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Now that it has a t on top, it's not bounded.
 
Now that it has a t on top, it's not bounded.
  
If you consider the constant signal x(t)=1, then <math>y(t) = \frac{{t }}{1+1^2} = \frac{{t }}{2}</math>, which is not bounded.
+
If you consider the constant signal x(t)=1, then <math class="inline">y(t) = \frac{{t }}{1+1^2} = \frac{{t }}{2}</math>, which is not bounded.
  
 
--[[User:Cmcmican|Cmcmican]] 19:26, 24 January 2011 (UTC)
 
--[[User:Cmcmican|Cmcmican]] 19:26, 24 January 2011 (UTC)
 
+
:<span style="color:green">Good! And what if there was no t on top? -pm </span>
 
===Answer 3===
 
===Answer 3===
 
Write it here.
 
Write it here.
 
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[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]
 
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]]

Revision as of 15:44, 24 January 2011

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

Share your answers below

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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

Write it here.


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