rlc circuit bandwidth? is this correct?

Discussion in 'Homework Help' started by ninjaman, Dec 13, 2014.

  1. ninjaman

    Thread Starter Member

    May 18, 2013

    I have simulated a rlc series circuit with values, 4.7k ohm, 151 mH, 1uF.
    I have set this up on multisim and used AC analysis to show what I think is the bandwidth.
    hopefully you should be able to see this clearly. I have 10vpk, making 20vpk-pk. I used 20 * 0.707 to get the -3db, half power point, giving 14.14 volts. I set the cursors at this value either end to show what I believe is the bandwidth.

    is this correct?
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
    Since the Bandwidth is measured between the -3db points, it is convenient to plot the response in db relative to the maximum which occurs at resonance. LTSpice makes this convenient. Note that by rearranging the network, the voltage across the resistor is proportional to the current, so we can plot either voltage or current. Since 10V is 20db, the half power points are at 17 db. Notice the phase of V(r) at the -3db points. LTSpice calculates the bandwidth as the Δf between cursors.


    Note that by rearranging the network again, the true usefulness of the RLC network is revealed. Look at the plot of V(out) vs frequency.

    Last edited: Dec 13, 2014
  3. ninjaman

    Thread Starter Member

    May 18, 2013

    this is a little confusing, I just wanted to know if I was on the right track. im still learning. going through things like decibels and LTspice is more than I want. I have tried LTspice and I know a lot of people on here like it but I have never been able to make it work. I have noticed that the values for frequency are the same. so im guessing that what I have done is ok. thanks for that!

  4. MrAl

    Distinguished Member

    Jun 17, 2014

    Mike suggested you use "db" because that is how these plots are usually made.
    db is simply the base ten log of the amplitude times 20, so we have:
    So you take the base 10 log then multiply the result by 20 and you get the amplitude in "db".

    If you dont want to use db then you need to find the points where the amplitude is 1/sqrt(2) down from the maximum normal output (not always from the peak, but from the passband gain). Since 1/sqrt(2) is simply 0.7071 you need to find the points where the gain is about 71 percent of what it is at the maximum point of the amplitude. If your max is 10 volts for example then the two points would be found at 7.1 volts, one to the right and one to the left of the 10 volt max, and this provides one high and one lower frequency. The bandwidth is then the higher frequency minus the lower frequency. If you have done this then you're ok :)
    Looking at your post again it looks like you did this ok.
    If you were to plot in db however you might see the two points more clearly.
    Just one question, why did you take the output from across the resistor (if that is what you really did) ?
    Last edited: Dec 13, 2014
  5. ninjaman

    Thread Starter Member

    May 18, 2013



    oh wait.....I have a few more things to figure out!?!

    oh no.....