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| + | [[Category:signals and systems]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:system invertibility]] | ||
[[Category:ECE301Spring2011Boutin]] | [[Category:ECE301Spring2011Boutin]] | ||
[[Category:problem solving]] | [[Category:problem solving]] | ||
| − | = Practice Question on System Invertibility= | + | |
| + | |||
| + | <center><font size= 4> | ||
| + | '''[[Signals_and_systems_practice_problems_list|Practice Question on "Signals and Systems"]]''' | ||
| + | </font size> | ||
| + | |||
| + | |||
| + | [[Signals_and_systems_practice_problems_list|More Practice Problems]] | ||
| + | |||
| + | |||
| + | Topic: System Invertibility | ||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
| + | |||
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 | ||
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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. | 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. | ||
| + | ---- | ||
---- | ---- | ||
==Share your answers below== | ==Share your answers below== | ||
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--[[User:Cmcmican|Cmcmican]] 17:08, 24 January 2011 (UTC) | --[[User:Cmcmican|Cmcmican]] 17:08, 24 January 2011 (UTC) | ||
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| + | *<span style="color:green">Good job! For some reason, this is a problem that a lot of students get stuck on. -pm </span> | ||
| + | *Why does z(t)=y(t-2)? | ||
| + | ::<span style="color:green">Instructor's answer: This is by the definition of the second system: it time delays its input. So if the input were x(t), then the output would be x(t-2). In this case, the input is called y(t), so the output is then y(t-2). You may want to look at this [[Video_Tutorial_on_How_to_Cascade_Transformations_of_the_Independent_Variable|video]] for more clarification. -pm </span> | ||
| + | *My question was poorly articulated. I should have asked rather how the inverse of the function was found. Once I have the inverse, I understand how to cascade; I did not understand how the inverse of y(t)=x(t+2) is y(t)=x(t-2). However, after reviewing it again, I see that since y(t)=x(t+2) then y(t-2)=x(t). Therefore the inverse of the signal is y(t)=x(t-2). | ||
| + | ::<span style="color:green">Instructor's comment: I see. But you know, in general, inverting an invertible system can be quite challenging. There is no method that works all the time. If you are lucky and figure out how to isolate x(t) in terms of y (e.g., y(t), y(t+1), t y(t), stuff like that), like you cleverly did above, then you are good to go. But sometimes isolating x(t) is hard. In such cases, thinking logically about what that the system does might tell you the answer. Like in this particular case: the system is a time delay, and once you realize that, it's a no brainer to invert it without doing any math. In other cases (think cryptography here...) finding the answer might even be NP hard! -pm </span> | ||
| + | ---- | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
| + | ---- | ||
===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
---- | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|More Practice Problems on "Signals and Systems"]] | ||
| + | |||
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
Latest revision as of 04:59, 21 September 2013
Practice Question on "Signals and Systems"
Topic: System Invertibility
Question
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)
- Good job! For some reason, this is a problem that a lot of students get stuck on. -pm
- Why does z(t)=y(t-2)?
- Instructor's answer: This is by the definition of the second system: it time delays its input. So if the input were x(t), then the output would be x(t-2). In this case, the input is called y(t), so the output is then y(t-2). You may want to look at this video for more clarification. -pm
- My question was poorly articulated. I should have asked rather how the inverse of the function was found. Once I have the inverse, I understand how to cascade; I did not understand how the inverse of y(t)=x(t+2) is y(t)=x(t-2). However, after reviewing it again, I see that since y(t)=x(t+2) then y(t-2)=x(t). Therefore the inverse of the signal is y(t)=x(t-2).
- Instructor's comment: I see. But you know, in general, inverting an invertible system can be quite challenging. There is no method that works all the time. If you are lucky and figure out how to isolate x(t) in terms of y (e.g., y(t), y(t+1), t y(t), stuff like that), like you cleverly did above, then you are good to go. But sometimes isolating x(t) is hard. In such cases, thinking logically about what that the system does might tell you the answer. Like in this particular case: the system is a time delay, and once you realize that, it's a no brainer to invert it without doing any math. In other cases (think cryptography here...) finding the answer might even be NP hard! -pm
Answer 2
Write it here.
Answer 3
Write it here.
