# System Properties; Linear, Causal, Time-Invariant, Stable?

I know the answers of the below question but I dont know why, except linear, could you explain the rest? Small tick: correct, small tick with dash incorrect answer.

• Have you tried to understand what is the meaning of time-invariance, causality, stability, etc., and attempted to apply any of your knowledge to the systems under discussion? If so, how about editing your question to include some of this work? (Don't bother answering this query as a comment: edit your question to include these responses) -1 pending such edits. – Dilip Sarwate Jan 25 '15 at 15:43

• time-invariance: if $y[n]$ is the response to input $x[n]$, $y[n-m]$ must be the response to $x[n-m]$ for any (integer) $m$. By inspection, this is obviously the case for systems A, B, and D. System C can get slightly confusing for beginners, but just use the criterion to see that for $x[n-m]$ the output is $x[-n-m]$ which does not equal $y[n-m]=x[-(n-m)]=x[-n+m]$. So system C is NOT time-invariant. For system E a simple substitution of the summation index shows you that the system is indeed time-invariant.
• causality: this is actually very simple. Just answer the question "does the output signal at any time depend on future values of the input signal?" If the answer is no, then the system is causal, otherwise it isn't. Systems A, B, and D only use the current input value to compute the output, so consequently they are all causal. System E uses the current and past input values, so it's also causal. Only system C is not causal because for negative values of $n$ the output depends on future values of the input.
• stability: usually the concept of bounded-input-bounded-output (BIBO) stability is used, which means that if the input signal is bounded, i.e. $|x[n]|\le M$, then also the output signal is bounded, i.e. $|y[n]|\le N$, for some arbitrary positive constants $M$ and $N$. For systems A, B, C, and D this is obviously the case, because the output can never be unbounded for bounded $x[n]$ (assuming that the constants $A$ and $B$ are finite, of course). For system E that's different, because you have an infinite sum which needs to converge for the output to be finite. But for a bounded input signal you can always construct a case where the output is unbounded. Just choose a constant input signal with any finite value, and the sum will grow without bounds. Hence, system E is not BIBO stable.
• @Anarkie: System C doesn't change the sign of the signal but of the time index. So e.g., $y[3]=x[-3]$, which is no problem for causality. But $y[-3]=x[3]$ is a problem because at time $n=-3$ we need to know the future input value at time $n=+3$, i.e. 6 samples later. – Matt L. Jan 25 '15 at 20:57