Bandwidth of Band Pass Filter

Discussion in 'General Electronics Chat' started by Mechatronical, Feb 18, 2014.

  1. Mechatronical

    Thread Starter Member

    Feb 15, 2014
    30
    0
    Last edited: Feb 18, 2014
  2. Mechatronical

    Thread Starter Member

    Feb 15, 2014
    30
    0
    It seems I've overlooked something.
    The equation states:
    S^2 + ω0/Q s0^2.
    If I solve this for b I get ω0/Q = (-ax^2-c) / x.
    (Where a is a value s^2, b = ω0/Q and c = ω0^2)

    However, I'm not sure if this is allowed.
     
  3. Veracohr

    Well-Known Member

    Jan 3, 2011
    551
    76
    You want to be solving for s, and the value of a is 1. B is given, it doesn't make sense to solve for it.
     
  4. Mechatronical

    Thread Starter Member

    Feb 15, 2014
    30
    0
    Let me guide you. This is the second grade equation:

    x = -b+-/sqrt(b^2 - 4ac) / 2a
     
  5. Mechatronical

    Thread Starter Member

    Feb 15, 2014
    30
    0
    Ok. So what if I calculate the value H(s) at a point s.
    When H(s) and s is given:

    H(s) = [ω0/Q s] / [s^2 + ω0/Q s + ω0^2]
    H(s) = [ω0 s] / [Qs^2 + ω0 s + Qω0^2]
    Qs^2 + ω0 s + Qω0^2 = [ω0s/H(s)]
    Q(s^2 + ω0^2) = [ω0s/H(s)] - ω0s
    Q = ([ω0s/H(s)] - ω0s) / (s^2 + ω0^2)
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Does this indicate you have answered your question or are you still seeking assistance?

    With respect to the relationship between Q and bandwidth one simply needs to keep in mind the point that the bandwidth will be twice the value of the real part of the complex roots of the transfer function 2nd order denominator.

    So if the complex roots are ...

    -\alpha \pm j\beta

    then the filter center (damped) frequency ωd would be given by

    \omega_d=\beta=\sqrt{\omega_o^2-\alpha^2}

    and bandwidth ...

    \text{\Delta\omega=2\alpha}

    Also in terms of Damping factor ζ & Quality factor Q

    \text{\alpha=\zeta\omega_o=\frac{\omega_o}{2Q}}
     
    Last edited: Feb 19, 2014
  7. LvW

    Active Member

    Jun 13, 2013
    674
    100
    Hi mechatronical,

    for a second-order bandpass function there is a simple relation between the quality factor Q=fo/B (fo=center frequency and B=3db bandwidth) and the pole location in the s-plane:
    The pole location can be described using the pole frequency wp (which is identical to wo=2*Pi*fo) and the pole quality factor Qp.

    Graphical interpretation:
    * wp is the length of the line between the pole (left half of the s-plane) and the origin;
    * Qp=1/[2*cos(phi)] with phi=angle between the wp line and the negative real axis. (Limitation: Double pole on the negative real axis with phi=0 and Qp=0.5)

    It can be shown that Qp=Q=fo/B.

    Edit:
    The last equation nicly shows how information from the complex frequency plane are related to real filter parameters which can be measured (bandwidth, pole frequency).
    And -of course, cos(phi)=σ/ωp with σ=real part of the pole.
     
    Last edited: Feb 19, 2014
Loading...