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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 13:56, 5 February 2011

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.


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

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.


Back to ECE301 Spring 2011 Prof. Boutin

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn