time constant vs frequency

Discussion in 'Homework Help' started by chaosmc, Aug 4, 2012.

  1. chaosmc

    Thread Starter New Member

    May 21, 2012
    17
    0
    In an underdamped response, with time constant for example τ=L/R and frequency that the oscillation happens is ω1 what is the relation between τ and ω1 ??

    In other words, increasing/decreasing the frequency of a transient, what behavior the attenuation of this transient will have?

    And to show my question in a formula:

    what is the relation between τ and ω1 if the transient has this form:

    e^(-t/τ)*sin(ω1*t)


    any suggestions will be appreciated..thanks!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Consider the series RLC circuit in under-damped mode - say in response to a voltage step input E.

    One can show that the current i(t) is given by

    i(t)=\frac{E}{\omega_d L}e^{-(\alpha t)}\sin{(\omega_d t)}

    where factor α is given by

    \alpha=\frac{R}{2L}

    and the damped frequency ωd is

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

    with the natural frequency ωo given by

    \omega_o=\frac{1}{\sqrt{LC}}

    Another term - usually referred to as damping factor ζ is defined by

    \zeta=\frac{\alpha}{\omega_o}=\frac{R}{2} \sqrt{ \frac{C}{L}}

    A damping factor ζ of 1 corresponds to a critically damped condition - at which point the response is no longer under-damped. For the series RLC circuit the critical damping condition is reached where R=2√(L/C).

    To change the damped frequency say, one may vary any of the elements R, L or C provided the damping factor ζ is kept below critical damping value of 1. You can play with the values to see what happens. For instance doubling the value of L will reduce the damping factor by a ratio of 1/√2. But doubling L also changes the natural frequency ωo and hence the value of the damped frequency ωd. Try substituting some simple numerical values to gain some understanding.

    For instance start with L=1H, C=4F and R=0.5Ω to find the response values then increase L to 2H to see what changes.
     
  3. mlog

    Member

    Feb 11, 2012
    276
    36
    T n k, you did a good job. I might add one item, which is simply an algebraic manipulation of one of your equations. It shows the relationship between the resonant frequency and the time constant for this 2nd order response.


    2\alpha=2\zeta\omega_{o}=1/\tau=R/L

    Is it appropriate to use the term "time constant" when talking of a 2nd order like this? I usually think of time constant \tau when working with a 1st order.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,736
    4,789
    I'd say yes, as long as it's understood (or stated) that the time constant is for the envelope.
     
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