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filter order

Discussion in 'Electronic Design' started by Candide Voltaire, Apr 14, 2009.

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  1. I was wondering if the following statement is always true

    The order of a passive filter consisting of only Ls,Rs and Cs can
    never be higher than the number of Ls and Cs

    Can anyone here confirm this or provide a counterexample

    regards,
    Candide
     
  2. MooseFET

    MooseFET Guest

    If you exclude all other devices that remember the past, it is true.
    If you allow crystals etc then the statement is false.
     
  3. Guest

    In this 'best of all possible worlds' you are correct. (As long as
    you count the stray inductance and capacitance in the circuit.)

    George Herold
     
  4. Guest

    As long as the filter is an "all pole" filter, the statement is
    correct. However, some filters, such as the elliptic (Cauer) also
    have real zeros, which require more Ls and Cs. By definition, the
    order of the filter is the order of the denominator polynomial of its
    transfer function. Consider, for example, a 3 pole elliptic filter.
    This filter has 3 poles (because the order of the denominator
    polynomial is 3) and 2 zeros. The total number of reactive
    components (inductors and capicitors) is 4, even though the order of
    the filter is 3. The same problem exists with inverse Chebyshev
    filters, since these filters also have real zeros, as well as complex
    poles, so the number of reactive components exceeds the order of these
    filters. For the other classical filter types (Butterworth,
    Chebyshev, Legendre, Bessel) the number of reactive components equals
    the order of the filter. Pardon my long-windeness :)
    Regards,
    Jon
     
  5. So the original statement is also correct in this case i.e. the order
    (3) is not higher than four
    The statement you make here differs from the one I posted originally
    and I think it is wrong to say the order equals the number of reactive
    components as for some networks e.g. the compensation network used for
    a 10:1 probe the order goes from 1 (for a network with two capacitors)
    to zero as the timeconstants are made equal
    Thanks for that, your reasoning helps us gain insights
     
  6. So the original statement is also correct in this case i.e. the order
    (3) is not higher than four
    The statement you make here differs from the one I posted originally
    and I think it is wrong to say the order equals the number of reactive
    components as for some networks e.g. the compensation network used for
    a 10:1 probe the order goes from 1 (for a network with two capacitors)
    to zero as the timeconstants are made equal
    Thanks for that, your reasoning helps us gain insights
     
  7. Guest

    Bzzzzz! wrong answer. The order of the denominator is the order of the
    filter.

    BTW, the original statement holds true even for eliptic filters. Look
    at it again:

    "The order of a passive filter consisting of only Ls,Rs and Cs can
    never be higher than the number of Ls and Cs "

    If you have some transmission zeros, it is true they add extra
    capacitors, but that isn't the issue. Even with the extra caps, the
    order of the filter won't be higher than the number of Ls and Cs. Now
    the sum of the Ls and Cs will not be the order, but that isn't what
    the theorem says.

    Here is another theorem of sorts. The transfer function of any point
    along the ladder filter will have the same denominator. I think this
    falls out of Mason's rule.

    Not that anyone does SCF these days, but if you did a SCF leapfrog
    filter, you generally got a better filter if you added transmission
    zeros. [This assumes you have filter software to build arbitrary
    responses.] The zeros didn't add any more op amps, so there was no
    noise penalty. Generally with any active filter, the attenuation of
    the input signal reaches the point where the signal falls in the noise
    floor. Often you could reduce the order of the filter using
    transmission zeros.
     
  8. As examples learn more than rules, could you supply a trivial example
    of this?

    regards,
    Candide
     
  9. That's indeed what most textbooks tell us, but the same network can
    have a different order depending on the values of its components, I
    already mentioned a simple example of this:

    r1
    u_i ______
    -----|______|----------
    | |-------u_ o
    |_____||________|
    || |
    c1 - -------|
    |
    ------
    | |
    _|_ |
    | | |
    | | __|__
    r2 |_| _____ c2
    | |
    | |
    | |
    ------
    |
    _|_


    when r1c1 <>r2c2 this is a first order network when r1c1=r2c2 this is
    a zero order network

     
  10. Maximum of numerator and denominator orders, no?


    Best regards,
    Spehro Pefhany
     
  11. nukeymusic

    nukeymusic Guest

    the original post was about passive time invariant linear systems with
    Rs Ls and Cs, mixers don't fit in that picture
     
  12. Mixers are passive. Nothing in the original post required the system to
    be linear and time invariant.


    Vladimir Vassilevsky
    DSP and Mixed Signal Design Consultant
    http://www.abvolt.com
     
  13. Umm.. .how do you make a mixer with ideal lumped L, R, C only?
     
  14. whit3rd

    whit3rd Guest

    A component is linear if its response is proportional to its input
    (double V,
    and the I doubles, like for R, L, C). A component is passive if it
    contains
    no amplifiers.

    The usual "passive" components of catalogs, however, are exactly the
    linear
    components. The original statement is correct, but mixers with
    transformers
    and diodes, also "passive" components, are NOT linear, and
    are not built with only R, L, C.
     
  15. Guest

    Yup, you're right. I misread your original post.
    Regards,
    Kral
     
  16. Guest

    But the limit is on the order of the filter. You can play games and
    get a lower order. All the theorem does is tell you the order can be
    no higher than a limit, but it could be lower. This is a case where
    you really need to read the statement carefully.
     
  17. nukeymusic

    nukeymusic Guest

    That's not correct, it's correct for an R but not for an L or C,
    e.g. if I double the voltage over an L the current will increase
    linearly
    for L and C there is a first order differential equation which
    describes the relation between voltage and current.
    An electronic network is linear when it obeys the rules of homogenity
    and superposition, when you only use ideal Rs,Cs and Ls you always get
    a linear network.

    regards,
    nukey
     
  18. whit3rd

    whit3rd Guest

    Look at that equation carefully: it specifies frequency dependence,
    NOT amplitude dependence. Double the amplitude of a
    voltage-source AC input signal, and the current through your
    L or C is going to double. The phase of the current remains
    the same as before the source changed.
     
  19. nukeymusic

    nukeymusic Guest

    But that's of course something different, in the preceding post you
    did't mention you were talking about an AC-source, you just mentioned
    doubling the voltage, that is a Heaviside step function. What you tell
    now is indeed another characteristic of linear circuits when excitated
    by a sinusoidal function (once the transient phenomena have
    disappeared).

    regards,
    nukey
     
  20. nukeymusic

    nukeymusic Guest

    But that's of course something different, in the preceding post you
    did't mention you were talking about an AC-source, you just mentioned
    doubling the voltage, that is a Heaviside step function. What you tell
    now is indeed another characteristic of linear circuits when excitated
    by a sinusoidal function (once the transient phenomena have
    disappeared).

    regards,
    nukey
     
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