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Homework Help: Impulse Response for a 2nd Order Diffy Q

  1. Jun 18, 2010 #1
    [tex]y'' + 6y' + 4y = x(t)[/tex]
    I found the eigenvalues to be -.764 and -5.24, no problems here. Next, since the order of the response is greater than the order of the excitation, I assumed the unit response took the form:

    [tex] h(t) = k_1e^{-.764t} + k_2e^{-5.24t}[/tex]

    So now I must find the values for k_1 and k_2. If I integrate both sides from -0 to +0, I should find one equation that will help in this endeavor. This is where I run into some problems.

    [tex]\int\limits_{-0}^{+0}h''(t)\, dx + 6\int\limits_{-0}^{+0}h'(t)\, dx + 4\int\limits_{-0}^{+0}h(t)\, dx = \int\limits_{-0}^{+0}\delta(t)\, dx[/tex]

    which becomes:

    [tex] h'(+0) - h'(-0) + 6[h(+0) - h(-0)] + 4\int\limits_{-0}^{+0}h(t)\, dx = u(+0) - u(-0)[/tex]

    Now, h(t) has no impulse in it. Therefore, an integral from -0 to +0 of h(t) = 0. Also, since the only excitation is the impulse, occuring at t = 0, and this system is causal, all h(t) and its derivatives evaluated at -0 evaluate to zero. Further, a step function at -0 is zero and at +0 is 1:

    [tex] h'(+0) + 6h(+0) = 1[/tex]

    I then substitute in h'(+0) and 6h(+0):

    [tex] h'(+0) + 6h(+0) = -.764K_1 - 5.24K_2 + 6k_1 + 6K_2 = 1[/tex]

    The problem is that the first equation generated in the solutions does not have the additional 6k_1 and 6K_2. Can someone explain to me why [tex]6\int\limits_{-0}^{+0}h'(t)\, dx[/tex] evaluates to zero?

    I've completed the problem, and some how got the same(right) answer? How can this be? When I solved these two systems of equations (the first, what I derived and the second what the solutions manual derived) i got the same answer.
    [tex]5.236k_1 + .760k_2 = 1[/tex]
    [tex]k_1 + k_2 = 0[/tex]

    [tex] -.760k_1 - 5.24k_2 = 1[/tex]
    [tex] k_1 + k_2 = 0[/tex]
    Last edited: Jun 18, 2010
  2. jcsd
  3. Jun 26, 2010 #2


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    The RHS is finite, so you need h'(0+) to also be finite. For the derivative h'(0+) to exist, h(t) has to be continuous at t=0, so you know that h(0+)=h(0-).

    You probably used this fact to say that h(0+)=0, which gave you your second equation k1+k2=0.
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