Second Order Circuit - Equation Decisions

Discussion in 'Homework Help' started by ghoti, Jun 13, 2010.

  1. ghoti

    Thread Starter New Member

    Jun 13, 2010
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    Hi,

    I am having difficulty understanding the decisions of variable selection in a second order circuit. (Time Domain)

    This is the problem I am working with.

    [​IMG]

    As this will be solved by hand I need to ensure an efficient solution.

    I am looking for a nodal equation with the cap and a mesh equation for the inductor.

    If the horizontal resistor is R1 and the vertical resistor R2

    My first equation is the nodal for the cap. node Vc.
    Current through the cap + current through R1 and
    the current through R1 is the current through R2 + current through the inductor

    So

     C_1\frac{dv_c}{dt} + i_l + L_1\frac{di_l}{dt}\frac{1}{R_2} = 0

    Now I am stuck with producing a satisfactory mesh equation that still allows me to solve for vc.

    The resultant characteristic equation will be m^2 + 2m + 2 = 0.


    Any help would be MUCH appreciated.
    Thanks,
    Alex
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Why not go for nodal analysis?

    You require a voltage answer so that would make more sense.

    Make the top node of the inductor V2 and that will give you the two nodes for setting up your equations in V1 & V2.
     
  3. ghoti

    Thread Starter New Member

    Jun 13, 2010
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    Hi,

    I am on the edge of my understanding here but.... if I describe V2 via nodal I end up with a solution for V2 which is fine however to solve for V1 I need to fully solve the V2 DE before I can sub it back into V1. I cannot however solve the V2 fully because V2(0-) and V2(0+) are not definable due to the discontinues nature of the voltage across the inductor.

    Essentially;

    (1) v_1' + 2v_1 = v_2
    (2) v1 = 2v_2 + \frac{1}{L}\int{v_2}
    Am I completely up the wrong tree somewhere?
     
  4. ghoti

    Thread Starter New Member

    Jun 13, 2010
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    I now believe my characteristic equation is also incorrect.

    LTspice confirms that ic(0+) is 1A (upward) so \frac{dv_c}{dt} must be equal to 2.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It's a little unclear what the current source value is - is it 1.u(t). In other words a constant 1A source from t=0+ ...?
     
  6. ghoti

    Thread Starter New Member

    Jun 13, 2010
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    on all t; it is 1 + a -negative step function at 0.

    f(-t) = 1
    f(0) = undef
    f(t) = 0
     
  7. ghoti

    Thread Starter New Member

    Jun 13, 2010
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    Just to clarify the result is also defined on all t;

    the 1+ outside the u(t) is the -t part

    and then the (f(t) -1)u(t) is +t part.
     
  8. Ghar

    Active Member

    Mar 8, 2010
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    This problem has 2 nodes and 2 meshes so it doesn't really matter which method you use but do stick to one. Doing both is a waste of time.

    V2(0-) is very well defined; it's 0.
    V2(0+) is also well defined; it's also 0. The reason is that the capacitor fixes the voltage across the resistors and the inductor fixes the current through it.

    Let's work through the initial conditions...
    We're looking at the DC solution so the inductor is a short and the capacitor is open. This means R2 is shorted by the inductor, you can ignore it. That also means V2(0-) is zero.
    If the capacitor is open all the current is going through R1, and since R2 has zero current all of it goes through L. That gives you IL(0-) = 1A.
    Since V2(0-) is 0, then V1(0-) must be 1A * 1 Ohm = 1V. This is equal to Vc(0-).

    Now the current source shuts off (t = 0+). You get that Vc(0+) = V1(0+) must still be 1V and IL(0+) must still be 1A.

    Solve the equations at t = 0+, by superposition you get:
    V_2(0^+) = V_c(0^+)\frac{R_2}{R_1 + R_2}- (R_1||R_2)I_L(0^+)<br />
\\=1\frac{1}{1+1} - (1||1) (1) = 0

    Edit:
    I should note that the discontinuity is not a problem. It happens exactly at t = 0 and only at t = 0.
    Discontinuities are only an issue if you try to differentiate them and you're not; in a capacitor you differentiate voltage which must be continuous and in an inductor it's current. The discontinuity is exactly how math says they must be continuous functions.
     
    Last edited: Jun 14, 2010
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    OK - understood.

    So for the purposes of analysis one can remove the current source from t=0+ and treat the circuit as if the capacitor was initially charged to 1V. This can be justified with some simple reasoning.

    As to the voltage on the inductor at t=0+ it would be a reasonable assumption that this is still zero - as it was at t=0-
     
  10. ghoti

    Thread Starter New Member

    Jun 13, 2010
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    Ghar,

    Thanks for your reply. I follow your derivation of the initial conditions and your note on discontinuity however I am wondering what your method is for producing the final equation. Perhaps I am very lost right now but;

    If I decide to select nodal and produce my two equations
    V_c' + \frac{1}{R_1C}V_c = \frac{1}{R_1C}V_l \\<br />
V_l'(\frac{1}{R_1} + \frac{1}{R_2}) + \frac{1}{L}V_l = \frac{1}{R_1}V_c'

    Getting rid of some of the variables to keep things clean.

    \frac{1}{2}(V_c' + 2V_c) = V_l \\<br />
2V_l' + 2V_l = V_c'

    Once sub one into the other I end up with a second order differential requiring I define v(0+) as well as v'(0+)

    V_c'' + V_c' + 2V_c = 0

    Now the only way I can think of to get Vc' is via ic = C\frac{dV_c}{dt} ?

    t_n_k:
    The voltage across an inductor can change instantaneously. It has been suggested to me that this is infact, the problem with my solution.

    In my initial nodal equation for VL, I included a \int{v}. As v is not a CTS function in an inductor this integral is in-fact undefined.

    Whilst my gut understands the logic of this I am a little lost as to its ramifications for this problem and how to get around it.
     
    Last edited: Jun 14, 2010
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I am able to find the following 2nd order DE with V1 as variable ....

    V_1^{''}+\[\frac{1}{(R_1+R_2)C}+\frac{R_1R_2}{(R_1+R_2)L}\]V_1^{'}+\frac{R_2}{LC(R_1+R_2)}V_1=0

    which leads to the same characteristic equation m^2+2m+2=0
     
  12. ghoti

    Thread Starter New Member

    Jun 13, 2010
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    That characteristic equation works but only if vc'(0) = 0 and vc(0) = 1

    if ic = c dvc/dt

    then for vc'(0) to be 0 ic must also be zero.

    I built the circuit in LTSpice and confirmed that the current in C rises instantly to 1A!

    To get the final result I am using:

     <br />
\sqrt{C^2 + D^2}e^{-\alpha} cos(\omega t - tan^{-1} \frac{D}{C})\\<br />
e^{-\alpha}[Csin(\omega t) + Dcos(\omega t)]<br />

    I am so very lost! appreciate your help on this one!

    Cheers,
     
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Consider what happens at t=0 for V1 (i.e. Vc)

    At t=0- the capacitor current will be zero and the voltage V1 will be 1V.

    At t=0 the current source jumps from 1A to zero, so the 1A current must then be supplied from the capacitor at that instant.

    Hence, at t=0 we have .....

    V_1(0)=1V

    i_c(0)=C\frac{dV_1}{dt}=-1A

    \frac{dV_1}{dt}=V_1^'=\frac{i_c(0)}{C}=\frac{-1}{0.5}=-2 \ Volt/sec
     
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  14. ghoti

    Thread Starter New Member

    Jun 13, 2010
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    t_n_k,

    Brilliant, I see it now and have reproduced it myself.

    A massive thankyou to both yourself and Ghar.

    I only just found this resource but plan on sticking around, maybe help out where I can as well.

    Once again,
    Thanks

    /passes t_n_k a beer.
     
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Cheers mate!
     
  16. Ghar

    Active Member

    Mar 8, 2010
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    My final equation is simply an instantaneous solution at t = 0+ and looks just like DC in terms of the math, since there's no more time dependence (because I made t = 0+)

    Integrals do work for finite discontinuous functions... since integrals are linear you just break the range apart. The discontinuity doesn't contribute anything because it exists for 0 time.
    i.e. if there's a discontinuity at 5 you can do the integral from 0 to 5- and from 5+ to +inf
    I'm sure there's some exceptions where it doesn't work but if you can see the area under the curve then the integral shouldn't have a problem.
     
    Last edited: Jun 14, 2010
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