My understanding of BIBO stability is that if a bounded input is applied to a discrete-time, linear, time-invariant BIBO system, the output will also be bounded. A sufficient condition for this is that the poles of the transfer function are within the unit circle.

Suppose I have a system with the transfer function $$ H(z)={(1−2z^{−1})(1−{\frac{1}{6}} z^{−1}) \over (1−\frac{1}{2}z^{−1})(1−\frac{1}{3}z^{−1})} $$ The poles of this system are within the unit circle and so is a BIBO stable system.

If I apply an input of $x[n]=(\frac{1}{3})^n$, which is bounded, the output would involve $H(\frac{1}{3})=\infty$, which is unbounded - i.e. a bounded input produced an unbounded output.

Obviously, there is something amiss in my understanding here, but what?


1 Answer 1


The signal $x[n] = (\frac{1}{3})^n$ is not bounded. As $n \to -\infty$, $x[n] \to \infty$; therefore, the system's response to your candidate signal $x[n]$ has no bearing on its BIBO stability.

A bounded, causal version of the signal (and likely what you were thinking) is $x_b[n] = (\frac{1}{3})^n u[n]$, where $u[n]$ is the unit step function. The response of the system to this signal is bounded as you would expect.

  • $\begingroup$ D'oh! D'oh! D'oh!!!!! $\endgroup$
    – Westerley
    Commented Aug 11, 2015 at 19:23
  • $\begingroup$ Also, you don't get the response of the system to $x[n]$ by evaluating $H(z)$ at some value of $z$... instead, you take the z-transform of $x[n]$, multiply it with $H(z)$, and call the result $Y(z)$. Now, you can do an inverse transform to get $y[n]$, which you will see is bounded. $\endgroup$ Commented Aug 11, 2015 at 21:53
  • $\begingroup$ I understood from this question that the output of a system when the input is a 'pure' exponential such as $\frac{1}{3}^n$ would involve evaluating $H(\frac{1}{3})$. If the input was $\frac{1}{3}^nu[n]$, then it would be as you say $\endgroup$
    – Westerley
    Commented Aug 12, 2015 at 1:28
  • $\begingroup$ @CMDoolittle: The OP is correct; the $z$ transform can be used in this way since exponential functions are the eigenfunctions of discrete-time LTI systems. The value $H(z)$ corresponds to the eigenvalue associated with the function $x[n] = z^{-n}$. $\endgroup$
    – Jason R
    Commented Aug 12, 2015 at 1:32
  • $\begingroup$ Wow. I did not know that. Thanks y'all! $\endgroup$ Commented Aug 12, 2015 at 3:28

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