for this example), we can see that φ⁢-1 φ -1 , φ⁢4 φ 4 , etc fall outside the support. Therefore these equations can be simplified. φ⁢0=2⁢h⁢0⁢φ⁢0 φ 0 2 h 0 φ 0 φ⁢1=2⁢(h⁢0⁢φ⁢2+h⁢1⁢φ⁢1+h⁢2⁢φ⁢0) φ 1 2 h 0 φ 2 h 1 φ 1 h 2 φ 0 φ⁢2=2⁢(h⁢1⁢φ⁢3+h⁢2⁢φ⁢2+h⁢3⁢φ⁢1) φ 2 2 h 1 φ 3 h 2 φ 2 h 3 φ 1 φ⁢3=2⁢h⁢3⁢φ⁢3 φ 3 2 h 3 φ 3 or in matrix form, φ⁢0φ⁢1φ⁢2φ⁢3=2⁢( h⁢0000 h⁢2h⁢1h⁢00 0h⁢3h⁢2h⁢1 000h⁢3 )⁢φ⁢0φ⁢1φ⁢2φ⁢3 φ 0 φ 1 φ 2 φ 3 2 h 0 0 0 0 h 2 h 1 h 0 0 0 h 3 h 2 h 1 0 0 0 h 3 φ 0 φ 1 φ 2 φ 3 Remember that φ⁢t φ t is zero outside the range 0 3 0 3 .

This matrix equation reveals that the vector of values of φ⁢t φ t on the integers 0…N−1 0 … N 1 is an eigenvector of the matrix P=2⁢( h⁢0000 h⁢2h⁢1h⁢00 0h⁢3h⁢2h⁢1 000h⁢3 ) P 2 h 0 0 0 0 h 2 h 1 h 0 0 0 h 3 h 2 h 1 0 0 0 h 3 Specifically, it is an eigenvector corresponding to the eigenvalue 11. However, not just any set of coefficients will give a matrix PP with 11 as an eigenvalue.

It turns out that PP will always have 11 as an eigenvalue whenever the coefficients h⁢n h n satisfy the condition

As a result, if the condition is satisfied, we can calculate the integer samples of φ⁢t φ t by calculating the eigenvector of PP corresponding to the eigenvalue 11.

We can also write the equation as follows: 2⁢( h⁢0000 h⁢2h⁢1h⁢00 0h⁢3h⁢2h⁢1 000h⁢3 )⁢φ⁢0φ⁢1φ⁢2φ⁢3−φ⁢0φ⁢1φ⁢2φ⁢3=0000 2 h 0 0 0 0 h 2 h 1 h 0 0 0 h 3 h 2 h 1 0 0 0 h 3 φ 0 φ 1 φ 2 φ 3 φ 0 φ 1 φ 2 φ 3 0 0 0 0 or (2⁢( h⁢0000 h⁢2h⁢1h⁢00 0h⁢3h⁢2h⁢1 000h⁢3 )−( 1000 0100 0010 0001 ))⁢φ⁢0φ⁢1φ⁢2φ⁢3=0000 2 h 0 0 0 0 h 2 h 1 h 0 0 0 h 3 h 2 h 1 0 0 0 h 3 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 φ 0 φ 1 φ 2 φ 3 0 0 0 0 (P−I)⁢φ=0 P I φ 0

Recall from the sum-integral equality that ∫φ⁢tdt=∑kφ⁢k t φ t k φ k . Therefore, if we want the normalization ∫φ⁢tdt=A t φ t A , say, we can get it by the normalization ∑kφ⁢k=A k φ k A Therefore, we can add this equation to the matrix equation we already have.

Notice that the matrix PP is a convolution matrix, but with every other row removed. It can be constructed by using the convmtx command and then removing every other row:

      H = convmtx(h(:),N);
      P = sqrt(2)*H(1:2:2*N-1,:);
    

Refinement

We can get more samples of φ⁢t φ t by using the dilation equation. φ⁢t=2⁢∑kh⁢k⁢φ⁢2⁢t−k φ t 2 k h k φ 2 t k To find the value of φ⁢t φ t on half integers n2 n 2 , for n∈Z n , let t=n2 t n 2 in the dilation equation.

φ⁢n2=2⁢∑kh⁢k⁢φ⁢n−k=2⁢h⁢n*φ⁢n φ n 2 2 k h k φ n k 2 h n φ n

(3)

According to the dilation equation, the value of φ⁢t φ t on the half-integers can be obtained by convolving the scaling filter h⁢n h n with the integer-samples of φ⁢t φ t , as illustrated in Figure 1.

Figure 1

Similarly, we can get samples of φ⁢t φ t on the quarter-integers by setting t=n4 t n 4 in the dilation equation.

φ⁢n4=2⁢∑kh⁢k⁢φ⁢n2−k=2⁢∑kh⁢k⁢g⁢n−2⁢k=2⁢[↑2]⁢h⁢n*g⁢n=2⁢[↑2]⁢h⁢n*φ⁢n2 φ n 4 2 k h k φ n 2 k 2 k h k g n 2 k 2 [↑2] h n g n 2 [↑2] h n φ n 2

(4)

where g⁢n≔φ⁢n2 ≔ g n φ n 2 . The notation [↑2]⁢x⁢n [↑2] x n represents an upsampled version of x⁢n x n . Recalling that the transfer function of [↑2]⁢h⁢n [↑2] h n is H⁢z2 H z 2 we get the following diagram (Figure 2).

Figure 2

The value of φ⁢t φ t for t=n8 t n 8 is found as

φ⁢n8=2⁢∑kh⁢k⁢φ⁢n4−k=2⁢∑kh⁢k⁢g⁢n−4⁢k=2⁢[↑4]⁢h⁢n*g⁢n=2⁢[↑4]⁢h⁢n*φ⁢n4 φ n 8 2 k h k φ n 4 k 2 k h k g n 4 k 2 [↑4] h n g n 2 [↑4] h n φ n 4

(5)

where g⁢n≔φ⁢n4 ≔ g n φ n 4 , with corresponding diagram (Figure 3),

Figure 3

Similarly, φ⁢n16=2⁢[↑8]⁢h⁢n*φ⁢n8 φ n 16 2 [↑8] h n φ n 8 Given the value of φ⁢t φ t on the integers, a simple method emerges for finding φ⁢t φ t on the dyadic rationals. In general, we have Equation 6 and Figure 4.

φ⁢n2j+1=2⁢ [ ↑ 2 j ] ⁢h⁢n*φ⁢n2j φ n 2 j 1 2 [ ↑ 2 j ] h n φ n 2 j

(6)

Figure 4

Let hh be a vector of the scaling filter coefficients. Let pp be a vector of samples of φ⁢t φ t on the integers. Then the samples of φ⁢t φ t on the dyadic rationals are obtained by the following simple loop.

	for k = 1:5
	   p = sqrt(2) * conv(h,p);
	   h = up(h,2);
	end
      

We also have the diagram in Figure 5,

Figure 5

Using the Z-transform, we can write

𝒵⁢φ⁢n2j=2j2⁢H⁢z⁢H⁢z2⁢⋯⁢H⁢zj−1⁢𝒵⁢φ⁢n 𝒵 φ n 2 j 2 j 2 H z H z 2 ⋯ H z j 1 𝒵 φ n

(7)

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Last edited by Charlet Reedstrom on Aug 12, 2005 7:55 pm -0500.