Passive bandpass center-frequency gain alteration

Discussion in 'Homework Help' started by Starhowl, Feb 13, 2014.

  1. Starhowl

    Thread Starter New Member

    Dec 3, 2013
    25
    0
    Task:
    Modify this bandpass circuit to change its center-frequency gain from 1 to 0.5 without changing \omega_0 or Q.
    [​IMG]
    The general transfer-function of a bandpass is
    T(s) = \frac{a_1 s}{s^2 + s \frac{\omega_0}{Q} (\omega_0)^2} where the natural modes can be found by finding the poles of any response-function without altering its natural structure.

    As the natural modes of the left-hand side are s^2 + s\frac{1}{RC} + \frac{1}{LC} and the natural structure isn't changed on the right-hand side of the alteration, the natural modes should stay the same as follows: = \frac{s\frac{1}{2CR}}{s^2+s\frac{3}{2}\frac{1}{CR}+\frac{1}{CL}}" alt="T(s) = \frac{\frac{1}{\frac{1}{R} + \frac{1}{sL} + sC}}{2R + \frac{1}{\frac{1}{R} + \frac{1}{sL} + sC}} = \frac{1}{1 + 2R({\frac{1}{R} + \frac{1}{sL} + sC})} = \frac{1}{1 + 2 + \frac{2R}{sC} + sC2R} = \frac{sL}{sL3 + 2R + s^2C2RL} = \frac{\frac{L}{C2RL}}{s^2 + s \frac{3}{2}\frac{1}{CR} + \frac{2R}{C2RL}}
    = \frac{s\frac{1}{2CR}}{s^2+s\frac{3}{2}\frac{1}{CR}+\frac{1}{CL}}" />

    As one can see, the center-frequency gain indeed has been halfed - but despite theory about natural modes \frac{\omega_0}{Q} has been altered.

    (Problem 16.37 from Microelectronic Circuits, 6th edition.)" alt="s^2 +s\frac{1}{(2R||2R)C}+\frac{1}{LC}.

    But by calculation
    T(s) = \frac{\frac{1}{\frac{1}{R} + \frac{1}{sL} + sC}}{2R + \frac{1}{\frac{1}{R} + \frac{1}{sL} + sC}} = \frac{1}{1 + 2R({\frac{1}{R} + \frac{1}{sL} + sC})} = \frac{1}{1 + 2 + \frac{2R}{sC} + sC2R} = \frac{sL}{sL3 + 2R + s^2C2RL} = \frac{\frac{L}{C2RL}}{s^2 + s \frac{3}{2}\frac{1}{CR} + \frac{2R}{C2RL}}<br />
= \frac{s\frac{1}{2CR}}{s^2+s\frac{3}{2}\frac{1}{CR}+\frac{1}{CL}}

    As one can see, the center-frequency gain indeed has been halfed - but despite theory about natural modes \frac{\omega_0}{Q} has been altered.

    (Problem 16.37 from Microelectronic Circuits, 6th edition.)" />
     
  2. Starhowl

    Thread Starter New Member

    Dec 3, 2013
    25
    0
    Whoooooops, I'm so sorry, I didn't see the R instead of 2R in my calculation - everything is fine! Thread can be closed! Or deleted.. Whatever. Can't I do this?
     
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