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Integration problem

  1. Jul 7, 2008 #1

    Defennder

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    1. The problem statement, all variables and given/known data
    I have no idea if this is a maths or EE problem, but is it possible to obtain the second line the expression for [tex]t_{PHL2}[/tex] from the first one, given the instructions for integration. I tried but didn't manage to get it.

    [​IMG]


    2. Relevant equations



    3. The attempt at a solution
    In particular I can't seem to make Kn disappear inside the ln expression.
     
  2. jcsd
  3. Jul 7, 2008 #2

    cristo

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    It should be true, since the passage claims it is! I have no idea what any of the things mean, so doubt I can help. Perhaps this is more of an electrical engineering type problem. I'll flag this up with engineering homework helpers to ensure that they don't miss it, since it's in the calculus forum.

    (You might want to show what you attempted, or how far you got).
     
  4. Jul 8, 2008 #3

    Defennder

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    Ok I'll try to explain what I think the passage means. The expression [tex]i_{DN} = 2K_n[(v_I - V_{th})v_0 - 0.5v_0^2] = -C_L \frac{dv_0}{dt}[/tex] can be integrated by separation of variables. So it should look like this after separating variables and inserting the limits of integration:

    [tex]\int^{V_{DD}/2}_{V_{DD}-V_{th}} \frac{1}{v_0^2 - 2K_n(v_I - V_{th})v_0} dv_0 = \int^{t_{PHL2}}_0 \frac{1}{C_L}dt[/tex]

    This is an integral of the form [tex]\int \frac{1}{x^2 - kx} dx = \frac{1}{k}ln|\frac{x-k}{x}|+C[/tex].

    So the definite integral works out to:

    [tex]\frac{1}{2K_n(v_I - V_{th})} \left[ln|\frac{v_0 - 2K_n(v_I - V_{th})}{v_0}|\right]^{V_{DD}/2}_{V_{DD}-V_{th}} = \frac{t_{PHL2}}{C_L}[/tex]

    As you can see this gets really confusing due to the number of algebraic expressions and subscripts, which is why I'm not sure if it's just algebra and simple integration or EE.

    Now the expression for the definite integral has [itex]K_n[/itex] and [itex]v_I[/itex] inside. I think I can account for [itex]v_I[/itex] by noting that [itex]V_{DD}[/itex] has been substituted for [itex]v_I[/itex] if I understood this part of my notes properly:
    [​IMG]

    But even after taking that into account I could only get this for the final expression after evaluating for the integral limits:

    [tex]\frac{1}{2K_n(v_I - V_{th})} ln\left|\frac{\frac{1}{2}V_{DD} - 2K_n(V_{DD}-V_{th})}{\frac{1}{2}(1-2K_n)V_{DD}}\right| = \frac{t_{PHL2}}{C_L}[/tex]

    The expression in the notes and the one I got doesn't match. How did they get rid of K_n inside the ln expression?

    As I said earlier I don't know if this is a simple algreba and integration problem or an EE problem where I would need extra information to work this out. I'm posting this here to see if anyone else managed to solve it. If they cannot, then clearly it's an EE problem and I'll request that it be moved to the Eng HW forums. Thanks.
     
  5. Jul 8, 2008 #4

    dynamicsolo

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    Defennder,

    back in this line

    [tex]
    \int^{V_{DD}/2}_{V_{DD}-V_{th}} \frac{1}{v_0^2 - 2K_n(v_I - V_{th})v_0} dv_0 = \int^{t_{PHL2}}_0 \frac{1}{C_L}dt
    [/tex]

    that should be

    [tex]
    \int^{V_{DD}/2}_{V_{DD}-V_{th}} \frac{1}{K_n v_0^2 - 2K_n(v_I - V_{th})v_0} dv_0 ...
    [/tex]

    from the K_n distributing over both terms on the left side of the differential equation. This will cause the K_n to cancel in the argument of the log function.

    I had been wondering if v_I had some particular relationship to V_DD , in order for the rest of the argument to simplify. I see this is the case...

    EDIT: I am able to get the expression to this point

    [tex]
    \frac{C_L}{2K_n(v_I - V_{th})} ln\left|\frac{V_{DD}-V_{th} }{V_{DD}} \cdot \frac{8(v__I - V_{th}) - 4(V_{DD}-V_{th})} {4(v_I - V_{th}) - V_{DD} }\right| = t_{PHL2}
    [/tex]

    Perhaps things can be carried from here if the relationship of v_I to V_DD is known...
     
    Last edited: Jul 8, 2008
  6. Jul 8, 2008 #5

    Defennder

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    Hey you're right. I somehow managed to omit that and was hence unable to get the answer. I could solve it now, thanks!

    P.S. [tex]v_I = V_{DD}[/tex] for this.
     
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