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Solving a circuit with Laplace

  1. Nov 30, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-11-30_0-44-27.png
    Assume zero initial conditions

    Step 1. Write the nodal equations to find i(t) in the time domain.
    Step 2.Solve the differential equation obtained in step 1 using laplace to obtain i(t).

    2. Relevant equations


    3. The attempt at a solution

    Convert to sdomain
    ##7e^{-6t}## becomes ##\frac{7}{s+6}##
    4H becomes 4s

    ##\frac{7}{s+6}+\frac{V_A}{5}+\frac{1}{4s}\int_{0}^{t}(V_A-V_B)=0##

    ##\frac{1}{4s}\int_{0}^{t}(V_B-V_A)+\frac{V_B}{3}=0##

    Am I doing this right?
     

    Attached Files:

  2. jcsd
  3. Nov 30, 2016 #2

    gneill

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    Staff: Mentor

    There should be no explicit integrals or derivatives required in the equations that you write. Laplace takes care of all that implicitly.

    The Laplace "variable" s is more of an operator than a variable. When s multiplies a Laplace domain function (such as sVa(s)) it is equivalent to differentiating the corresponding time-domain function. When a Laplace domain function is divided by s, it is equivalent to integrating in the time domain. So (ignoring initial conditions):

    ##sF(s) ⇔ \frac{d}{dt}f(t)##

    ##\frac{1}{s} F(s) ⇔ \int_o^t f(\tau)~d \tau##

    A nice thing about the Laplace domain is that reactive components have "impedances" that automatically take care of their own differentiation and integration. Thus Laplace domain "impedances" sL for inductors and 1/(sC) for capacitors carry with them their own operators that automatically handle the calculus. All you have to do is write your circuit equations with these "impedances", manipulating them with standard algebra, and it's as though all the calculus were being done for you behind the scenes.
     
  4. Nov 30, 2016 #3
    ##\frac{7}{s+6}+\frac{V_A}{5}+\frac{V_A-V_B}{4s}=0##
    ##\frac{V_B-V_A}{4s}+\frac{V_B}{3}=0##
    Does this look better?
     
  5. Nov 30, 2016 #4

    gneill

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    Staff: Mentor

    Yes, that looks better. Just check the sign on the first term of the first equation. As written you have the source current flowing out of node A (judging by how you've written the other terms).
     
  6. Nov 30, 2016 #5
    ##-\frac{7}{s+6}+\frac{V_A}{5}+\frac{V_A-V_B}{4s}=0##
    ##\frac{V_B-V_A}{4s}+\frac{V_B}{3}=0##

    ##\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A-V_B=0##
    ##\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A=V_B##

    ##\frac{\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A-V_A}{4s}+\frac{\frac{-28s}{s+6}+\frac{4sV_A}{5}+V_A}{3}=0##

    ##\frac{-7s}{s+6}+\frac{V_A}{5}+\frac{-28s}{3s+18}+\frac{4sV_A}{15}+\frac{V_A}{3}=0##


    ##V_A(\frac{1}{5}+\frac{1}{3}+\frac{4s}{15})=\frac{49s}{3s+18}##
    ##V_A(\frac{8}{15}+\frac{4s}{15})=\frac{49s}{3s+18}##
    ##V_A=\frac{49s}{3s+18}*\frac{15}{8+4s}##

    Am I on the right track?
     
    Last edited: Nov 30, 2016
  7. Nov 30, 2016 #6

    gneill

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    Staff: Mentor

    Check the first term on the last line
     
  8. Nov 30, 2016 #7
    ##\frac{-7}{s+6}+\frac{V_A}{5}+\frac{-28s}{3s+18}+\frac{4sV_A}{15}+\frac{V_A}{3}=0##
    ##V_A(\frac{1}{5}+\frac{4s}{15}+\frac{1}{3})+\frac{-28s-21}{3s+18}=0##
    ##V_A(\frac{4s+8}{15})=\frac{28s+21}{3s+18}##
    ##V_A=\frac{28s+21}{3s+18}*\frac{15}{4s+8}##
     
    Last edited: Dec 1, 2016
  9. Dec 1, 2016 #8
    Trying to get Vb instead of Va. Could you check my work? i(t) = VB/3, correct?
    ##\frac{V_B-V_C}{4s}+\frac{V_B}{3}=0##
    ##V_B-V_A+\frac{4sV_B}{3}##
    ##V_A=V_B+\frac{4sV_B}{3}##
    sub into eqA
    ##-7e^{-6t}+\frac{V_B(1+\frac{4s}{3})}{5}+\frac{V_B(1+\frac{4s}{3})}{4s}##
    simplify
    ##-7e^{-6t}+\frac{V_B}{5}+\frac{4sV_B}{15}+\frac{V_B}{4s}+\frac{V_B}{3}=0##
    ##-7e^{-6t}+\frac{8V_B+V_B4s}{15}+\frac{V_B}{4s}=0##
    ##\frac{8V_B+V_B4s}{15}+\frac{V_B}{4s}=7e^{-6t}##
    mult everything by 60s
    ##32sV_B+16s^2V_B+15V_B=420se^{-6t}##
    ##e^{-6t} = \frac{1}{s+6}##
    ##32sV_B+16s^2V_B+15V_B=\frac{420s}{s+6}##

    ##V_B=\frac{420s}{(4s+3)(4s+5)(s+6)}##

    Partial fraction decomp:
    upload_2016-12-1_2-34-14.png

    Inverse Laplace:
    VB= upload_2016-12-1_2-40-35.png

    i(t)=VB/3 =
    upload_2016-12-1_2-41-53.png

    is this correct?
     
  10. Dec 1, 2016 #9

    gneill

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    Staff: Mentor

    Yes, that's correct.
    Problem: You can't mix time-domain and Laplace domain functions. ##7e^{-6t}## doesn't belong here.
     
  11. Dec 1, 2016 #10
    I ended up getting VB= upload_2016-12-1_3-23-6.png
    i(s) = VB(s)/3 = upload_2016-12-1_3-25-13.png

    i(t)= upload_2016-12-1_3-25-26.png

    However, the answer should be:
    upload_2016-12-1_3-24-58.png

    Any idea what I'm doing wrong?

    EDIT: I think I see my mistake. I will rework and post results
     
    Last edited: Dec 1, 2016
  12. Dec 1, 2016 #11

    gneill

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    Staff: Mentor

    All I can tell is that your Laplace expression for VB has not turned out correctly. There's a lot of tedious algebra involved so you'll need to be very careful at each step.

    I 'll admit that reviewing every line of algebra is not my favorite thing to do, so instead I'm going to give you a goal to work towards. The solution for VB(s) should be able to be expressed in the form:

    ##VB = A \frac{1}{(s+6)(s+2)}##, where A is a numeric constant.
     
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