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About filters and cut-off frequencies

A good question was brought up by my students in the lab exercise in the course Electronics I yesterday: What actually makes the cut-off frequency of a filter circuit (1st order) so special? Why do we look out for a point which happens to be located \SI{3}{\dB} or a factor of \sfrac{1}{\sqrt{2}} under the amplitude of the pass band? Why not \sfrac{1}{3}, \sfrac{1}{\sqrt{5}}, \SI{5}{\dB}?

The first answer, which has some historical relevance, would be to look at a resistor R and a signal with an amplitude V_0 and another signal with an amplitude V_1=V_0\cdot\sfrac{1}{\sqrt{2}} while calculating the electrical power:

    \[ P_0 = \frac{V_0^2}{R}\qquad  P_1 = \frac{V_1^2}{R} = \frac{\left(V_0\,\sfrac{1}{\sqrt{2}}\right)^2}{R} = \frac{1}{2}\,P_0 \]

The power has been reduced to half the original power, it could be the electric power which is converted to sound waves in a speaker, or the power converted to electromagnetic waves in an antenna.

But already here we can make one important observation: 20\log_{10}\left(\sfrac{1}{\sqrt{2}}\right)=\SI[output-decimal-marker={.}]{-3.0103}{\dB}\approx\SI{-3}{\dB}

Is this everything then? It still sounds random to choose exactly this point. If we for example look at a simple RC filter then we can show another reason which makes this point’s frequency so special. In a low-pass filter we have a constant amplitude \sfrac{v_{ut}}{v_{in}}=1 for low frequencies while we observed mathematically and experimentally that the amplitude drops with \SI{-20}{\dB} per decade of frequency for high frequencies. If we draw out these two asymptotic lines in a Bode diagram, then we will find exactly one point where the lines intersect – you guessed it: it’s at the cut-off frequency (the same holds of course true for a high-pass filter). We can see this if we zoom into the area around the cut-off frequency in a Bode diagram:

Bode diagram of a low-pass filter - zoomed into the area around the cut-off frequency.

Bodediagram of a low-pass filter – zoomed into the area around the cut-off frequency.

Then was this everything now? Is there more about the cut-off frequency?

As we also saw we have a phase angle of \SI{+45}{\degree} (high-pass) or \SI{-45}{\degree} (low-pass) between the input- and the output signal at the cut-off frequency. In the complex notation of the j\,\omega-method this correlates to the fact that the real and the imaginary part of the transfer function A_v=\sfrac{v_{ut}}{v_{in}} have the same magnitude at this point. The relation between real part, imaginary part, amplitude and phase angle I have plotted below for both the low-pass and the high-pass filter.

It is interesting to observe that the imaginary part actually reaches a minimum at this point (=\num{-0.5} for the low-pass filter) or a maximum (=\num{+0.5} for the high-pass filter), respectively.

You can see that the cut-off frequency actually is a remarkable point for a filter circuit.

Low-pass filter

Low-pass filter: RC and RL.

Low-pass filter: RC and RL.


High-pass filter

High-pass filter: RC and RL.

High-pass filter: RC and RL.


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