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Determining r and c values by use of bode plot

  1. Dec 17, 2004 #1

    I have got difficulties with the following:a simple RC circuit (scheme of the circuit and measurements attached).
    I measured out the bode plot and now I have to find the values of R and C ,but I don't know how.
    Can someone help me please.

    Attached Files:

  2. jcsd
  3. Dec 17, 2004 #2


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    Staff Emeritus

    There are two things you need: The bode plot and the circuit. From the circuit you can derive the transfer function using the laplace domain based on the resistors and capacitors. From the bode plot you can do the inverse bode and figure out what the actual values are of the transfer function. You'll need to play around with the bode equation that you get so it matches the transfer function you derived from the circuit.

    If you don't know how to do the inverse bode, there are two things you need to know: the slope of the line in dB and the point the lines crosses the zero. From the slope you'll know if you have repeated roots and if the term is in the numerator or denominator. From the crossover point you'll know the denominator of the omega term.
  4. Dec 17, 2004 #3

    Sorry could not open your document -- if by Bode plot you mean the loss curve of the RC circuit -- this cannot tell you the values of r or c
    only their impedance ratio at some frequency.
    If you know one then you can find the other.
    A significant point which requires no calculus is the 3Db down point i.e. where the output voltage has fallen by sqrt(2) , at this point the r,c impedances are equal i.e. where R = 1/ (C.w ) w=2.pi.f

    Hope this helps Ray.
    Last edited: Dec 17, 2004
  5. Dec 19, 2004 #4
    you have to download the file and unzip it than you can open it.
    I've found the transfer function,but I've got to find two values out of it. (R1 is knowk,R2 and C1 aren't)
    I am unable to find the inverse function.
    Can you give me some more info please?

    thank's very much
  6. Dec 20, 2004 #5
    What is the theoretical definition of the phase shift?
  7. Dec 20, 2004 #6

    Phase applies to the time relationship between two wave forms which are both repetitions of some basic shape over a unit period T.
    These waveforms do not have to have the same shape -- but to have a constant phase -- or time relation they must have either the same frequency ( cycles (of the shape) per second OR be in an integer relation.
    The simplest situation is two sinewaves of the SAME frequency but which are shifted with respect to each other in time, after shifting by one wavelength they will again be in phase -- so phase 0 to 2.pi is equivalent to one basic period T.
    This definition can perhaps be best seen with a rotating vector or rod
    --- if I take a rod fixed at one end and rotate it in a circle -- then if I plot the vertical (y) height against the angle (x) in a graph I will generate a sinewave.
    But I can do the same with a second rod of equal or different length and generate a second sinewave .
    If now I rotate them both togeter -- BUT at different starting angles

    then they both generate sinewaves but are clearly shifted along the (x) axis
    by some angle (0-2.pi) which IS the phase.

    in maths we say v1 = A. sine (w.t) where w = 2.pi.f f=Hertz
    v2 = B sine (w.t + b) t = time
    the phase difference is b ( a constant )
    from this it can be seen that if b=2.pi there is no difference in the time relation.

    By the way I had tried to unzip your image but it would not open correctly.
    If other people had the same problem try a compressed small Jpeg image
    at 50 dpi or less.


    Attached Files:

    Last edited: Dec 20, 2004
  8. Dec 21, 2004 #7
    now you should be able to open the file
    I would really appreciate your help because I've got to find the solution very soon!

    Attached Files:

  9. Dec 21, 2004 #8
    by the way,R1=1000 ohm
  10. Dec 21, 2004 #9


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    Staff Emeritus

    I can tell you by looking at the bode plot that C*R1 = 1/1000 and (R1+R2)*C = 1/100

    I know this by doing the inverse bode of the amplitude. You can estimate a pole at 100 and a zero at 1000. That means the transfer function is this: H(jw) = (jw/1000+1)/(jw/100+1). Finally, from the equation you derived you can relate the values to the resistors and capacitor.
  11. Dec 21, 2004 #10
    Sorry Gotilio can download file but cannot open it in wordpad -- maybe my end problem
    or the diagram content ---- I found a small diagram in zipped BMP format ends up at 1Kb -- I dont need the plot just circuit and known values you can sketch the r's as boxes and the c's as aline pair.
  12. Dec 21, 2004 #11
    thanks very much!
    Is it possible that C*R2 = 1/1000 en not C*R1 = 1/1000 ?
  13. Dec 21, 2004 #12


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    Homework Helper

    No. dduardo gave the correct answer.

    Just to double check, the frequency on your bode plot is in rad/s right? If so then dduardo answer, should give you the right solution.

    If it's in Hz, then I believe that C*R1 = 1/(1000*2*pi) and (R1+R2)*C = 1/(100*2*pi)
  14. Dec 21, 2004 #13
    My equation for A was wrong ,the nominator should be 1+jWCR2,
    so in this case C*R2=1/1000,right?
    by the way the plot is in hertz
    If I solve it like this my values are :111 ohm for R1 and 9 microF for C
  15. Dec 21, 2004 #14


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    Oops sorry, the equations should be:

    C*R2 = 1/(1000*2*pi)
    (R1+R2)*C = 1/(100*2*pi)

    It seems like you solved without using 2pi. You are given R1=1000 right?

    If R1=1000 ohm, I get R2=111 ohm and C=1.43 microfarad.
  16. Dec 22, 2004 #15
    is the cut-off frequency equal to the frequency of the pole?
  17. Dec 22, 2004 #16

    Only a small diagram but it may help -- real frequency w = 2.pi.f
    runs up the vertical axis the vectors drawn go from that frequency point to each pole and represents the magnitude and phase of each pole in the transfer function shown.
    Ray This is a zipped BMP image open with paint or similar after unzipping.

    Attached Files:

    • try.zip
      File size:
      32.7 KB
  18. Dec 23, 2004 #17


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    Homework Helper

    Yes they're the same.
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