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The input x(t) and the output y(t) of a system are related by the equation  
 
The input x(t) and the output y(t) of a system are related by the equation  
  
<math>y(t)=\frac{1}{1+x^2(t)}.</math>
+
<math>y(t)=\frac{ {\color{red} t }}{1+x^2(t)}.</math>
  
 
Is the system stable? Answer yes/no and ustify your answer.  
 
Is the system stable? Answer yes/no and ustify your answer.  
 +
:<span style="color:red">OOPS, I actually meant to put a "t" on top of the fraction (now in red). -pm</span>
 
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==Share your answers below==
 
==Share your answers below==
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--[[User:Cmcmican|Cmcmican]] 17:44, 24 January 2011 (UTC)
 
--[[User:Cmcmican|Cmcmican]] 17:44, 24 January 2011 (UTC)
 +
 +
:<span style="color:green">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. </span>
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:<span style="color:green"> Hint for this case: Look at the constant signal x(t)=1. -pm </span>
  
 
===Answer 2===
 
===Answer 2===

Revision as of 14:07, 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

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

Write it here.

Answer 3

Write it here.


Back to ECE301 Spring 2011 Prof. Boutin

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Ryne Rayburn