Is System Stability of LTI System Proven?

[the_amateur]

Is the following system stable. If so how.

y(t)= $\frac{d}{dt}$ x(t)I have tried the following proof but i think it is wrong.

PROOF:

1. The System is LINEAR
2. The system is time invariant

So on applying the stability criterion for LTI systems

ie . $\int^{\infty}_{-\infty} h(t) dt$ < $\infty$ --------- 1 For the above system h(t) = $\delta^{'}(t)$

so on applying h(t) = $\delta^{'}(t)$ in eq. 1
$\int^{\infty}_{-\infty} \delta^{'}(t) dt$ = $\delta(0)$

So the system is not stable.
I think the above proof is way off the mark.
please provide the correct proof. thanks

 

[jashua]

The system (BIBO) stable if the impulse response is absolutely integrable, that is,

$\int^{\infty}_{-\infty} |h(t)| dt < \infty$

For the differentiator, we have (as you have found in your question)

$h(t) = \delta (t)'$.

Since

$\int^{\infty}_{-\infty} | \delta (t)' | dt= \infty$,

the system is not stable.The system is LTI as

$\frac{d}{dt} (a_1x_1(t) + a_1x_1(t)) = a_1 \frac{d}{dt} x_1(t) + a_2 \frac{d}{dt} x_2(t) = a_1y_1(t) + a_2y_2(t)$

and

$\frac{d}{dt} x(t-\tau) = y(t-\tau)$.

 

[the_amateur]

Thanks for your reply but i am trying to find if it is stable or not. i think my derivation is wrong.

 

[Larrytsai]

here let x(t) be input that is bounded. As you let t-> inifinity, y(t) remains real and bounded, therefore the system is Stable.

if we had y(t) = tx(t), then if t->infinity x(t) is still bounded, but it is being multiplied by a factor of infinity so your output y(t) -> infinity therefore is unstable.

 

[the_amateur]

@Larrytsai

you mentioned the following

here let x(t) be input that is bounded. As you let t-> inifinity, y(t) remains real and bounded, therefore the system is Stable.

if we had y(t) = tx(t), then if t->infinity x(t) is still bounded, but it is being multiplied by a factor of infinity so your output y(t) -> infinity therefore is unstable.

where x(t) can be any input signal. You say that in the system that i have posted the system is stable since x(t) is stable but i find by intuition that this is not the case if the input signal has discontinuities. At the point of discontinuity the output would then be a DIRAC delta $\delta(t)$ whose value is infinite at t = 0 (http://en.wikipedia.org/wiki/Dirac_delta_function" ). So by intuition the system is not stable.

The problem is I am unable to provide a rigorous proof to back my intuition. I believe my proof is wrong.

Any help would be appreciated. thanks!