Given a signal with a list of discrete values 1,2,3 (t=0),0,1,

Whether the signal is causal when the arrow is in 3 and why? I know the definition of causal is that the output will not depend on future input, but how to apply the definition in this question?

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    $\begingroup$ A system is called causal if the system output $y[n]$ at time $n$ does not depend at all on the input values $x[n+1], x[n+2], x[n+3], \cdots$ all of which are still in the future at time $n$. A signal does not have an input or output and so it cannot be said to be causal. $\endgroup$ Commented Feb 15, 2015 at 22:14
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    $\begingroup$ @DilipSarwate: I agree with you. Nevertheless it is quite common to speak of causal signals as defined here. $\endgroup$
    – Matt L.
    Commented Feb 15, 2015 at 22:25
  • $\begingroup$ Thanks. I understand now by check the link which @DilipSarwate mentioned above. $\endgroup$
    – Joe
    Commented Feb 15, 2015 at 22:41
  • $\begingroup$ @MattL.I have a following question. How did you use the theory which defines the system causal to explain the definition of causal function in the link? Thank you! $\endgroup$
    – Joe
    Commented Feb 15, 2015 at 22:51

1 Answer 1


A signal x[n] is said to be causal if it could be the impulse response of a causal LTI system. That is, a signal is causal if x[n] = 0 when n < 0.

  • $\begingroup$ So the signal is h[n] = [4,0,1,0,2] I was hoping someone can answer this question by doing z transform and then judging from ROC $\endgroup$
    – Joe
    Commented Feb 15, 2015 at 22:22
  • $\begingroup$ If the signal is different from 0 for n < 0 then it isn't causal. It isn't necessary to calculate the z-transform. But the z-transform of a finite causal signal would not have terms with positive exponents in z (check the definition). Then the ROC of a finite duration causal signal would be the whole z-plane except for 0 (including infinity). $\endgroup$ Commented Feb 15, 2015 at 22:31

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