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.

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