When analyzing an image signal as a series of Dirac impulses in the continuous spatial domain, I applied continuous low pass filters (to reconstruct or smooth it) with area equal to 1 to preserve the low frequency amplitudes in the signal spectrum. When using a kernel (which, correct me if I'm wrong, is just the "sampled" discrete version of the original continuous filter in the spatial domain) I was wondering what is the meaning of the sum of the kernel components, and when I should care?

I often see examples with kernels with components which sum to one, stating they preserves the original image brightness. But if the original continuous signal integrates to one, it does not mean the discrete version will sum to one, so this left me a bit confused.

  • $\begingroup$ Hi! So is this an analog image processing problem? Or are you after a digital image processing eventually. Or are you not sure of ? Also please provide an example of such kernels... $\endgroup$
    – Fat32
    Commented Oct 8, 2018 at 9:25
  • $\begingroup$ @Fat32 Hello! I am dealing with digital images. One example of kernel which sums to 1 is the box blur ( 3x3 kernel with all components set to 1/9) $\endgroup$
    – yggdrasil
    Commented Oct 8, 2018 at 14:57

1 Answer 1


Considering an LTI filter impulse response (aka filter kernel) $h[n_1,n_2]$, then its frequency response will be

$$ H(\omega_1,\omega_2) = \sum_{n_1} \sum_{n_2} h[n_1,n_2] e^{-j \omega_1 n_1}e^{-j \omega_2 n_2} $$

When an image is filtered, its average brightness is mostly affected by the DC response of $h[n_1,n_2]$, which is given by $H(0,0)$. When you want the brightness not to change after filtering, then you set $H(0,0) = 1$, which means that

$$ H(0,0) = \sum_{n_1} \sum_{n_2} h[n_1,n_2] e^{-j 0 n_1}e^{-j 0n_2} = \sum_{n_1} \sum_{n_2} h[n_1,n_2] = 1 $$

Which shows that the sum of impulse response (aka filter kernel) samples is one.


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