Signals & Systems - Stable or Not?

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Discussion Overview

The discussion revolves around the stability of a system characterized by an impulse response involving delta functions, as well as various integration techniques related to impulse responses and other functions. Participants explore concepts of bounded input-bounded output (BIBO) stability, integration of delta functions, and integration of exponential and trigonometric functions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests that the system is stable if the coefficients of the delta functions are within the range -1 to 1, proposing that this would ensure bounded output for bounded input.
  • Another participant asks for the output of the system given an arbitrary input and questions whether the output remains bounded if the input is finite.
  • A participant inquires about the integration of the impulse response, specifically the integral of a delta function, and proposes a relationship involving the unit step function.
  • One response clarifies that the integral of a delta function over an interval that includes its argument is equal to one.
  • A participant expresses confusion about the abstract nature of the module and recalls a related lecture on the integral of the delta function.
  • Another participant presents a new integration problem involving an exponential and sine function, discussing potential approaches including integration by parts and the use of trigonometric identities.
  • One participant suggests expressing sine as a complex exponential to facilitate the integration process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the stability of the system, and multiple viewpoints regarding integration techniques and their applications are presented. The discussion remains unresolved on several points, particularly regarding the integration of the sine function and the implications for system stability.

Contextual Notes

Some participants express uncertainty regarding the assumptions needed for stability analysis and the conditions under which the integrals are evaluated. There are unresolved mathematical steps in the integration problems presented.

Angello90
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Homework Statement


The system with impulse response [tex]h(t) = a_{1}\delta(t-t_{1}) + a_{2}\delta(t-t_{2})[/tex]. Is this system stable or not?

Homework Equations


The Attempt at a Solution


I drew the h(t) response and I got something like that:
attachment.php?attachmentid=30358&stc=1&d=1291478969.jpg

but I don't really know how to prove it. There is no input give, only input response.
I would say that it is iff [tex]-1\leq a_{1} \leq 1[/tex] & [tex]-1\leq a_{2} \leq 1[/tex], since for this values input (if it was a unit step, say) would be bounded, thus producing bounded output. I.E. BIBO stable. Is that right?

Thanks guys!

Homework Statement


Homework Equations


The Attempt at a Solution

 

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Write down the output of this system for an arbitrary input x(t). Is the output bounded if the input is finite?
 
Ok so I think I have the main idea but one question.
How one may integrate impulse response?
[tex]\int_{-\infty}^{\infty}\delta(\tau-t)d\tau[/tex]

I know that for step unit, integral would be:
[tex]\int_{-\infty}^{\infty}u(\tau-t)d\tau = \int_{t}^{\infty}d\tau[/tex]
since step unit is 0 before t.

Would it be true to say that:
[tex]\int_{-\infty}^{\infty}\delta(\tau-t)d\tau = \int_{0}^{t}d\tau[/tex]

Thanks
 
No, the integral of a delta function is one so long as the interval includes the point at which its argument is zero. In the case you gave, the delta function is zero except at tau=t, but this is in the interval -inf to +inf so
[tex] \int_{-\infty}^{\infty}\delta(\tau-t)d\tau =1[/tex]
 
Thanks mate, I remember that from a lecture except it was
[tex]\int_{-\infty}^{\infty}\delta(\tau)d\tau =1[/tex]

Anyway thanks! This module is so confusing. Almost too abstract as for engineering course!
 
I got another problem though.
How should I integrate:
[tex]\int_{-\infty}^{\infty}e^{-(t-\tau)}sin(t-\tau) d\tau[/tex]

I know I can bring one e to the front:
[tex]e^{-t}\int_{-\infty}^{\infty}e^{\tau}sin(t-\tau) d\tau[/tex]
but what do I do with sin()? I used trigonometric property on it:
[tex]sin(A+B) = sin(A)cos(B)-cos(A)sin(B)[/tex]
and got:
[tex]e^{-t}(sin(t)-cos(t))\int_{-\infty}^{\infty}e^{\tau}(cos(\tau)-sin(\tau)) d\tau[/tex]

Should I expand it into two integrals and use integration by parts? Or am I wasting my time?
 
Angello90 said:
Thanks mate, I remember that from a lecture except it was
[tex]\int_{-\infty}^{\infty}\delta(\tau)d\tau =1[/tex]

Anyway thanks! This module is so confusing. Almost too abstract as for engineering course!
If you are integrating from -inf to +inf, the offset is irrelevant. Change variables u=tau-t and evaluate the integral, to prove it to yourself.
Angello90 said:
I got another problem though.
How should I integrate:
[tex]\int_{-\infty}^{\infty}e^{-(t-\tau)}sin(t-\tau) d\tau[/tex]

I know I can bring one e to the front:
[tex]e^{-t}\int_{-\infty}^{\infty}e^{\tau}sin(t-\tau) d\tau[/tex]
but what do I do with sin()?
I would express sin as a complex exponential. You need to think a little about what the resulting integrals evaluate to.
 
Last edited:

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