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Could someone explain the difference between the following responses of a state variable filter? It looks as though the band pass response is approximately 3dB lower in the second image.

Assuming a state variable filter:

enter image description here

I have seen frequency responses like this:

enter image description here

and like this:

enter image description here

enter image description here

  • Are they both correct?
  • Can they be achieved from the same state variable filter?
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    \$\begingroup\$ I think they are all valid and can all be achieved with the same circuit. Remember for a band-pass filter the frequency responce at f0 is determined by the Q-factor, if Q=1 then the frequency responce at f0 is 0dB, if Q<1 it is <0dB and if Q>1 it is >0dB. The bandwidth is also determined by the Q: BW=f0/Q. I looked up "state variable filter" and found this site electronics-tutorials.ws/filter/state-variable-filter.html I believe that's where you found the first graph? This page has equations for Q and A (DC) etc. so you can just calculate the resistors and caps you need to get.. \$\endgroup\$
    – Vinzent
    Jan 5, 2018 at 9:33
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    \$\begingroup\$ ..The frequency responce that you see on either graph. basically just take any of the three graphs you have and work your way backwards from gain/BW to resistor/cap values, using the equations on the website. \$\endgroup\$
    – Vinzent
    Jan 5, 2018 at 9:35
  • \$\begingroup\$ Thanks for your helpful comments. Yes, the first graph is from that website, however when I went on to design a state variable filter, the frequency response I achieved was similar to the second one. That's why I was wondering if the freq response I obtained was indeed correct, and if so how come I did not achieve the first response type. \$\endgroup\$
    – rrz0
    Jan 5, 2018 at 9:38
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    \$\begingroup\$ ..Another reason for the difference between the graphs could be that (remember dB is just relative log-gain, relative to something we call 0dB) The first graph might just be showing the frequency responce of all three filters with their amplitude at f0 =0dB /normalized to 0dB. while the other three graphs might show the ouput of each filter relative to the input. The advantage with the first graph is ofcause that it allowes you to easily find the -3dB point because the peak is placed at 0dB, with the other two graphs in order to find the band-width for instance you would have to look for the.. \$\endgroup\$
    – Vinzent
    Jan 5, 2018 at 9:44
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    \$\begingroup\$ ..point on the graph where the frequency responce of the BP-filter has droped -3dB below the peak which is at -3dB (so -6dB) to find the -3dB point. Personally I prefer the first graph because it focuses on the relative output of each filter. the following three graphs however show what you are most likely going to see in the real world if you are measuring the frequency responce of all three filters relative to the input. \$\endgroup\$
    – Vinzent
    Jan 5, 2018 at 9:47

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The bottom two graphs are correct, for what would be a Butterworth response (Q=0.707) You can set the Q equal to one (Q=1) and get no attenuation of the BP response at the center frequency. But then there will be a gain peak at the center frequency of both the LP and HP response.

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  • \$\begingroup\$ Indeed, in other words the first graph is just representing the relative dampening of each output with its amplitude at f0 defined as 0dB would you agree? \$\endgroup\$
    – Vinzent
    Jan 5, 2018 at 15:22
  • \$\begingroup\$ How did you arrive to the conclusion that Q = 0.707 for the bottom graphs? Indeed, the state variable filter I designed has a Q factor of 0.707 and that response. \$\endgroup\$
    – rrz0
    Jan 5, 2018 at 17:27
  • \$\begingroup\$ @Rrz0, After you've done a few of these they become old friends. The BP gain at max is equal to the Q. (And no bumps in the HP an LP response.) \$\endgroup\$ Jan 6, 2018 at 3:31

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