\begin{definition}
A system $F$ is causal $\iff$ $\forall T\in\R, \forall x_1,x_2 \in X$ such that $x_1(t) = x_2(t)$ $\forall t \leq T$ produce  $y_1(t) = y_2(t)$ $\forall t \leq T$. In general, we say that a causal system doesn't use future values. A causal system may use present or past values only. In other words, a causal system is \emph{nonanticipitory}. In general, to process information in real-time, a system must be causal.

\end{definition}

\begin{example}
Here is an example of a system defined by its output signal:
$
y(n) = 
\begin{cases}
0 & \mbox{if $n$ is odd} \\
1 & \mbox{if $n$  is even} \\
\end{cases}
$

The system must be causal. The method here is to do a brute force proof.

\begin{proof}
Let $x_1(n),x_2(n) \in X$ be arbitrary. In this case, we have $y_1(n) = y_2(n)$ $\forall n$.
\end{proof}

\end{example}

\begin{example}
Here we have a system given by
$$ \delta(n) \to \fbox{F} \to \delta(n+1) + \delta(n) $$

This is only a partial characterization of $F$, thus, the system can be causal, but doesn't have to be. The method here is to show two examples of why both parts of the previous statement are true.

\begin{enumerate}
\item Let $F$ be defined by $y(n) = \begin{cases} 1 & \mbox{if $n\in\{-1,0\}$} \\ 0 & \mbox{otherwise} \\ \end{cases}$. Then this system is causal, because any two arbitrary functions will have the same output.
\item Let $F$ be defined by $y(n) = x(n) + x(n+1)$. This system is not causal since it ``looks into the future''.
\end{enumerate}

\end{example}

\begin{example}
Here we have a system given by
$$ \delta(n) \to \fbox{G} \to \delta(n+1) + \delta(n) $$
where $G$ is linear. Here, the system cannot be causal. The method here is to compare with another signal. If we pass in the zero signal, then in order for the system to be causal, the outputs should be identical up to $n=0$. But the outputs are not the same at $n=-1$, therefore the system is not causal. We can use this property because of the ZIZO property of linear systems. Another method we can use is to scale the input by $k$. Then $x_1(n) = x_2(n)$ $\forall n \leq -1 \wedge y_1(-1) \neq y_2(-1) \Rightarrow$ the system is not causal.
\end{example}

Let $P$ be the statement ``the output of the system precedes the input''

Let $L$ be the statement ``the system is linear''

Let $C$ be the statement ``the system is causal''

Then given a system that is linear:
\begin{align*}
&L \wedge C \Rightarrow \lnot P \\
&P \Rightarrow \lnot \left( L \wedge C \right) \\
&P \Rightarrow \lnot L \vee \lnot C \\ 
&\lnot C \quad \mbox{since we know $L$ is true}
\end{align*}

\begin{example}
Here we have a system given by
$$ \delta(n) \to \fbox{H} \to \delta(n+1) + \delta(n) $$
where $H$ is time invariant. The system cannot be causal. Here, the method is to compare with a time-shifted version of the input signal. 
\begin{proof}
Let $x_2(n) = x_1(n-1)$. Then $x_1(n) = x_2(n)$ $\forall n \leq -1 \wedge y_1(-1) \neq y_2(-1)$. 
\end{proof}
\end{example}

Let $TI$ be the statement ``the system is time invariant''

Then given $TI$ is true:
\begin{align*}
&TI \wedge C \Rightarrow \lnot P \\
&P \Rightarrow \lnot TI \vee \lnot C\\
&\lnot C \quad \mbox{ since we know $TI$ is true}
\end{align*}

Note that the output preceding the input doesn't preclude causality unless additional information is given, such as time invariance or linearity.


Here are some more examples:
\begin{enumerate}
\item $y(t) = cos(x(t))$ is casual because it only depends on the current time $t$.
\item $y(t) = x(cos(t))$ is not causal, since $y(0)  = x(1)$.
\end{enumerate}

Causality