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.
Linearity
- 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.
Scaling
- 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.
It is correct. You could also find out about the nonliterary by looking at the Taylor series expansion of $\sin(x(t))$:
$$\sin(x(t))=x(t)-\frac{x^3(t)}{3!}+\frac{x^5(t)}{5!}+\cdots$$
the higher degrees of $x(t)$ are all nonlinear.