Memorylessness of simple delay system

Question

As usual, $y(t)$ is the output signal of a system, and $x(t)$ is the input signal. I'm wondering whether or not a certain system has memory.

It's easy for me to see that the system

$y(t) = \int_{t-T}^{t} x(t) dt$ where $T > 0$

has memory, because an output at time $t_0$ depends on all values of input from $[t_0 - T, t_0]$.

But what about the following system, which just delays input by some constant $C$:

$y(t) = x(t - C)$ where $ C > 0 $

Upon first glance, this doen't feel like it has memory because there is no integral. But the output of a delay system does depend upon inputs from the past. Does that mean a simple constant delay system has memory?

Answer 1

A system is memoryless if its output ($y(t)$) for each value of the independent variable ($t$ in this case) at a given time is dependent only on the input at that same time ($x(t)$).

Every system that consists of a delay (like the one in your example) or an accumulator, for example, are systems with memory.

This can be seen just by replacing $t$ by some number. Take, for instance, $t=0$. In that case, the output for that instant ($y(0)$) depends on the input for another value of the independent variable ($x(C),C>0$); that means that the system has memory.

Answer 2

Yes, simple delay system has memory. In fact simple delay system has to store (remember) infinite no: of input values. It has to store all the inputs 'separately' from (t-C) to present instant.

Whereas the integrator block, even though the o/p still depends on infinite no: of past inputs, doesn't have to store each input separately. Only the sum of previous inputs needs to be stored, which is a single quantity.

Hence the amount of memory required is less in an integrator compared to a simple delay block.