Talk:Low-pass filter

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Please add new talk topics as new sections at the bottom.

1 Old talk not previously in any section
2 2nd order filters all the same?
3 Filter circuit question
4 Slope
5 math tags
6 Why only electronc types?
7 Confusing Paragraph
8 Attenuates versus reduces
9 Realizable for finite, or for infinite, signals?
10 Passive electronic realization
11 Electronic low-pass filters

[edit] Old talk not previously in any section

What about Bandwidth?

Can someone add an example and disscusiuon on the transfer function of a low pass filter? mickpc

I have added the Low Pass Filter Image, however am not sure how to scale it, anyone feel free to do so.

this page should have some dicussion of LPFs' uses in electronic music. - mhjb

What do you want to know? 213.64.153.109 22:04 Dec 10, 2002 (UTC)

I want to know if it's the high-pass or the low-pass which filters tape hiss.

Tape hiss is very highfrequent hence you need too block out these high frequensies. Therefor you need a lowpass filter. /same guy as before

The following content was at Lowpass filter by 195.145.245.249:

A filter used, amongst others, in sound synthesis, that only lets pass waves below its cutoff frequency. With analog realizations of lowpass filters, the cutoff of higher frequencies is gradual, with frequencies being dampened increasingly the higher they get. Typical values for this slew rate are 12 dB or 24 dB per octave, meaning that a signal one octave above the cutoff frequency will be dampened by 12/24 dB.

Dori | Talk 17:06, Dec 2, 2003 (UTC)

slew rate is wrong and everything else is already included. - Omegatron 14:31, Dec 31, 2004 (UTC)

"The Bode plot for this type of filter resembles that of a first-order filter, except that it falls off quadratically instead of linearly."

That depends on the type of filter. A Butterworth of any order will look like a straight line, too. - Omegatron

Moved from article, awaiting clarification from contributor (see also, Talk:High-pass filter). --Lexor|Talk 13:10, 24 Jun 2004 (UTC)

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

Talk:Electronic_filter#Ideal_filters

removed from article:

Capacitors naturally resist changes in voltage.

It is this natural resistance (not to be confused with Ohmic resistance) that the functionality of the low-pass filter is realized.

With low-frequencies, the voltage to the capacitor changes slowly and provides sufficient time for the capacitor to change voltage through the current-voltage relationship $I = C\frac{dV}{dt}$ .
For high-frequencies, the voltage to the capacitor changes too fast for sufficient charge to build up in the capacitor to change the voltage.

This understanding is rooted in the concept of reactance where the capacitor will naturally block DC but pass AC.

Taking a more fluidic vision of this passive circuit, then if the capacitor blocks DC then it must "flow out" the path marked

V o u t

(analogous to removing the capacitor).

If the capacitor passes AC then it "flows out" the path where the capacitor effectively short circuiting

V o u t

with ground (analogous to replacing the capacitor with just a wire).

[edit] 2nd order filters all the same?

Moved to Wikipedia talk:WikiProject Electronics#1st-_and_2nd-order_filters

[edit] Filter circuit question

FTA: "At higher frequencies the reactance drops, and the capacitor effectively functions as a short circuit." -- shouldn't this be an "open" circuit? A short would tie Vout to ground, thus eliminating all frequencies. (129.42.208.182)

No, it's a short circuit; at high frequencies, the output is tied to ground. So it doesn't pass high frequencies. That's why it's a low-pass filter, right? Pfalstad 22:36, 24 October 2005 (UTC)

[edit] Slope

Why does the text show a cutoff of 6 dB per octave while the bode plot shows 20 dB per octave? Shouldn't it be 20 dB per octave in the text also? If the frequency doubles, there is a change of 6 dB, but an octave is a factor 10 change in frequency.

The bode plot shows 20 dB per decade, not octave. An octave is a doubling in frequency; a decade is 10x frequency change. It would be nice if the text said something about 20 dB per decade, since the diagram emphasizes that. Pfalstad 12:40, 25 January 2006 (UTC)

[edit] math tags

I've edited the "Passive digital realization" section to use math tags. It would be swell if someone could check my work, then delete the text-only version if I've got it right.

I know the intention of the section is to show that the new output is determined by the previous output, but this isn't made clear by the representation, even in the text-only version. Right now, $y$ appears on both sides of the equation. What's the best way to fix this? -- Mikeblas 15:36, 28 July 2006 (UTC)

Fixed... Anyway i'm not so sure about the correctness of the first (alpha) formula. Two sources report that the correct value is $\alpha = e^{-2 \pi f_c}$ where $f_c=\frac{f_{-3\,\mathrm{dB}}}{f_{sample}}$ ... I'll investigate more. Anyway, why the first formula is definitely bigger? (solved, was just the second, simpler, one wasn't rendered in png 10:11, 16 September 2006 (UTC)) Danilo Roascio 17:39, 14 September 2006 (UTC)

[edit] Why only electronc types?

There are mehanical and acoustic (optical?) low pass filters. Should this page mention them? I think so.--Tugjob 15:25, 22 June 2007 (UTC)

Yes, maybe a section on "other domains" or something. Dicklyon 16:56, 22 June 2007 (UTC)

I think a very general description of low pass filtering (in all its forms) and then the different types would be appropriate. Any volunteers?--Tugjob 23:26, 12 July 2007 (UTC)

[edit] Confusing Paragraph

If someone could enumerate a little more on this paragraph:

However, this filter is not realizable for practical, real signals because the sinc function extends to infinity. The filter would therefore need to predict the future and have infinite knowledge of the past in order to perform the convolution. It is effectively realizable for pre-recorded digital signals (by padding the ends of the signal with zeros to the point that the error after filtering is less than the quantization error), or perfectly cyclic signals that repeat for infinity.

It is interesting, but confusing to someone with only slight background of the area.

See if my edits helped. Dicklyon 19:08, 26 June 2007 (UTC)

[edit] Attenuates versus reduces

See definitions of attenuate. They usually involve "the level of" (or the strength, power, magnitude, etc.). The "reduces" currently there in parens is not a great synonym, because it leads to edits such as I just reverted with "reduces the frequencies". The trouble with this is that frequencies are not what is changed. What is reduced is the level of the frequency components. An edit that reflects this understanding would be welcome. Dicklyon 23:05, 12 July 2007 (UTC)

[edit] Realizable for finite, or for infinite, signals?

BrianWilloughby, I concur that "this second paragraph seems to have a history of misinterpretation," but I'm not sure I agree with you on what the right statement should be. Seems to me that when the signal is of finite duration, then the filtering with the infinite impulse response is possible, or realizable; whereas if the input signal is of infinite duration, then it would require both an infinite amount of computation and an infinite delay, and the ideal filter is therefore NOT realiable for such signals. What's your interpretation? Dicklyon 22:46, 12 August 2007 (UTC)

I'm glad you questioned this because the edit caught my eye too. It occurs to me that Brian must be speaking to the infinite duration of the sinc signal required for the convolution while the orginal version of the paragraph was speaking to the finite duration of the signal to be filtered. Alfred Centauri 23:06, 12 August 2007 (UTC)

As far as my understanding goes an ideal filter is *perfectly* realizable in the frequency domain on a digital signal of finite length but not *desirable* due to the gibbs effect.

Since I am involved primarily with vision systems I will be talking about length and the spatial-domain instead of duration and the time-domain.
Take for example a snapshot of a square wave across a horizontal scanline in the spatial-domain (for simplicity assume only 1-Dimension). In this scenario the pixel intensity is a function of the pixel coordinate and since we are talking about a square wave this would be a dark segment followed by a bright segment followed by a dark segment. There are two sharp transitions (edges), dark to bright and bright to dark. Say we want to smooth these transitions (soften the edges) with the help of an ideal low-pass filter constructed in the frequnecy domain. Its frequnecy response would look like a step function and its phase response would be zero at all all frequencies. Assume that the length of our scanline is 1024 pixels then we would need exactly 513 frequencies (due the redundancies in the output of a DFT on real input) in order to fully reconstruct the scanline in the spatial domain with absolutely no no loss of information.

If we take the FFT of our scanline and apply the ideal low-pass filter by multiplication in the frequency domain and then do the inverse FFT to go back to the spatial domain the resulting scanine will have overshoots and ringing arround the edges of the square wave. This happens because we have thrown away alot of the high frequencies that would be otherwise needed for its full reconstruction. As said earlier we need exaclty 513 unique frequencies to fully reconstruct this wave but we are now effectively using much less. Of course, our purpose is not reconstruction but smoothing. The ideal low-pass filter does exactly what it is supposed to do and it is 100% accurate. The problem is that the ideal low-pass is not suitable for this application. We instead need to keep some higher frequencies to reduce overshoot and ringing at the edges. In this case a gaussian low-pass would do a much better job.

Please refer to the "Gibbs phenomenon" article for more information. Also forgive me and correct me if I am wrong or if the above explanation is not applicable to other fields.erm 13:50, 5 October 2007 (UTC)

[edit] Passive electronic realization

In the section "Passive electronic realization" it states that DC can not flow through the capacitor and AC can. It is to my understanding , excluding leakage, that no current flows through the capacitor. The charge is stored on one of the plates depending on the direction of the current. When the current changes direction the charge is released on the respective plate and built up on the opposite plate.

Any Comments?

Rrace001 15:05, 25 October 2007 (UTC)

That's technically correct. Current does not actually flow from one plate of the capacitor, across the dielectric, and onto the other plate. But for all practical purposes a capacitor can be analyzed as if the current did flow in this way. It's long-standing practice to state that AC current flows through a capacitor... though it's definitely confusing to those who are just learning about electronics. Mathematically the reactance of a capacitor in an AC circuit is 1/(2*pi*f*C), so at very high frequencies it's essentially a short circuit. ǝɹʎℲxoɯ (contrib) 15:15, 25 October 2007 (UTC)

I'd say it's not technically correct. As with other two-terminal devices, the current flowing through a device is the current flowing into one terminal and out the other. What happens inside is just a distraction for you. And DC current can flow throught a capacitor; but it takes a linear voltage ramp to make that happen, and it may be hard to sustain that for a long time; there certainly are applications, such as integrators in photodetector cells, where the voltage that represents the integral of a DC current is what you care about. Dicklyon 22:14, 25 October 2007 (UTC)

I think we basically agree. In standard electronics usage, the current through a device is defined as you express it. However, this definition is somewhat contradictory to the everyday meaning of "through" which means "in one side, across the middle by some path, out the other side." ǝɹʎℲxoɯ (contrib) 00:46, 26 October 2007 (UTC)

Are you concerned that the electrons that come out are not the same ones that went in? Are you not believing that electrons are indistinguishable, and hence that question has no meaning? Dicklyon 01:30, 26 October 2007 (UTC)

Well... I work in a nanoelectronics lab where I try to understand novel semiconductors, so I'm well-aware that electrons are indistinguishable :-) I'm just saying that the conventional, everyday notion of flow "through" something doesn't intuitively match up with what goes on in a capacitor. I remember having trouble with that notion when playing with my first electronics kit when I was 10 or 11 and all the way till I learned about circuit design in college. This article isn't necessarily the best place to clarify that, but maybe the capacitor article could? ǝɹʎℲxoɯ (contrib) 02:38, 26 October 2007 (UTC)

[edit] Electronic low-pass filters

I have removed my question Karlwalton (talk) 15:23, 30 March 2008 (UTC)