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Transfer function challenge #3

Discussion in 'Electronic Design' started by The Phantom, Sep 10, 2004.

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  1. The Phantom

    The Phantom Guest

    A modern active filter building block is the GIC (generalized
    immittance converter). Descriptions can be found in many textbooks
    with expressions for the relevant properties, but these usually assume
    the amplifier gains are infinite.

    For the circuit shown, find the input impedance for finite amplifier
    gains. There are two amplifiers of gain A1 and A2, as shown. There
    are five impedances, Z1 through Z5. Assume the nodes are numbered 1
    through 5, with node 1 being the input, node 2 between Z1 and Z2, node
    3 between Z2 and Z3, etc.

    Obtain a final result in the form of a fraction with the numerator
    consisting of a sum of products with no fractions, and the denominator
    likewise. Leave the numerator and denominator unfactored so that
    comparison with other people's results is facilitated. The numerator
    and denominator should look *something* like this:

    Z1 Z2 Z3 A1 + Z5 A1 A2 + Z2 Z4 Z5 A2 + Z3 Z4 Z5 A1 + ...
    ---------------------------------------------------------------------
    Z2 Z4 Z5 A2 + Z1 Z2 Z3 Z4 A1 A2 + Z2 Z3 Z4 + Z1 Z5 + A1 Z1 Z2 +...


    Since you have the general expression with finite amplifier gains, you
    can then substitute an expression for amplifier gain with rolloff and
    see what the effect of amplifier imperfection is.

    Or, let the amplifier gains go to infinity, and find the limiting
    input impedance for that case.

    You might use a different method to solve this problem if you have
    access to a computer algebra system (CAS) such as Mathcad, than if you
    do it all by hand. That would be an interesting topic for discussion:
    how best to solve this kind of problem given modern computer
    mathematical assistants.


    Vin ----o-----------.
    | |
    .-. |
    Z1 | | |
    | | |
    '-' .-----. |
    .---------| | | |
    | | | | |
    | .-. | .-----.
    | Z2 | | | \- +/
    | | | | \ / A1
    | '-' | V
    A2 /^\ .---|---' |
    /+ -\ | | |
    '-----' | .-. |
    | | | | | Z3 |
    | | | | | |
    | '----' '-' |
    | |----------'
    | |
    | .-.
    | Z4 | |
    | | |
    | '-'
    '----------|
    |
    .-.
    Z5 | |
    | |
    '-'
    |
    GND ---
    -
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
     
  2. The Phantom wrote...
    Or go to Adel Sedra and Peter Brackett's 1978 book and get the answers.
     
  3. Jim Thompson

    Jim Thompson Guest

    [snip]
    As has been discussed here (and A.B.S.E), this gyrator configuration
    is non-optimum. This version I developed moves nodes slightly and
    yields better performance per GBW... see my website, SED/Schematics
    page.

    ...Jim Thompson
     
  4. Jim Thompson

    Jim Thompson Guest

    Anyone heard from Peter since the first hurricane? He lives in
    Indialantic-by-the-Sea, Florida.

    ...Jim Thompson
     
  5. Prof. Sedra liked it so much that it's featured on the cover..

    Best regards,
    Spehro Pefhany
     
  6. xray

    xray Guest

    FWIW My cousin does too and they escaped without any signinfcant damage
    (leeward side). Hopefully Peter was as fortunate. Oh, it would have been
    the 2nd one that was challenging in Indiatlantic.
     
  7. The Phantom

    The Phantom Guest

    I picked it because it has a pleasing symmetry, not because of any
    performance advantage or disadvantage. That would be an interesting
    topic for another thread. This one's purpose is, as you said:

    "...I'd like to have the young-bucks here actually do the grunt work
    and get proficient."

    This version I developed moves nodes slightly and
     
  8. Dr. Neutron

    Dr. Neutron Guest

    That wouldn't be playing the game. Kind of like the physics test question that says,
    "You are standing on the street in front of a tall building with a mercury barometer. How
    can you determine the height of the building?"

    The street-wise kid says, "I'd find the building super and tell him, 'I'll give you
    this excellent mercury barometer if you'll tell me the height of the building'".

    (Then there's the fellow who says, "I'd go to the top of the building and drop the
    barometer off and time how long it takes to hit the sidewalk. From that I can determine
    the height of the building".)

    Look it up only to verify your own work.
     
  9. Ban

    Ban Guest

    I remember a famous audio A/D converter manufacturer showing reconstruction
    filters in his data sheets with this gyrator based circuit. But they have
    magically disappeared and were replaced with simple Sallen-Key type
    lowpasses. The idea of Cauer (elliptic) filters doesn't go very well with
    high-end audio, and anyway with oversampling have become useless.
    I couldn't get in my simulation programm a stable output with this gyrator,
    when simulating a "super-capacitor" with Z1 and Z5 being capacitors. Bad
    alternating bursts were showing up, now I understand why AD thew that
    structure out.
    There are better ways to form BiQuad filters, I use a 5opamp approach, which
    lets me directly and independently implement the coefficients or a 3 opamp
    state variable filter, which has 3 resistors to programm "f" and "Q" for
    numerator and denominator. There are also solutions with 1 and 2 opamps, but
    they do not separate the coefficients and are difficult to adjust.
     
  10. Jim Thompson

    Jim Thompson Guest

    [snip]

    Unfortunately it rarely happens :-(

    ...Jim Thompson
     
  11. Roy McCammon

    Roy McCammon Guest

    Easy enough.
    V5 = V3 = Vin
    I5 = V5/Z5
    I4 = I5
    V4 = V5+Z4*I4
    I3 = (V3-V4)/Z3
    I2 = I3
    V2 = V3 + Z2*I2
    I1 = (Vin - V2)/Z1
    Zin = Vin/I1
     
  12. Jim Thompson

    Jim Thompson Guest

    [snip]

    Not only are the gains finite, but they have, over most of their
    useful frequency range, a phase shift of 90°.

    ...Jim Thompson
     
  13. The Phantom

    The Phantom Guest

    This is only true if the amplifier gains are infinite. You lose points for not
    following directions. As I said above:
    "For the circuit shown, find the input impedance for finite amplifier gains."

    Not as easy.

    But this does raise an interesting additional question. If the amplifier gains are
    finite, what additional condition(s) would be required so that V5=V3, or so that V3=Vin.
    Would it be possible at all?
    Again, as I said above:
    Obtain a final result in the form of a fraction with the numerator consisting of a sum of
    products with no fractions, and the denominator likewise.
     
  14. The Phantom

    The Phantom Guest

    How did you model the limited GBW of the amps? A single pole
    rolloff? Can you post the model and the numbers for the parameters of
    the rolloff?

    .. see my website, SED/Schematics
     
  15. Jim Thompson

    Jim Thompson Guest

    See "Op-Amp-Config.zip" on the Subcircuits & Symbols page of my
    website. It is modeled just like a real op-amp: slewed pole-splitting
    capacitor plus an excess phase all-pass to emulate the delay in the
    devices.

    ...Jim Thompson
     
  16. Ban wrote...
    You have to have proper models for the components (capacitor esr, opamp
    series output inductance etc.), and moreover, mature complete designs
    add a single stabilizing part for robustness.
    Very good, please enlighten us.
     
  17. Roy McCammon

    Roy McCammon Guest

    I've got several 10's of thousnads of notch filters based
    on that topology out in the field for 20+ years. My first
    prototype oscilated, but I redid it with lower impedences
    and it worked fine ever sense. Though, I have had a very
    good analog designer tell me that he had had problems with
    that topology.
     
  18. Roy McCammon wrote...
    Excellent, I'd really like to have a go at the rejected
    version, do you have any around or can you give us the
    exact circuit and part values?
     
  19. Roy McCammon

    Roy McCammon Guest

    the best I can do is show the working version and try to guess at the
    old values.
     
  20. Rich Grise

    Rich Grise Guest

    You left out, measure your own shadow and the shadow of the building at
    the same time, and do similar triangles. Use the barometer to figure out
    where to go have a pina colada. %-}

    Cheers!
    Rich
     
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