Active Filter Design

Discussion in 'Homework Help' started by EEPenguin, Apr 24, 2009.

  1. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    I just want to be pointed in the right direction, not looking for somebody to do the problem for me (that's why I'm not posting the actual problem)

    So, Given a transfer function, I need to design an Active Filter using only two resistors two capacitors and one op amp. I really don't know where to begin.

    Please give me some guidance. Thanks
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    What are the design parameters of your filter? (e.g. Low-pass, high-pass)

    Are you looking for Butterworth, Bessel, or other response?

    hgmjr
     
  3. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    First, what do you know about transfer functions?

    Are you familiar with the transfer function of any filters, i.e. first order, second order, low pass, high pass, bandpass,...?
     
  4. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    As far as I understand it would be a butterworth? We never really went over different types of responses like that, but I googled it and thats what my Matlab output looked like. It has a high frequency roll off.

    I know I'm not helping much, It's not very descriptive, but literally I was just given the transfer function, and those restrictions.

    This is a Sophomore level college class, if that might give you any idea on the scope of what I would know. Thanks
     
  5. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    Stay at home...

    I don't know a whole lot, I guess that's where my problem lies. I know first orders have only one energy storage element in the filter and seconds have two. A capacitor is 1/s and inductor is Ls (doesn't apply here though)

    And that's about all I can think of off the top of my head
     
  6. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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  7. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    Ok I understand the impedances for Vo/Vi and then subbing in Wb to clean it up. But I really don't get how to work it backwards. I need the Frequency and such, I can stuble through it this way.

    Or am I missing the point you're trying to make here?

    That was a good link though, made sense out of everything. I needed stuff like that a while ago lol
     
  8. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Is the transfer function that you are given an equation or a graph? If it is an equation, you need to manipulate it into the forms where you get the Wb = 1/RC type of substitution. The defines the circuit elements. If you are doing it with 2 resistors and two capacitors you will eventually get two of these terms.

    If it is a graphical transfer function, then the poles in the graph define the 1/RC points.

    Also note:

    Impedance of C -> Zc = 1/jwC or 1/sC L -> Zl = jwL or sL
    it is all linked together.
     
  9. hgmjr

    Moderator

    Jan 28, 2005
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    Can you post the transfer function you were given?

    hgmjr
     
  10. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    Howdy:

    Funny you should mention this, because I'm working on the E-book chapter on active filters.

    In a nutshell, here are the difference between the different types of polynomials. By the way, Butterworth and Chebyshev are just two of the most common ones. There are also Cauer Elliptical filters, various hybrids of all three, and a few I haven't even heard of yet. :)

    Butterworth filters are also known as maximally flat filters. They have the smoothest response IN the passband, but have rather gentle slopes outside the passband, which makes them rather "weak" filters for some applications.

    Chebyshev filters (about three different spellings, by the way!) have the steepest slopes for a given number of poles, but at the cost of a bumpy passband. These are more suitable for radio frequency circuits where a "strong" filter action is needed.

    A Cauer elliptical filter has INFINITE attenuation at one point outside the passband, but then the response RISES again outside that limit...which may or may not be a problem in some applications.

    All these types of filters can be implemented in either a passive or active configuration.

    Hope this helps get you started.

    Eric
     
  11. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    I didn't want to post the function because I didn't want to look like I was looking for handouts. But here it is

    S^2/[S^2 + (2.4x10^4)S + (5.76x10^8)]
     
  12. hgmjr

    Moderator

    Jan 28, 2005
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    Is this what your equation looks like?

    \frac{s^2}{s^2+2.4\times10^4s+5.76\times10^8}

    hgmjr
     
  13. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    Yea, sorry I don't know how to type it like that.
     
  14. hgmjr

    Moderator

    Jan 28, 2005
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    The transfer function contains what is referred to as a "zero" at the origin due to the s^2 term in the numerator. That would appear to indicate that the filter will not pass a DC signal.

    hgmjr
     
  15. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    And the Poles are when the denominator is zero.

    But, what does this mean? What can I draw from this exactly that will help me design this thing?
     
  16. hgmjr

    Moderator

    Jan 28, 2005
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    The challenge as I understand it, is for you to determine the active filter topology that corresponds to the equation that you have been given. The filter topology must involve the use of two capacitors, two resistors and an opamp.

    hgmjr
     
  17. hgmjr

    Moderator

    Jan 28, 2005
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    I believe you should be looking for the topology of a second-order, high-pass, unity-gain, Sallen-Key active filter.

    hgmjr
     
  18. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    I looked those up as well, and the only type of Topology we've seen is The Sallen Key. I have a pretty good idea of the topology this thing will have. Most likely a cap, then the feedback node, another cap, With a feedback resistor and a resistor going to ground before the input.


    Once again I don't know how to imbed images so I'll post this link

    http://ourworld.compuserve.com/homepages/Bill_Bowden/opamp.htm#2Ofilter.gif
     
  19. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
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    Try to simplify the denominator to the form

    (s + a)(s + b)

    or

    (1 + 1/sa)(1 + 1/sb)

    and then review the transfer function material again.
     
  20. EEPenguin

    Thread Starter Member

    Apr 11, 2009
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    Ok, my ti-89 gives me one answer and I worked it out by hand with the quadratic formula and got something completely different.

    I really wish I knew how to write in that math formula font, because this is ugly. Ha.
     
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