Talk:High-pass filter

Irritation sentence

I moved this sentence from the article, because I don't understand what it is trying to say, irritation of whom (the listener?), could the contributor please clarify the meaning and it can be added back. --Lexor|Talk 13:07, 24 Jun 2004 (UTC)

There is often irritation when you actively want to cut low frequencies with a high pass....

High-pass filter image

I have added the High Pass Filter Image, however am not sure how to scale it, anyone feel free to do so. — Preceding unsigned comment added by Dragoljub (talk • contribs) 03:08, 15 November 2004 (UTC)

Alpha code

I have edited the formula for alpha in the code section. —Preceding unsigned comment added by 194.138.39.53 (talk) 09:59, 15 April 2008 (UTC)

Computer algorithm

With the given equations, if the time interval is the same as the RC time constant, alpha calculates as 0.5. But this is incorrect, after one time constant of time the output should reduce to 1/e, not 1/2. Surely the correct equation is e^(-dt/RC). Richard whitehead (talk) 20:48, 13 June 2008 (UTC)

That depends on how you choose to discretize the continuous-time filter. I think your method probably corresponds to the impulse invariance method, but that's not the only sensible choice. Dicklyon (talk) 21:11, 13 June 2008 (UTC)

I didn't realise I was being that clever! I just imagined the input signal varying as a sequence of steps. In this case, the current decays between steps according to exponential decay, and then a new lot of current is added on when the step happens (both ends of the capacitor changing voltages by the same amount because the step is instantaneous, so the new current is delta-V / R). I expect the result is very similar unless the simulation time step is too long (too similar to RC). Richard whitehead (talk) 13:09, 14 June 2008 (UTC)

See the new comments about the relationship between α and RC. As you can see, the decay truly is exponential, but the mapping from "continuous-time frequencies" to "discrete-time frequencies" is not linear, and so the "time constant" isn't going to match especially for α that large. Keep in mind that whenever you go from CT to DT, you're wrapping an infinite line (i.e., from 0 Hz to ∞ Hz) to a finite line (i.e., the semicircular circumference from 0 radians/sample to π radians/sample). There's no way to do that linearly, and so the frequencies only "match" in a small region. That's why we have different CT-to-DT methods; sometimes we need them to match in different ways. For example, we might need to make sure all CT poles land inside the unit circle, or we might need to make sure that the match is very close near the corner frequency... or lots of other things. There's always going to be mismatch, and that's why it's always best to do your design in your target context. If it's going to be realized in discrete-time, it should be designed there too. —TedPavlic (talk) 20:46, 8 January 2009 (UTC)

One more thing. As you can see on the new sections of the page, this method approximates the derivative by taking the slope over one sampling period. Each CT-to-DT method approximates the derivative in a different way. Put another way, each CT-to-DT method deals with the space "in between" samples in different ways. As you can see, there are tradeoffs. Lots of people like approximating the derivative as the slope between time points. A consequence of that is that your impulse response doesn't have the same time constant (however, there is a qualitative relationship between the DT and CT time constants; as one gets bigger, so does the other). —TedPavlic (talk) 21:17, 8 January 2009 (UTC)

High-pass vs. 1st order LTI high-pass

The title of this article is high-pass filter. You can implement a discrete-time high-pass filter without attempting to match the characteristics of any particular continuous time high-pass filter. So the emphasis on matching the 1st-order continuous time version seems misplaced to me.Ccrrccrr (talk) 17:03, 28 May 2009 (UTC)

For example, the statement, "the effect of a high-pass filter can be simulated on a computer," seems misplaced. You can implement a high-pass filter in discrete time, and it actually is a high-pass filter, not an approximation of one. Ccrrccrr (talk) 17:05, 28 May 2009 (UTC)

See recent changes (as a start). —TedPavlic (talk) 23:04, 28 May 2009 (UTC)

Nice! Thanks. The article is still a bit of an odd collection of facts but you fixed the worst of it very nicely.Ccrrccrr (talk) 11:54, 29 May 2009 (UTC)

More information for non-experts

I'd like to see a summary for less technical people. It would be nice to include a sample audio clip showing what something sounds like before and after a high pass filter is applied. Kevink707 (talk) 18:44, 5 February 2010 (UTC)

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Requests for more information

Please add a section covering resonance. Also, please explain the difference between a single-pole filter and a 2-pole filter (is the latter simply two of the former in series?) 174.106.34.169 (talk) 14:11, 26 January 2023 (UTC)