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===Answer 1=== | ===Answer 1=== | ||
| − | + | Yes, this system is invertible. The inverse is <math>y(t)=x(t-2)</math> | |
| + | |||
| + | Proof: | ||
| + | |||
| + | <math>x(t) \to \Bigg[ system 1 \Bigg] \to y(t) = x(t+2) \to \Bigg[ inverse \Bigg] \to z(t) = y(t-2) = x((t-2)+2) = x(t)</math> | ||
| + | |||
| + | --[[User:Cmcmican|Cmcmican]] 17:08, 24 January 2011 (UTC) | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 13:08, 24 January 2011
Contents
Practice Question on System Invertibility
The input x(t) and the output y(t) of a system are related by the equation
$ y(t)=x(t+2) $
Is the system invertible (yes/no)? If you answered "yes", find the inverse of this system. If you answered "no", give a mathematical proof that the system is not invertible.
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
Yes, this system is invertible. The inverse is $ y(t)=x(t-2) $
Proof:
$ x(t) \to \Bigg[ system 1 \Bigg] \to y(t) = x(t+2) \to \Bigg[ inverse \Bigg] \to z(t) = y(t-2) = x((t-2)+2) = x(t) $
--Cmcmican 17:08, 24 January 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.
