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Historically when speaking about lets say an analog low-pass filter, the cut-off frequency of that filter is defined as -3dB.

So this is the frequency where the output sinusoid signal's amplitude attenuates to 0.7 of the input(Voltage-ratio). When squared this ratio we obtain %50 which is the Power-ratio.

It is very obvious that once upon a time it was decided the cut off-frequency should be the frequency which causes a %50 power loss of a particular frequency sinusoidal signal input.

My question is: what could be practical reason defining it %50? Isn't %50 still big amount and how could it be associated with being filter. I could make sense if it were chosen %95 for example.

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  • \$\begingroup\$ The -3dB point is traditionally regarded as the end of the useful passband, rather than the beginning of the useful stopband. The latter is too dependent on specific needs to have a single universally applicable definition. \$\endgroup\$ – Brian Drummond Nov 15 '15 at 13:25
  • \$\begingroup\$ How could it be defined useful just a %50 attenuation. That is I dont get. What is the story and reasons behind it I meant. \$\endgroup\$ – user16307 Nov 15 '15 at 13:30
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    \$\begingroup\$ One aspect that makes it most useful is reciprocity : because the R and C components (or L and R) of impedance are in quadrature, making their magnitudes equal gives you sqrt(2) voltage loss, 0.5 power. Any other definition would not have this property. For example, interchanging R and C gives you a high pass filter with the same nominal cutoff frequency. Any other definition of cutoff frequency would give you a different frequency for the transposed filter! Thus -3dB is a uniquely useful definition. \$\endgroup\$ – Brian Drummond Nov 15 '15 at 13:37
  • \$\begingroup\$ So lets say in an RC or RL low-pass filter assuming same oppostions(impedances), if the R and C, or R and L are interchanged; the circuit will be a high-pass filter with still the same attenuation. But what is useful about it? \$\endgroup\$ – user16307 Nov 15 '15 at 13:44
  • \$\begingroup\$ To be more precise: the 0.7 you use comes from 1/sqrt(2), which is actually ~0.707. \$\endgroup\$ – Gabriel Staples Jun 3 '18 at 7:54
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First, calling it the "cutoff frequency" leads to misconceptions. "Rolloff frequency" is a better name that gives you a more accurate mental picture of what is really happening.

Using the -3 dB point is not arbitrary. It falls out from the math naturally. For a R-C filter:

  ω = 1 / RC

where ω is the frequency in radians/second, R is in Ohms, and C in Farads. For the frequency in Hz, use:

  f = 1 / 2\$\pi\$RC

If you plot the Log(amplitude) as a function of Log(frequency), such as in a Bode plot, then the -3 dB frequency is where the asymptotes for the pass band and stop band meet. Put another way, at frequencies well into the pass band, the filter looks like a horizontal line. At frequencies well into the stop band, the filter is a line with a slope of 20 dB per decade (+ or - depending on high or low pass). If you draw those two lines and extend them to where they meet, it will be at the -3 dB rolloff frequency.

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  • \$\begingroup\$ ok so in log-log scale the horizontal line becomes a line with a slope around -3dB. that explains the practical use of it. \$\endgroup\$ – user16307 Nov 15 '15 at 14:36
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    \$\begingroup\$ @user16307 It doesn't become a line with a slope of -3 dB, it becomes an asymptotic line with a slope that's some integer multiple of 20 dB/decade. Instead, the distance at the cutoff frequency between the asymptotes and true plot is about 3 dB (or some integer multiple for the case of repeated poles/zeros). \$\endgroup\$ – IPoiler Nov 16 '15 at 0:50
  • \$\begingroup\$ I know I meant "with a slope at around -3dB" at that -3dB point.. \$\endgroup\$ – user16307 Nov 16 '15 at 8:59
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Got too long for comments.

The -3dB point is traditionally regarded as the end of the useful passband, rather than the beginning of the useful stopband. The latter is too dependent on specific needs to have a single universally applicable definition.

One aspect that makes it most useful is reciprocity : because the R and C components (or L and R) of impedance are in quadrature, making their magnitudes equal gives you sqrt(2) voltage loss, 0.5 power. Any other definition would not have this property. For example, interchanging R and C gives you a high pass filter with the same nominal cutoff frequency. Any other definition of cutoff frequency would give you a different frequency for the transposed filter! Thus -3dB is a uniquely useful definition.

It's a single reference point. Given the 3dB point and a little more info (filter order, type e.g. 4th order Butterworth) you can tell where other characteristic points : 1dB flatness, 60dB stopband etc are.

Or you can work backwards : if you need 40dB attenuation at 1 kHz from a 2nd order Butterworth LPF for example, you know from filter design sources that a 2nd order filter has an ultimate slope of 40 dB/decade, and in the special case of a Butterworth filter, intercepts the 0dB line at the 3dB point, so the 3dB point will be 1 decade from the start of the stopband, i.e. 1000Hz / 10, or 100Hz. If you need 60dB attenuation, the -3dB point becomes 1.5 decades below 1000Hz, or about 30Hz. If that's too low a frequency, you need a steeper filter, such as a higher order one.

Filter design by scaling and similarity, with reference to the -3dB point, has a long history, and accumulation of experience and literature behind it.

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Quote: "Can you explain why it is perfectly reasonable to use? That is what I'm asking actually."

Let me try another explanation: Based on the general second-order transfer function it is common practice to desribe the various lowpass filter responses (Butterworth, Bessel, Chebyshev,...) using the pole location in the complex s-plane (pole: Zero locations of the denominator). This is because it can be shown that the factors in the denominator D(s) of the general transfer function can be simply expressed by two parameters: Pole frequency wp and pole quality factor Qp.

When we apply this nomenclature also to to a first order lowpass function it is easy to show that, in this case, the pole frequency wp is identical to the 3dB angular frequency (and Qp=0.5). That means: We have a single real pole on the neg. real axis.

In summary: With the aim to describe the various lowpass responses (first and second order) by the same parameters (wp and Qp) we automatically arrive - for a 1st order function - at a frequency wc=wp (3dB angular frequency).

Remark:Perhaps it is interesting to note that the second-order Butterworth lowpass also has a pole frequency wp that is identical to the 3dB threshold.

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If we are going to describe a filter with a single figure, then half power has got a nice ring to it.

In a single pole RC filter, the RC time constant gives the -3dB bandwidth directly, as f3dB = 1/2piRC, whether it's highpass or lowpass. So it's a perfectly reasonable figure to use. In a Butterworth filter, the -3dB bandwidth similarly drops out of the filter equations.

However, nobody who actually wants to use a filter, to a specification, relies on the -3dB bandwidth as more than a rough idea of the type of filter. -3dB is a long way down for anyone who wants a non-distorted passband, and it doesn't say anything about the delay flatness of the filter.

In a Chebychev designed filter, the usual 'cut-off' frequency is when the passband is the ripple amplitude down, rather than 3dB down. The ripple is often 1dB or even 0.1dB.

If you are interested in the stopband of a filter, different applications will demand anything from -30dB to -100dB, so a single figure will be useless for any specific design.

If I compare filters with -3dB bandwidths of 1kHz, 1MHz and 10GHz, I will have a pretty good idea from those numbers that the first will be built out of opamps and RCs, the second from Ls and Cs, and the third from patches of transmission line. But will know nothing about the passband flatness or stopband attenuation.

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  • \$\begingroup\$ you wrote "So it's a perfectly reasonable figure to use." Can you explain why it is perfectly reasonable to use? That is what I'm asking actually. \$\endgroup\$ – user16307 Nov 15 '15 at 13:52
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    \$\begingroup\$ No, I can't. You either get it, or you don't. As a filter needs a host of figures to describe it, it cannot be described by one alone. But people want one for general use, and have picked one. What is unreasonable about picking one that is somewhere between the usable passband and stopband, and drops out of the equations of the simplest types of filter? \$\endgroup\$ – Neil_UK Nov 15 '15 at 15:07
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For a butterworth filter the transfer function is

$$|H(j\Omega)|=\frac{1}{1+ \Omega^{n2} }$$

where Omega is the normalized frequency $$\Omega = \frac{\omega}{\omega_0}$$

so when Omega equals 1 the transfer function equals a half, thus giving us our famous -3dB or half power point.

Edit: Sorry I should have written s as jOmega fix it up

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If you want to have an overall frequency response of magnitude 1 when splitting a signal into two frequency bands (like with a two-way loudspeaker box), you choose the same roll-off frequency for low- and highpass. It would seem that adding two amplitudes of sqrt(2)/2 would result in a magnitude of more than 1, but since the respective filters turn by 45° in opposing directions, the overall amplitude actually stays straight (for orthogonal signals, powers rather than amplitudes add).

Of course that means that one has to choose loudspeakers with the same sort of phase response at roll-off frequency...

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