$\delta[n+1]$ equals to one only $n+1 = 0$ ?
If the 1. is true, so what does a signal $x[n+1] $ stand for? Is this signal causal or anticausal? I mean, if 1. is true, $n = -1$, so is this signal causal? (because it just happens before the time I want). But Oppenheim says that it is anticausal. Why?
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4$\begingroup$ Signals cannot be causal or anti-causal, only systems. $\endgroup$– Peter K. ♦Commented Apr 25, 2016 at 11:23
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2 Answers
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A discrete signal can be defined at different integer time indices. Generally, one talks about causal or anticausal systems. But since a signal can, when convolved, be interpreted as a system, some say that:
Causal signals are signals that are zero for all negative time, while anticausal are signals that are zero for all positive time.
You can read that in the Classification of signals. Accordingly to that definition, your signal would be anticausal. Oppenheim sounds correct. One interpretation is: if you convolve a signal $x$ by an anticausal signal, the result will start "sooner" that $x$, a breach in causality.
answered Apr 25, 2016 at 11:26
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$\begingroup$ Oh, it seems that anticausal is absolutely noncausal. I say a wrong concept.. What does $x[n+1]$ stand for? It is the signal 1s before or after the original signal? $\endgroup$ Commented Apr 25, 2016 at 12:57
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$\begingroup$ If you have $x$, and write $y[n] = x[n+1]$, then $x[n+1]$ is $1$ when $n=-1$, so $y[-1] =1$, before the original. $\endgroup$ Commented Apr 25, 2016 at 13:21
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$\begingroup$ **This system is not causal, since the current value of the output depends on a future value of the input. ** This sentence is copied from Oppnhiem. He says that $x[n+1]$ is the future value. It seems so weird.. $\endgroup$ Commented Apr 25, 2016 at 14:29
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$\begingroup$ Exactly, because to get $y[-1]$ you need $x[0]$, a future value regarding index $-1$ $\endgroup$ Commented Apr 25, 2016 at 15:21
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The signal $x(n+1)$ just means that it is a signal based on $x(n)$ but with the samples at other time instances, in particular, all the samples are at the previous time instance. For example, the value that $x(n)$ had at $n=0$, now would be at $n=-1$, because to obtain $x(0)$ with the new signal $x(n+1)$ you have to put the time index $n=-1$, so that $x(-1+1) = x(0)$.
Based on the definition that was given in Laurent Duval's answer, if your signal had a non zero value at $n=0$ then after the shift it would become non causal.
Hope that helps.
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$\begingroup$ What you mean is that the signal is based on a stanard original time$( t= 0)$. Is it right? $x(n+1)$ stands for the time before original signal? $\endgroup$ Commented Apr 25, 2016 at 13:00
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$\begingroup$ I wouldn't call it original time, but sure if it makes you understand it better. It just means that $x(n+1)$ is the same signal (same shape and values of the samples) but all samples are shifted to the LEFT in the time axis. $\endgroup$– boneCommented Apr 25, 2016 at 17:08