0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Your code should be s^2+4*s+5. $\endgroup$ – msm Dec 3 '16 at 21:03
  • 1
    $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Dec 4 '16 at 2:37
  • 1
    $\begingroup$ $Y(s)$ is the output of the system, so its fine as written. $\endgroup$ – Batman Dec 4 '16 at 5:28
1
$\begingroup$

For these calculations, it is better to give the Wolfram Alpha answers:

inverse Laplace transform 1/(s*(s^2 + 4*s + 5))

Which gives the correct expression, consistent with the Final Value expression of 1/5:

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

This is very different from the complex (phasor) expression, which could inexplicably have a 1/10, but when evaluated it is still 1/5:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.