About filters and cut-off frequencies

Post author:uwezi
Post published:2013-09-18
Post category:Electronics / Everything / Science
Post comments:0 Comments

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 $V_1=V_0cdotsfrac{1}{sqrt{2}}$ 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: $20log_{10}left(sfrac{1}{sqrt{2}}right)=SI[output-decimal-marker={.}]{-3.0103}{dB}approxSI{-3}{dB}$

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.