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

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.

ringo
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1 Answers1

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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 $x_2(t)<0$.

Roger V.
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  • Why is the result of negative α f(αx(t)) ≠ αf(x(t))? Please tell me in more detail. –  Oct 08 '21 at 16:14
  • @ringo because $\alpha x(t)>0$, when $x(t)<0$ and vice versa, so different cases apply. – Roger V. Oct 08 '21 at 16:51

  • How is αf(x) expressed as a formula when alpha is negative?

  • 'we get' In the following formula, isn't the range αx(t), not x(t)? (such as f(αx(t)) = 0, αx(t)<0 )

    • –  Oct 08 '21 at 17:20
  • @ringo define $x_1(t)=\alpha x(t)$ and evaluate $f[x_1(t)]$. It is rally about working through the math. – Roger V. Oct 08 '21 at 18:45
  • OK, Thanks for your answer. – ringo Oct 09 '21 at 04:10