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The original question is given as

$$\frac {d^3y}{dt^3}+y=u=(1-t)e^{-2t}$$

The initial value y(0) = 0 and the same for all derivatives of y.

  1. Determine Y(s)
  2. What happens to u(t) and y(t) when $t\rightarrow \infty$ ?

Through Laplace transforms and partial fractions we arrive at (also given in the answer for 1.)

$$Y(s)=\frac{1}{(s+2)^2(s^2-s+1)}$$

Now, using the final value theorem, we could evaluate

$$\lim_{t\rightarrow \infty}y(t)=\lim_{s\rightarrow 0}sY(s)$$

Using Wolfram alpha, I get

$$\lim_{s\rightarrow 0}\frac{s}{(s+2)^2(s^2-s+1)}=0$$

But the given answer to question 2. is that $|y(t)|\rightarrow \infty$ when $t\rightarrow \infty$

What am I doing wrong?

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1 Answer 1

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For $u(t)$, $U(s) = \frac{s+1}{(s+2)^2}$, therefore $\lim_{s \to 0} s U(s) = 0$. However, for $y(t)$ the Final Value Theorem does not hold, because $y(t)$ does not converge. For FVT to work, $\lim_{t \to \infty} y(t)$ must exists, i.e. finite. But this is not the case since the system is not stable, i.e. it has roots at right half plane. Therefore $\lim_{t \to \infty} |y(t)| = \infty$ for all $u(t) \not\equiv 0$ (assuming all initial conditions are 0).

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