Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For my intro to DSP class, the TA gave us this discrete system and asked to prove whether or not it's BIBO-stable:

$$T \left\{ x[n] \right\} = \sum \limits_{k=\text{min}(n, n_0)}^{\text{max}(n, n_0)} {x[k]}$$

for some integer constant $n_0$.

Intuitively, it's unstable, and it can be easily proven by a counterexample if $x[n]$ is the unit step and $n_0 = 0$.

However, the TA tried to use a general proof by invoking the triangle inequality. Assuming $x[n]$ is bounded (or $|x[n]| \leq M < \infty$), he said:

$$\left| T \left\{ x[n] \right\} \right| = \left| \sum \limits_{k=\text{min}(n, n_0)}^{\text{max}(n, n_0)} {x[k]} \right| \leq \sum \limits_{k=\text{min}(n, n_0)}^{\text{max}(n, n_0)} {|x[k]|} \leq (|n-n_0|+1)M$$

Obviously the right-hand side is unbounded, as it goes to infinity with increasing $n$, but to me it doesn't seem to imply that the system on the left-hand side is unbounded (because of the inequality).

My question is, can his attempt be augmented to show that the left-hand side is also unbounded? Or is a counter-example the only way to prove it?

share|cite|improve this question
The triangle inequality only gives an upper bound. An example (its not really a counterexample) showing unbounded output with a bounded input is sufficient. – copper.hat May 12 '12 at 0:37
up vote 1 down vote accepted

Since $n$ is increasing, you can assume that $n>n_0$. Then you can write the above system as the linear, time-invariant discrete time system: $$ \sigma_{n+1} = \sigma_n + x_n, \; \; \; \sigma_{n_0} = 0.$$ ($T(x)(n) = \sigma_n$ in this case.) A system of this form is BIBO stable iff all poles lie inside the unit circle. Since this system has a pole at $1$, it is not BIBO.

share|cite|improve this answer

This can only really be done by counterexample since you can think of plenty of bounded signals which give a bounded output (in fact a sufficient condition is for the sequence to be summable). Indeed your TA has effectively shown an upper bound which does tend to infinity and will not tell you that your function is BIBO. Perhaps he was trying to say that the worst case scenario is for a signal of the form $x[n]=M$ so that all other signals $|y[n]|\leq M$ are not as big in output (even though they still might give unbounded output).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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