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:

${\begin{aligned}\left(f*h\right)(t)&=\int _{-\infty }^{\infty }h(\tau )\,f(t-\tau )\,\mathrm {d} \tau \\&={\frac {1}{5}}\,\int _{0}^{5}f(t-\tau )\,\mathrm {d} \tau \\&={\frac {1}{5}}\,\int _{t-5}^{t}f(\sigma )\,\mathrm {d} \sigma \end{aligned}}$

Where $\sigma =t-\tau$

3 Cases:

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

Problem 4

$f_{2}$ defined by $a^{2}=\left(3,1,-2,0,-3,9,-3\right)$ ($a_{k}^{2}=0$ for $k<0$ or $k>7$

Haar Wavelet Decomposition into $b^{1}$, $a^{1}$, and $a^{0}$ for $f_{2}=f_{0}+w_{0}+w_{1}$

${\begin{aligned}a_{k}^{j-1}&={\frac {1}{2}}\,\left(a_{2k}^{j}+a_{2k+1}^{j}\right)\\b_{k}^{j-1}&={\frac {1}{2}}\,\left(a_{2k}^{j}-a_{2k+1}^{j}\right)\end{aligned}}$

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

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

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:

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

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

$\phi (x)=\sum _{k\in \mathbb {Z} }p_{k}\,\phi (2x-k)$

$p_{k}=2\,\int _{-\infty }^{\infty }\phi (x)\,\phi (2x-k)\,\mathrm {d} x$

wavelet

$\psi (x)=\sum _{k\in \mathbb {Z} }\left(-1\right)^{k}\,p_{1-k}\,\phi (2x-k)$

decomposition

$a_{k}^{j-1}={\frac {1}{2}}\,\sum _{m\in \mathbb {Z} }p_{m-2k}\,a_{m}^{j}$

$b_{k}^{j-1}={\frac {1}{2}}\,\sum _{m\in \mathbb {Z} }p_{1-m+2k}$

reconstruction

$a_{k}^{j}=\sum _{m\in \mathbb {Z} }p_{k-2m}\,a_{m}^{j-1}+\sum _{m\in \mathbb {Z} }\left(-1\right)^{k}p_{1-k+2m}\,b_{m}^{j-1}$

are these $p_{k}$ indexes correct?

filters

low-pass decomposition $\ell _{k}={\frac {1}{2}}\,p_{-k}$

high-pass decomposition $h_{k}={\frac {1}{2}}\,\left(-1\right)^{k}\,p_{k+1}$

low-pass reconstruction ${\tilde {\ell }}_{k}=p_{k}$

high-pass reconstruction ${\tilde {h}}_{k}=\left(-1\right)^{k}\,p_{1-k}$

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:

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

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

${\begin{aligned}{\mathcal {F}}\left[\phi (x)\right]={\hat {\phi }}(\xi )&={\mathcal {F}}\left[\sum _{k\in \mathbb {Z} }p_{k}\,\phi (2x-k)\right](\xi )\\&=\sum _{k\in \mathbb {Z} }p_{k}\,{\mathcal {F}}\left[\phi (2x-k)\right](\xi )\end{aligned}}$

Let's look at that inner Fourier transform:

${\begin{aligned}{\mathcal {F}}\left[\phi (2x-k)\right](\xi )&={\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }\phi (2x-k)\,\mathrm {e} ^{-i\,\xi \,x}\,\mathrm {d} x\\&={\frac {1}{2}}\,{\frac {1}{\sqrt {2\pi }}}\phi (u)\,\mathrm {e} ^{-i\,\xi \,\left({\frac {u+k}{2}}\right)}\,\mathrm {d} u\\&={\frac {1}{2}}\,\mathrm {e} ^{-{\frac {i\,\xi \,k}{2}}}\cdot \underbrace {{\frac {1}{\sqrt {2\pi }}}\,\int _{-\infty }^{\infty }\phi (u)\,\mathrm {e} ^{-i\,{\frac {\xi }{2}}\,u\,\,\mathrm {d} u}} _{{\hat {\phi }}\left({\frac {\xi }{2}}\right)}\end{aligned}}$

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

${\begin{aligned}{\hat {\phi }}(\xi )&=\sum _{k\in \mathbb {Z} }p_{k}\,{\mathcal {F}}\left[\phi (2x-k)\right](\xi )&=\sum _{k\in \mathbb {Z} }p_{k}\,\left({\frac {1}{2}}\,\mathrm {e} ^{-{\frac {i\,\xi \,k}{2}}}\,{\hat {\phi }}\left({\frac {\xi }{2}}\right)\right)\end{aligned}}$

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

${\begin{aligned}{\hat {\phi }}(\xi )&=P\left(\mathrm {e} ^{i\,{\frac {\xi }{2}}}\right)\,{\hat {\phi }}\left({\frac {\xi }{2}}\right)\\\end{aligned}}$

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

${\begin{aligned}{\hat {\phi }}(\xi )&=\left(\prod _{r=1}^{n}P\left(\mathrm {e} ^{-{\frac {i\,xi}{2^{r}}}}\right)\right)\,{\hat {\phi }}\left({\frac {\xi }{2^{n}}}\right)\end{aligned}}$

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

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

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

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