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Difference Equation

  1. Feb 10, 2016 #1
    I have a difference equation which is given as:

    ΔP = e^P [1]

    where we can re-write ΔP as: Δ P = P_2 - P_1, where the subscripts indicate two distinct discrete time indices.

    What I would like to do: is to convert this into a continuous time expression and solve it, if possible.

    In order to help give some insight, I will solve a similar type of problem where I know the solution.

    ΔP = c_1 [2]

    Note here, that in all cases we are running the recursive algorithm at a fixed data rate. Therefore, I can rewrite equation [1] as:

    Δ P = P_2 - P_1 = c_2 ⋅ Δ t

    where c_1 = c_2 ⋅ Δ t

    This allows me to divide both sides by [equation] \Delta t [/equation]:

    ΔP /Δt = c_2

    And in the limit:

    dP/dt = c_2

    which then becomes:

    P(t) - P(0) = c_2⋅(t - t_0)

    And so the result is that this recursive equation [2] gives us a linear ramp if we were to implement it. What I am trying to do for equation [1] is figure out what this expression will look like.
  2. jcsd
  3. Feb 10, 2016 #2
    To which discrete time index is e^P refering?
  4. Feb 10, 2016 #3


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    You can approximate it by a "continuous derivative" with different results depending upon what "continuity" assumptions you make.
  5. Feb 10, 2016 #4


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    looks explosive if I try a few numbers:
    Code (Text):

    n   P1           exp(P1)         P2
    0   0.0001       1.0001      1.0002
    1   1.0002       2.718826    3.719026
    2   3.719026    41.2242     44.94323
    3  44.94323      3.3E+19     3.3E+19
    I find it hard to believe this is what's intended ... ?
  6. Feb 10, 2016 #5
    Sorry all I forgot it's e^p1
  7. Feb 10, 2016 #6
    Good point, it's actually e^(p1 - c_3)

    Which is a negative exponential, but we can rewrite that as e^p1 / e^c_3 so I left out the denominator since it was a constant value. c_3 is just a constant
  8. Feb 10, 2016 #7
    I'm open to any solutions that are simple with some reasonable assumptions. :-)
  9. Feb 10, 2016 #8
    Hey look who's back :smile:
  10. Feb 11, 2016 #9


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    Still explodes, at some point P1 > c3.
    Let C3 = 100 and P1 start at 1:

  11. Feb 11, 2016 #10


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    It looks like you have, for a fixed data rate
    ##\Delta P = e^P,## so ##P_2 - P_1 = e^{P_1 } \Delta t##?
    In the same way you simplified the first problem,

    ##\frac{\delta P}{\delta t} = \frac{ P(t+\delta t) - P(t) }{ \delta t} = e^{ P(t) }.##
    In the limit, this will give you ## P' = e^{P(t)} ##

    As BvU has pointed out, no matter what your starting P_1 is, your function will eventually blow up. Is there some other context for this problem? Maybe some initial values?
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