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Derivative of the Wavelet Scaling Function

Module by: Ivan Selesnick

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The derivative of φt φ t (if it exists!) also satisfies a dilation equation. To see this, take the derivative of both sides of the dilation equation. ddtφt=ddt2∑nhnφ2t−n t φ t t 2 n h n φ 2 t n or φt=2∑nhnddtφ2t−n φ t 2 n h n t φ 2 t n As ddtφ2t−n=2φ′2t−n t φ 2 t n 2 φ 2 t n we have

φ′t=2∑n2hnφ′2t−n φ t 2 n 2 h n φ 2 t n (1)
Therefore, φ′t φ t satisfies a dilation equation with coefficients 2hn 2 h n . We can find values of φ′t φ t on the integers t=n t n using the procedure described already. We will get the matrix equation 2P φ ′= φ ′ 2 P φ φ As a result, we see that 11 must be an eigenvalue of the matrix 2P 2 P ; otherwise, the derivative of φt φ t does not exist on the integers, meaning that φt φ t is not differentiable. Equivalently, 1/2 12 is an eigenvalue of PP.

It turns out that if hn h n satisfies the sum rule

∑n-1nnhn=0 n 1 n n h n 0 (2)
in addition to this sum rule, then 1/2 12 is an eigenvalue of PP.

We will not show that if hn h n satisfies the sum rules, this equation from the module on the dilation wavelet equation and Equation 2, then 1/2 12 is an eigenvalue of PP. (We already know that 11 is an eigenvalue of PP.) We will do this by showing that 1/2 12 is an eigenvalue of PT P .

Define e 0 ≔111⋮1 ≔ e 0 1 1 1 ⋮ 1 e 1 ≔012⋮N−1 ≔ e 1 0 1 2 ⋮ N 1 We already saw that e 0 P= e 0 e 0 P e 0 which proves that 1 is an eigenvalue of PP. Now let's looks at e 1 P e 1 P . e 1 P=2∑nnh2n∑nnh2n−1∑nnh2n−2⋮∑nnh2n−(N−1) e 1 P 2 n n h 2 n n n h 2 n 1 n n h 2 n 2 ⋮ n n h 2 n N 1 We now simplify the right hand side. First notice that if hn h n satisfies this equation, then ∑n-1nnhn=0⇔∑n2nh2n−∑n2n−1h2n−1=0 ⇔ n 1 n n h n 0 n 2 n h 2 n n 2 n 1 h 2 n 1 0 ⇔∑n2nh2n−∑n2nh2n−1+∑nh2n−1=0 ⇔ n 2 n h 2 n n 2 n h 2 n 1 n h 2 n 1 0 ⇔∑n2nh2n−∑n2nh2n−1+12=0 ⇔ n 2 n h 2 n n 2 n h 2 n 1 1 2 0 ⇔∑nnh2n−∑nnh2n−1+122=0 ⇔ n n h 2 n n n h 2 n 1 1 2 2 0 ⇔∑nnh2n−1=∑nnh2n+122 ⇔ n n h 2 n 1 n n h 2 n 1 2 2 The last line says that the second term of e 1 P e 1 P is equal to the first term plus the constant 122 1 2 2 . Let's find an expression for the third term ∑nnh2n−2 n n h 2 n 2 .

∑nnh2n−2=∑kk+1h2k=∑kkh2k+∑kh2k=∑nnh2n+12=∑nnh2n+222 n n h 2 n 2 k k 1 h 2 k k k h 2 k k h 2 k n n h 2 n 1 2 n n h 2 n 2 2 2 (3)
That means that the third term of e 1 P e 1 P is equal to the first term plus twice the previous constant. In general, it can be similarly shown that
∑n-1nnhn=0⇔∑nnh2n−k=∑nnh2n+k22 ⇔ n 1 n n h n 0 n n h 2 n k n n h 2 n k 2 2 (4)
Therefore
e 1 P=C e 0 +12 e 1 e 1 P C e 0 1 2 e 1 (5)
with C=2∑nnh2n C 2 n n h 2 n We can show that 1/212 is an eigenvalue of PP by constructing the vector e 0 +β e 1 e 0 β e 1 . For the right choice for ββ, this vector turns out to be an eigenvector with eigenvalue 1/212.
e 0 +β e 1 P= e 0 P+β e 1 P= e 0 +βC e 0 +12 e 1 =1+βC e 0 +β2 e 1 e 0 β e 1 P e 0 P β e 1 P e 0 β C e 0 1 2 e 1 1 β C e 0 β 2 e 1 (6)
Therefore, e 0 +β e 1 P=12 e 0 +β e 1 e 0 β e 1 P 1 2 e 0 β e 1 provided 1+βC=12 1 β C 1 2 . Solving for ββ gives β=-12C β 1 2 C Therefore, we have the following result ∑nhn=2∧∑n-1nhn=0∧∑n-1nnhn=0⇒1  and  1/2  are eigenvalues of  P n h n 2 n 1 n h n 0 n 1 n n h n 0 1   and   12   are eigenvalues of   P

If hn h n satisfies these sum rules, then we can use the procedure described before to calculate φ′t φ t on the dyadic rationals. If hn h n does not satisfy these sum rules then φt φ t is not differentiable.

Moments

The sequence hn h n satisfies the sum rules

∑n-1nhn=0 n 1 n h n 0 (7)
∑n-1nnhn=0 n 1 n n h n 0 (8)
if and only if Hz H z can be written as Hz=Qzz-1+12 H z Q z z 1 2 To see this, differentiate Hz H z , H′z=Qz-2z-1z-1+1+Q′zz-1+12 H z Q z -2 z z 1 Q z z 1 2 and set z=-1 z -1 . Then H-1=0∧H′-1=0 H -1 0 H -1 0 We already saw that H-1=0 H -1 0 gives Equation 7. Writing out H′-1 H -1 we get
Hz=∑nhnz-n H z n h n z n (9)
H′z=-∑nhnnz-n+1 H z n h n n z n 1 (10)
H′-1=∑nhnn-1n H -1 n h n n 1 n (11)
This gives sum rule Equation 8. The result above can therefore be stated as H1=2∧H-1=0∧H′-1=0⇒1  and  1/2  are eigenvalues of  P H 1 2 H -1 0 H -1 0 1   and   12   are eigenvalues of   P or as H1=2∧Hz=Qzz-1+12⇒1  and  1/2  are eigenvalues of  P H 1 2 H z Q z z 1 2 1   and   12   are eigenvalues of   P

For the existence of higher derivatives, the matrix PP must have further eigenvalues of the for 12k 1 2 k .

In general,

∀,0≤k≤K−1:∑nhn=2∧∑n-1nnkhn=0⇒11/21/4…12K−1  are eigenvalues of  P 0 k K 1 n h n 2 n 1 n n k h n 0 1 12 14 … 1 2 K 1   are eigenvalues of   P (12)
Or in terms of Hz H z :
H1=2∧Hz=Qzz-1+1K⇒11/21/4…12K−1  are eigenvalues of  P H 1 2 H z Q z z 1 K 1 12 14 … 1 2 K 1   are eigenvalues of   P (13)
For signal processing applications, it is desirable that φt φ t be relatively smooth, and differentiability indicates the smoothness of the function. However, φt φ t can not be infinitely differentiable when hn h n is finite in length. When hn h n is finite in length, then φt φ t will be finite in length, and the matrix PP will have a finite number of eigenvalues. In particular, PP will have a finite number of eigenvalues of the form 12k 1 2 k . Therefore, φt φ t can not be infinitely differentiable. The scaling functions that are the most differentiable are the spline scaling functions. But they are not orthogonal, as we will see later.

Normalization for φ'(t)

We can write the equation 2P φ ′=φ 2 P φ φ as 2P−I φ ′=0 2 P I φ 0 When we computed samples of φt φ t we used the normalization ∑nφn=A n φ n A in this equation. What normalization should we use when we compute samples of φ′t φ t ? We will see that ∑nφ′n n φ n is not the right quantity to normalize, but that ∑nnφ′n n n φ n is.

Begin with the dilation equation for the derivative of φt φ t . φ′t=2∑k2hkφ′2t−k φ t 2 k 2 h k φ 2 t k Set t=n∈ℤ t n , φ′n=2∑k2hkφ′2n−k φ n 2 k 2 h k φ 2 n k Sum over nn,

∑nφ′n=∑n2∑k2hkφ′2n−k=2∑n∑k2hkφ′2n−2k+2∑n∑k2h2k−1φ′2n−2k+1=2∑k2h2k∑nφ′2n−2k+2∑k2h2k−1∑nφ′2n−2k+1=2∑k2h2k∑nφ′2n+2∑k2h2k−1∑nφ′2n−1=2∑k2h2k∑nφ′2n+2∑k2h2k−1∑nφ′2n−1=22∑nφ′2n+22∑nφ′2n−1=2∑nφ′2n+2∑nφ′2n−1=2∑nφ′n n φ n n 2 k 2 h k φ 2 n k 2 n k 2 h k φ 2 n 2 k 2 n k 2 h 2 k 1 φ 2 n 2 k 1 2 k 2 h 2 k n φ 2 n 2 k 2 k 2 h 2 k 1 n φ 2 n 2 k 1 2 k 2 h 2 k n φ 2 n 2 k 2 h 2 k 1 n φ 2 n 1 2 k 2 h 2 k n φ 2 n 2 k 2 h 2 k 1 n φ 2 n 1 2 2 n φ 2 n 2 2 n φ 2 n 1 2 n φ 2 n 2 n φ 2 n 1 2 n φ n (14)
That is, ∑nφ′n=2∑nφ′n n φ n 2 n φ n or ∑nφ′n=0 n φ n 0 This assumes of course that φ′t φ t is defined on the integers, for which we need the sum rule ∑n-1nnhn=0 n 1 n n h n 0 which we saw before. Similar to what we saw before, if we define S j ≔12j∑nφ′n2j ≔ S j 1 2 j n φ n 2 j and use the same procedure used before, we find that S j =2 S j − 1 S j 2 S j − 1 Because S 0 =∑nφ′n=0 S 0 n φ n 0 we find that S j =0 S j 0 for all j≥0 j 0 . Taking the limit j→∞ j gives ∫φ′tdt=limj→∞ S j =0 t φ t j S j 0 or
∫φ′tdt=∑nφ′n=0 t φ t n φ n 0 (15)
Therefore, we can not use the quantity ∑nφ′n n φ n to normalize φ′t φ t .

Let us consider now instead the quantity T j ≔12j∑kk2jφ′k2j ≔ T j 1 2 j k k 2 j φ k 2 j Notice that limj→∞ T j =∫tφ′tdt j T j t t φ t Set t=k2j t k 2 j in the dilation equation, φ′k2j=2∑n2hnφ′2k2j−n φ k 2 j 2 n 2 h n φ 2 k 2 j n Multiply by k2j k 2 j and sum over kk,

∑kk2jφ′k2j=∑kk2j2∑n2hnφ′2k2j−n=2∑nhn∑kk2j−1φ′k2j−1−n=2∑nhn∑ll2j−1+nφ′l2j−1=2∑nhn∑ll2j−1φ′l2j−1+2∑nhn∑lnφ′l2j−1=2∑ll2j−1φ′l2j−1∑nhn+2∑nnhn∑lφ′l2j−1=2∑ll2j−1φ′l2j−12+0=2∑ll2j−1φ′l2j−1 k k 2 j φ k 2 j k k 2 j 2 n 2 h n φ 2 k 2 j n 2 n h n k k 2 j 1 φ k 2 j 1 n 2 n h n l l 2 j 1 n φ l 2 j 1 2 n h n l l 2 j 1 φ l 2 j 1 2 n h n l n φ l 2 j 1 2 l l 2 j 1 φ l 2 j 1 n h n 2 n n h n l φ l 2 j 1 2 l l 2 j 1 φ l 2 j 1 2 0 2 l l 2 j 1 φ l 2 j 1 (16)
because ∑lφ′l2j−1=0 l φ l 2 j 1 0 . Dividing both sides by 2j 2 j gives 12j∑kk2jφ′k2j=12j−1∑kk2j−1φ′k2j−1 1 2 j k k 2 j φ k 2 j 1 2 j 1 k k 2 j 1 φ k 2 j 1 or T j = T j − 1 T j T j − 1 Taking the limit j→∞ j gives
∫tφ′tdt=∑nnφ′n t t φ t n n φ n (17)
This is similar to this equation.

If we choose to normalize φt φ t such that ∫φtdt=A t φ t A , then what value of ∫tφ′tdt t t φ t does this imply? By the product rule, tφt ′=φt+tφ′t t φ t φ t t φ t Integrating both sides, ∫-∞∞ tφt ′dt=∫-∞∞φtdt+∫-∞∞tφ′tdt t t φ t t φ t t t φ t (This is just integration by parts.) If φt φ t is of finite support (or decays sufficiently fast), then ∫-∞∞ tφt ′dt=tφt|t=-∞∞=0 t t φ t t t φ t 0 Therefore we get

∫-∞∞tφ′tdt=-∫-∞∞φtdt t t φ t t φ t (18)
So if we choose to normalize φt φ t so that ∫φtdt=A t φ t A , then we also have ∫tφ′tdt=-A t t φ t A . Combining this with Equation 17 indicates that we should normalize the eigenvector when we compute samples of φ′t φ t by ∑nnφ′n=-A n n φ n A Appending this normalization to the equations we already have for the integer samples of φ′t φ t gives the following matrix system.
2P−I012…N−1 φ ′=0-A 2 P I 0 1 2 … N 1 φ 0 A (19)
We can compute samples of φ′t φ t on a finer grid by the same refinement method used for φt φ t .

Exercise

Develop the procedure for the second derivative φt′′ 2 φ t . Give a condition that ensures that φt′′ 2 φ t exists on the integers, and show why the condition works. Show how to compute φt′′ 2 φ t on the integers. Show what normalization on φt′′ 2 φ t is induced by the normalization ∫φtdt=A t φ t A . Using Matlab, make plots of φ′t φ t and φt′′ 2 φ t for some examples hn h n that satisfy the appropriate conditions.

Splines

Given KK, the shortest hn h n for which Hz=Qzz-1+1K H z Q z z 1 K has Qz Q z equal to a constant chosen so that H1=2 H 1 2 . The result is Hz=2z-1+12K H z 2 z 1 2 K It turns out that the scaling function φt φ t corresponding to this is a square pulse repeatedly convolved with itself. The result is a spline - a function that is piecewise polynomial.

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