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!

RC Circuit Bode Plot Help

  1. Dec 1, 2014 #1
    1. The problem statement, all variables and given/known data
    Make a blode plot of a series RC high pass filter (log(gain) vs. f)...

    2. Relevant equations

    log(Gain)=log(Vout/Vin)
    I'm pretty sure the "Gain" is just the transfer function, for a highpass it should be the V_in = Vsupply, and V_out = V_capacitor.

    3. The attempt at a solution

    I have an RC circuit hooked up and I'm going to make the bode plot taking data from the oscilloscope, I'm just going to vary the frequency and collect voltages, but I'm not sure how to go about doing it because of the phase difference between the input voltages (Ch. 2 blue line) and the output voltage (over the capacitor, Ch.1 yellow line).

    Can I just measure peak-to-peak? I'm not 100% sure if that's correct because of the phase difference between peaks. Do I need to take voltages over the same point on the "time" axis or is peak-to-peak ok even if at different points with respect to time axis?
     

    Attached Files:

  2. jcsd
  3. Dec 1, 2014 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    There are two kinds of Bode plots - |gain| and phase. If all you need is the gain, pick the amplitudes of input & output voltages and disregard the phase shift.
     
  4. Dec 1, 2014 #3

    gneill

    User Avatar

    Staff: Mentor

    Peak-to-peak voltages will work fine. You can always convert them to peak values (divide by 2).

    You might want to think about where you take the Vout measurement. That is, what's the placement of the resistor and capacitor for a high-pass filter? It's essentially a voltage divider. How does the impedance of a capacitor vary with frequency?

    A Bode plot generally displays the power gain in dB. When you work with voltage ratios (Vout/Vin) take 20 Log(Vout/Vin) to get the equivalent power gain in dB.
     
  5. Dec 1, 2014 #4
    Actually what I have hooked up is a low pass filter, I made a mistake there... I'm asked to make a high pass and a low pass and put them in series to see what happens. But the circuit I have now is just the low pass, ...a resistor and capacitor in series, it's the capacitor connected to the low end (ground) so this should be low pass.

    So now I need to measure the voltages over the resistor and the supply (entire circuit). So in my excel spreadsheet I should just make a column of "V_supply - V_capacitor" to have my " V_resistor" and use that as my V_out when I make the Bode plot for the low pass.

    For a high pass I'll swap the capacitor and resistor and measure the voltage over the capacitor.
     
  6. Dec 1, 2014 #5
    Also I think I see why it doesn't matter if the peak values are at different "times" because the transfer function is a function of "s" which is a complex variable dependent on frequency, so there is no time dependency.
     
  7. Dec 2, 2014 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

    No, Bode gain plots are typically voltage gain plots with dB = 20log10(Vout/Vin).
     
  8. Dec 2, 2014 #7

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You could have used a voltmeter in lieu of your oscilloscope in which case you would not have been aware of time at all. Time has nothing to do with the gain measurement.

    Just to be sure: your Vin is hooked up to the generator high and low sides and your 'scope (Vout) to the capacitor high and low (low being ground) for the lowpass. Yes, for the high-pass you swap the R and C.
     
  9. Dec 2, 2014 #8

    gneill

    User Avatar

    Staff: Mentor

    There are 10 decibels in a bel.

    $$dB = 10 log_{10}\left( \left(\frac{V_{out}}{V_{in}}\right)^2\right) = 20 log_{10}\left( \frac{V_{out}}{V_{in}}\right)$$
     
  10. Dec 2, 2014 #9

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Thanks g I had that part figured out already :)
    A Bode plot graphs 20log(Vout/Vin) = 20log (voltage gain).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: RC Circuit Bode Plot Help
  1. Bode Plot (Replies: 5)

Loading...