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:Why does z(t)=y(t-2)? | :Why does z(t)=y(t-2)? | ||
::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 | ::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 | ||
| + | :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). | ||
===Answer 2=== | ===Answer 2=== | ||
Revision as of 12:56, 5 February 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)
- 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)?
- 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).
Answer 2
Write it here.
Answer 3
Write it here.
