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I have calculated the transfer function for a given RC circuit, and determined that it was high pass. I was wondering if you could cancel its affect with an inverse circuit, and how this circuit would look?

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  • \$\begingroup\$ If I consider your question from a physical point of view, it seems clear that every resistor transforms part of the energy of the signal in heat, ant that transformation is an irreversible process. If you consider negative resistance and capacitance, then your question makes more sense. \$\endgroup\$
    – CasaMich
    Commented Sep 25, 2015 at 6:48
  • \$\begingroup\$ @CasaMich, whether or not an inverse operation exists depends on the transfer function of the original filter. For example, whilst not practically advisable, an op-amp integrator can be followed by an op-amp differentiator. It should be noted, though, that if a given frequency (or band) has been removed, the information is lost and can't be re-created. For example, a DC component may be removed from a signal by an RC filter. \$\endgroup\$
    – Chu
    Commented Sep 25, 2015 at 7:28
  • \$\begingroup\$ I guess your question is about if adding a low pass filter will attenuate the high pass filter... \$\endgroup\$
    – R Djorane
    Commented Sep 25, 2015 at 7:50
  • \$\begingroup\$ Show us your circuit. \$\endgroup\$
    – Chu
    Commented Sep 25, 2015 at 7:56

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For specific cases, you can. For others, you can't. So we can't answer regarding your specific filter without knowing its circuit or transfer function.

For an example where it's possible, look at pre-emphasis and de-emphasis networks, used to boost a range of frequencies before a noisy transmission system, and reduce them (also reducing the noise) at the receiver.

For an example where it isn't, look at your transfer function : is there any frequency - say, DC for a HP filter - where the output is zero? ( a so-called "zero" in the transfer function) If so, you cannot design a matching filter that covers that frequency. Having eliminated DC, you must use some other technique such as a "black level clamp" in analog video to restore it.

However you may be able to design a matching filter for a spectral range that excludes the zero - such as a frequency range down to 20Hz, considered "good enough" for most audio purposes.

With more complex transfer functions involving second order filter sections, you may be able to undo the amplitude response but not the phase response.

If you need to cancel changes to the phase response, the best you can do is add a "group delay equaliser" usually implemented as an "all pass filter". This does not undo the phase changes (you would need to wind time backwards for that!) but adds further delays such that the entire signal is delayed by a constant time.

A constant time delay appears in phase measurements as a phase shift linearly increasing with frequency (30 degrees at 1 kHz is the same as 60 degrees at 2 kHz) and is often called a "linear phase" system.

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Yes, a low pass filter can be used to offset a high pass filter.

However, while the frequency response can be made flat in theory, that theory doesn't take into account signal to noise ratio. The low frequencies coming out of the high pass filter may be so attenuated that there is little signal to noise ratio left, or may even be below the noise floor. In that case they are effectively permanently gone. You can amplify the low frequency range, but all you get will be louder noise.

One thing that helps is that in this situation the signal to noise ratio isn't the ratio of signal to all the noise, but only the signal to the noise frequencies in that signal. There is often little noise in the low frequencies. With the lower noise floor at low frequencies, those signal frequencies have a higher signal to noise ratio, and therefore more chance of being recovered in a useful way.

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  • \$\begingroup\$ Perhaps it should be added that such an "offset" requires that both stages (lowpass and highpass) are in parallel. Only in this case, both transfer functions are added. \$\endgroup\$
    – LvW
    Commented Sep 27, 2015 at 8:57
  • \$\begingroup\$ @lvw: What the OP asked about and I answered about was two filters in series. Their transfer functions multiply. That's why one attenuating some frequencies below the noise floor becomes unrecoverable. \$\endgroup\$ Commented Sep 27, 2015 at 12:11
  • \$\begingroup\$ O.Lathrop: I only spoke about the transfer functions - not the noise properties. A highpass function can be "compensated" (neutralized) by a lowpass function only if both are added (not multiplied). For a series connection we need a PI block. \$\endgroup\$
    – LvW
    Commented Sep 27, 2015 at 14:06
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Let´s assume you have a simple first order highpass: H1(s)=sT/(1+sT)

If we want to find another series block which is able to "cancel" the high pass effect, we need to fulfill the equation H=H1*H2=1.

That means: we have to realize the function H2=1/H1=(1+sT)/sT.

This is a simple PI block (proportional-integrate).

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