1. Not finding help here? Sign up for a free 30min 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!

Series RLC filter (trap filter)

  1. May 5, 2006 #1
    Hey! I need to solve the following problem:

    a) Show the trap filter below acts to reject signals at a frequency
    [tex] \omega = \frac{1}{\sqrt{LC}} [/tex]


    b) How does the width of the frequency band rejected depend on the resistance R?

    For a) I thought that at resonance (when

    [tex] \omega = \frac{1}{\sqrt{LC}} [/tex] )

    the reactance of L & C = 0 and therefore Z= R. In a series circuit the current is the same everywhere, but the voltage divides itself according to V= IR. Therefore the voltage on L & C is 0 (R=0) and the voltage on R= IR. But is that correct? I think Vout is measured on R and NOT L & C, but how would I know? Btw what is the function of that grounded symbol, what does it mean and is it relevant for this question?

    As for b) I know that
    [tex] Q= \frac{\omega_0}{\Delta \omega} [/tex]


    [tex] \Delta \omega = \frac{\omega_0}{Q} [/tex]


    [tex] Q= \frac{\omega_0 L}{R} [/tex]

    Substitution gives:

    [tex] \Delta \omega = \frac{\omega_0}{\frac{\omega_0 L}{R}} [/tex]

    [tex] =\frac{R}{L}[/tex]

    [tex] \Delta f= \frac{\Delta \omega}{2 \pi}[/tex]

    [tex] = \frac{\frac{R}{L}}{2 \pi} [/tex]

    [tex]= \frac{R}{2 \pi L}[/tex]

    Though my textbook says that

    [tex] \Delta \omega = \frac{R}{2L} [/tex]

    Could anybody please tell me what I'm doing wrong?!
    Last edited: May 5, 2006
  2. jcsd
  3. May 5, 2006 #2


    User Avatar

    Staff: Mentor

    How are they defining delta-omega? Are they maybe defining it as one side of the width of the stop band, and you are defining it as the full width?

    BTW, in your first part, since the reactive impedance of the series LC goes toward zero at resonance, think of the R as the top part of a voltage divider and the LC as the bottom part of the voltage divider. What do you get for the transfer function of a voltage divider as the bottom impedance goes towards zero?
  4. May 5, 2006 #3
    In my book you have this graph that shows the average power supplied by the generator to the series combination as a function of generator frequency. The resonance width (delta omega) is the frequency difference between the two points on the curve where the power is half its maximum value.... so I guess they are defining it as the full width?

    So... where did I go wrong in part b?

    Do you mean this equation:

    Well that one will equal 1, so Vout= Vin :grumpy: ?
    So the voltage on the resistor (the Vout) = the whole signal, so the whole signal gets transferred at resonance? But that doesn't match the question, I mean it is given that at resonance the whole signal is blocked....
    Last edited: May 5, 2006
  5. May 5, 2006 #4


    User Avatar

    Staff: Mentor

    The impedance of a parallel LC goes to infinity at resonance, and the impedance of a series LC goes to zero at resonance. So | Vo/Vi | goes to zero at the resonance of the circuit you show. The equation that you wrote is for when the LC is at the Vin spot, and the R is from Vout to ground.
  6. May 6, 2006 #5
    Ah I see, thanks!
    But what about this delta omega? Is my thinking correct and the answer in my book wrong or not?
  7. May 8, 2006 #6


    User Avatar

    Staff: Mentor

    Well, to check the delta-omega, you can write the equation for the input impedance at Vi, and solve for the two omega values where the input impedance goes to SQRT(2) of the minimum impedance (which is R). Subtract the two omegas to see if you get your previous answer or the book's answer.
  8. May 8, 2006 #7
    This is what I've done. Please tell me if I've done it okay, cause I need to deliver it in 2 hours....:

    V= IR, so Vin= IZ = I [tex] \sqrt{R^2 + (wl - \frac{1}{wc})^2} [/tex]

    Vin= IR [tex] \sqrt{1 + (\frac{wl - 1/wc}{R})^2} [/tex] so

    [tex]\frac{(wl - 1/wc)}{R}[/tex] = + or - 1 to give sqrt(2)

    [tex]\frac{wl}{R} - \frac{1}{Rwc}[/tex] = + or - 1

    [tex]\frac{w^2 l C}{RwC} - \frac{1}{Rwc} + or - \frac{RwC}{RwC} =0 [/tex]


    [tex]w^2 l C + or - RwC - 1 = 0 [/tex]

    using the abc formula
    a= LC
    b= + or - RC
    c= -1


    [tex] w = \frac{+ or - RC + or - \sqrt{R^2 C^2 + 4 LC}}{2LC} [/tex]


    [tex] w = \frac{-RC + or - \sqrt{R^2 C^2 + 4 LC}}{2LC} [/tex]


    [tex] w = \frac{+ RC + or - \sqrt{R^2 C^2 + 4 LC}}{2LC} [/tex]


    [tex] w= \frac{2 RC}{2 LC} [/tex]

    which is:

    [tex] w= \frac{R}{L} [/tex]

    So may I conclude that the book is wrong? Or did I make a mistake somewhere?!
    Last edited: May 8, 2006
  9. May 8, 2006 #8


    User Avatar

    Staff: Mentor

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?