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Spring 2018 Math 994: Comp. Harmonic Analysis

The next theorem shows that the analytic wavelet transform is invertibleso long as the wavelet satisfies a weak admissibility condition given by (22).Recall that fa(t) is the analytic part of a real valued signal f(t).

Theorem 27. Let f 2 L2(R) be real valued and let 2 L1(R)\L2(R) be ananalytic wavelet. Then

Wf(u, s) =1

2Wfa(u, s) (21)

Furthermore, if

C =

Z +1

0

| b (!)|2

!d! < 1 (22)

then

f(t) =2

C <

Z +1

0

Z +1

�1

Wf(u, s) s(t� u) duds

s2

�(23)

and

kfk22 =

2

C

Z +1

0

Z +1

�1

|Wf(u, s)|2 duds

s2(24)

Proof. First note that the L1(R) norm of s is:

k sk1 =psk k1

which is obtained through a simple change of variables.We now prove (21) first. Recall that e s(t) =

⇤(�t) and consider

fs(u) = Wf(u, s) = f ⇤ e s(u)

Young’s inequality (14) proves that fs 2 L2(R):

kfsk2 = kf ⇤ e sk2 kfk2ke sk1 =

psk k1kfk2 < 1

The Fourier transform of fs is:

bfs(!) = bf(!)be s(!) = bf(!)ps b ⇤(s!)

Since b (!) = 0 for ! 0, and bfa(!) = 2 bf(!) for ! > 0, we derive that

bfs(!) =1

2bfa(!)

ps b ⇤(!)

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which is the Fourier transform of (1/2)fa ⇤ e s(u).Now let us prove (24). We use the Plancheral formula and Tonelli:

2

C

Z +1

0

Z +1

�1

|Wf(u, s)|2 duds

s2=

1

2C

Z +1

0

Z +1

�1

|fa ⇤e s(u)|

2du

ds

s2

=1

C 4⇡

Z +1

0

Z +1

�1

| bfa(!)|2| b (s!)|2 d!ds

s

=1

4⇡

Z +1

0| bfa(!)|2

1

C

Z +1

0| b (s!)|2ds

sd!

(25)

Now make the change of variables ⇠ = s!, which induces the change ofmeasure ds = (s/⇠)d⇠. We have:

(25) =1

4⇡

Z +1

0| bfa(!)|2

1

C

Z +1

0

| b (⇠)|2

⇠d⇠ d!

=1

4⇡

Z +1

0| bfa(!)|2 d!

=1

2kfak

22 = kfk

22

Finally we prove (23). To do so we prove that

fa(t) =1

C

Z +1

0

Z +1

�1

Wfa(u, s) s(t� u) duds

s2(26)

The result will then follow since Wfa(u, s) = 2Wf(u, s) and <[fa(t)] = f(t).To prove (26) we write the right hand side of (26) as:

h(t) =1

C

Z +1

0

Z +1

�1

Wfa(u, s) s(t� u) duds

s2=

1

C

Z +1

0fa ⇤

e s ⇤ s(t)ds

s2

We compute the Fourier transform of h(t), which gives:

bh(!) = 1

C

Z +1

0

bfa(!)be s(!) b s(!)

ds

s2

=bfa(!)C

Z +1

0

ps b ⇤(s!)

ps b (s!) ds

s2

=bfa(!)C

Z +1

0| b (s!)|2ds

s

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But this is exactly the same integral we encountered in (25), where we provedthat it equals C . Thus we obtain bh(!) = bfa(!). But then we must haveh(t) = fa(t) and the theorem is proved.

The condition (22) is the admissibility condition of the wavelet . Notethat to be finite, we must have b (0) = 0, or equivalently

R = 0. This

condition is nearly su�cient. If b 2 C1(R) as well, then one can verify thatthe admissibility condition is satisfied. By Theorem 11, if is su�cientlylocalized in time such that

| (t)| C

1 + |t|n+1+✏

then we must have b 2 C1(R).

Exercise 42. Read Section 4.3.2 of A Wavelet Tour of Signal Processing.

4.4.2 Wavelet Ridges

Section 4.4.3 of A Wavelet Tour of Signal Processing.

Motivated by the hyperbolic chirp example and the poor performance of thewindowed Fourier ridges for this example, we define and study wavelet ridges.We utilize an approximately analytic “wavelet” (t) of the form:

(t) = g(t)ei⌘t

where the window function g(t) satisfies the same assumptions as in thewindowed Fourier case; namely:

• supp g = [�1/2, 1/2]

• g(t) � 0 so that |bg(!)| bg(0) for all ! 2 R

• kgk2 = 1 but also bg(0) =Rg(t) dt = kgk1 ⇡ 1

Let �! be the bandwidth of bg. If ⌘ > �! then

b (!) = bg(! � ⌘) ⌧ 1, 8! 0

Thus (t) is not strictly a wavelet nor is it strictly analytic, but it nearlysatisfies both conditions.

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Notice that dilated and translated wavelets can be written as:

u,s(t) =1ps

✓t� u

s

◆= gs,u,⌘/s(t)e

�i(⌘/s)u

where

gs,u,⇠(t) =1psg

✓t� u

s

◆ei⇠t

The resulting wavelet transform use time frequency atoms similar to thoseof the windowed Fourier transform. However, in this case the scale s variesover (0,+1) and ⇠ = ⌘/s:

Wf(u, s) = hf, u,si = hf, gu,s,⌘/siei(⌘/s)u

Theorem 26 computes hf, gu,s,⇠i when f(t) = a(t) cos ✓(t). Applying thistheorem to the wavelet transform gives:

Wf(u, s) = hf, gu,s,⌘/siei(⌘/s)u =

ps

2a(u)ei✓(u) [bg(s[⌘/s� ✓

0(u)]) + "(u, ⌘/s)]

A normalized scalogram computes

ePWf(u, ⌘/s) =|Wf(u, s)|2

s=

1

4a(u)2 |bg(s[⌘/s� ✓

0(u)]) + "(u, ⌘/s)|2

If the error term "(u, ⌘/s) is negligible, ePWf(u, ⌘/s) obtains its maxima at(u, ⌘/su) where

⌘

su= ✓

0(u) =) su =⌘

✓0(u)

The corresponding points (u, ⌘/su) are called wavelet ridges.Recall the error term "(u, ⌘/s) is broken into four components:

|"(u, ⌘/s)| "a,1(u, ⌘/s) + "a,2(u, ⌘/s) + "✓,2(u, ⌘/s) + sup!�s✓0(u)

|bg(!)|| {z }

(iv)

At the ridge points (u, ⌘/su) the first error term "a,1 and the fourth errorterm can be made negligible if the bandwidth �! satisfies

�! su✓0(u) =) �! ⌘

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but this was assumed from the start in order to make an approximatelyanalytic wavelet, so these two error terms are guaranteed to be small by thechoice of the wavelet. Using Theorem 26, the second order terms at the ridgepoints are bounded as:

"a,2(u, ⌘/su) sup|t�u|su/2

s2u|a

00(t)|

|a(u)|= sup

|t�u|⌘/2✓0(u)

⌘2

✓0(u)2|a

00(t)|

|a(u)|

and

"✓,2(u, ⌘/su) sup|t�u|su/2

sup|t�u|su/2

s2u|✓

00(t)| = sup|t�u|⌘/2✓0(u)

⌘2

✓0(u)2|✓

00(t)|

Thus since ✓0(u) is in the denominator, we see that if the instantaneous

frequency is small, a0(u) and ✓0(u) must have slow variations (i.e., a00(u) and✓00(u) need to be small), but a0(u) and ✓0(u) are allowed to vary much morequickly when the instantaneous frequency is large.

Now turn to our more general signal model:

f(t) =KX

k=1

ak(t) cos ✓k(t)

Recall that to separate the K instantaneous frequencies we require that

bg(sku[✓0k(u)� ✓0

l(u)]) ⌧ 1, 8 k 6= l,⌘

sku

= ✓0

k(u)

which can be obtained if

�! sku|✓

0

k(u)� ✓0

l(u)| =⌘|✓

0

k(u)� ✓0

l(u)|

✓0k(u), k 6= l

Under the assumptions on the window g, the primary free parameter onehas is the frequency ⌘. There is a tension between on the one hand wantingto make ⌘ large relative to the bandwidth, so that the wavelet is nearlyanalytic, the error terms "a,1 and (iv) are small, and so multiple instantaneousfrequencies are separated; however, the second order error terms "a,2 and "✓,2may blow up if ⌘ is made too large.

Let us now return to the examples of the linear and hyperbolic chirps.We start with the sum of two hyperbolic chirps, which the windowed Fouriertransform had trouble analyzing:

f(t) = a1 cos

✓↵1

�1 � t

◆+ a2 cos

✓↵2

�2 � t

◆

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In this case ✓k(t) = ↵k/(�k�t) and ✓0k(t) = ↵k/(�k�t)2. Since the amplitudesa1 and a2 are constant, the second order term "a,2(u, su) = 0. The othersecond order error term is bounded as:

"✓,2(u, su) maxk=1,2

sup|t�u|⌘/2✓0k(u)

⌘2 |✓

00

k(t)|

✓0(u)2

maxk=1,2

sup|t�u|⌘(�k�u)2/2↵k

⌘2 2↵k

(�k � t)3(�k � t)4

↵2k

maxk=1,2

sup|t�u|⌘(�k�u)2/2↵k

⌘22(�k � t)

↵k

This error term will be small if

⌘2⌧

↵k

�k � t

This will be the case if, for example, t 2 [0, �k) and ⌘ ⌧

p↵k/�k. Figure

16 illustrates how the wavelet ridges successfully follow the instantaneousfrequencies of the two hyperbolic chirps.

Now let us go back to the two linear chirps signal

f(t) = a1 cos(bt2 + ct) + a2 cos(bt

2)

which has frequencies ✓1(t) = bt2 + ct and ✓2(t) = bt

2. We thus have

|✓0

1(u)� ✓0

2(u)|

✓01(u)=

|c|

2bt! 0 as t ! +1

Thus for some finite t we will not be able to separate the instantaneous fre-quencies because of interferences. Figure 17 illustrates this phenomenon, asfor large t the ridges follow the interference patterns rather than the instan-taneous frequencies.

The take home message is that better is more sparse. This is true of coursefor compression, where sparser representations require less memory to store.But the linear and hyperbolic chirp examples show that sparsity also meanswe have found a time frequency transform that has a resolution adapted tothe time frequency properties of the signal, in which case the number of ridgepoints is small. Conversely, if signal structures do not match our dictionaryof time frequency atoms, then their energy will di↵use over many such atomswhich produces more ridge points.


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Figure 16: Analysis of a signal consisting of two hyperbolic chirps. (a) Nor-malized scalogram ePWf(u, ⌘/s); (b) Wavelet ridges.

79


Figure 17: Analysis of a signal consisting of two linear chirps. (a) Normalizedscalogram ePWf(u, ⌘/s); (b) Wavelet ridges.

80


4.4.3 Real Wavelets

Section 4.3.1 of A Wavelet Tour of Signal Processing.

We now shift our focus to real wavelets. As we shall see, real valued waveletsare good for measuring sharp single transitions, and in particular measuringthe the regularity of f(t) at a specific point t = u. Indeed, since

R = 0,

the wavelet transform Wf(u, s) = hf, u,si measures the variation of f ina neighborhood of u proportional s. “Zooming in” on these variations willallow us to measure the regularity of f at u.

For now we show that like the analytic wavelet transform, the real wavelettransform is invertible and preserves the energy of f . We collect these resultsin the next theorem.

Theorem 28. Let 2 L1(R) \ L2(R) be a real function such that

C =

Z +1

0

| b (!)|2

!d! < +1

Then, for any f 2 L2(R),

f(t) =1

C

Z +1

0

Z +1

�1

Wf(u, s)1ps

✓t� u

s

◆du

ds

s2

and

kfk22 =

1

C

Z +1

0

Z +1

�1

|Wf(u, s)|2 duds

s2

The proof is nearly identical to the proof of Theorem 27, and so is omitted.An example of a real valued wavelet, which we shall use later and which alsosatisfies the conditions of Theorem 28, is the so called “Mexican hat wavelet.”The Mexican hat wavelet is the second derivative of a suitably normalizedGaussian function g�(t) with mean zero and standard deviation �:

(t) = g00

�(t) =2

⇡1/4p3�

✓t2

�2� 1

◆e�t2/2�2

(27)

The Fourier transform of (t) is

b (!) = �!2bg�(!) = �

p8�5/2⇡1/4p3

!2e��2!2/2

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Figure 18: Mexican hat wavelet for � = 1 and its Fourier transform.

Figure 18 plots the Mexican hat wavelet and its Fourier transform.Figure 19 computes the wavelet transform of the signal from Figure 13

using the Mexican hat wavelet. The maximum scale is smaller than one be-cause the support of f(t) is limited to t 2 [0, 1]. In all numerical calculations,including this one, the minimum scale is limited by the sampling interval ofthe discretized signal due to aliasing e↵ects. When the scale decreases, thewavelet transform Wf(u, s) has a rapid decay to zero in the regions where thesignal is regular. Isolated singularities create cones of large amplitude waveletcoe�cients that converge to the locations of the isolated singularities, as onthe left hand side of the signal. The right hand side of the signal is singularalmost everywhere. If this part can be modeled as a multifractal, then undercertain assumptions the distribution of singularities can be estimated fromWf(u, s), which can characterize the underlying signal generation process.

In numerical applications, both the minimum and maximum scale arelimited. We now examine the wavelet transform when we compute Wf(u, s)only for s < s0. In this case we lose the low frequency components of f(t),since it is the supports of e s(!) =

ps b ⇤(s!) as s ! +1 that collapse in

around ! = 0; see Figure 20 for an illustration.In order to recover this lost low frequency information, we introduce a

single scaling function � that is an aggregation of all wavelets at scales largerthan one. The modulus of its Fourier transform is defined as:

|b�(!)|2 =Z +1

1| b (s!)|2 ds

s=

Z +1

!

| b (⇠)|2

⇠d⇠ (28)

The complex phase of b�(!) can be arbitrarily chosen; in particular we canset it to zero so that � is real valued. One can verify that k�k2 = 1. The

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Figure 19: Real wavelet transform Wf(u, s) computed with a Mexcian hatwavelet. The vertical axis represents log2 s, Black, gray and white pointscorrespond, respectively, to positive, zero, and negative wavelet coe�cients.

Figure 20: Scaled Fourier transform | b (2j!)|2 for 1 j 5 and ! 2 [�⇡, ⇡].Notice the gap around ! = 0 due to the limitation of the largest scale s = 25.

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definition of b�(!) and the admissibility condition yields:

|b�(!)|2 |b�(0)|2 = lim⇠!0

|b�(⇠)|2 = C , 8! 2 R

Denote by �s(t) the scaling of �(t) by s:

�s(t) =1ps�

✓t

s

◆, e�s(t) = �

⇤

s(�t)

The low frequency approximation of f at scale s is

Lf(u, s) = f ⇤ e�s(u)

In this case, the wavelet inversion formula becomes:

f(t) =1

C

Z s0

0Wf(·, s) ⇤ s(t)

ds

s2+

1

C s0Lf(·, s0) ⇤ �s0(t) (29)

For the Mexican hat wavelet defined in (27), the Fourier transform of thescaling function is:

b�(!) = 2�5/2⇡1/4p3

r!2 +

1

�2e��2!2/2

See Figure 21 for a plot.

Figure 21: Scaling function associated to a Mexican hat wavelet and itsFourier transform.


Exercise 45. Read Section 4.5 of A Wavelet Tour of Signal Processing.

84


Exercise 46. Let � be the scaling function defined by (28). Prove thatk�k2 = 1.

Exercise 47. Prove the reconstruction formula given in (29), which can berewritten as:

f(t) =1

C

Z s0

0f ⇤ e s ⇤ s(t)

ds

s2+

1

C s0f ⇤ e�s0 ⇤ �s0(t)

Exercise 48. Adapt your windowed Fourier ridge code from Exercise 38 tocompute the normalized scalogram ePWf(u, ⌘/s) and corresponding waveletridges. Test your code on the sum of two linear chirps and the sum of twohyperbolic chirps. Turn in plots of your wavelet ridges. Do you get somethingsimilar to the plots in Figures 16 and 17?

Exercise 49. Implement the real wavelet transform Wf(u, s) using the Mex-ican hat wavelet. Write down a signal f(t) similar to the one from Figure 13(does not have to be exactly the same!) and compute Wf(u, s) numerically.Turn in a plot of your signal and a plot of Wf(u, s), as in Figure 19.

5 Wavelet Zoom

Chapter 6 of A Wavelet Tour of Signal Processing

Wavelet transforms can zoom in and characterize the regularity of a functionf(t) at individual points t = v.

5.1 Lipschitz Regularity

Section 6.1 of A Wavelet Tour of Signal Processing

5.1.1 Lipschitz Definition and Fourier Analysis

Section 6.1.1 of A Wavelet Tour of Signal Processing

We begin by defining the spaces C↵(R) and C↵(R) for any ↵ � 0. Firstnote that previously we defined Cn(R) as the space of functions f(t) with n-continuous derivatives; in fact we should have called this space Cn(R), wherethe extra · indicates that these functions (and their derivatives) do not have

85


to be bounded. The space Cn(R) should have then been defined as the spaceof functions with n continuous, bounded derivatives, i.e.,

Cn(R) = {f 2 Cn(R) : 8 k = 1, . . . , n, kf(k)k1 < +1}

Let 0 ↵ < 1. A function f 2 C↵(R) if f is continuous and there existsa constant K > 0 such that

8 t, u 2 R, |f(t)� f(u)| K|t� u|↵

This enforces a global Lipschitz regularity on f . The space C↵(R) consistsof those functions f 2 C↵(R) that are also bounded, i.e.,

C↵(R) = {f 2 C↵(R) : kfk1 < +1}

These definitions are extended to arbitrary ↵ � 0 in the following way. Letn < ↵ < n + 1 for some n 2 N. Then f 2 C↵(R) if and only if f 2 Cn(R)and f

(n)2 C↵�n(R). Similarly, f 2 C↵(R) if and only if f 2 Cn(R) and

f(n)

2 C↵�n(R).We can link these definitions to Taylor’s theorem. Suppose that f 2

C↵(R) for n < ↵ < n + 1. Let v 2 R and let Jvf(t) be the jet of f at v,which is the n-degree Taylor polynomial of f around v:

Jvf(t) =nX

k=0

f(k)(v)

k!(t� v)k

Denote the residual by Rvf(t), i.e.,

Rvf(t) = f(t)� Jvf(t)

Then using Taylor’s theorem and the fact that f 2 C↵(R),

|Rvf(t)| K|t� v|↵

Conversely, suppose that f(t) is a continuous function for which there exists auniversal constant K and such that for each v 2 R, there exists a polynomialpv(t) of degree at most n such that

|f(t)� pv(t)| K|t� v|↵

Then f 2 C↵(R).

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References

[1] Stephane Mallat. A Wavelet Tour of Signal Processing, Third Edition:The Sparse Way. Academic Press, 3rd edition, 2008.

[2] Yves Meyer. Wavelets and Operators, volume 1. Cambridge UniversityPress, 1993.

[3] Elias M. Stein and Rami Shakarchi. Fourier Analysis: An Introduction.Princeton Lectures in Analysis. Princeton University Press, 2003.

[4] Richard L. Wheeden and Antoni Zygmund. Measure and Integral: AnIntroduction to Real Analysis. Marcel Dekker, 1977.

[5] Elias M. Stein and Guido Weiss. Introduction to Fourier Analysis onEuclidean Spaces. Princeton University Press, 1971.

[6] John J. Benedetto and Matthew Dellatorre. Uncertainty principles andweighted norm inequalities. Contemporary Mathematics, 693:55–78, 2017.

[7] Karlheinz Grochenig. Foundations of Time Frequency Analysis. SpringerBirkhauser, 2001.

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