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 ?


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

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    $\begingroup$ 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
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    $\begingroup$ 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

(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.

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    $\begingroup$ 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

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