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Help deriving a circuit equation

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Question is here


    2. Relevant equations

    Shown in the image, but just to restate:

    Vc(t) = A+Be^(-t/τ)

    n/m=e^-1

    OR

    p/m=1-e^-1

    3. The attempt at a solution

    First thing I tried to do was solve the Vc(t) equation for e, which comes out to e=(-A/B)^(-τ/t). I'm not even sure if that's relevant, though, or where to take it from there if it is. I'm a little lost on how to proceed with this.
     
  2. jcsd
  3. Nov 7, 2013 #2

    berkeman

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    Staff: Mentor

    The key is to use the equation for the V(t), and compare V(t) to V(t+τ)...
     
  4. Nov 7, 2013 #3
    So try setting V(t) equal to V(t+τ), which would be A+Be^(-t/τ)=A+Be^(-t+τ/τ)? That still doesn't relate n, m, and p to each other, though. From looking at the graph, it looks like there's some sort of relationship between m/n/p and A or A+B, but I can't quite tell what it is.
     
  5. Nov 7, 2013 #4

    berkeman

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    Staff: Mentor

    No, don't set them equal because they are not.

    The key in the figure (it is small and hard to see on the figure) is that the two vertical lines are a time τ apart...
     
  6. Nov 7, 2013 #5
    I see that the m line is time τ apart from the n/p line. I also see that m is the difference between A (the maximum possible voltage) at time t and V(t), and that n is the difference between A and V(t+τ), and that p is the difference between m and n. I can't quite tell how to relate them though. Maybe...A+Be^(-t/τ)+m=A+Be^(-t+τ/τ)+n?

    EDIT: Wait, no. m=A-(A+Be^(-t/τ)) and n=A-(A+Be^(-t+τ/τ)), but I'm still not sure how to relate them.
     
    Last edited: Nov 7, 2013
  7. Nov 7, 2013 #6

    berkeman

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    Staff: Mentor

    Now that I look closer at the diagram, it is pretty confusing. It's like they drew it upside down, with A+B at the bottom, and A at the top of the vertical axis. Weird.

    Try this as a first way to prove the ratios, and then you can see if you can prove it for this upside down version.

    Flip the drawing about its middle horizontal axis, so A is at the bottom, and A+B is at the top. Now set A=0, so the waveform starts with initial value B and decays exponentially to zero. Now since A=0, the equation for the waveform becomes V(t) = B e^-t/τ

    Now take the ratio of V(t+τ) / V(t), and see what you get...


    EDIT -- I fixed the V(t) equation -- I had left out the minus sign in the exponent.
     
  8. Nov 7, 2013 #7
    So in that case V(t+τ) = A+Be^(-t+τ/τ). I put that over the equation you gave me for V(t), but I'm not sure what it represents or what it's equivalent to. I assume it's equivalent to one of the ratios it shows in the problem statement, I just can't tell which or how to prove it.
     
  9. Nov 7, 2013 #8

    berkeman

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    Staff: Mentor

    Did you miss where I said to set A=0?
     
  10. Nov 7, 2013 #9
    I think that example confused me.

    But I understand what's going on now a little bit better.

    I think n=V(t+τ)=A+Be^(-t+τ/τ) and m=A+Be^(-t/τ), and that setting them over each other and reducing that should come out to e^-1. But it comes out to e, not e^-1. m/n does come out to e^-1 though, so I think I'm just missing one last component of the problem.

    EDIT: I also played around with the flipped graph a little more, and I get e instead of e^-1 for that one as well. I can't figure out why that is or what I should do to fix it.
     
    Last edited: Nov 7, 2013
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