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Consider a system: $$\dot x = \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}x+ \begin{bmatrix}0\\1 \end{bmatrix}u$$

$$y = \begin{bmatrix}1 & 1\end{bmatrix}y$$

Its transfer function is:

$$\frac{Y(s)}{U(s)} = H(s) = \frac{1}{s+1}$$

  1. Its BIBO stable.
  2. However, since there is an uncontrollable mode which is unstable, this system is not stabilizable.
  3. By simulink, if plotting '$y$', it will approach infinity as $t$ goes to infinity if $x(0)$ is not $0$.

My question is:

If $u(t)$ is a step function, and simulate $y(t)$ by simulink, it will be a step-like response.
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However, if using state-space form, $y(t)$ will approach infinity if $x(0)$ is not $0$.

enter image description here

Why are both different? Hope for a detailed explanation. What is the meaning of each of them?

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1 Answer 1

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Short answer: This is because transfer function model assumes that all initial conditions are $0$.

See that $y(t) = x_1(t) + x_2(t)$ where $x_1(t)=e^t x_1(0)$. The system is unstable, so $y(t)$ will go to infinity if $x_1(0) \neq 0$. In the transfer function model, you lose $x_1(t)$, because it is uncontrollable and the initial conditions.

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