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Textbook states Initial value theorem as follows:

if $x[n]$ is equal to zero for $n < 0$, the initial value, $x[0]$, may be found from $X(z)$ as follows:

$$ x[0] = \lim_{\ z \rightarrow \infty}\ X(z) $$

Ok...no problem there... then, I was working problems 4.16 in Schaum's outline for DSP.

$$ X(z) = \frac{z}{(1-0.5z^{-1})(1-0.6z^{-2})}\ \ \ \ |z|< 0.5 $$

The textbook gives function $X(z)$ and says applying "initial value theorem" to $X(z)$, we see that the limit doesn't exist therefore the system is non-casual.

Is applying the "initial value theorem" to a system sufficient to determine if a system is non-casual? or are they inferring this information in some other way?

Then, the book goes on to say, we can modify the system by adding delay to make "initial value theorem" converge to a value.

function is modified as follows:

$$ X(z) = \frac{1}{(1-0.5z^{-1})(1-0.6z^{-2})}\ \ \ \ |z|< 0.5 $$

Applying initial value theorem to this problem, we see that it converges to a finite value therefore the system is causal.

Again is the "initial value theorem" sufficient to prove a system is casual?

I guess i'm just confused because the only rule that the textbook gives that links casual systems to the initial value theorem is as a pre-requisite of the system being right-sided before applying the initial value theorem.

I was wondering what's the missing rule if there is one? I can kind of see that if the limit isn't met, that might imply the system is not casual, on the other hand, why can't a causal system have an infinite initial value, mathematically speaking... or is that something that just would never happen in the real world therefore it's always non-casual. and what about the other way around?

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    $\begingroup$ Your textbook definition is incomplete. It should read "If $x[n]=0$ for $n<0$, then $\lim_{z\to\infty}X[z]$ exists and is equal to $x[0]$, the initial value of the sequence." Note that $X[0] = \lim_{z\to\infty}X[z]$ as you write is nonsensical: some people might fudge notation a little bit and write $X(\infty)$ as short-hand or the more formal $\lim_{z\to\infty}X[z]$ but even such benighted folks will not use $X[0]$ as short-hand for the limit. Check your book! It probably says $$x[0]=\lim_{z\to\infty}X[z]$$ and not $$X[0]=\lim_{z\to\infty}X[z].$$ Upper and lower case matter.... $\endgroup$ – Dilip Sarwate Jan 4 at 21:22
  • $\begingroup$ sorry, typo with the uppercase character. $\endgroup$ – Bill Moore Jan 5 at 0:14
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    $\begingroup$ So how about editing your question to clean it up instead of just acknowledging it in a comment which many people won't even bother reading? $\endgroup$ – Dilip Sarwate Jan 5 at 3:23
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One important thing that hasn't been mentioned is the case when the limit does exist. If the limit exists, you cannot conclude that $X(z)$ is necessarily the transform of a causal sequence. Take as an example the function

$$X(z)=\frac{1}{(1-az^{-1})(1-bz^{-1})}\tag{1}$$

Clearly,

$$\lim_{z\to\infty}X(z)=1\tag{2}$$

yet, $X(z)$ as given by $(1)$ is not necessarily the transform of a causal sequence. There exist three sequences with different ROCs with $X(z)$ as their transform: one of them is causal, one is left-sided, and one is double-sided.

In sum, if the limit $\lim_{z\to\infty}X(z)$ does not exist, we can conclude that there is no causal sequence corresponding to the transform $X(z)$. If the limit exists, we know that there's at least one causal sequence corresponding to $X(z)$. However, there are usually also other, non-causal sequences with the same transform.

In your example, the fact that the ROC is given by $|z|<0.5$ clearly indicates that the corresponding sequence is left-sided, so there's no need to apply the initial value theorem to check causality.

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Basing on the definition of the z-transform, the initial value theorem can be written and proved as follows: $$ \lim_{\ z \rightarrow \infty}\ X(z) = \lim_{\ z \rightarrow \infty}\ \sum_{k=-\infty}^{\infty} f[k]z^{-k} $$

In the case that $f[k]=0\forall k < 0$ (causal), obviously this limit will converge to $f[0]$ (assuming the signal doesn't contain infinite values). If there exists any negative $k$ for which $f[k]\neq 0$, it doesn't converge.

Now concerning the values of the signal (disclaimer: I'm more of an engineer than a mathematician), in the discrete domain, I've never seen any case where a signal took infinite values, not even theoretically (unlike in the continuous domain, where the Dirac delta takes care of those cases). Having a signal taking infinite values in discrete time would be very problematic, as everything is defined as sums (instead of integrals).

To conclude, from my point of view, assuming $f[k]\in \mathbb{C}$, $$\lim_{\ z \rightarrow \infty}\ X(z)\in \mathbb{C} \Leftrightarrow f[.]\text{ is causal}$$ meaning that if the limit $X(z)$ is finite (may as well be complex), then $f[.]$ is causal.

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  • $\begingroup$ thanks! The part I was missing was "non-convergence if negative k present with value", that makes sense now. $\endgroup$ – Bill Moore Jan 5 at 0:22
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Begin with the power series expansion of the Z-transform of a sequence $x[n]$ :

$$ \begin{align} X(z) &= \sum_{n=-\infty}^{\infty} x[n] z^{-n} \\ &= ~... +~ x[-1] z^2 + x[-1] z + x[0] + x[1] z^{-1} + x[2] z^{-2} + ...\\ \end{align} $$

Now if the sequence $x[n]$ is causal then those samples of $x[n]$ for $n < 0$ are by definition zero and the Z-transform becomes:

$$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} = x[0] + x[1] z^{-1} + x[2] z^{-2} + ... $$

Now it's very easy to see that when $z$ is sent to infinity, $X(z) = x[0]$. If the signal is known to have all finite samples, then this limit will be finite for a causal signal. But if thes sequence is not causal, then there exist at least one nonzero sample $x[n]$ for $n < 0$ and the Z-transform includes positive power of $z$ and, therefore, now if you set the limit of $z$ to infinity the result will be infinite (or does not converge) which indicates that there is a nonzero sample to the left of $n=0$ hence the sequence is nancausal.

Yes it's sufficient to conclude (provided all samples are finite)

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