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So I have a system with the following inputs and outputs:

\begin{align} x[n]&=\left( \frac12 \right)^{n}u[n] + 2^{n}u[-n-1]\\ y[n]&=6\left( \frac12 \right)^{n}u[n] - 6\left( \frac34 \right)^{n}u[n] \end{align}

The transfer function $H(z)$ can be found to be:

$$\frac{Y(z)}{X(z)}=\frac{1-2z^{-1}}{1-\frac{3}{4}z^{-1}}$$

with a zero at $2$, and a pole at $\frac34$. Now how can we find the ROC of this system given that one of the sequences ($x[n]$) has an anti-causal term?

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  • $\begingroup$ I see the point of your question now-- the input x[n] does not have a z-transform clearly due to no ROC overlap between causal and anti-causal parts, but given a transfer function that can have x[n] as an input (or any other input as the transfer function is independent of the input provided) can we also say the ROC does not exist? (I am not confident that it can or can't, just elaborating on your question). Does the stated input which is not the system then imply that the ROC of the system must use the bilateral transform where no ROC would exist? Not sure, will watch for a better answer. $\endgroup$ – Dan Boschen Mar 26 '17 at 14:00
  • $\begingroup$ So no answers yet but really interested in hearing what the experts have to say. Without input from others that may know more, I would say that the Z-transform must be defined for the transfer function given (whether it is unilateral or bilateral) in order to give the ROC, as the transfer function is independent of the input. However perhaps there is a convention that given the entire problem statement is defined with causal and non-causal inputs that the bilateral Z-Transform must be assumed. If no one answers please give us an update on what your professor says once you get the test back. $\endgroup$ – Dan Boschen Mar 26 '17 at 16:51
  • $\begingroup$ @DanBoschen: The input signal does have a Z-transform; its region of convergence is the annulus $\frac12<|z|<2$. $\endgroup$ – Matt L. Mar 26 '17 at 17:13
  • $\begingroup$ @MattL where I am getting confused is: Isn't $\sum_{-\infty}^0 u(-n)z^{-n} = \sum_0^{\infty}u(1/k)z^{-k}$ using the substitution that n = -1/k? Also in general do we need to assume bilateral transform for ROC of the transfer function itself given the input was non-causal or does that need to be explicitly stated? $\endgroup$ – Dan Boschen Mar 26 '17 at 17:50
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    $\begingroup$ @DanBoschen: that's right; the bilateral Z-transform naturally reduces to the unilateral transform if $x[n]=0$ for $n<0$, so I don't usually regard them as two "different Z-transforms". $\endgroup$ – Matt L. Mar 26 '17 at 19:39
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Let's look at the pole locations and the corresponding regions of convergence (ROCs) of the $\mathcal{Z}$-transforms of the input and output signals:

  1. $X(z)$: poles at $z=\frac12$ and $z=2$ with ROC $\frac12<|z|<2$ (two-sided).
  2. $Y(z)$: poles at $z=\frac12$ and $z=\frac34$ with ROC $|z|>\frac34$ (causal).

Now we know that the ROC of the multiplication of two $\mathcal{Z}$-transforms equals the intersection of the ROCs of the individual $\mathcal{Z}$-transforms, unless there is pole-zero cancellation, in which case the resulting ROC can be bigger than the intersection.

$H(z)$ has a zero at $z=2$ and a pole at $z=\frac34$. Since the zero cancels the pole of $X(z)$, the restriction $|z|<2$ for the ROC of $Y(z)$ is removed. The two theoretically possible ROCs of $H(z)$ are $|z|<\frac34$, corresponding to an anti-causal and unstable system, and $|z|>\frac34$, corresponding to a causal and stable system. In the first case, the ROC of $Y(z)$ would need to be $\frac12<|z|<\frac34$, which is not the case, and which would correspond to a two-sided unstable signal. $^1$ Consequently, the ROC of $H(z)$ must be $|z|>\frac34$, which means that the system described by $H(z)$ is causal and stable.


1. I use the term unstable signal for a signal that, if interpreted as the impulse response of an LTI system, would correspond to an unstable system (i.e., the ROC does not contain the unit circle).

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    $\begingroup$ On the second line in the last paragraph, don't you mean the restriction is removed for the ROC of $X(z)$ and not $Y(z)$? $\endgroup$ – Valentin Mar 26 '17 at 17:32
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    $\begingroup$ @Valentin: No, I meant the restriction for the ROC of $Y(z)$, because the ROC of $X(z)$ is determined by $x[n]$ and can't be changed, whereas the ROC of $Y(z)$ is determined by the ROCs of $X(z)$ and $H(z)$, and is changed by pole-zero cancellations. The pole-zero cancellation changes the restriction for the ROC of $Y(z)$ initially imposed by the ROC of $X(z)$. $\endgroup$ – Matt L. Mar 26 '17 at 18:57

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