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I've got the filter that's given in the picture below, they calculate the pole and the zero of the transfer function but what do these two values for omega say, is it kind of a cut-off frequency? (kantelpunt is Dutch and if I translate it literally it means 'tipping point')

Notice that, for the omega:

  • $$|\text{s}_{\text{z}}|=\omega_{\text{z}}=4\cdot10^4\text{rad/s}$$
  • $$|\text{s}_{\text{p}}|=\omega_{\text{p}}=8\cdot10^3\text{rad/s}$$

enter image description here

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First note that R2 is smaller than R1, for this reason we can first ignore R2 to get an idea what the circuit does.

For very low frequencies the impedance of the capacitor is larger than the resistance R2. R1 and C form a low pass filter, the corner frequency is given by the first frequency (the pole).

For very high frequencies the capacitor is a short, the circuit acts approximately as an ohmic voltage divider.

The point where the low pass behavior turns into voltage divider behavior is given by the second frequency (the zero).

The behavior is shown in the plot below.

enter image description here

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'kantelpunt' could also mean 'transition points'. These are frequencies where there is a marked transition in the behavior of the circuit.

Up to \$8.10^3 rad/s\$ the output voltage is approximately equal to the input voltage.

From \$8.10^3 rad/s\$ to \$4.10^4 rad/s\$ the output voltage is decreasing by 20 db for every 10 fold increase in frequency.

Above \$4.10^4 rad/s\$ the output voltage is approximately 14db lower than the input voltage.

enter image description here

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  • \$\begingroup\$ Does these points have something to do with the -3dB points (the cutt-off frequencies)? \$\endgroup\$ – Jeans Boss Jun 24 '16 at 20:19
  • \$\begingroup\$ These are called the corner frequencies. The maximum error in the approximations occurs here and it is -3dB. \$\endgroup\$ – Suba Thomas Jun 24 '16 at 20:25
  • \$\begingroup\$ Corner frequency is also called cutoff frequency. It is also called break frequency. \$\endgroup\$ – Suba Thomas Jun 24 '16 at 20:34
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Think about what the amplitude transfer of this network will look like.

What will this network do for very low frequencies ? Hint: then the capacitor behaves as an open.

What will this network do for very high frequencies ? Hint: then the capacitor behaves as a short.

Now that you know what happens at the frequency extremes, you can estimate what the overall shape will be. So there will a frequency where the amplitude will start to drop over frequency and one where it will do the opposite.

These points are the poles and zeros and they correspond with the frequencies at which the numerator and denominator of H(s) become zero.

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  • \$\begingroup\$ For the pole and zero I cannot use real part transfer function = imaginary part transfer function but why? \$\endgroup\$ – Jeans Boss Jun 24 '16 at 17:48
  • \$\begingroup\$ You need the imaginary part to represent the phase change due to the filter. If you would only use the real part there's no way to take the phase into account and it is important to account for the phase as you will not get the proper transfer function without it. With more complex filters or an LC resonator you cannot even get the proper transfer without using imaginary numbers. It is needed to take into account the energy exchange between L and C which causes resonance which drastically changes the transfer curve. \$\endgroup\$ – Bimpelrekkie Jun 24 '16 at 20:40

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