# $y(n) = x(-n)$ , causal or not , memory or memoryless?

$y(n) = x(-n)$ , causal or not , memory or memory-less ?

it's a question in digital signal processing course .

My guess it's memory less , causal because $x(-n)$ is only the inverse of the function ?

Memoryless System A system is memoryless if the output y[n] at every value of n depends only on the input x[n] at the same value of n

Causality A system is causal it’s output is a function of only the current and previous samples

• It helps (you and the others) that the problem statement has a subject and a predicate. Eg: Is the filter/system defined by the relation $y[n]=x[-n]$ causal or not... ? ($x[n]$ is the input and $y[n]$ the output? Do you know the definition of a memory-less filter? What is the output at time, say $n=5$? – leonbloy Jun 1 '12 at 23:39
• I got it in the sheet like , tell whether the system is memory or memory less, causal or non – xsari3x Jun 1 '12 at 23:56
• well, a system is memoriles if output y[n] for every value of n (say, for n=5, y) depends only of the input at the same instant (x) But y does depend rather on x[-5], so... – leonbloy Jun 2 '12 at 1:02

• It is not causal. Consider $n=-1$. The output $y(-1)$ depends on the input at $n=1$, which is only available in future.
• It is not memoryless. Consider any $n > 0$. The output $y$ depends on the input at $-n$ which was applied in the past. Thus the system has memory.