| Line 49: | Line 49: | ||
=== Answer 3 === | === Answer 3 === | ||
| − | If there is not a t on top (i.e it is back to being a '1'), then the signal is bounded. | + | If there is not a t on top (i.e it is back to being a '1'), then the signal is bounded*. |
Considering the case where <math>|x(t)| \le \infty</math> then <math>0<\frac{{1}}{1+x^2(t)}\le1</math>. | Considering the case where <math>|x(t)| \le \infty</math> then <math>0<\frac{{1}}{1+x^2(t)}\le1</math>. | ||
| − | <math>\therefore y(t)</math> is bounded by <math> | + | <math>\therefore y(t)</math> is bounded by <math>M = \pm 1</math> |
| − | '''Addendum''': This only works for <math>x(t) \in \Re</math> as there are imaginary values that cause it to be unstable. | + | |
| + | '''*Addendum''': This only works for <math>x(t) \in \Re</math> as there are imaginary values that cause it to be unstable. | ||
--[[User:Darichar|Darichar]] 14:05, 26 January 2011 (UTC) | --[[User:Darichar|Darichar]] 14:05, 26 January 2011 (UTC) | ||
Revision as of 11:34, 26 January 2011
Contents
Practice Question on System Stability
The input x(t) and the output y(t) of a system are related by the equation
$ y(t)=\frac{ {\color{red} t }}{1+x^2(t)}. $
Is the system stable? Answer yes/no and ustify your answer.
- OOPS, I actually meant to put a "t" on top of the fraction (now in red). -pm
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
This system is stable. I'm actually not sure how to show this, does the following logic work?
$ \lim_{x(t) \to 0}\frac{1}{1+x^2(t)} = 1 $ and $ \frac{1}{1+x^2(t)} < 1 $ for all x(t), thus the system is stable.
I'm not sure that the justification works here...
--Cmcmican 17:44, 24 January 2011 (UTC)
- Unfortunately no. Here is how you should go about answering such questions. If you think it is stable,
- then assume that x(t) is bounded (i.e., |x(t)|<m ) and then try to show that y(t) is also bounded (|y(t)<M ).
- If you think it is not stable, then try to think of a bounded signal x(t) for which y(t) would not be bounded.
- Hint for this case: Look at the constant signal x(t)=1. -pm
Answer 2
Now that it has a t on top, it's not bounded.
If you consider the constant signal x(t)=1, then $ y(t) = \frac{{t }}{1+1^2} = \frac{{t }}{2} $, which is not bounded.
--Cmcmican 19:26, 24 January 2011 (UTC)
- Good! And what if there was no t on top? -pm
Answer 3
If there is not a t on top (i.e it is back to being a '1'), then the signal is bounded*.
Considering the case where $ |x(t)| \le \infty $ then $ 0<\frac{{1}}{1+x^2(t)}\le1 $.
$ \therefore y(t) $ is bounded by $ M = \pm 1 $
*Addendum: This only works for $ x(t) \in \Re $ as there are imaginary values that cause it to be unstable.
--Darichar 14:05, 26 January 2011 (UTC)
