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Plotting Impedance of Parallel RLC Circuit

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Plot |z| vs. f(Hz) of the circuit. R=100, C=600 pf = 6E-12 F, L=10E-6 H . All elements of the circuit are in parallel. Also identify the resonant frequency.


    2. Relevant equations
    ZR=R
    ZL=jwL
    ZC=1/(jwC)=-j/(wC)
    w=2(pi)f
    w0=1/sqrt(LC)
    f0=1/(2(pi)sqrt(LC))
    G=1/R

    3. The attempt at a solution
    Z1=ZR
    Z2=ZL
    Z3=ZC
    Zeq= 1/((1/R)+1/(jwL)+jwc)

    Switching to admittance and skipping a few steps, it can be found that
    Y=G + j * 2 * pi * f * C * (1 - (f0 / f) ^ 2)
    Z=R + 1 / (j * 2 * pi * f * C * (1-(f0 / f) ^ 2))
    Z=100 + 1/(j * 2 * pi * f * 600E-12 *(1 - (2054681.48 / f) ^ 2)

    To plot, i used a few lines in matlab:

    f=linspace(0,4000000,100000);
    x=1i.*2.*pi.*(600.^(-12));
    z=100+(1./(x.*f.*(1-((2054681.48./f).^2))));
    plot=semilogx(z);

    The plot shows me a vertical line at f=100

    My professor told us that the graph should be bell shaped, and i have no idea where i've gone wrong. Anyone see a mistake?
     
  2. jcsd
  3. Mar 30, 2010 #2

    vela

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    For one thing, 1/(a+b) isn't equal to 1/a + 1/b, which is what you did when you went from the admittance back to the impedance.
     
  4. Mar 30, 2010 #3
    Ok, change it to this then:

    Z=1/(100 + j * 2 * pi * f * 600E-12 *(1 - (2054681.48 / f) ^ 2))

    The plot is still a vertical line.
     
  5. Mar 30, 2010 #4

    vela

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    Well, that's not quite right either. I'll leave it to you, however, to fix your algebra mistakes.

    I don't know Matlab, but I'm guessing you're letting f start from 0. Two problems with that. At f=0, you're dividing by zero in your expression. Also, when f=0, Z=0 since the inductor is a short, so the log of Z is undefined.
     
  6. Mar 31, 2010 #5

    The Electrician

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    In this expression:

    x=1i.*2.*pi.*(600.^(-12));
    ....^

    is that lower case i that I've pointed to supposed to represent SQRT(-1)?

    If it is, that's a problem. You can't plot expressions that have an imaginary component. You have to plot the magnitude, or modulus, of the impedance expression.

    Some mathematical packages have a command such as Abs[] that can do it. Or you can do it yourself by plotting SQRT(Z * Conjugate(Z)).
     
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