# Math Help - Orthonormal Conjugate Quadrature Filter, Fourier Transform

1. ## Orthonormal Conjugate Quadrature Filter, Fourier Transform

This question is strange enough so I'm beginning to suspect a typo/misinterpretation.

Let $w$ be defined by
$\hat{w}=\{ \begin{array}{c}
1 \text{ if } \xi \in (0.5,1] \\
-1 \text{ if } \xi \in [-1,-0.5) \\
0 \text{ otherwise }\end{array} \}$

Show that $w$ is admissible and compute its normalization constant.

First of all it is very strange to think of the fourier transform to be a function on the reals as opposed to a sequence, right? (I suspect this might be a typo)

Anyway, I interpret admissible to mean: satisfies the Orthonormal Conjugate Quadrature Filter Conditions.

So first I actually find $w$ by taking the inverse fourier transform:

$w=\check{\hat{w}}=\sum_{\xi=-\infty}^{\infty}\hat{w(\xi)}e^{2\pi i \xi x}$
$=e^{2\pi i x}-e^{-2\pi i x}=-2isin(2\pi x)$

and this is zero at all integer values, which makes it a very strange trivial case (which is not orthonormal).

Any thoughts?
Thanks.

2. Originally Posted by verdi
First of all it is very strange to think of the fourier transform to be a function on the reals as opposed to a sequence, right? (I suspect this might be a typo)
You may be confusing Fourier transform and Fourier series. The Fourier transform of a function is a function on the reals.

3. Good point.

That gives me

Edit: I spotted a miscalculation...I'll get back to this

$w(\xi)=\check{\hat{w(\xi)}}=\int_{-\infty}^{\infty}\hat{w(x)}e^{2\pi i \xi x} dx=\int_{\frac{1}{2}}^{1}e^{2\pi i \xi x} dx-\int_{-1}^{-\frac{1}{2}}e^{2\pi i \xi x} dx$

$=(\frac{1}{2\pi i \xi})((e^{2\pi\xi}-e^{\pi\xi})-(e^{-\pi\xi}-e^{-2\pi\xi}))$

$=(\frac{1}{\pi \xi})(sin(2\pi \xi)+ i cos(\pi \xi))$

which takes on $\frac{-i}{\pi \xi}$ at odd $\xi$, $\frac{i}{\pi \xi}$ at even $\xi$

and at 0, we get $2+\infty$.

Is that really a good result. It's technically non-zero at infinitely many integers and I don't like what's going on at 0.

Does this all make sense?

4. $
=(\frac{1}{2\pi i \xi})((e^{2\pi\xi}-e^{\pi\xi})-(e^{-\pi\xi}-e^{-2\pi\xi}))
$

$
=(\frac{1}{\pi i \xi})(cos(2\pi \xi)- cos(\pi \xi))
$

$
\xi \text{ even } \rightarrow 0$

$\xi \text{ odd } \rightarrow \frac{2}{\pi i \xi}$
$\xi \text{ 0 (special case of even)}\rightarrow 0$

I don't believe that this counts as finite support. Correct me if I am wrong.

5. Originally Posted by verdi
$
=(\frac{1}{2\pi i \xi})((e^{2\pi\xi}-e^{\pi\xi})-(e^{-\pi\xi}-e^{-2\pi\xi}))
$

$
=(\frac{1}{\pi i \xi})(cos(2\pi \xi)- cos(\pi \xi))
$

$
\xi \text{ even } \rightarrow 0$

$\xi \text{ odd } \rightarrow \frac{2}{\pi i \xi}$
$\xi \text{ 0 (special case of even)}\rightarrow 0$

I don't believe that this counts as finite support. Correct me if I am wrong.
Remember $\xi$ is real, not just integer. The support certainly isn't finite. I don't know however what "admissible" and "normalizing constant" mean in this context.

6. Well my impression based on the context is that we're trying to construct something analogous to the Haar filter:
$h(k)=\frac{1}{\sqrt{2}} \text{ if k=0 or k=1}$
=0 if $k\neq{0}$ and $k\neq{1}$

Which we at least only care about on integer values, even if it could be extended to a larger domain.

7. I think I might have cleared some things up for myself.
Continuous wavelet transform - Wikipedia, the free encyclopedia