Signals & Systems - Stable or Not?

AI Thread Summary
The discussion centers on determining the stability of a system with an impulse response defined by h(t) = a1δ(t-t1) + a2δ(t-t2). It is suggested that the system is BIBO stable if the coefficients a1 and a2 are within the range of -1 to 1, ensuring bounded output for bounded input. Participants discuss the integration of the delta function and its properties, confirming that the integral of δ(τ-t) over all time equals one. Additionally, there are inquiries about integrating more complex expressions involving exponential and trigonometric functions, with suggestions to use variable changes and complex exponentials for simplification. The conversation highlights the challenges of understanding these concepts within an engineering context.
Angello90
Messages
65
Reaction score
0

Homework Statement


The system with impulse response h(t) = a_{1}\delta(t-t_{1}) + a_{2}\delta(t-t_{2}). 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 -1\leq a_{1} \leq 1 & -1\leq a_{2} \leq 1, 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

 

Attachments

  • Untitled-1.jpg
    Untitled-1.jpg
    9.2 KB · Views: 1,885
Physics news on Phys.org
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?
\int_{-\infty}^{\infty}\delta(\tau-t)d\tau

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

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

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
<br /> \int_{-\infty}^{\infty}\delta(\tau-t)d\tau =1<br />
 
Thanks mate, I remember that from a lecture except it was
\int_{-\infty}^{\infty}\delta(\tau)d\tau =1

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

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

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
\int_{-\infty}^{\infty}\delta(\tau)d\tau =1

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:
\int_{-\infty}^{\infty}e^{-(t-\tau)}sin(t-\tau) d\tau

I know I can bring one e to the front:
e^{-t}\int_{-\infty}^{\infty}e^{\tau}sin(t-\tau) d\tau
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:
Back
Top