# System that has derivative of input is non causal

Consider a system $$y(t) = \dot{x}(t)$$ where $$y$$ is the output and $$x$$ is the input. Given an initial condition $$x_0$$ and two inputs $$x_1$$ and $$x_2$$ such that $$x_1(t)=x_2(t) , 0 \le t < t_0$$ the outputs $$y_1,y_2$$ corresponding to $$x_1,x_2$$ respectively should satisfy $$y_1(t)=y_2(t) , 0 \le t < t_0$$ since $$x_1(t)=x_2(t) , 0 \le t < t_0 \implies \dot{x}_1(t)=\dot{x}_2(t) , 0 < t < t_0$$ How is this system non causal then?

• What is the question? – Cherny Oct 29 '18 at 7:15
• Added the question. – Anant Joshi Oct 29 '18 at 11:00
• (1) Why do you think the system is non-causal? (2) How is the analysis you did related to causality? (double check the definition of causal system) – MBaz Oct 29 '18 at 13:49

Just for example, we can take 2 cases: $$x_1(t) = -t , x_2=|t| \Rightarrow \dot x_1(t)=-1, \dot x_2(t) =undefined$$ $$\Rightarrow y_1(t)\neq y_2(t)$$ despite $$\forall t\leq 0 : x_1(t)=x_2(t)$$