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: On determining the -3dB point in an RLC circuit

  1. Oct 4, 2014 #1
    1. The problem statement, all variables and given/known data
    OK, I have the response function for a notch filter, and I want to find out what the frequency where it has the -3dB response.

    2. Relevant equations
    [tex]H(\omega) = \frac{1-\omega^2LC}{1-\omega^2LC+i\omega RC}[/tex]

    This is my response function. My [itex]\omega_0 = \frac{1}{\sqrt{LC}}[/itex]

    3. The attempt at a solution
    OK, so I figure that what I am looking for is where [itex]20\log_{10}(H(\omega)) = -3[/itex]. This should be simple enough, but I am having some difficulty and I suspect I am messing up something stupid and obvious.

    I am pretty sure that all I need is to have [tex]H(\omega) = \frac{1-\omega^2LC}{1-\omega^2LC+i\omega RC} = \frac{1}{\sqrt{2}}[/tex]. But it's getting there that's the problem and figuring out what my omega ought to be. I tried simply saying that the absolute value of my response function squared should be 1/2, and thereby getting rid of the imaginary component. This is kind of a silly algebra question I suppose, I feel like there is some ridiculously simple thing I am not seeing here.

    Looking at it I need something where plugging in [itex]i \omega RC = i \frac{1}{\sqrt{LC}} RC[/itex] yields a square root of 2, or plugging in R/L as my change in omega and multiplying things out, but I feel like I am missing something even simpler than that. Anyhow, any help is appreciated.
  2. jcsd
  3. Oct 4, 2014 #2


    User Avatar

    Staff: Mentor

    You were on the right track before. It's the magnitude of the transfer function that you want to set to ##1/\sqrt{2}##. So first you'll need to deal with obtaining the magnitude expression.

    If it's a notch filter you should expect to find more than one value of ##\omega## that yields a -3 dB gain.
  4. Oct 4, 2014 #3
    OK, and it should be no problem if I get imaginary numbers for omega, though I am still having a little trouble solving the expression. My instinct is to plug in omega +/- R/L and see what that gets me. Is that the right Idea? (delta-omega is R/L IIRC)
  5. Oct 4, 2014 #4


    User Avatar

    Staff: Mentor

    I think you you find two real roots to the resulting expression (a notch has two sides after all). There may be another pair of imaginary roots that can be ignored. Were you given actual values for the components?

    You'll probably find a quartic expression with ##\omega^4## and ##\omega^2## terms. A substitution will make that a quadratic.

    Unless you know something in particular about the notch filter in question, plugging in likely values based on combinations of components is unlikely to be fruitful.
  6. Oct 4, 2014 #5
    no actual values, i think we're just supposed to say what the omega is for the -3dB spots.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted