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Mathematical Model of a Circuit.

  1. Apr 23, 2005 #1
    Before i start i may point a few things:

    I don't know if this is college level in other places, but around here it is.
    And, English is not my Native language, so bear with my mistakes...

    Anyway, my class is "Mathematical Models" AKA Control Systems. It is part of physics and math, anyways, while they give me a circuit kinda looking like THIS:

    http://img94.echo.cx/img94/7059/circuito0gc.png

    Well, that's the circuit. It is a Direct Current circuit, and i must find the "Mathematical Model" of this circuit,


    As you can see, Entrada (Input) is V(t) and the output is VR(1). While i know acording Kirkchof laws,

    V(t) = VR(1) + VR(2) or

    V(t) = VR1 + VC

    And in this case, the R1 and R2 Resistance are Equal, so R1 = R2 = R.

    In order to write a math model i only need to express it on V(t) and VR(1) (Meaning, just the Output and Input) so i need to express VR(2) or VC in terms of VR1 ONLY and not V(t) cause that doesn't make sense.

    This is when i go ballistic. While if the output is V(C) or VR(2) is piece of cake, here i cannot do nothing. All i can remember and investigate is that

    IR1 = IR2 + IC

    VR1/R = VR2/R + C . dVC/dt (Derivate) and while is in this position, i cannot put it into terms of VR1.

    I KNOW i am missing something, but it was a long time ago when i saw physics... please, you are my last line of defense, i tried all my other exercises and this one is bothering me.

    The meaning of doing that is finding the "Transference Function" i need to...

    Laplace Output (In this circuit pic, Salida :D )
    ---------------
    Laplace Input (In this circuit, Entrada)


    Thx beforehand, S'-
     
  2. jcsd
  3. Apr 23, 2005 #2

    Andrew Mason

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    If Vt is constant (DC circuit), the current through the capacitor is 0 so it is just:

    It = Vt/(R1+R2)

    VR1 = ItR1

    VR2 = ItR2

    VC = VR2

    AM
     
  4. Apr 23, 2005 #3

    OlderDan

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    You are talking about an input voltage that varies with time in a DC circuit. I can only assume that what you are talking about is a consant voltage that is turned on at some time we can call t = 0 and looking for the response of the circuit from that time forward. This is a classic problem in circuit theory, so I'll bet this is it.

    Let E be the magnitude of the applied voltage at t = 0

    At t = 0 the capacitor will have been fully discharged through R2. Now the circuit sees the constant applied voltage E, but the potential difference accross the capacitor is proportional to the charge on the capacitor. It will take time for that charge to build up. The voltage on the capacitor will reach a maximum when the current through R1 is the same as the current in R2, which will be E/2 if R1 = R2. Between t = 0 and any later time, the potential on the capacitor will approach E/2 asymptotically.

    The mathematical description of this is a differenctial equation that includes the current flowing into the capacitor as dq/dt where q is tha capacitor's charge. The voltage on the capacitor is qC. Can you take it from there?
     
  5. Apr 23, 2005 #4
    Thanks for the attention.

    Actually i kinda know the theory. What i really need is an equation that is expresed as

    V(t) = THE DIFERENCIAL EQUATION

    The only variables that could go there is the Output, as VR1 on this case.

    THis is where i can't advance. The idea is to make a general equation, and i really take it from there.

    What equations can i get from Kirkchof laws?

    V(t) = VR1(t) + VC(t)
    IR = IR + IC ................(1)
    VR2 = VC

    If the output where in case, VC(t) the problem is easily solved.

    From equation (1)

    VR1/R = VR2/R + C . dVC/dt

    VR1/R = VC/R + C . dVC/dt

    VR1 = VC + RC . dVC/dt

    Then i replace.

    V(T) = VC + RC . dVC/dt + VC =>

    V(T) = 2VC + RC . dVC/dt

    As you see, this is the math model from the same circuit but the output is really VC. If it's VR2 it's easy using the same equations. Maybe i am missing some equations i cannot remember? I just need the equation...
     
  6. Apr 23, 2005 #5

    OlderDan

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    How's this? Be sure to check my work, but I think this is it

    V(t) = V1(t) + V2(t) = I(t)R1 + Vc(t)
    I(t) = I1(t) = I2(t) + Ic(t)
    I2(t) = Vc(t)/R2
    Ic(t) = dq(t)/dt = CdVc(t)/dt
    I(t) = Vc(t)/R2 + CdVc(t)/dt
    V(t) = R1*[Vc(t)/R2 + CdVc(t)/dt] + Vc(t)
    V(t) = Vc(t)*(R1+R2)/R2 + (R1*C)dVc(t)/dt

    This is the same as yours for R1 = R2 = R

    This is all you can do until you put in the driving voltage V(t). Then solve the DE for Vc(t)
     
    Last edited: Apr 23, 2005
  7. Apr 23, 2005 #6
    :smile: Thanks for your help, but it isn't enough...

    All those equations i knew them, all of them, and i can't really put the general V(t) equation in terms of VR1 or V1 as you put it only... all those equations i tried them, played with them and i cannot see another else... :uhh:

    Well, i am going to keep researching till i get this done, and if you remember something up please post!
     
  8. Apr 24, 2005 #7

    OlderDan

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    There is no avoiding the dependence of V1(t) on Vc(t). If that is what you are trying to do, it will not happen.

    V(t) = V1(t) + Vc(t)
    V1(t) = V(t) - Vc(t)
    dV1(t)/dt = dV(t)/dt - dVc(t)/dt

    If you know the driving function V(t) then you can find the first term, but you are stuck with the dVc(t)/dt term. You can also replace dV(t)/dt by taking the derivative of the result already obtained.

    V(t) = Vc(t)*(R1+R2)/R2 + (R1*C)dVc(t)/dt
    dV(t)/dt = [(R1+R2)/R2]*dVc(t)/dt + (R1*C)d^2Vc(t)/dt^2

    dV1(t)/dt = [(R1+R2)/R2]*dVc(t)/dt - (R1*C)d^2Vc(t)/dt^2 - dVc(t)/dt
    dV1(t)/dt = [R1/R2]*dVc(t)/dt - (R1*C)d^2Vc(t)/dt^2
     
  9. Apr 24, 2005 #8

    Andrew Mason

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    I think Dan is right.

    [tex]V(t) = V_c(t)\frac{(R1+R2)}{R2} + (R1*C)\frac{dV_c(t)}{dt}[/tex]

    If R1=R2:

    [tex]V(t) = 2V_c(t) + (RC)\frac{dV_c(t)}{dt}[/tex]

    [tex]\frac{dV_c(t)}{dt} + \frac{2}{RC}V_c(t) = \frac{V(t)}{RC}[/tex]

    You have to solve that for V_c. I get:

    [tex]V_c = \frac{V(t)}{2}(1-e^{-2t/RC}) - \frac{1}{RCe^{2t/RC}}\int_0^t (V'(t)\frac{RC}{2}(e^{2t/RC}-1)dt[/tex]

    If V(t) is constant = V:

    [tex]V_c = \frac{V}{2}(1 - e^{-2t/RC})[/tex]

    AM
     
  10. Apr 24, 2005 #9
    Yes, i cannot find the way of killing Vc.

    Tomorrow if i can i will try to solve this in a test, and one way or another i will bring the answer here (Maybe i am not expressing my problem correctly)

    Thanks guys.

    S'
     
  11. Apr 24, 2005 #10

    Andrew Mason

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    Just substitute the expression for V_c into the equation for total current:

    [tex]I_t = V_1/R_1 = \frac{V(t)-V_c}{R_1}[/tex]

    [tex]I_tR_1 = V(t) - (\frac{V(t)}{2}(1-e^{-2t/RC}) - \frac{1}{2e^{2t/RC}}\int_0^t (V'(t)(e^{2t/RC}-1)dt)[/tex]

    AM
     
  12. Apr 26, 2005 #11
    Well i was wrong, it didn't have to be expresed into Vt =... just Input terms = Output. Hopfully i won the test.
     
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