Clasify the output $y(t)$ knowing that $x(t)=e^{-t}$, $t>0$

Question

Given the transfer function $$G(s)=10\frac{s+1}{s^2-2s+5}$$ clasify the output $y(t)$ given the input $x(t)=e^{-t}$, $t>0$.

---

I know that $$y(t)=\mathscr{L}^{-1}[Y(s)]=\mathscr{L}^{-1}[X(s)G(s)].$$ We know that $$X(s)=\mathscr{L}^{-1}[e^{-t}]=\frac{1}{s+1},$$ so $$Y(s)=\frac{1}{s+1}10\frac{s+1}{s^2-2s+5}=10\frac{1}{s^2-2s+5}=10\frac{1}{(s-\frac{3}{2})^2+\frac{41}{9}},$$ so $$\boxed{y(t)=\mathscr{L}^{-1}[Y(s)]=\frac{30}{\sqrt{41}}\sin\left(\frac{\sqrt{41}}{3}t\right)e^{\frac{2}{3}t}}.$$ So to clasify $y(t)$ we need to study the following limit: $$\lim_{t\to\infty}\frac{30}{\sqrt{41}}\sin\left(\frac{\sqrt{41}}{3}t\right)e^{\frac{2}{3}t}.$$ I have studied that I can calculate its value in two ways:

1. I have seen that $\sin(x)$ is bounded and $e^x$ is an increasing function, to when $x\to\infty$ the product diverges (and hence $\boxed{\text{the output is oscillatory undamped}}$).
2. Using the Final value theorem (or Initial value theorem): $$\lim_{t\to\infty}y(t)=\lim_{s\to0}sY(s).$$ But if I find the last limit I end up with: $$\lim_{s\to0}sY(s)=\lim_{s\to0}\frac{10s}{s^2-2s+5}=0,$$ so $\lim_{t\to\infty}y(t)=0$, but that IS NOT TRUE, since $\not\exists\lim_{t\to\infty}\frac{30}{\sqrt{41}}\sin\left(\frac{\sqrt{41}}{3}t\right)e^{\frac{2}{3}t}$.

What am I missing in (2)? Why this theorem fail?

I can only use these 2 properties, I do not know another one, as someone mentioned in a recent question: Show that $\lim_{x\to\infty}\sin(x)e^{x}$ diverges.

Answer 1

You cannot use the final value theorem in this case. The proper statement of the theorem is as follows:

Suppose that the following conditions are satisfied:

1. $f$ is continuously differentiable and $f'$ has a Laplace transform
2. $f'$ is absolutely integrable, that is, $\int_0^\infty|f'(\tau)|\mathrm{d}\tau$ is finite,
3. $\lim_{t\to\infty}f(t)$ exists and is finite,

then,

\begin{equation} \lim_{t\to \infty}f(t) = \lim_{s\to 0^+}sF(s). \end{equation}

If $F$ is a rational function (the quotient of two polynomial functions) or a rational function multiplied by an exponential, that is \begin{equation} F(s) = e^{-as}\frac{P(s)}{Q(s)} \end{equation} and

1. the (complex) non-zero roots of $Q$ (called the poles of $F$), say $s_1,\ldots,s_n$, are all in the open left hand side half-plane of the complex plane (i.e., $\Re(s_i)<0$ for all $s_i\neq 0$), and
2. $F$ has at most one pole at the origin,

then the final value theorem applies (under these assumptions, all three requirements of the theorem are satisfied).

In your case, the poles of $G$ are not in the open left complex plane (this is clear because there is a sign change in the coefficients of the denominator), so the FVT cannot be used. You can confirm that the poles of $G$ are $1\pm 2i$.

Answer 2

For functions that are increasing exponentially in $t$, the initial value theorem holds, but the final value theorem that is quoted, breaks down. Let's revisit the derivation of the initial/final value theorem,

If $|f(t)|<e^{ct}$ and for simplicity we assume that $s$ is real and positive:

$$|sF(s)|=|\int_{0}^{\infty}sf(t)e^{-st}dt|=|\int_{0}^{\infty}f(\frac{y}{s})e^{-y}dy|<\int_{0}^{\infty}e^{-(1-c/s)y}dy=\frac{1}{1-\frac{c}{s}}$$

This estimate converges only if $0<s<c$. This proves that $\lim_{s\to 0}sF(s)=0$. However this is in clear contradiction with the fact that for $f(t)$ unbounded as in the example provided by the OP, $\lim_{t\to\infty}f(t)=\infty$.

This happens because we can no longer use the dominated convergence theorem that allows us to interchange the order of limit and integration to prove the FVT statement. $$\lim_{s\to 0}sF(s)=\lim_{s\to 0}\int_{0}^{\infty}f(\frac{y}{s})e^{-y}dy\neq\int_{0}^{\infty}(\lim_{s\to 0}f(\frac{y}{s}))e^{-y}dy$$

and the two limits are generally unequal, since $f$ diverges at infinite time. Look here for an explicit example where the FVT fails.

Generally, physical systems exhibit bounded response at final times and thus the FVT is true for them. These systems happen to have poles only in the left hand plane and the particular system under examination fails this requirement.