Can anyone explain how a shelving filter is able to maintain a flat frequency 
response both above and below its center frequency?

I have a sense of first-order low-pass filters that have analogs in the 
physical world - it somehow makes sense that the frequency response continues 
to drop off as the frequency increases above the cutoff. However, while a high 
shelving filter has a similar drop in frequency response just above its cutoff, 
eventually the response levels off and remains flat as frequency increases. For 
example, a high shelf set for -10 dB would have a 0 dB response for frequencies 
significantly below the cutoff, a transition around the cutoff with first order 
response, and then once the response reaches -10 dB for high frequencies, the 
response remains at -10 dB as frequency increases.

I'm trying to understand how this is possible. The Wikipedia article for audio 
EQ has a section that seems to hint at how this is achieved.

https://urldefense.proofpoint.com/v2/url?u=https-3A__en.wikipedia.org_wiki_Equalization-5F-28audio-29-23Shelving-5Ffilter&d=DwIFAg&c=009klHSCxuh5AI1vNQzSO0KGjl4nbi2Q0M1QLJX9BeE&r=TRvFbpof3kTa2q5hdjI2hccynPix7hNL2n0I6DmlDy0&m=upGIdDKIRsFXJZNFq2HDaJZ3o2g4X_MUDEt6vyMkQUWMkMzq2MaNL8TGk2mG7h3U&s=GWUlmLkjr9OLMrPKS-R0jAEZJ5TrrXSRi1VVGxm20Dc&e=
 

This section mentions that there is both a pole and a zero in a shelving 
filter. Is that how the response becomes flat in that region? Is it because the 
first-order slopes of both the pole and zero add together and end up being flat 
as frequency increases, albeit at a loss of 10 dB in the example above?

By the way, I tried to search the internet for an answer, but 99% of the hits 
are articles about how to use a shelving filter, or what it does, but none on 
how it does it. The remaining 1% of the articles are about the detailed 
mathematical transfer functions for shelving filters, without any simple 
overview of how they work, or what they're doing mathematically at a high level 
rather than formula level.

Thanks for any insight,

Brian Willoughby

Reply via email to