Why is Corner Frequency always -3db

During my electronic tutorials on filters, I was told that the corner frequency of RL filters and RC filters is where the gain deviates by -3db.

however, there was no stated reason why it had to always be -3db.

I would love to know the reason why it has to be -3db

Hi,

a frequently asked question. There there are already discussion here in theis forum and even more in the internet.
Use the links in the box "Similar threads" below.

In short:
It´s not always -3dB.

- 3dB means half the power:
Imagine a 2 way speaker box. There is a crossover installed. One is a HPF the other is a LPF. Say both filters have a cutoff at 2.5kHz. Now at exactly this frequency: half of the power comes from the low frequency speaker, the other half of power comes from the high frequency speaker (in ideal case).

Another cause:
Draw a chart (log X-axis = frequency, linear dB axis) of an LPF (or HPF) there is a flat horizontal line from 0Hz close to fc...then there is a bent line around fc .. then it becomes again a straight line with constant slope.
Now extend the horizontal straight line in direction fc ... and extend the other straight line in direction to fc.
Where these both lines cross is fc (on most linear filters) ... and when you check the attenuation at this frequency: it will be -3dB. (on most linear filters)
(while the Y axis in dB is linear, the amplitude is logarithmic because of the mathematical calculation from amplitude to dB)

Klaus

Excellent explanation by KlausST.

Just to add a tiny bit ... it's calculated from Log10( 1/ 2) = -0.3010299 dB

however, there was no stated reason why it had to always be -3db...

Click to expand...

Actually it is related to logs and exponents.

It is typical to consider 1/2 value, e.g., half life, half-power or half of some signal that decays exponentially.

For an exponentially decaying signal, one interesting property is that half-signal value is constant; it does not depend on the original reference point.

If you have an exponentially decaying signal, and you plot that on a half-log paper (x-lin and y-log), you get a straight line.

If you also convert the y-axis scale, 1/2 point will be log(1/2)=-log(2)=-0.3010 or -3db. That is where the 3db comes in the picture.

Hence, -3db point is the one where the signal becomes 1/2; at -6db point the signal becomes 1/4; at -12db point the signal becomes /16th of the original.

Next question will be the order of the filters. Well, that is another story.

c_mitra said:

Actually it is related to logs and exponents.

It is typical to consider 1/2 value, e.g., half life, half-power or half of some signal that decays exponentially.

For an exponentially decaying signal, one interesting property is that half-signal value is constant; it does not depend on the original reference point.

If you have an exponentially decaying signal, and you plot that on a half-log paper (x-lin and y-log), you get a straight line.

If you also convert the y-axis scale, 1/2 point will be log(1/2)=-log(2)=-0.3010 or -3db. That is where the 3db comes in the picture.

Hence, -3db point is the one where the signal becomes 1/2; at -6db point the signal becomes 1/4; at -12db point the signal becomes /16th of the original.

Next question will be the order of the filters. Well, that is another story.

Click to expand...

Try not to obfuscate the issue. Nobody is asking about exponential decay or about filter theory.

Try not to obfuscate the issue...

Click to expand...

Are you confused? Don't worry, you are not alone!

c_mitra said:

Are you confused? Don't worry, you are not alone!

Click to expand...

No I am not c_mitra. I am just requesting you not to post confusing responses filled with irrelevant information.