Signals & Systems - Stable or Not?

In summary, the conversation discusses a system with impulse response h(t) = a_{1}\delta(t-t_{1}) + a_{2}\delta(t-t_{2}) and whether it is stable or not. The conversation also discusses how to integrate an impulse response and how to integrate the function \int_{-\infty}^{\infty}e^{-(t-\tau)}sin(t-\tau) d\tau. The suggestion is to express sin as a complex exponential and evaluate the resulting integrals.
  • #1
Angello90
65
0

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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  • #2
Write down the output of this system for an arbitrary input x(t). Is the output bounded if the input is finite?
 
  • #3
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
 
  • #4
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]
 
  • #5
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!
 
  • #6
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?
 
  • #7
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.
 
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FAQ: Signals & Systems - Stable or Not?

What is the definition of a stable signal?

A stable signal is one that does not grow infinitely or oscillate uncontrollably. In other words, it remains bounded and does not exhibit any unpredictable behavior.

How can we determine if a system is stable?

To determine stability, we can analyze the impulse response or step response of the system. If the response is bounded and approaches zero as time goes to infinity, the system is stable. Another method is to check the poles of the transfer function - if all poles are in the left half of the complex plane, the system is stable.

Can a system be stable for some inputs but not for others?

Yes, a system can be conditionally stable. This means that it is stable for certain inputs but unstable for others. It is important to analyze the system's stability for all possible inputs to ensure reliable performance.

What are the consequences of having an unstable system?

An unstable system can lead to unpredictable and undesirable behavior. It may cause the system to amplify or attenuate signals uncontrollably, resulting in distorted output. It can also cause the system to become unstable and fail to perform its intended function.

How can we ensure stability in a system?

To ensure stability, we can use techniques such as feedback control, filtering, and adjusting system parameters. It is also important to properly design and analyze the system before implementation to identify and address any potential stability issues.

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