I'm a bit befuddled by noise's effect on derivative filters.

I've always 'known' that straightforward first difference derivative filters of discrete signals amplifies noise, but I'm struggling to prove that fact.

I thought it was caused by a greater gain for higher frequencies, but when I take the z-transform for a first difference filter I get $$ H(Z) = 1 - Z^{-1} $$

and if I restrict the magnitude of $Z$ to 1 to get the DTFT and take $|H(Z)|$ the result is $$ 2 |\sin(\frac{\omega}{2})|$$

which isn't increasing or constant in frequency like a high pass filter.

Could you explain why the noise increases if not for this effect?


Any filter whose magnitude is above unity at a given frequency $\omega$ amplifies that freuency.

Typical white noise is a wideband discrete-time signal whose spectrum ranges from $-\pi$ to $\pi$.

The digital first difference filter is an LTI system with the impulse response $h[n] = \delta[n] - \delta[n-1]$ and frequency response $H(\omega) = 1 - e^{-j\omega}$, whose magnitude is:

$$ |H(\omega)| = \sqrt{ 2 - 2 \cos(\omega) } = \sqrt{ 4 \sin^2(\omega/2) } = 2|\sin(\omega/2)|$$

as you can see as $\omega \to \pi$, then the magnitude goes to $2$ which is larger than unity and therefore amplifies any noise content found at those frequencies...

Note that in any digital system, there will always be quantization noise, and therefore it will always be amplified using the given first difference filter.


Let $y[n] = x[n] + v[n]$ be the signal you want to filter, with $x[n]$ the useful part of the signal and $v[n]$ some additive white Gaussian noise of variance $\sigma^2$ in the bandwidth of interest.

After filtering with your first order difference filter $h[n] = \delta[n] - \delta[n-1]$, you get

$$z[n] = (h \ast y)[n] = y[n] - y[n-1] = \underbrace{x[n] - x[n-1]}_{\text{useful part}} + \underbrace{v[n] - v[n-1]}_{\text{noise } v'[n]}.$$ The variance of the noise $v'[n]$ you obtain after filtering has thus doubled and is given by $2\sigma^2$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.