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LTI System Property

  • #1
Is the following system stable. If so how.

y(t)= [itex]\frac{d}{dt}[/itex] 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 . [itex]\int^{\infty}_{-\infty} h(t) dt[/itex] < [itex]\infty[/itex] --------- 1


For the above system h(t) = [itex] \delta^{'}(t)[/itex]

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

So the system is not stable.



I think the above proof is way off the mark.
please provide the correct proof. thanks
 
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Answers and Replies

  • #2
43
0
The system (BIBO) stable if the impulse response is absolutely integrable, that is,

[itex]\int^{\infty}_{-\infty} |h(t)| dt < \infty [/itex]

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

[itex] h(t) = \delta (t)' [/itex].

Since

[itex] \int^{\infty}_{-\infty} | \delta (t)' | dt= \infty[/itex],

the system is not stable.


The system is LTI as

[itex]\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)[/itex]

and

[itex]\frac{d}{dt} x(t-\tau) = y(t-\tau)[/itex].
 
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  • #3
Thanks for your reply but i am trying to find if it is stable or not. i think my derivation is wrong.
 
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  • #4
228
0
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.
 
  • #5
@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 [itex]\delta(t)[/itex] whose value is infinite at t = 0 (http://en.wikipedia.org/wiki/Dirac_delta_function" [Broken]). 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!
 
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