state-variable and biquad filters

Discussion in 'Homework Help' started by PG1995, Jun 2, 2012.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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  2. mlog

    Member

    Feb 11, 2012
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    It's called a biquad filter because it is developed from a biquadratic equation of the form ax^{4}+bx^{2}+c=0.
     
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  3. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Thank you, mlog.

    Could someone please help me with the Q1?

    Regards
    PG
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    As for Q1 figure 15-12 show the answer to your question
     
  5. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    Hi Jony

    I think you referred to FIGURE 15-21. Please have a look on the attachment. Thanks.

    Regards
    PG
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    OK I see why you get confuse.
    We set each integrators to have Fc = Fo = 7.23KHz and this Fc is -3dB for LPF and for HPF. But not for BH filter. The maximum output in BH output we get at frequency equals to Fc = 7.23KHz. So for the BH output this voltage is our 0dB not -3dB.
    This imagine show everything

    [​IMG]
     
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  7. steveb

    Senior Member

    Jul 3, 2008
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    For Q1, the answer is that you can't just think blindly about parameter definitions. The idea of "bandwidth" is a concept that is defined relative to the total response, or final output. You need to define new 3dB points in the final output response, then the numerical value of the bandwidth will be clear.

    Think about it, what if you shifted the highpass filter cutoff above the lowpass filter cutoff? Would you then have negative bandwidth? No, the response would still have a peak, and there would be points on either side of the peak that are 3 dB down from the peak. It's all relative.
     
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