The cutoff frequency is defined as the frequency where the amplitude of \$H(j\omega)\$ is \$\frac{1}{\sqrt{2}}\$ times the DC amplitude (approximately -3dB, half power point).

I wish to understand why it is \$\frac{1}{\sqrt{2}}\$, and not \$\frac{1}{\sqrt{3}}\$ or some other value. Can we derive the cutoff frequency in any other way?


marked as duplicate by nidhin, PeterJ, David, Daniel Grillo, Olin Lathrop Feb 24 '16 at 12:21

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    \$\begingroup\$ The cutoff frequency is chosen to be the half-power point mostly by convention. It is convenient, and easy to calculate for RC filters. If you would like to know why the half-power point corresponds to 1/sqrt(2), well, it just does. From an intuitive perspective, it is because power is V^2/R. If you say V1^2 / R = 2(V2^2)/R, and then solve for V1/V2, you will get the sqrt(2) ratio. \$\endgroup\$ – mkeith Feb 24 '16 at 5:28
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    \$\begingroup\$ HARI, don`t forget it is a DEFINITION only. And as such it is NOT applied for any kind of filter. It is common to use the 3dB criterion for first-order filters and higher-order filters with BUTTERWORTH response. However, for CHEBYSHEV and CAUER type filter functions we have a different definition (for practical reasons). \$\endgroup\$ – LvW Feb 24 '16 at 8:30

This -3dB point is actually half the power point. It is logical to consider a mid-way point where power of amplitude/voltage goes up or down! If power reduces by half, the voltage (with unit load) reduces by 1/root(2) or 0.707

10*Log(0.5) = -0.30102999566 dB

Even if voltage plot is considered, the corresponding half point is

20*Log(0.707) = -3.01161172406 dB

We can examine this in a while. But let's identify this.

Transfer function of any filter is actually two plots. One is for amplitude, and one for phase. A filter on a PCB/Chip/Model/DSP requires insight into both these components.

One can actually construct a hardware setup to plot these in real world for a real filter. Like a sweep of sinewave inputs with increasing frequency and measuring the output amplitude and phase w.r.t input.

For simplicity engineers leaves out phase in the transfer function. (however there are phase sensitive applications, like LVDS based serial standards, where clock and data are embedded on the same signal)

When one ignores phase, you need to consider "Power of signal". V*I, I^2*R or V^2/R (But with complex impedance, no need to leave out phase)

With normalized load P = V^2

With 5V signal power is 25W

-3dB voltage is 5*0.707 = 3.535V Corresponding power is 3.535*3.535 = 12.5W (Which is half of the original)

This has remained an interesting question, as it deserves multiple angles based on applications.

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    \$\begingroup\$ "This -3dB point is actually half amplitude point" typo? \$\endgroup\$ – biggvsdiccvs Feb 24 '16 at 7:06

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