A system is memoryless, according to our definition, if $y(t) = f(x(t))$. If $y(t) = f(t,x(t))$, then the system has memory. Memoryless systems are \emph{instantaneous systems}

\begin{example}
Consider an AM Modulator defined by $y(t) = \cos(\omega_0 t)x(t)$. This system has memory because it must keep track of time to know what the value of $\cos(\omega_0 t)$ is. 
\end{example}

\begin{example}
Consider a resistor where the voltage across the circuit is given by $y(t) = Rx(t)$. This system is memoryless. If the voltage was given by $y(t) = R(t)x(t)$, then the system has memory.
\end{example}

If we consider an LTI system, what must be true of $h(n)$ for the system to be memoryless? Let's look at an output signal $y(n)$, whose value is the convolution of an arbitrary input $x(n)$ and $h(n)$.

\begin{align*}
y(n) &= \sum \limits_{k=0}^{\infty} h(n)x(k-n) \\
y(n) &= \cdots + h(-1)x(n+1) + h(0)x(n) + h(1)x(n-1) + \cdots \\
\end{align*}

Note that in order to have an instantaneous system without memory, the only term should be $h(0)x(n)$. So
$$y(n) = \sum \limits_{\ell=0}^{\infty} h(n)x(\ell-n) = kx(n)$$
where $k \in Z$. Therefore, $h(n) =k\delta(n)$ in order for the system to be memoryless.


Let $M$ be the statement ``the system is memoryless''.


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


Is every memoryless system causal? In other words, is the following statement a tautology?
$$M \Rightarrow C$$
Since we define a memoryless system as a system whose output can be given by $y(t) = f(x(t))$, then we can prove this by using the definition of causality. Let $y_1(t) = f(x_1(t))$ be a memoryless system, and let $y_2(t) = f(x_2(t))$ be a memoryless system. It is clear that if $x_1(t) = x_2(t)$ $\forall t \leq T$, then $y_1(t) = y_2(t)$ $\forall t \leq T$, therefore, every memoryless system is causal, and the statement $M \Rightarrow C$ is a tautology.


Let $M$ be the statement ``the system is memoryless''.


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


Is every memoryless system time-invariant? In other words, is the following statement a tautology?
$$M \Rightarrow TI$$
We can also prove this with the definition of time-invariance. Let $y(t) = f(x(t))$, and let $\hat{x}(t) = x(t-T)$. Then we have $\hat{y}(t) = f(\hat{x}(t)) = f(x(t-T)) = y(t-T)$. Therefore, every memoryless system is time-invariant, and the statement $M \Rightarrow TI$ is a tautology.


Note that other textbooks say that this system is time-invariant: $y(t) = x(t)g(t)$. But note that $\hat{x}(t) = x(t-T) \Rightarrow \hat{y}(t) = \hat{x}(t)g(t) \neq y(t-T)$. Therefore, the system is not time-invariant. We can also deduce that the system is not memoryless since $\lnot TI \Rightarrow \lnot M$ is always true.


Given that a system is memoryless and $x(t_0) = x(t_1)$, we can deduce that $y(t_0) = y(t_1)$. In other words, $M \wedge \left( x(t_0)=x(t_1) \right) \Rightarrow y(t_0) = y(t_1)$.


To disprove memorylessness of a system, you can simply find a known memoryless signal and compare the signals where they have identical inputs at some time $t_0$. If $x_1(t_0) = x_2(t_0)$, then $y_1(t_0) \neq y_2(t_0) \Rightarrow$ the system has memory.  

Memory in Systems