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.

System Properties

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    $\begingroup$ 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. $\endgroup$ Jan 25, 2015 at 15:43

1 Answer 1


Such questions have been asked here before, and the concepts are simple and important. Dilip Sarwate is right to ask you to add details as to where you get stuck, so we can give better answers (instead of just solving the problem for you).

I'll try to explain the basics in a clear way so that this answer may also be useful for other users, and to help you understand and solve the problem yourself. Assuming you understand linearity, let's move on to the other properties:

  • 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.
  • $\begingroup$ Thank you very much, understood the terms very well but I didn't understand why System is non Causal. Could you elaborate that part please? Because the system just negates the sign, currently say we have input 9 it will come out as -9, will a future input, say 7 can change this -9? or how... $\endgroup$
    – Anarkie
    Jan 25, 2015 at 20:53
  • $\begingroup$ @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. $\endgroup$
    – Matt L.
    Jan 25, 2015 at 20:57
  • $\begingroup$ Nice explanation and clearly understand these concepts. But i have a question about the part E stability. when we consider the summation on a finite interval and say 0 to 10. then the system E is stable? $\endgroup$
    – user17873
    Oct 19, 2015 at 6:12
  • $\begingroup$ @user17873: Yes, because then the output will always be bounded for a bounded input signal. $\endgroup$
    – Matt L.
    Oct 25, 2015 at 14:57

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