In image processing we may use derivatives to help us detect the edges. While at mean time, this operation would also make the data noisier. But why do we have this effect?

My intuition is that if we denote the image data as

$I_{actual}(x, y) = I_{ideal}(x, y) + I_{noise}(x, y)$.

Then apparently when we do $I_{actual}(x_2, y) - I_{actual}(x_1, y)$, the noises will cancel out with each other? But how come they end up adding up to each other?

  • $\begingroup$ First of all why do assume the noises will cancel each other out? The only case where that would be true would be with very low-frequency noise such as "DC" offset noise. How about you do a frequency response of Iactual(x2,y)−Iactual(x1,y) ? Basically it is an FIR with 2 taps -1 and + 1 $\endgroup$
    – Ben
    Commented Mar 5, 2018 at 19:48

2 Answers 2


Differentiation in one axis will amplify high-frequency components. Assuming noise at Fs/2 (i.e maximum frequency), the gain at this frequency will be 6 dB. Intuitively, if you have a sequence alternating between +1 and -1 i.e (1, -1, 1, - 1), Then the difference will be +2, - 2, +2, -2, a gain of 6 dB

Another reason is that by differentiating, you remove the low-frequency components, so your eyes see mostly the high-frequency components giving you the impression of having more noise.


Let's have a look at:

$$\Delta I (x_1,x_2) = I_{actual}(x_2, y) - I_{actual}(x_1, y) = I_{ideal}(x_1, y) + I_{noise}(x_1, y) - I_{ideal}(x_2, y) - I_{noise}(x_2, y)$$

If you assume that for close $x_1$ and $x_2$ the $I_{ideal}$ doesn't change much so that $$ I_{actual}(x_2, y) \approx I_{ideal}(x_2, y)$$ then $$\Delta I (x_1,x_2) = I_{noise}(x_1, y) - I_{noise}(x_2, y)$$

Now we have to make some assumptions about the noise. One standard assumption is that noise is independent and identically distributed across the image. Independence means that the noise is uncorrelated.

If $$ I_{noise}(x, y) \sim N(0, \sigma^2)$$ then $$ \Delta I (x_1,x_2) \sim N(0, 2\sigma^2)$$

This means that the variance of the difference noise will be twice the variance of the original noise (similar to what Ben suggests in his answer).

  • 1
    $\begingroup$ I prefer this approach to Ben's. Ben doesn't separate the noise and the signal. Instead of saying the ideal portion doesn't change much, you could just break the delta into ideal and noise components, where the ideal portion is the derivative you are trying to calculate. $\endgroup$
    – David
    Commented Mar 6, 2018 at 18:41

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