T { x [ n ] } </mrow> = <munderover> <mo movable

Question

T {x [n]} = \sum_{k = min (n, n_{0})}^{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:

| T {x [n]} | = | \sum_{k = min (n, n_{0})}^{max (n, n_{0})} x [k] | \leq \sum_{k = min (n, n_{0})}^{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?

Answer 1

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:

σ_{n + 1} = σ_{n} + x_{n}, σ_{n_{0}} = 0.

(

T (x) (n) = σ_{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.

Answer 2

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).

T { x [ n ] } </mrow> = <munderover> <mo movable

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Answers (2)

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