Complex RLC circuit impedance and resonance?

Discussion in 'Homework Help' started by electronice123, Sep 15, 2014.

  1. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    I am trying to understand how to calculate impedance and resonance in a complex RLC circuit.
    The circuit (attached) consists of: A 100V RMS AC source, 1.3H inductor, 1k resistor, and a 250pF cap in series, In parallel with the 1.3H inductor and 1k resistor is a 44.2pF cap.

    I am trying to calculate and understand how the impedances and resonance work in this circuit.
    The calculated Fres is 8,828Hz, but multisim shows the Fres is 8,032Hz (going by Vpk across 250pF cap)
    XL=65,606
    R1=1K
    XC(C1)=448,305
    XC(C2)=79,250

    It seems to me that the parallel capacitance is shifting the resonance, but I can't seem to get the numbers right. The series capacitance has an XC of 79,250, so I would think at resonance the parallel RLC portion of the circuit should have the same impedance but opposite in phase.

    The problem is, the parallel RLC portion of the circuit at 8,032 Hz is capacitive, with a phase angle of -89.85 degrees and a impedance of 56,732 ohms. Here's how I calculated it.
    First I took the sum of the XL and R1, then combined it with the XC of C2 using the standard formula for parallel impedances (1/ 1/Z + 1/Z =Zp) = 56,732 ohms.
    I calculated the phase angle using: tan^-1((XL-XC)/R) and came up with -89.85 degrees

    Can anyone explain what I am doing wrong here?
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    What is the source of your circuit?

    on edit ...

    Can you link to the homework assignment?
     
    Last edited: Sep 15, 2014
  3. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello,

    The first thing is that the total impedance of two parallel impedances is not exactly what you typed there, it is:
    Zp=1/(1/Z1+1/Z2)=Z1*Z2/(Z1+Z2)

    The second form above is harder to mess up when actually doing the calculation.

    Second and more importantly, you should really learn to combine complex impedances using complex algebra if you dont already. It is almost as simple as combining resistors except we use complex numbers instead of regular numbers, and for the inductor we use s*L and for the capacitor we use 1/(s*C) and then later replace all the 's' with 'j*w' and then simplify. The result is one complex number which is the complex impedance in w, and for any w will be just one complex number.

    For this network it appears that there are two critical frequencies:
    f1=8138Hz
    f2=20996Hz

    The total impedance is either highest or lowest at these two points.
    The impedance at f1 is:
    1384.828853273425-7.366151421609984*j
    and gets higher as we move away so f1 gives us a local min.
     
    electronice123 likes this.
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As we have seen discussed elsewhere it's a matter of how one defines resonance.
    Mr. Al's f1 is the natural frequency of the circuit.
    His f2 is the value at which source voltage & current are in phase
     
    Last edited: Sep 15, 2014
  5. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    Thanks everyone for the replies, especially MrAl.
    I guess I need to go back and understand complex numbers. That's what I will do. Thank you.
     
  6. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    So I have one more question about the above complex rlc circuit. ... Is there an actual formula to calculate resonance?
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Each circuit requires it's own separate analysis. To come up with the formula, you analyze the circuit for current or voltage and then find the point where you see the type of resonance you want to see.

    For a parallel RLC where L and C are ideal and R is just in parallel with L and C, the resonance is always w=1/sqrt(LC), but once series resistance enters the picture for either L or C or both, we must calculate the resonant frequencies for that particular circuit. This is also true if the circuit is totally different than the typical parallel circuit as your circuit is.

    I can show you how to do this analysis if you understand math with complex numbers.
     
  9. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    That would be great.
    I studied complex numbers and they don't seem too difficult so I think I'll understand. I just want to know how to calculate the series resonance freq L1/C2 which changes because of C1 & R1, do both C1 and R1 change it?
     
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Have a look here:
    http://en.wikipedia.org/wiki/RLC_circuit
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    C1 connects to the top of C2, and the other side of C1 connects to a voltage source that has impedance of zero, and the bottom of the voltage source connects to the bottom of C2, so impedance wise C1 is in parallel with C2. This means the resonant frequency should go down because the sum of the two caps is greater than C2 and a larger value cap produces a lower resonant frequency.

    Without C1 we would have a series circuit:
    w=1/sqrt(L1*C2)

    With C1 if we only consider it in parallel with C2 we get:
    w=1/sqrt(L1*(C2+C1))

    so you see C1 changes the resonant frequency, and it will change it by quite a bit unless it is very small compared to C2.
    In fact, without C1 we get around 8.8kHz, and with C1 we get around 8.1kHz.
    So C1 definitely changes the resonant frequency.

    R1 probably changes it to a lesser effect, but since it will appear in a formula for the resonant frequency it will change it. How much it changes it depends on the other component values too. To find out exactly, we would do an analysis of the circuit and find at least one of the resonant frequencies and see how much R1 changes that frequency.

    To start with, the impedance of the circuit is:
    Zs=(s*C2*R1+s*C1*R1+s^2*C2*L1+s^2*C1*L1+1)/(s*C2*(s*C1*R1+s^2*C1*L1+1))

    This equation comes from a general circuit analysis where we simply combine all the impedances to get the impedance presented to the voltage source. It contains all the elements, and the variable 's' which we will later change to "j*w" in order to solve for the resonant frequency. Here w is the angular frequency and 'j' is the imaginary operator.

    If you dont know how to get that equation yet then we'd have to go over that first. If you do, then we could proceed. When we proceed we will do the very simple calculation where we replace every 's' with "j*w" and then solve for various things.

    If this sounds like what you would like to we can go ahead and do that next.
     
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  12. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    This circuit will actually have two resonant frequencies, one series and one parallel. In either case, only one of the C's is relevant.
     
  13. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    Yes please keep going. Your explanation so far is really good.

     
  14. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Ok starting with the equation for Zs we can expand the top and bottom and get:
    Zs=(s*C2*R1+s*C1*R1+s^2*C2*L1+s^2*C1*L1+1)/(s^2*C1*C2*R1+s^3*C1*C2*L1+s*C2)

    So the numerator and denominator are:
    Ns=s*C2*R1+s*C1*R1+s^2*C2*L1+s^2*C1*L1+1
    Ds=s^2*C1*C2*R1+s^3*C1*C2*L1+s*C2

    Next, we replace all 's' in Ns and Ds with "j*w" and get:
    Njw=j*w*C2*R1+j*w*C1*R1+j^2*w^2*C2*L1+j^2*w^2*C1*L1+1
    Djw=j^2*w^2*C1*C2*R1+j^3*w^3*C1*C2*L1+j*w*C2

    Now using the rules for exponents of the imaginary operator j we replace every power of j with the equivalent nad get:
    Njw=j*w*C2*R1+j*w*C1*R1-w^2*C2*L1-w^2*C1*L1+1
    Djw=-w^2*C1*C2*R1-j*w^3*C1*C2*L1+j*w*C2

    Now factor both into real and imaginary parts:
    Njw=-w^2*C2*L1-w^2*C1*L1+1+(w*C2*R1+w*C1*R1)*j
    Djw=-w^2*C1*C2*R1+(w*C2-w^3*C1*C2*L1)*j

    which can be written as:
    Njw=a+b*j
    Djw=c+d*j

    Now what we want is the imaginary part, because we will set that equal to zero to find one type of resonant frequency, the one where the imaginary part of the response or impedance is equal to zero. There are a couple ways to get this, but in formulaic form we have:
    imagpart(Njw/Djw)=(b*c-a*d)/(d^2+c^2)

    Note that is the imaginary part only. Now we use the parts from Njw and Djw and insert them into that formula numerator and we get:
    N=-w^2*C1*C2*R1*(w*C2*R1+w*C1*R1)-(-w^2*C2*L1-w^2*C1*L1+1)*(w*C2-w^3*C1*C2*L1)
    D=(d^2+c^2)

    Next, we set N/D=0 and then multiply both sides by D so we get:
    -w^2*C1*C2*R1*(w*C2*R1+w*C1*R1)-(-w^2*C2*L1-w^2*C1*L1+1)*(w*C2-w^3*C1*C2*L1)=0

    Noting we can factor out one w to begin with, we get:
    -w^2*C1*C2^2*R1^2-w^2*C1^2*C2*R1^2-w^4*C1*C2^2*L1^2-w^4*C1^2*C2*L1^2+w^2*C2^2*L1+2*w^2*
    C1*C2*L1-C2=0

    We now note that this is a fourth degree equation in w, but is just a quadratic in w^2, so we solve for W=w^2 and insert all the component values and get:
    -2.19761516*10^-20*W^2+4.3990699636000005*10^-10*W-1=0

    Solving for W, we get two solutions:
    W1=2.6147571329917808*10^9
    W2=1.7402714732282398*10^10

    Taking the square root we get:
    w1=51134.69598024203
    w2=131919.3493475555

    and dividing both by 2*pi we get:
    f1=8138.33962875679
    f2=20995.61653813006

    If you wanted to you could solve the fourth degree equation above symbolically and that would give you an explicit formula, but it's more work and comes out somewhat long:


    w1^2=
    -(sqrt((C1^2*C2^2+2*C1^3*C2+C1^4)*R1^4+(-2*C1*C2^2-6*C1^2*C2-4*C1^3)*L1*R1^2+C2^2*L1^2)+
    (C1*C2+C1^2)*R1^2+(-C2-2*C1)*L1)/((2*C1*C2+2*C1^2)*L1^2)

    w2^2=
    (sqrt((C1^2*C2^2+2*C1^3*C2+C1^4)*R1^4+(-2*C1*C2^2-6*C1^2*C2-4*C1^3)*L1*R1^2+C2^2*L1^2)+
    (-C1*C2-C1^2)*R1^2+(C2+2*C1)*L1)/((2*C1*C2+2*C1^2)*L1^2)

    There is another type of resonant frequency where the response peaks or dips, that is harder to solve. Instead of setting the imaginary part to zero we set the first derivative equal to zero, and this will give us two slightly different frequencies. In this circuit they are probably almost the same, just not exactly the same.
     
    Last edited: Sep 23, 2014
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  15. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
    302
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    Amazing explanation. Thank you so much, is exactly what I needed. Thanks
     
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