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As far as I have grasped the concept is that if we are given the expression

$$y[n]=(2x[n]-x^2[n])^2\,,$$

it is a memory less system because even if we give negative values of $n$, we still get the overall result in the positive sign, right?

And this expression is memory system because we get the negative answer for negative values of $n$, right?

$$y[n]=(2x[n]-x^2[n])\,.$$

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  • $\begingroup$ Do you need more details before validating an answer? $\endgroup$ – Laurent Duval Nov 28 '18 at 19:48
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A confusion may arise, for causal systems, from mistaking "having negative signal values (amplitudes)" and "depending on negative time indices".

A strict memory-less system does depend neither on past (for the causal case) nor on future values, only the current ones, to determine the current value of the output. The output at $n$ only depends on inputs at $n$. For a strict memory-less and causal system, you could not depend on former (negative, in a way relative to the current index) time indices.

By extension, some allow memory-less systems to allow a constant delay, i.e. to hold only one value at a given lag: the output at $n$ only depends on inputs at $n-k$ of a fixed integer $k$: So, globally, memory-less system can afford the following shape, with $k$ any integer:

$$y[n] = f(x[n-k])\,.$$

For realizable causal systems, that cannot look ahead in the future, on:y $k\ge 0 $ are allowed.

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    $\begingroup$ Hi! is it, "...it does not depend either on A or on B..." OR " it depends neither on A nor B..." $\endgroup$ – Fat32 Sep 18 '18 at 9:08
  • $\begingroup$ Thx, my leaking English... $\endgroup$ – Laurent Duval Sep 18 '18 at 9:10
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The memoryless system means a system which does not use past values of the input. There is nothing to do with the sign of your output. That just means there isn't any storage element in your system.

Although reverse is not necessarily true every time, that implies also casuality.

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