Suppose $\sin(x(t))$. I want to prove whether $\sin(x(t))$ is both linear and time-invariant or not. Is this proof flawed, or is it sound and correct?

Time Invariance:

  • Suppose $x_1(t)$ is a particular input signal to the system.
  • Now, suppose $x_2(t)$ is $x_1(t)$ but shifted by $T$ time units such that $x_2(t) = x_1(t - T)$.
  • Suppose $y_1(t)$ and $y_2(t)$ are the responses of the input signals of the system.
  • To prove time-invariance, we must determine whether $y_1(t - T) = y_2(t)$ holds or not.

Therefore, we expand both sides:

$$LHS: y_1(t - T) = \sin(x_1(t - T))$$

$$RHS: y_2(t) = \sin(x_2(t))$$ $$sin(x_1(t - T))$$

Lemma 1: Therefore, $\sin(x(t))$ is a time-invariant system.


  • Suppose $x_1(t)$ and $x_2(t)$ represent two distinct input signals, and $y_1(t)$ and $y_2(t)$ represent the two responses of the input signals into the system, respectively. Therefore...

$$y_1(t) = \sin(x_1(t))$$ $$y_2(t) = \sin(x_2(t))$$

  • To prove linearity, we must prove two properties: the scaling property and the additive property.


  • To prove the scaling property, we must determine whether the following equation is true:

$$Ay(t) = \sin( Ax(t))$$

such that $A \in \mathbb{R}$.

  • After substitution...

$$A \sin(x(t)) \neq \sin(Ax(t))$$

  • Therefore, the system does not satisfy the scaling property.

  • Therefore, the system is not linear.

  • Therefore, the system is not both linear and time-invariant.

  • 1
    $\begingroup$ it is time-invariant, since $\sin( x(t+a)) = \sin(T_a x(t)) = T_a \sin(x(t)) = \sin(x(t+a))$ but not linear (since $\sin(x_1(t))+\sin(x_2(t)) \ne \sin(x_1(t)+x_2(t))$ $\endgroup$ – reuns Sep 30 '16 at 10:48

It is correct. You could also find out about the nonliterary by looking at the Taylor series expansion of $\sin(x(t))$:


the higher degrees of $x(t)$ are all nonlinear.


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