2
$\begingroup$

How do I check the following system $$ y[n]=u[n] $$ is BIBO stable or not ?

$u[n]$ is the unit step function

My Attempt:

For the BIBO stability, the necessary and sufficient condition is $$\sum_{n=-\infty}^{+\infty} |h[n]|<\infty$$.

Let $x[n]=\delta[n]$, so $y[n]=h[n]=u[n]$ $$ \sum_{n=-\infty}^{+\infty} |h[n]|<\infty=\sum_{n=0}^{+\infty} |u[n]|=1+1+1+......=\infty $$

which proves it is an unstable system. Is it the right way to approach the problem or do I need to first write $u[n]$ in terms of the delta function ?

$\endgroup$

closed as unclear what you're asking by Matt L., A_A, jojek Apr 14 '17 at 21:40

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ What do you mean by "the system" $y[n]=u[n]$? Is the output always a step function, regardless of the input signal? Or do you mean a system with impulse response $h[n]=u[n]$? $\endgroup$ – Matt L. Mar 9 '17 at 13:14
  • $\begingroup$ @MattL. The actual problem is to test stability of the following system, $y[n]=u[n]$. So i think it means $\mathcal{H}x[n]=y[n]=u[n]$ $\endgroup$ – ss1729 Mar 9 '17 at 13:36
  • 2
    $\begingroup$ But $y[n]=u[n]$ doesn't make much sense if $y[n]$ denotes the output signal, and if $u[n]$ denotes the step function. In some texts, $u[n]$ denotes the input. Make sure you understand what they mean, and clarify your question, otherwise we can't help. $\endgroup$ – Matt L. Mar 9 '17 at 13:39
  • $\begingroup$ @MattL. in the problem $u[n]$ denotes the unit step function for sure. $\endgroup$ – ss1729 Mar 9 '17 at 13:41
  • $\begingroup$ Are you sure it is $y[n]=u[n]$ and not $h[n]=u[n]$? In the first case that system would be rather useless as it would have a step function at its output, regardless of what signal you are putting ad the input. Maybe you could post the original problem instead of transcribing it. $\endgroup$ – Tendero Mar 9 '17 at 14:04
-2
$\begingroup$

(Long Comment) Sorry guys, I did not get why that system created so many conceptual problems to you. For instance, consider the unforced system $y(n)=0$. If I define $\pi:\{0\}\times\mathbb Z\to\{0\}$, $\pi(y,n)=0$ then $(\{0\},\mathbb Z,\pi)$ defines a dynamical system on the state space $\{0\}$. In fact $\pi$ is continuous, and it satisfies the identity axiom ($\pi(x,0)=x$ for all $x\in\{0\}$) and the group axiom ($\pi(\pi(x,n_1),n_2) = \pi(x,n_1+n_2)$ for all $(x,n_1),(x,n_2)\in\{0\}\times\mathbb Z$).

I agree that this is a degenerate case. I guess I can extend the state space to $\mathbb R$ by defining the system in a different way. For instance, let $$ x(n+1)=0\qquad (1) $$ be a ordinary difference equation and define $\pi$ as the flow of (1). Then $(\mathbb R,\mathbb N,\pi)$ defines a dynamical system on $\mathbb R$. Take the forced system \begin{align*} x(n+1) &= 0\cdot x(n) + 0\cdot u(n)\\ y(n) &= 0\cdot x(n) + 1\cdot u(n)\qquad\qquad (2) \end{align*} That makes sense to me. It is a LTI system with zero relative degree. To check BIBO stability let take a reference input $\bar u:\mathbb N\to\mathbb R$ and a perturbation $\delta u:\mathbb N\to\mathbb R$. Let $u(n)=\bar u(n)+\delta u(n)$. Let $\bar y:\mathbb N\to\mathbb R$ be the solution corresponding to the input $\bar u$ and let $y:\mathbb N\to\mathbb R$ be the one corresponding to the input $u$. Then \begin{align*} \|y(n)-\bar y(n)\| = \|u(n)-\bar u(n)\|=\|\delta u(n)\| \end{align*} So trivially, for any $\epsilon>0$ there exists $\delta_\epsilon>0$ such that $\|\delta u(n)\|\le \delta_\epsilon\implies \|y(n)-\bar y(n)\|\le\epsilon$. You can see this also as a continuity property of the solutions. In fact, you can define the space $Y$ of all the bounded output trajectories $y:\mathbb N\to\mathbb R$ and the space $U$ of all the bounded input ones. Let endow $Y$ with the norm $\|y\|:=\sup_{n\in\mathbb N}|y(n)|$ and $U$ with the norm $\|u\|:=\sup_{n\in\mathbb N}|u(n)|$. Then the system (2) definest an operator $T:U\to Y$, and asking for BIBO stability is the same as asking for the continuity of $T$ in the sup norms.

$\endgroup$
  • 1
    $\begingroup$ This is complete gobbledegook for the question asked; and it doesn't answer the question that I can fathom: -1. $\endgroup$ – Peter K. Apr 10 '17 at 13:22

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