Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

I have a question which may sound strange, but still I'd like to ask community to explain that. Hope other people may find it interesting aswell.

As we know, cutoff frequency is the one frequency harmonics of which is being reduced by 3 db (or Vpp is sqrt(2) times lesser than harmonics Vpp before the filter).

This image gives more or less clear picture of that idea (consider the red line, cutoff frequency marked as fcp):

enter image description here

What that means: the filter passes all harmonics of the input signal through, but some of them are being reduced in power. Harmonics of some frequency are being reduced by 3 db, this frequency is a cutoff frequency of that particular filter.

This is the defininition and the explanation of my understanding of the term. I'd be grateful for corrections, if any :)

Now, the question itself: the word cutoff means that something is being cut. So, we may say that harmonics of particular frequency are filtered, if Vpp reduced by 3 db or more. The question is - why exactly 3 db? Saying exactly I mean

enter image description here

share|improve this question
You are using the word "harmonics" a lot, but you don't need it at all to describe the cutoff frequency. At the cutoff frequency, any signal component is attenuated by 3 dB, whether it's a harmonic of some other component or not. –  The Photon Sep 4 at 22:13
@ThePhoton Well I use word "harmonics" because DC signal has no oscillations and thus no frequency. You mean that any alternating signal might have cutoff frequency, meander wave for example? –  Alexey Malev Sep 4 at 22:15
A filter has a cut-off frequency, regardless of whether the signal input to it is a periodic signal or not. If the signal is not periodic, then you wouldn't describe it's components as harmonics. For example, the input signal might be white noise, and the transfer function of the filter will still tell you how that signal is affected by the filter. –  The Photon Sep 4 at 22:17
@ThePhoton No objections on that, cutoff frequency is definetely a filter (circuit) property which doesn't depend on input signal. –  Alexey Malev Sep 4 at 22:19
I learned this stuff from a Control Systems perspective and always viewed the -3 dB point as a convention. In essence a common vocabulary among engineers. You knew that a 2nd order system would drop off by 20 db per decade above the cutoff (or we called it bandwidth) frequency of the filter. So someone could describe the filter or controller as a first/second order filter/controller with a bandwidth of 100 Hz. From there we knew it was more or less flat below 100 Hz and dropped off according to the order of the filter/controller above it. –  gabe Sep 5 at 7:11

6 Answers 6

up vote 5 down vote accepted

The 3-dB cutoff is just one commonly used way to describe a filter. For some applications you might want to specify the 10-dB cutoff or 60-dB cutoff instead. It is convention that if someone says "cutoff frequency" without being more specific, they are talking about the 3-dB cutoff. You should think of the 3-dB cutoff as the frequency where the filter begins to roll off, not where it is strongly attenuating the signal.

There are a couple of practical reasons to use the 3-dB point as the conventional cut-off frequency.

  1. In a one-pole RC filter, the 3-dB frequency is conveniently found at a radian frequency of \$\dfrac{1}{RC}\$.

  2. If you draw a piecewise linear approximation to your one-pole transfer function plot, using a horizontal line segment for the low frequencies, and a 20-dB per decade sloped line for the frequencies well above cut-off, these two approximate lines will meet at the 3-dB frequency. Often when we just want to sketch a frequency response we make a Bode plot, which uses exactly this approximation.

enter image description here

(image courtesy Wikimedia)

share|improve this answer
That is exactly what I wanted to know, thank you. –  Alexey Malev Sep 4 at 23:02
Thanks for the upvote. But feel free to wait 24 hours to give people all around the world a chance to answer your question before accepting an answer. –  The Photon Sep 4 at 23:04
Thanks for fairness ;) I do monitor all the activity, so have no fear on that. –  Alexey Malev Sep 4 at 23:06

The question is - why exactly 3 db?

When any signal reduces by 3dB (or more precisely 3.0103 dB as you pointed out) the power that that signal can drive into a resistor halves.

share|improve this answer
So this is a kind of widely accepted agreement that a frequency considered filtered out or "been cut off" when it is attenuated by 3 db or by half of its initial power? My question is acually - why one half and not 2/3 or some other ratio for instance.. –  Alexey Malev Sep 4 at 22:23
Half is a more fundamental ratio than two-thirds or three-quarters. Other than that I can't give any other reasoning. –  Andy aka Sep 4 at 22:29
I thought about it from the filter application point of view, but lack of hands-on experience makes a decision whether 3db attenuation is enough or not a little bit complicated. Anyway, you clarified things, thanks a lot :) –  Alexey Malev Sep 4 at 22:33
3 dB is hardly any attenuation - if you were talking about an audio signal, the perceived volume roughly drops by half every 10 dB. You be hard-pressed to notice a 1 dB drop in volume unless you were listening intently. 3dB is what I call "noticeable but not extreme by any means"! –  Andy aka Sep 4 at 22:39
The 3-dB point is also conveniently at 1/RC when describing a one-pole RC filter in angular frequency units. But it is just one commonly used way to describe a filter. For some applications you might want to specify the 10-dB cutoff or 60-dB cutoff instead. It is convention that if someone says "cutoff frequency" without being more specific, they are talking about the 3-dB cutoff. You should think of the 3-dB cutoff as the frequency where the filter begins to roll off, not where it is strongly attenuating the signal. –  The Photon Sep 4 at 22:48

Contrary to seemingly popular belief, the cutoff frequency isn't the frequency in the passband where power falls by 3dB, it's the frequency where the amplitude of the signal falls by 3dB.

Thereafter, in the stop band, the amplitude of the signal will drop at a rate depending on the order of the filter.

For a first order filter, the rate will be 6dB per octave, for a second order filter it'll be 12dB per octave, and so on.

share|improve this answer

In something like a crossover network, sounds above a certain frequency will be directed to one speaker, while those below will be directed to another. If the "cutoff" point is defined as attenuating half the power, then a high-pass filter and a low-pass filter at the same frequency will each pass half the power. At frequencies sufficiently far away from the cutoff, one filter's output will approach full power while the other's will approach zero. In each case, the combined output of the two filters will be roughly equal to full power. Note that for the frequency response to be flat near the cutoff frequency, the filters must be properly matched to each other, but it's still much easier to say that a clean crossover will occur when the cutoff frequency for a high-pass filter should be set to match that of a low-pass one, than to specify the cutoff frequencies in way that would require them to differ.

With regard to the fact that 3.00000dB isn't "exactly" half, think of "3dB" not as a precise 3.00000dB ratio, but rather as a shorthand for "half-power", in much the way as a "3.58Mhz" crystal is a shorthand for 3,579,545Hz [most such crystals may not be accurate to seven decimal places, but 3,5800,000Hz would be out of tolerance].

share|improve this answer
"....while the other will pass nothing" Really nothing??? –  LvW Sep 5 at 11:20
@LvW: The further one goes above or below the cutoff frequency, the closer filters will come to passing either "full power" or "nothing"; assuming ideal components, many filters' attenuation may be made arbitrarily close to 100% attenuation by making the frequency high or low enough. –  supercat Sep 5 at 17:10
@LvW: Further, if at some particular frequency, one side attenuates its input by at least 95%, and the other side by at most 5%, then the total power output by both sides at that frequency will be between 95% and 105%; for many purposes, a 5% variation in power is too small to worry about. –  supercat Sep 5 at 17:16
This sounds a bit more realistic than "nothing". My comment was because - according to my experience - some students/newcomers think that a lowpass is something like a "brickwall" with infinite damping within the so called "stop region". I think, in a forum like this an exact description of the real circuit behaviour is appropriate. –  LvW Sep 6 at 8:05
@LvW: Do you like the edit above? –  supercat Sep 8 at 15:31

You have to use some threshold to measure the frequency, so we usually choose half-power frequency. That is,

$$10\log_{10}{1\over2} = 20\log_{10}{1\over{\sqrt2}}$$

share|improve this answer
Sorry, I didn't understand the point. Can you please explain it in a few more words, particular unclear thing is threshold refererence.. –  Alexey Malev Sep 4 at 22:21
In your example, the frequency response is a curve. How to characterize it with a single number? We simply choose a place where the curve crosses. -3 dB is a reasonable choice. It is where the power the voltage can drive is diminished by half. –  markrages Sep 4 at 22:24
Thanks, now idea is clear. The only remaining question is why 3 db (half power reduction)? Wrote that also under another answer. I guess this level of attenuation is ok for some basic (or maybe widely spread) applications, but not sure where the sqrt(2) number came from. –  Alexey Malev Sep 4 at 22:28
Because power is proportional to voltage squared, so voltage is proportional to the square root of power. –  markrages Sep 4 at 22:29

Alexey - you should know that the so-called "3dB-cutoff" is an arbritary point on the filters magnitude response with the aim to distinguish between the path region and the "stop" region of the filter ("stop" means: increasing attenuation). It is common practice to use the 3dB point for first order circuits - however, in case of circuits of higher order following Chbeyshev or Cauer lowpass approximation we usually have "cutoff frequencies" corresponding to attenuation values of, for example, 0.1/0.5/1/2/3 db.

As another argument: System theory intensively makes use of pole frequencies which are defined in the complex frequency area "s" (The magnitude of the transfer function at pole frequencies approaches infinity because the denominator is zero). And it appears that the magnitude of a 1st order filter at its pole frequency is -3dB down (and the phase is -45 deg). For a 2nd order filter the phase at the pole frequency is exactly -90 deg.)

Summary: It is common practice to use the concept of pole frequencies (in the complex frequency domain) to describe the magnitude and phase response of filters. And for 1st order filters it appears that the magnitude at a frequency identical to the pole frequency is 3 dB down.

**EDIT:**Here comes another argument supporting the 3dB concept(for first order responses): If the bandwidth BW for a bandpass (lowest order) is defined between the two 3dB points the resulting quality factor Q=fo/BW is identical to the "quality" factor of the corresponding pole Qp (which is defined based on the pole location in the complex frequency plane), that means: Q=Qp.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.