How Is the Bandwidth of an Image Reduced by an Octave in Gaussian Pyramid Procedure?

Question

Reference paper 1. For arguments sake, let's say that the maximum and minimum frequency in an image are 256Hz and 1024Hz. When I perform filtering using the gaussian function, the "bandlimit" reduces by an octave (1/2). Does this mean the new freq. range will be from 128Hz to 512Hz? Or does it mean that it will be 256Hz to 512Hz, but the bandpass signal (While calculating the Laplacian pyramid) will now have a freq. range 512Hz to 1024Hz (which is 1 octave)?

2. Also, apart from the low pass filtering, since downsampling is also performed, shouldnt the image bandwidth reduce by more than an octave? I'm guessing that the LPF operation is what causes the reduction in bandwidth

Answer 1

The model is simple.
Assume we have data I0 on the frequencies range of $\left[ 0, 1000 \right] $ [Hz].

We have 2 Black Boxes:

• BlackBox 1 - LPF which always has a cut off at the middle of the spectrum.
• BlackBox 2 - Decimation by 2.

What happens if we apply BlackBox 1 on the data?
We'll get an LPF version of it we'll call it L0 with data on the range $ \left[0, 500 \right] $ [Hz]..
We can also have H0 = I0 - L0 which is vasically all data on the range $ \left[500, 1000 \right] $ [Hz].

Now, be decimating I1 = decimate(L0) we have new data on the range $ \left[ 0, 500 \right] $ [Hz].
Applying the same tricks as above (Using the same black boxes which is what's nice about it) can yields the following series:

• L# - Always contains the lower half frequency of the # Level.
• H# - Always contains the higher half frequency of the # Level.

If you see the amount of data is reduced by factor of 2 each level (Octave).
Another nice property that you can reproduce the data using the lowest level LPF and the HPF of all other levels.