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Thread: differential eq problem

  1. #1
    Mar 2008

    differential eq problem

    Could anyone please help with this?

    A radar controlled tracking system is described by the following diff eqn:

    X''(o)+100X'(o)+10,000X(o)= 10,000X(d)

    X(d)=angular position of the target (in radians)
    X(o)=angular position of the tracker (in rads)

    If the target is moving at a rate of 0.5rad/s i.e. X(d) = 0.5t rads
    what will the steady state error E be, where E=X(d) - X(o)

    Thanks in advance
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  2. #2
    Super Member Aryth's Avatar
    Feb 2007
    First, we split it into two equations:

    X''(o)+100X'(o)+10,000X(o)= 10,000X(d)

    Now, I'm not sure if you're familiar with D notation, but:

    D = \frac{d}{dx}

    Dy = \frac{dy}{dx}


    D^n = \frac{d^n}{dx^n}

    So therefore:

    (D^2 + 100D + 10,000)X(o) = 0

    And the second equation:

    (D^2 + 100D + 10,000)X(o) = 10,000X(d)

    The Initial Steps

    1. D(e^{kx}) = ke^{kx} and D^n(e^{kx}) = k^ne^{kx}

    2. Hence (D^2 + 100D + 10,000)e^{\alpha x} = \alpha^2e^{\alpha x} + 100\alpha e^{\alpha x} + 10,000e^{\alpha x}

    = (\alpha^2 + 100\alpha + 10,000)e^{\alpha x}

    = \left(\alpha + \left(-50 + \frac{173.21i}{2}\right)\right)\left(\alpha + \left(-50 - \frac{173.21i}{2}\right)\right)e^{\alpha x}

    So the roots of the polynomial \alpha^2 + 100\alpha + 10000 are 50 \pm \frac{173.21i}{2}

    Hence (D^2 + 100D + 10000)e^{\left(50 + \frac{173.21i}{2}\right)x} = 0 and (D^2 + 100D + 10000)e^{\left(50 - \frac{173.21i}{2}\right)x} = 0

    3. From 1 and 2 we deduce that for any quadratic function of D with constant coefficients that: (aD^2 + bD + c) = e^{kx}(ak^2 + bk + c). This equals 0 if k is a root of ak^2 + bk + c.

    First we deal with the Homogeneous Equation, for the solution we expect:

    1. Two linearly independent solutions, X_1(o) and X_2(o).
    2. Two arbitrary constants c_1 and c_2.
    3. The general solution will be X_c(o) = c_1y_1 + c_2y_2.

    First, we see that a possible solution is X(o) = Ae^{kx}

    From the initial work, we can see that both roots are in fact complex and in the form k = \alpha + i\beta or k = \alpha - i\beta, therefore, the general solution is of the form:

    X(o) = e^{\alpha x}[Pcos(\beta x) + Qsin(\beta x)]

    We have the roots:

    k_1 = 50 + \frac{173.21}{2}i

    k_2 = 50 - \frac{173.21}{2}i

    So, therefore we arrive at the solution:

    X_c(o) = e^{50x}\left[Pcos\left(\frac{173.21}{2}x\right) + Qsin\left(\frac{173.21}{2}x\right)\right]


    P = A + B


    Q = i(A - B)

    Where A and B are constants.


    X_c(o) = e^{50x}\left[(A + B)cos\left(\frac{173.21}{2}x\right) + i(A - B)sin\left(\frac{173.21}{2}x\right)\right]

    Nonhomogeneous Equation

    The general solution of this is:

    y(x) = X_c(o) + X_p(o)

    Where X_c(o) is the general equation we had above and X_p(o) is a particular solution of the differential equation.

    Ok, to solve this we need to go back to the formula:

    (D^2 + 100D + 10000)X(o) = 10000X(d)

    Got to go to class. If anyone else would like to solve it they can.
    Last edited by Aryth; Mar 10th 2008 at 03:56 PM.
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