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I have got two different answers for the ROC of the signal. In that PIC, I have solved it in 2 methods, but I'm getting different answer. Which one is correct? Also please explain how to find the ROC of shifted signal?

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  • $\begingroup$ Sorry ,in that first case x(z)=1+0.5 z^-1 $\endgroup$ – Ajay vishwanath Nov 1 '15 at 12:34
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The $\mathcal{Z}$-transform of your signal is

$$X(z)=1+\frac12 z^{-1}\tag{1}$$

(as you figured out by yourself).

The ROC is obviously the whole $z$-plane except for $z=0$, because there's a pole at $z=0$. Note that the ROC of the $\mathcal{Z}$-transform of any finite length sequence must be the whole $z$-plane, possibly except for $z=0$ (for sequences with non-zero values for $n>0$), and $z=\infty$ (for sequences with non-zero values for $n<0$). So by merely looking at the sequence you could determine its ROC.

The reason why your second method of determining the ROC doesn't work is because when you add the two $\mathcal{Z}$-transforms you get a pole/zero cancellation:

$$X(z)=\frac{1}{1-\frac12 z^{-1}}-\frac{\frac14 z^{-2}}{1-\frac12 z^{-1}}=\frac{(1-\frac12 z^{-1})(1+\frac12 z^{-1})}{1-\frac12 z^{-1}}=1+\frac12 z^{-1}$$

So the restricted ROCs of both functions caused by the pole don't show up in the final result due to this pole/zero cancellation.

Note that you could express the unit impulse by $\delta[n]=a^nu[n]-a^nu[n-1]$ and obtain arbitrary ROCs of the separate $\mathcal{Z}$-transforms of the two terms on the right-hand side, depending on the choice of $a$. However, the ROC of the $\mathcal{Z}$-transform of $\delta[n]$ (which is $1$) is obviously the whole $z$-plane, including $z=0$ and $z=\infty$.

In general, if you have a $\mathcal{Z}$-transform $X(z)$ and you shift the corresponding sequence to the right by $k$ samples, $X(z)$ gets multiplied by $z^{-k}$. So all poles and zeros of $X(z)$ remain unchanged, and a pole at $z=0$ and a zero at $z=\infty$ are added, both with multiplicity $k$. Of course they can cancel with already existing poles or zeros. Analogously, if you shift the sequence to the left by $k$ samples, $X(z)$ gets multiplied by $z^k$, which adds a $k$-fold pole at $z=\infty$ and a $k$-fold zero at $z=0$. As an example, take the sequence $a^nu[n]$ with $|a|<0$. Its $\mathcal{Z}$-transform is

$$X(z)=\frac{z}{z-a}$$

with a pole at $z=a$ and a zero at $z=0$. If you shift the corresponding sequence to the right by one sample, the resulting $\mathcal{Z}$-transform is $z^{-1}X(z)$. The original zero at $z=0$ gets cancelled by the new pole at $z=0$, and you're left with the original pole at $z=a$ and a new zero at $z=\infty$. If you shift the sequence to the left by one sample you get $zX(z)$, which gives a double zero at $z=0$, the original pole at $z=a$, and an additional pole at $z=\infty$.

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  1. Your first answer does not include the 0.5 coefficient on the $z^{-1}$ term.

  2. Your second answer needs to look at both together, not separately.

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