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questions about Basics of active filters (using op amp), please help/

Discussion in 'Electronic Basics' started by Liu Ju, May 4, 2004.

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  1. Liu Ju

    Liu Ju Guest

    Dear members:

    I am studying about applications of op amps in designing the active
    filter in a book.
    I have some confusion and would like to ask for explaination.

    1. In the book they analyze an active low pass filter consisting of "
    1 R, 1 C, and 1 op amp"
    The R and C create a passive low pass RC filter and are connected to
    the positive pin of the op amp in series.

    The book provides the transfer function of the low pass filter as
    K=1/(1+P*(omega break)*R*C)
    Where (omega break) is the corner frequency (at which the
    K=1/SQRT(2)).

    I don't understand this function. In my understanding P=j*omega
    (Omega: frequency of the input) and thus the transfer function should
    be: K=1/(1+P*R*C).
    Why is there the (omega break) in the function? Please explain it to
    me!

    2. In the book they said we can get the transfer function of a high
    pass filter by changing somethings from the transfer function of a low
    pass filter.

    I guess that in the transfer function of a low pass filter, we replace
    R by 1/C, C by 1/R, (omega break) by 1/(omega break) and P by 1/P, and
    thereby we have the transfer function of the corresponding high pass
    filter.


    Is it correct?

    By the way, would anyone, who know the basic theory of the active
    filters and op amp, give me the contact email address. I have some
    complex confusion and would like to ask but there are some figures
    that I cannot present in the form of text in GOOGLE group. Thank you
    very much.


    Sincerely
    LiuJu
     
  2. Liu Ju wrote:
    (snip)
    This is a basic 1 pole RC filter followed by a buffer amplifier that
    keeps operation of the RC filter independent of any downstream load.
    The break or corner frequency is the frequency at which the response
    is switching from little attenuation versus frequency to an
    attenuation proportional to frequency. If you draw a graph of the
    amplitude response versus frequency on log log paper, the response is
    a hyperbola that approaches constant (flat, horizontal) amplitude in
    the low frequency side, and constant down slope to the high frequency
    side with the intersection of those two asymptotes at the corner or
    break frequency.

    This frequency is most naturally described in radians per second,
    instead of cycles per second. R*C describes that radian frequency.
    Since there are 2*pi radians in a complete cycle, you have to multiply
    by those factors to change the units of frequency from radians per
    second to cycles per second.
    If you interchange the capacitor and resistor, the right (higher
    frequency side) asymptote of the response graph becomes horizontal
    (flat response versus frequency) and the left (lower frequency side)
    slopes down proportional to frequency. The corner or break frequency
    is still R*C radians per second.

    Both the high and low pass filters are just voltage dividers made up
    of one fixed impedance (the resisters impedance is not a function of
    frequency) and one variable impedance (the impedance of a capacitor is
    1/(2*pi*F*C), if F is is hertz or cycles per second, or 1/(omega*C) if
    the frequency is in units of radians per second) with a 90 degree
    phase shift. The attenuation versus frequency response is just the
    result of dividing the voltage between these two impedances.
    Not quite. The break frequency still remains at R*C because that is
    the frequency at which the magnitudes of the resistive and capacitive
    impedances are equal.
    You are welcome to email me if this post has been at all helpful to
    you.
     
  3. CFoley1064

    CFoley1064 Guest

    Subject: questions about Basics of active filters (using op amp), please
    If you're going to be working with active filters, you should look at "Active
    Filter Cookbook" by Don Lancaster. All the basics of a tough subject are
    covered there, in a manner which is slightly unorthodox but accessible to just
    about everybody. The best part is that it has "instant design" sections which
    walk you through component specification quickly. It's really oriented toward
    actually making and using active filters, as well as analysis. I found the
    book to be helpful, and would highly recommend it to any newbie.

    The book is still in print, and is available from the library, amazon, or Mr.
    Lancaster's website.

    http://www.tinaja.com/
    http://www.tinaja.com/books/bkdons.asp

    Good luck
    Chris
     
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