MATH 414 Lecture 37

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Exam Discussion

Problem 2

Frequency response: $\sqrt{2\pi }\hat{h}(\omega )=\frac{1}{5}\frac{1-{\mathrm{e}}^{-5i\omega }}{i\omega }$

Convolution filter:

$\begin{array}{rl}(f*h)(t)& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}h(\tau )f(t-\tau )\mathrm{d}\tau \\ & =\frac{1}{5}\underset{0}{\overset{5}{\int }}f(t-\tau )\mathrm{d}\tau \\ & =\frac{1}{5}\underset{t-5}{\overset{t}{\int }}f(\sigma )\mathrm{d}\sigma \end{array}$

Where $\sigma =t-\tau$

3 Cases:

1. $t<0$, then $\sigma \in [t-5,0)$: thus $(f*h)=0$.
2. $t\in (0,5)$, then $\sigma \in [t,t-5]$: thus $(f*h)=\frac{1}{5}\cancel{\underset{t-5}{\overset{0}{\int }}f(\sigma )\mathrm{d}\sigma }+\frac{1}{5}\underset{0}{\overset{t}{\int }}f(\sigma )\mathrm{d}\sigma =\frac{1}{15}(1-{\mathrm{e}}^{-3t})$
3. $t>5$, then $\sigma >0$: thus $(f*h)=\frac{1}{15}({\mathrm{e}}^{-3(t-5)}-{\mathrm{e}}^{-3t})$

Problem 4

$f_2$ defined by $a^2=(3,1,-2,0,-3,9,-3)$ ($a_k^2=0$ for $k<0$ or $k>7$

Haar Wavelet Decomposition into $b^1$, $a^1$, and $a^0$ for $f_2=f_0+w_0+w_1$

$\begin{array}{rl}{a}_k^{j-1}& =\frac{1}{2}(a_{2k}^j+a_{2k+1}^j)\\ b_k^{j-1}& =\frac{1}{2}(a_{2k}^j-a_{2k+1}^j)\end{array}$

First decomposition level $a^1$, $b^1$ gives:

$\begin{array}{rl}{a}^1& =(2,-1,3,-\frac{3}{2})\\ b^1& =(1,-1,-6,-\frac{3}{2})\end{array}$

Corresponding to $k=0,1,2,3$ (and $a_k^1=b_k^1=0$ for $k<0$ or $k>3$)

Second decomposition level $a^0$, $b^0$ gives:

$\begin{array}{rl}{a}^0& =(\frac{1}{2},\frac{3}{4})\\ b^0& =(\frac{3}{2},\frac{9}{4})\end{array}$

Corresponding to $k=0,1$ (and $a_k^0=b_k^0=0$ for $k<0$ or $k>1$)

MRA Summary

So far, we've assumed that the $p_k$'s are real. Let's keep that assumption and see what we have so far:

Multi-Resolution analysis:

scaling function

$\varphi (x)=\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k\varphi (2x-k)$

$p_k=2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (x)\varphi (2x-k)\mathrm{d}x$

wavelet

$\psi (x)=\sum _{k\in \mathbb{\mathbb{Z}}}{(-1)}^k{p}_{1-k}\varphi (2x-k)$

decomposition

$a_k^{j-1}=\frac{1}{2}\sum _{m\in \mathbb{\mathbb{Z}}}{p}_{m-2k}{a}_m^j$

$b_k^{j-1}=\frac{1}{2}\sum _{m\in \mathbb{\mathbb{Z}}}{p}_{1-m+2k}$

reconstruction

$a_k^j=\sum _{m\in \mathbb{\mathbb{Z}}}{p}_{k-2m}{a}_m^{j-1}+\sum _{m\in \mathbb{\mathbb{Z}}}{(-1)}^k{p}_{1-k+2m}{b}_m^{j-1}$

are these $p_k$ indexes correct?

filters

low-pass decomposition ${\ell }_k=\frac{1}{2}{p}_{-k}$

high-pass decomposition $h_k=\frac{1}{2}{(-1)}^k{p}_{k+1}$

low-pass reconstruction ${\tilde{\ell }}_k=p_k$

high-pass reconstruction ${\tilde{h}}_k={(-1)}^k{p}_{1-k}$

Properties of $p_k$'s

The entire scheme of an MRA depends on the $p_k$'s, which, in turn, may very well depend on the scaling function definition

The $p_k$'s must satisfy the following properties:

• $\sum _{k\in \mathbb{\mathbb{Z}}}{p}_{k-2\ell }{p}_k=2{\delta }_{\ell ,0}$
• $\sum _{k\in \mathbb{\mathbb{Z}}}{p}_{2k}=\sum _{k\in \mathbb{\mathbb{Z}}}{p}_{2k+1}=1$

If we have $\varphi (x)=\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k\varphi (2x-k)$, then

$\begin{array}{rl}{ℱ}[\varphi (x)]=\hat{\varphi }(\xi )& ={ℱ}[\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k\varphi (2x-k)](\xi )\\ & =\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k{ℱ}[\varphi (2x-k)](\xi )\end{array}$

Let's look at that inner Fourier transform:

$\begin{array}{rl}{ℱ}[\varphi (2x-k)](\xi )& =\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (2x-k){\mathrm{e}}^{-i\xi x}\mathrm{d}x\\ & =\frac{1}{2}\frac{1}{\sqrt{2\pi }}\varphi (u){\mathrm{e}}^{-i\xi \left(\frac{u+k}{2}\right)}\mathrm{d}u\\ & =\frac{1}{2}{\mathrm{e}}^{-\frac{i\xi k}{2}}\cdot \underset{\hat{\varphi }\left(\frac{\xi }{2}\right)}{\underbrace{\frac{1}{\sqrt{2\pi }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (u){\mathrm{e}}^{-i\frac{\xi }{2}u\mathrm{d}u}}}\end{array}$

Plugging this back in to our Fourier transform of $\varphi$ gives:

$\begin{array}{rlr}\hat{\varphi }(\xi )& =\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k{ℱ}[\varphi (2x-k)](\xi )& =\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k\left(\frac{1}{2}{\mathrm{e}}^{-\frac{i\xi k}{2}}\hat{\varphi }\left(\frac{\xi }{2}\right)\right)\end{array}$

Let $P(z)=\frac{1}{2}\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k{z}^k$. Then

$\begin{array}{rl}\hat{\varphi }(\xi )& =P\left({\mathrm{e}}^{i\frac{\xi }{2}}\right)\hat{\varphi }\left(\frac{\xi }{2}\right)\\ \end{array}$

We can expand the $\hat{\varphi }\left(\frac{\xi }{2}\right)$ in the RHS $n$ times to get

$\begin{array}{rl}\hat{\varphi }(\xi )& =\left(\prod _{r=1}^{n}P\left({\mathrm{e}}^{-\frac{ixi}{2^r}}\right)\right)\hat{\varphi }\left(\frac{\xi }{2^n}\right)\end{array}$

If we let $n\to \mathrm{\infty }$, then $\hat{\varphi }\left(\frac{\xi }{2^n}\right)\to \hat{\varphi }(0)=\frac{1}{\sqrt{2\pi }}$. Hence

$\begin{array}{rl}\hat{\varphi }(\xi )& =\frac{1}{\sqrt{2\pi }}\prod _{r=1}^{\mathrm{\infty }}P\left({\mathrm{e}}^{-\frac{i\xi }{2^r}}\right)\end{array}$

For the Haar wavelet, $\hat{\varphi }(\xi )=\frac{1}{\sqrt{2\pi }}\prod _{r=1}^{\mathrm{\infty }}\left(\frac{1+{\mathrm{e}}^{-\frac{i\xi }{2^r}}}{2}\right)$

Now we can construct new scaling functions and wavelets by just working with $\hat{\varphi }$.