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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:

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

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

    $\displaystyle D = \frac{d}{dx}$

    $\displaystyle Dy = \frac{dy}{dx}$


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

    So therefore:

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

    And the second equation:

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

    The Initial Steps

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

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

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

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

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

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

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

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

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

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

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

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

    We have the roots:

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

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

    So, therefore we arrive at the solution:

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


    $\displaystyle P = A + B$


    $\displaystyle Q = i(A - B)$

    Where A and B are constants.


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

    Nonhomogeneous Equation

    The general solution of this is:

    $\displaystyle y(x) = X_c(o) + X_p(o)$

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

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

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

    Got to go to class. If anyone else would like to solve it they can.
    Last edited by Aryth; Mar 10th 2008 at 02:56 PM.
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