Find if a function is bounded through its Laplace transform

[libelec]

Homework Statement

If f(t) transforms into F(s), so that $$\[ F(s) = \frac{{s + 1}}{{s^2 + as + 1}},a \in \]$$, 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.

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).

Help?

 

[JSuarez]

Start by computing the roots of the denominator (as a function of a). What is the inverse transform for each of the cases?

 

[libelec]

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 $$\[ F(s) = \frac{{s + 1}}{{s^2 + as + 1}} = \frac{{s + 1}}{{\left( {s + \frac{{a - \sqrt {a^2 - 4} }}{2}} \right)\left( {s + \frac{{a + \sqrt {a^2 - 4} }}{2}} \right)}} \]$$.

I find no way to anti-transform this.

 

[JSuarez]

For what values of a are the roots real and distinct? A double real root? Complex conjugates?

 

[libelec]

If a² > 4, then those are real roots. Otherwise, complex and conjugate (since a $$\in$$R). If a = 2, then the roots are double (1). Still, no way to find the anti-transform.

 

[JSuarez]

Well, if $$a^2>4\Leftrightarrow a>2 \vee a<-2$$. These are distinct real roots, say $$\alpha$$ and $$\beta$$, which means that you can factor
$$F(s) = \frac{{s + 1}}{{s^2 + as + 1}}$$

As:

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

Where A and B are constants. How do you invert that? What's the behaviour of the inverses?

 

[libelec]

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.e^ax, which are not bounded.

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

Thanks.

 

[JSuarez]

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.