On 9/14/2010 8:40 AM, Invisible wrote:

> No, I'm fairly sure JPEG uses the Discrete Cosine Transform. Now
> JPEG2000 really *does* use some manner of wavelet transform (although I
> don't remember which one).

Ah.. You're right, it is JPEG2000.

>> And is rather clever, when you think about it.
>>
>> Take a signal, repeatedly break it up into pairs of low and high
>> frequencies, subsample those segments, and place them back into the
>> buffer.
>
> I'm not at all clear on exactly how it does that.
>

Using a pair of filters, low pass and high pass in a bank,

so you filter low frequencies, this becomes the first stage, which is 
downsampled by 2.

Let see, this is a nice diagram:

http://en.wikipedia.org/wiki/Discrete_wavelet_transform#Cascading_and_Filter_banks

Apparently, once the filters are run over it, you can downsample each of 
the filtered components. The low pass can then be filtered again How 
reducing the sample rate of the high pass data doesn't discard data I'm 
not sure, yet.

>
> Most transform coding methods work not by *discarding* points, but by
> *quantinising* them according to how "important" they are deemed to be.
> This one seems a little unusual in that respect.

Right. *But* you can drop every other sample because of the Nyquist limit!

>
>> Another use of wavelets: Fourier analysis of signals.
>
> It's not Fourier analysis if it's not a Fourier transform. ;-)
>

ERRRR... Frequency domain analysis. In the end you're doing Fourier 
transforms on each portion you've separated out in the decomposition.

> Apparently there's a limit to how much precision you can get in time and
> frequency. Increasing the resolution of one necessarily decreases the
> resolution of the other. This is apparently due to the Heisenberg
> uncertainty principle. (Which is interesting, since I thought that
> applies only to quantum mechanics, not to general mathematical
> phenomena...)

Interesting. It makes sense, though. Low frequencies change relatively 
little in the time domain. You can tell when it occurs, when exactly a 
peak is, but now how it changes if delta-t is less than the frequency of 
the wave.

-- 
~Mike

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