Is this system linear?

Question

$$y(t)=\begin{cases}0 \hspace{4.86cm}x(t)<0\\ x(t)+x(t-2)\hspace{2cm} x(t)\geq0\end{cases}$$

Assuming $C>0$. If $x(t) = C$ is an input, the output will be $2C$, and the output of $-x(t)=-C$ must be $-2C$ to satisfy scaling. But when the input is negative, the output is 0.

Therefore, since scaling is impossible, it is not linear, right?

If I'm wrong, please let me know why.

Answer 1

No, this system is not linear.

Linearity requires:

1. additivity: $f(x_1+x_2)=f(x_1)+f(x_2)$
2. homogenuity: $f(\alpha x)=\alpha f(x)$ for all $\alpha$

Now, if we take $$ f[x(t)]=\begin{cases} 0, x(t)<0,\\x(t)+x(t-2), x(t)\geq 0\end{cases} $$ then for positive $\alpha$ we do have $f(\alpha x)=\alpha f(x)$, but for negative alpha we get $$ f(\alpha x(t)) = \begin{cases}0, x(t)<0,\\ \alpha x(t)+\alpha x(t-2), x(t)\geq 0 \end{cases}\neq \alpha f(x) $$

One could also check the additivity condition, which becomes tricky, if, e.g., $x_1(t)>0$ and $x_2(t)<0$.