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Homework Help: Find if a function is bounded through its Laplace transform

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    If f(t) transforms into F(s), so that [tex]\[
    F(s) = \frac{{s + 1}}{{s^2 + as + 1}},a \in
    [/tex], prove that if a < 0, the function f(t) isn't bounded, and if a >= 0, it is bounded. Prove that if -2 < a < 2, f(t) oscilates.

    3. The attempt at a solution

    I honestly have no idea how to do this. I think I have to use the final value property, but that gives me that f(t) has a finite limit when t approaches to infinity wheather a is positive or not (actually, if a is 0, it gives me that f(t) diverges).

  2. jcsd
  3. Feb 1, 2010 #2
    Start by computing the roots of the denominator (as a function of a). What is the inverse transform for each of the cases?
  4. Feb 1, 2010 #3
    The roots are terrible to look at:

    s^2 + as + 1 = \left( {s + \frac{{a - \sqrt {a^2 - 4} }}{2}} \right)\left( {s + \frac{{a + \sqrt {a^2 - 4} }}{2}} \right)

    Then I find that [tex]\[
    F(s) = \frac{{s + 1}}{{s^2 + as + 1}} = \frac{{s + 1}}{{\left( {s + \frac{{a - \sqrt {a^2 - 4} }}{2}} \right)\left( {s + \frac{{a + \sqrt {a^2 - 4} }}{2}} \right)}}

    I find no way to anti-transform this.
  5. Feb 1, 2010 #4
    For what values of a are the roots real and distinct? A double real root? Complex conjugates?
  6. Feb 1, 2010 #5
    If a2 > 4, then those are real roots. Otherwise, complex and conjugate (since a [tex]\in[/tex]R). If a = 2, then the roots are double (1). Still, no way to find the anti-transform.
  7. Feb 1, 2010 #6
    Well, if [tex]a^2>4\Leftrightarrow a>2 \vee a<-2[/tex]. These are distinct real roots, say [tex]\alpha[/tex] and [tex]\beta[/tex], which means that you can factor

    F(s) = \frac{{s + 1}}{{s^2 + as + 1}}




    F(s) = \frac{A}{{s-\alpha}}+\frac{B}{{s-\beta}}


    Where A and B are constants. How do you invert that? What's the behaviour of the inverses?
  8. Feb 1, 2010 #7
    Oh, right, partial fractions, I forgot about them. Well the, I will get to exponential functions, A.e-ax, where a>0, A is a constant, and those are bounded functions. If the roots where negative, then the exponentials would be of the form A.eax, which are not bounded.

    And if the roots are complex, then I would get sines or cosines, right?

  9. Feb 1, 2010 #8
    Exactly. For 0<a<2 (complex conjungate roots, with negative real part), you have sines and cosines (oscillatory behaviour); for a>=2, you have negative real roots. Therefore, in both cases, the inverse will go to 0, as t goes to infinity (bounded functions).

    If -2<a<=0, you will have oscillatory behaviour (complex roots again, with nonnegative real parts), which means sines and cosines again; if a<=-2, then you will have real roots, but at least one of them will be positive, so the inverse is unbounded.
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