Talk:Low-pass filter

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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).

1 2nd order filters all the same?
2 Filter circuit question
3 Slope
4 math tags

[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)