Hi Brian,
In Laplace terms:
Hshelf(s)=(1-k)*2*pi*fc/(s+2*pi*fc)+k=(k*s+2*pi*fc)/(k+2*pi*fc)
so you have one zero and one pole and
Hshelf(0)=1
Hshelf(inf)=k
which is definitely a 1st-order shelving filter, hence what I suggested
was correct.
Whether this is intuitive or not depends on your intuition. I just feel
like suggesting to base your reasoning on math, as analogies and
intuitions have limited scope and usefulness in engineering and science.
Best,
Stefano
Il 14/12/23 00:34, brianw ha scritto:
Hi Stefano,
What you describe seems neither intuitive nor correct.
Mixing a scaled amount of the input, which is presumably flat in response, with
a lowpass in parallel would still produce a falling amplitude with rising
frequency. Intuitively, I am assuming no phase shift that would cause
constructive or destructive interference between the parallel audio paths.
Adding two copies of a signal, each of different amplitudes at different
frequencies, still results in an output amplitude that is relative to both
inputs. i.e. The output would not be flat above the cutoff unless both signals
are flat in the frequency range above the cutoff.
Even if there would be a convenient phase shift that would somehow maintain a
flat response above the high shelf cutoff, it's certainly not intuitive that
the specific phase shift required would be automatic. When phase shift is
introduced by accident, it's often vastly different than expected.
I will seek out the texts referenced by the other responses to this thread.
Brian
On Nov 26, 2023, at 1:31 AM, Stefano D'Angelo wrote:
Hello,
Intuitively (I hope) you could simply think of a low shelving filter as a
parallel of a lowpass + a scaled amount of the input, and conversely of a high
shelving filter as a parallel of an highpass + a scaled amount of the input.
E.g., in case of a low shelving whose output = (1 - k) * lowpass output + k * input,
with 0 < k < 1, at DC you have gain = 1 and at freq=infty you have gain = k.
(Actually, you have to scale the lowpass/highpass too if you want gain=1 at DC or
infinity).
Cheers,
Stefano
Il 26/11/23 07:32, brianw ha scritto:
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
