# Causal non-causal sequences

This answer says that $Y = zX$ is a simplest example of non-causal system because it corresponds to $y_n = x_{n+1}$ and current output depends on future input. Yet, it is causal because both $x_n$ and $y_n$ are 0 for $n < 0.$ How causal can be non-causal? I am struggling to resolve this absurd and want to ask about the relationship between these two notions of causality and also the 3rd one, which seems to be related to pole values and ROC rather than number of poles?

First, we can speak of a causal filter or a causal function (or sequence, or signal).

In general, a filter is causal if its output at present time ($n$) never depends on the input at future times ($n+m$, with $m>0$). Let us restrict to LTI filters (and we assume discrete time) - so that the filter is fully specified by an impulse response function $h(n)$. In that case, the above property can be concisely stated as follows:

• A LTI discrete-time filter is causal iff $h(n)=0$ for $n<0$

This motivates the definition of a causal signal.

• A discrete-time function (signal, sequence) $g(n)$ is causal iff $g(n)=0$ for $n<0$

Notice that this later definition does not involve filters (it's just motivated by them). And notice that the two can be combined in:

• A LTI discrete-time filter is causal iff its response function $h(n)$ is a causal function.

In your assertion regarding the second link ("it is causal because both $x_n$ and $y_n$ are $0$ for $n<0$") you seem to be confusing both meanings. To determine that a filter is causal, one does not look for the causality of inputs or ouputs ($x_n$ and $y_n$) but for the causality of the response function $h_n$

Further, instead of $h(n)$ we can work with its Z-transform $H(z)$; we have $y[n] = x[n] \star h[n] \implies Y(z)=H(z) \, X(z)$. But, remember the relation "signal" $\leftrightarrow$ "Z transform" is not one-to-one unless the ROC (region of convergence) is also specified. A single $H(z)$ can have several corresponding $h[n]$("anti-transform"), for different ROCs. Alternatively, instead of giving a ROC, we might be given a causal (or anticausal) condition. In particular, if we are given a (rational) $H(z)$ and we are told that $h(n)$ is causal, then the ROC must extend outwards from the biggest pole. For an explanation of this, see any Signal Processing textbook, or here.

In your example, $Y(z)= z X(z)$, so $H(z)=z$. To analyze this you can reason in two ways:

1) In terms of zeros and poles. $H(z)$ has a zero at $z=0$ and a pole at infinity. Hence, because there is a single ROC (all the plane), and it extends inwards from the pole, hence there can be only one valid $h(n)$ which must be anti-causal.

(I insist: here you could deduce from $H(z)$ that the filter was anti-causal; but normally you can't; say, if $H(z)=z/(z-1)$ you'd have two possible $h[n]$, one causal, one anticausal).

2) Explicitly, formally. By inspection, $H(z)=\sum_{n} h(n) z^{-n} = z \implies h(n)=\delta(n+1)$ Which is anticausal.

• Thanks. I just did not get how the non-causal Y = zX is causal, how reading data from the future is reconciled with being causal filters/signals? – Val Jul 31 '14 at 14:38
• Y = zX is not causal – leonbloy Jul 31 '14 at 14:39
• Really? Do you understand my question? Or you are saying that there is a principal difference between not and non in English? – Val Jul 31 '14 at 14:48
• I just don't understand why you insist on saying "$Y = zX$ is causal". That's false. – leonbloy Jul 31 '14 at 15:08
• @leonboy It is explained several times. At first in the referenced answer and then repeated in the first sentence of my question. I can say once more that Y = zX corresponds to $y_n = x_{n+1}$, which means that you compute current output based on the future input, which means non-casualty. But, I see no reason why you will notice this after one more repetition. – Val Jul 31 '14 at 15:49