# Causality of z-transform $a^nu[n+1]$

To preface, this is not a homework related question but purely for self-study purposes.

I'm try to do the analyse of z-transform of $$a^nu[n+1]$$. It is clearly a non-causal signal, I try to explain it by using the definition.

Based on the causality of z-transform definition: If x(n) is a infinite duration causal sequence, ROC is exterior of the circle with radius a. i.e. $$|z| > |a|$$.

I am not sure if my explanation is correct: $$\mathcal{Z}\big\{a^nu[n+1]\big\} = a^{-1}z+\sum_{n=0}^{\infty} (a/z)^{n}$$

The ROC of this function in z-plane is $$|z| > |a|$$ except at $$z = ∞$$.

It does not obey the causality definition because the boundary exists ( $$|a|<|z|< ∞$$), i.e. the ROC pattern is a ring, it is a non-causal signal.

Am I correct? I am kinda confused with the causality definition.

• Signals just are. Systems can be non-causal. That signal description just says that it starts before $n = 0$ and remains non-zero to $n = +\infty$. Because, in signal processing convention, $n = 0$ means "now" and not the moment of the big bang, it's perfectly acceptable (if mathematically awkward) to have a signal that's nonzero for $n < 0$. Generally, if it happens (and it's not an impulse response, which does indicate a non-causal system), then it's most convenient to just change your time index. – TimWescott Jan 18 at 18:13

• Technically, you are absolutely right. Signals just are and systems can be characterized as causal or not. However, it is very common to call signals as causal if $$x(t) = 0, t < 0$$ and anti-causal if $$x(t) = 0, t> 0$$ The term is obviously borrowed from systems and their impulse response. – GKH Jan 18 at 21:20