In the paper The Laplacian Pyramid as a Compact Image Code,

it says The pyramid algorithm reduces the filter band limit by an octave from level to level. How is the conclusion drawn? Is this related to weighting function and downsampling width and height by 2? How does weighting function and downsampling operation affect the filter band limit?

Edit: From one document, it shows the frequency response of binomial filters enter image description here


1 Answer 1


It's due to the weighting function used in the Gaussian pyramid downsampling.

Lets take a 1D patch(1D patch because it makes our calculation easy and also the kernel is separable) with p10,p11,p12,p13,p14,p15 in the Level 1.

Assume maximum gradient possible is Max - Min between any two pixels.

The pixel p20 in level 2 is formed by influence of pixels p10,p11,p12,p13,p14 in level1 and pixel p21 if formed by influence of pixels p12,p13,p14,p15,p16. The corresponding image(I'm not not able to upload image).

The weights of the kernel are

w(0) = a,

w(-1) = w(1) = 1/4

w(-2) = w(2) = 1/4 -a/2.

For Gaussian pyramid a = 0.4(approximately)

w(0) = 0.4

w(-1) = w(1) = 0.25

w(-2) = w(2) = 0.05

The equation for new pixel p20

p21 = w(-2)*p10 +w(-1)*p11+w(0)*p12 +w(1)*p13+w(2)*p14

Substituting the values for the weights we get,

p20 = 0.05*p10 + 0.25*p11 + 0.4*p12 + 0.25*p13 + 0.05*p14

Similarly value of pixel p21

p21 = 0.05*p12 + 0.25*p13 +0.4*p14 +0.25*p15 + 0.05*p16

Now the difference between pixels p20 and p21 is,

p20 - p21 = 0.05*p10 + 0.25*p11 +0.35*p12 -0.35*p14 -0.25*p15+0.05*p16

So for the difference to be maximum p10 = p11 =p12 = Max and p14=p15=p16 = Min

p20 -p21 = 0.7(Max-Min)

p20- p21 = (Max -Min)/√2

Since the filters are separable, applying the kernel in another direction over the image again will reduce the pixel difference by a ratio of √2. The maximum gradient in x and y direction in next level of Gaussian pyramid is (Max - Min)/2

The maximum gradient possible is reduced by half or in frequency domain reduces the frequency by half(an octave).

  • $\begingroup$ Thanks for your reply. In frequency domain reduces the frequency by half(an octave). Is the reason that the width and height are reduced by half? I do one experiment. I resize one image by half in width and height, then perform FFT for original image and resized image and compare their magnitude spectrum. They looks like similar. Why is it called the frequency is reduced by half? Because FFT result of resized image is reduced by half comparing with the FFT result of original image? If so, it is not related to weighting function? $\endgroup$ Nov 11, 2016 at 9:07
  • $\begingroup$ Simple resizing won't affect the frequency. Instead of resizing, apply a mean filter over the image and compute FFT. Compare it with the FFT of original image. You will see that high frequency components will be missing.This is the same in case of Gaussian Pyramid.The frequency reduction is because of the smoothing. In Gaussian Pyramid, the weights of kernel are responsible for a frequency reduction factor of 1/2. $\endgroup$ Nov 11, 2016 at 10:36
  • $\begingroup$ Thanks for valuable clarification. Does frequency reduction mean the bandwidth of original image is reduced by 2? If I choose 3 tap smoothing filter, the frequency reduction factor is still 1/2? I don't understand image size won't affect the spectrum. If image resizing occurs, the frequency of small image will be affected due to aliasing. $\endgroup$ Nov 11, 2016 at 12:46
  • $\begingroup$ Yes. Bandwidth will be reduced by 2. Aliasing artifact occurs only if the sampling frequency is greater than the spatial frequency so there is no guarantee every image resize will get affected by aliasing. If you try to make a 3 tap filter you will end up with kernel similar to a = 0.5 in 5 tap filter I guess. So the frequency reduction factor will be less than 2. $\endgroup$ Nov 11, 2016 at 14:36
  • $\begingroup$ From one document, it gives frequency response of several binomial filters. Is it possible to draw the conclusion that the frequency reduction factor is 1/2 from frequency response? $\endgroup$ Nov 14, 2016 at 9:11

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