Determining whether or not a system is BIBO stable

Question

I'm trying to determine whether or not a system with impulse response

$$ h(t) = \sum_{n=-\infty}^{\infty} \delta(t-2n) $$

is BIBO stable. I haven't touched this material for a very long time -- could anyone lend a helping hand? I recall needing to show that

$$ \int_{-\infty}^{\infty} |h(t)| \, dt < \infty, $$

but $h(t)$ is defined in a very weird way, and calculating the integral is proving to be difficult.

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EDIT: Does the following argument seem reasonable? Since were summing an infinite number of $\delta$ functions, each will have an argument of $0$ for some unique value of $t$. Thus, there are infinitely many values of $t$ for which $\delta(t-2n)=1$, so the integral diverges.

Answer 1

In your case

$$\int_{-\infty}^{\infty}|h(t)|dt=\int_{-\infty}^{\infty}h(t)dt= \int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}\delta(t-2n)dt= \sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty}\delta(t-2n)dt$$

And since

$$\int_{-\infty}^{\infty}\delta(t-2n)dt=1$$

(as correctly pointed out by Dilip Sarwate in his comment to Phonon's answer), the above sum is

$$\sum_{n=-\infty}^{\infty}1=1+1+1+\ldots=\infty$$

I believe that this is basically the argument you gave in your question.

Answer 2

Hint:

Simplify the expression from the inside outward.

First, look at $|h(t)|$. Does $h(t)$ ever go negative? Can you simplify the absolute value expression?

Once you do that, your argument works. There are several ways of evaluating this integral. One of them is to show that

$$\lim_{N\rightarrow \infty} \int^N_{-N}\left( \sum_{n=-\infty}^{\infty}\delta(t-2n) \right)dt \rightarrow \infty.$$

As $N$ grows large, you're capturing more and more impulses under the integral.