# How to calculate the steady state response $y_{ss}(t)$ of a LTI system given the Laplace transform $Y(s)$?

I am given the Laplace transform of the output of a LTI system: $$Y(s) = \frac{1}{s((s+2)^2+1)}$$ Asked is what the steady state response $y_{ss}(t)$ would be. I think that $y_{ss}(t) = \lim_{t\to\infty} y(t)$, since after waiting infinit long, the system should be in steady state. (Right?)

I thought to use the final value theorem:

$$\lim_{t\to\infty}y(t)=\lim_{s\to 0}sY(s)$$ This gives $$\lim_{s\to 0} sY(s)=\lim_{s\to 0}\frac{1}{(s+2)^2+1} = \frac{1}{5}.$$

This is different from $\frac{1}{10}$. When I let a computer algebra system calculate $\mathscr{L}^{-1}[Y(s)] \bigg{|}_{t=\infty}$ I get $\frac{1}{10}$. (I'm using wxMaxima and used limit(ilt(1/(s*(s^2 + 2*s + 10)), s, t), t, inf);.)

What am I doing wrong? Thanks in advance.

• Your code should be s^2+4*s+5. – msm Dec 3 '16 at 21:03
• unlike transient response, which is with zero input but some states are non-zero at $t=0$, the steady state response needs a defined input. – robert bristow-johnson Dec 4 '16 at 2:37
• $Y(s)$ is the output of the system, so its fine as written. – Batman Dec 4 '16 at 5:28

$$\frac{1}{5} - \frac{1}{5} e^{-2 t} (2 sin(t) + cos(t))$$
$$\frac{1}{10} i (e^{(-2 - i) t} ((2 + i) e^{2 i t} + (-2 + i)) - 2 i)$$