1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Circuit with 2 resistors, inductor, and capacitor

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://img534.imageshack.us/img534/228/helps.png [Broken]

    I have trouble completing (b), as I do not know how to solve for the damping factor in this circuit, and hence cannot characterize this system. I have the poles of the system, where s ~= -3048.059 and s~= -8201.941. I also have the transfer function, which I used to find the poles.

    2. Relevant equations
    I found the transfer function to be:

    [tex]G(s) = \frac{1}{CLs^2 + \frac{L+R^2C}{R}s + 2}[/tex]

    3. The attempt at a solution
    I just don't know how to solve for zeta.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 12, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I suggest you use Laplace transforms to solve the differential equation for the damped harmonic oscillator and compare what you get there with G(s). Hopefully, you can then see how to solve for [itex]\zeta[/itex].
     
  4. Jun 12, 2010 #3
    I don't understand how I would apply that.

    [tex]\ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega_0^2 x = 0[/tex]

    After Laplace transform (assuming initial conditions are zero):

    [tex]X(s)s^2 + 2\zeta\omega_0X(s)s + \omega_0^2X(s)[/tex]

    Are you suggesting to equate this to the denominator? Your reply is ambiguous, I can't relate this to the original transfer function unless I use the reciprocal of it or some other method.
     
  5. Jun 12, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You actually want to look at the forced damped harmonic oscillator.

    [tex]
    \ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega_0^2 x = f(t)
    [/tex]

    Depending on what f(t) is, you get a different response x(t) from the system. In the s-domain, you'll find X(s)=H(s)F(s) for some function H(s), which is what I'm suggesting you find. With the circuit, ei plays the role of the forcing function, and the response of the system is eo. In the s-domain, you have Eo(s) = G(s)Ei(s).
     
  6. Jun 12, 2010 #5
    I understand that part, but I still don't know what you mean by applying that second order differential equation to this. Are you implying that I need to know the actual values for the forcing function to characterize this circuit?
     
  7. Jun 12, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No. I'm saying compare H(s) to G(s). The transfer function G(s) describes a forced damped harmonic oscillator, but it's not written in terms of the quantities [itex]\omega_0[/itex] and [itex]\zeta[/itex] that characterize the oscillator. By comparing H(s) and G(s), you can identify how the coefficients in G(s) relate to those quantities.
     
  8. Jun 12, 2010 #7
    Hmm ok, you edited the previous post.

    H(s) = G(s)
    Then:
    [tex]\frac{X(s)}{F(s)} = \frac{1}{CLs^2 + \frac{L+R^2C}{R}s + 2}[/tex]
    [tex]\frac{s^2 + 2\zeta\omega_0s + \omega_0^2}{F(s)} = \frac{1}{CLs^2 + \frac{L+R^2C}{R}s + 2}[/tex]

    But I'm left with the F(s) which I cannot do anything with?

    I found the poles to both be unique, real roots, and thus characterized the system as overdamped. I would still like to find zeta regardless, though.
     
  9. Jun 12, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I didn't say set them equal to each other, and X(s) isn't equal to that polynomial.

    First, take the Laplace transform of

    [tex]\ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega_0^2 x = f(t)[/tex]

    to find the oscillator's transfer function, H(s)=X(s)/F(s).
     
  10. Jun 12, 2010 #9
    Sorry, I typed that wrong.

    [tex]X(s)s^2 + 2\zeta\omega_0X(s)s + \omega_0^2X(s)=F(s)[/tex]
    [tex]H(S) = \frac{X(S)}{F(S)} = \frac{1}{s^2 + 2\zeta\omega_0s + \omega_0^2}[/tex]
     
  11. Jun 13, 2010 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Now, if you pull a factor of 1/LC out front, you have

    [tex]G(s) = \frac{1}{LC} \left[\frac{1}{s^2+(L/R+RC)s+2/LC}\right][/tex]

    The factor of 1/LC in front is just a constant; it only affects the overall gain of the circuit. Comparing what's in the square bracket and what you found for H(s), you should be able to identify what [itex]\omega_0^2[/itex] and [itex]2\zeta\omega_0[/itex] are equal to and then solve for [itex]\zeta[/itex].
     
  12. Jun 13, 2010 #11
    Okay, that's initially what I thought I had to do when I read this question. The constant in the denominator (the 2) was the part that I was confused about. I realize now that the numerator, when constant, will have no effect on the denominator when finding a value for [itex]\omega_0[/itex]. Thanks for your help.
     
  13. Jun 13, 2010 #12
    I believe your s term is wrong, it seems that it should be the reciprocal of that.

    [tex]\frac{1}{RC} + \frac{R}{L}[/tex]

    Just in case anybody else sees this post.
     
  14. Jun 13, 2010 #13

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I just started from what you had posted for G(s) earlier, assuming you had gotten it right, but yes, you're right otherwise the units don't work out correctly.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook