2nd Order All-Pass Filter Design

Discussion in 'The Projects Forum' started by circuitmayne, Oct 5, 2010.

  1. circuitmayne

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    Oct 5, 2010
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    I've been try to design a second order all pass filter but I'm having a lot of trouble finding a source that actually explains how to calculate the right component values. I especially am looking for a formula or equation which I can use to modify Q with so that I can play around with my circuit till I get the result I want. Does anybody know where to find a circuit schematic that is also accompanied by some equations and formulas?
    I greatly appreciate any help I can get.
     
  2. marshallf3

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    all pass filter ?
     
  3. Ghar

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  4. SgtWookie

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    If you want to design a passive LC filter, google Elsie.
     
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  5. Ron H

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  6. JoeJester

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    I doubt it's an all pass because after choosing C, you need to identify f0 to compute k.

    So, if this is an all pass, why must fo be in the calculation?
     
  7. Ron H

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    IIRC, f0 is the frequency where the phase shift is 180 degrees, in a 2nd order allpass. The Q determines how rapidly the phase shifts as a function of frequency.
    If you didn't need f0, you could use a piece of wire instead of a filter.
     
    Last edited: Oct 6, 2010
  8. gootee

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  9. circuitmayne

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    i want to thank you guys for your help. i'm going to try a couple of these options out and see which one works. i will post and update soon. if anyone else has any suggestions, please feel free to share. thanks
     
  10. JoeJester

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    True. However, if there are limits on the pass, it can not be an all pass. Of course if your definition of all pass is all I need for the application, then your playing loose with the language.
     
  11. Ron H

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    If I understand your point, that is like saying that a voltage follower isn't really a voltage follower because it doesn't have unity gain from DC to blue light.
    An active allpass is basically bandwidth limited by the GBW of the op amp(s). If you had ideal op amps, the allpass would have unity gain from DC to infinite frequency.

    Maybe I'm missing your point.

    All I was trying to explain is that an allpass filter intentionally changes the phase shift (group delay) of the signal as a function of frequency, and f0 defines the region where the change occurs.

    To explain one application of allpass filters:
    My experience with was designing group delay correction circuits for frequency response shaping networks, which were used to flatten the response of tape recorders which recorded and played back digital signals.
    Without group delay correction, the signal zero crossings (eye pattern) would give very poor error rates.
     
  12. JoeJester

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    Ron,

    The name allpass is a misnomer since it restricts the signals to a specific bandwidth to phase shift the selective frequency or frequencies. That was my point.

    Prior to this thread, if you said all pass, I would have thought of a piece of wire, although that too would have an upper limit.
     
  13. Ron H

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    I don't understand what you mean. Allpass means that the amplitude response is flat over the frequency range where the phase shift is changing.
    You were not aware of what an allpass network does. I'm not sure you do now.
     
  14. JoeJester

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    Low pass, high pass, bandpass, and notch are all descriptive terms with respect to filters. Now add all pass to that list.

    Semantics. That is what I am saying, semantics.

    I'll get over the misnomer.
     
  15. R!f@@

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    All pass....what's the use of All pass.

    I see Hi pass, low pass, band pass, but all pass.. I am totally lost
     
  16. Ron H

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    Did you read post #11, last paragraph?
     
  17. R!f@@

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    Ya but all what....what...pass what ?
     
  18. JoeJester

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    Ron H,

    All I know is if I design a 2nd order all pass according to page 8.107, the response is not flat. I am using unity gain and I am not expecting an outrageous flat response to over 1 MHz. My first attempt was between 10 and 1kHz, or an fo of 100.

    When I get back from dinner, I'll post the circuit and results.

    On edit:

    After thinking about the problem, I realized I "picked" the incorrect value for C. Although the instructions stated "pick C", it meant do the calculation. So, after making that adjustment, it certainly works better and as advertised.
     
    Last edited: Oct 9, 2010
  19. Ron H

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    I think there is an error in the PDF. It says that R4=Q/2. Did you start with Q=1, which would yield R4=499m, and calculate the rest of the values from there? 499m is way to low for anything other than an ideal op amp, and the load on the source is 1Ω at low frequency, and 0.5Ω at high frequency. This is way too low for real-world sources to handle. You're not an analog hardware guy, are you?;) No disrespect intended - just an observation.
    "Pick C" means choose a value for C. I started with C=100nF, Q=1, and F0=100. The frequency response comes out to be bandpass. I have Active Filter Design Handbook, Moschytz and Horn, 1981. If you juggle their equations (p.48), the correct value for R4 is Q^2*R3. The value of R3 is actually arbitrary, since it is only the ratio of R3 to R4 that affects the transfer function. I picked R3=10k.
    Attached are my simulation results.
    The graphic on the left is with R4=Q^2*R3 (correct).
    The graphic on the right is with R4-Q/2 (wrong).
     
  20. JoeJester

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    That error was costly to me. Not in a monetary fashion thankfully.

    Anyways, here's the page where I found the formulae that led me astray.
     
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