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While solving a problem, there is a problem that I don't understand. $$ Y(t) = \{X(t)\}^3 $$ A system has a response that is the cube of its excitation.It is a problem of determining the Invertibility of the system in question in continuous time.

The answer is "The cube root operation is multiple valued. Therefore the system is not invertible"

Are functions with multiple valued absolutely not invertible?

According to the answer from the professor in the subject, if a function like $X(t) = t^2$ is included in $X(t)$, it cannot be used because it becomes a counterexample.

Doesn't that mean all systems are not-invertible?

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The answer is "The cube root operation is multiple valued. Therefore the system is not invertible"

Maybe. That depends a bit on how you define it: if constrain $x(t)$ and $y(t)$ to be real it's perfectly invertible sine any real number has exactly one cube root.

So $$y(t) = x^3(t)\,\; x,y \in \mathbb{R} \tag{1}$$ is invertible but $$y(t) = x^3(t)\, \; x,y \in \mathbb{C} \tag{2}$$ is not.

For the same reason $y(t) = x^2(t)\,\; x,y \in \mathbb{R}$ is NOT invertible since square roots of positive real numbers are not unique and square roots of negative real numbers are complex.

According to the answer from the professor in the subject, if a function like $X(t) = t^2$ is included in $X(t)$, it cannot be used because it becomes a counterexample.

Sorry, I have no idea that sentence is supposed to mean. What exactly does "include in $X(t)$ " mean and a counterexample of what ?

Doesn't that mean all systems are not-invertible?

Off course not. There are plenty of invertible systems. Equation (1) above is invertible. Every LTI systems that's minimum phase is invertible (and causal and stable).

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  • $\begingroup$ Thank you very much, I understand it, According to the answer from the professor in the subject, if a function like X(t)=t2 is included in X(t), it cannot be used because it becomes a counterexample. What I was trying to say is, If put a function called $t^2$ in a function called $x(t)$, thinking that $x(t) = t^2$, it becomes one-to-one($\because y(t) = (x(t))^3 = t^6 $), so I heard that this is a counterexample to the idea that $y(t) = (x(t))^3 $ is invertible. $\endgroup$
    – quele
    Commented Oct 16, 2023 at 13:08

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