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Time Reversal Property of Z-Transform
Z-Transform
The Z-transform is a mathematical tool which is used to convert the difference equations in discrete time domain into the algebraic equations in z-domain. Mathematically, if $\mathrm{\mathit{x\left ( n \right )}}$ is a discrete time function, then its Z-transform is defined as,
$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\, }X\left ( z \right )\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$
Time Reversal Property of Z-Transform
Statement – The time reversal property of Z-transform states that the reversal or reflection of the sequence in time domain corresponds to the inversion in z-domain. Therefore, if
$$\mathrm{\mathit{x\left ( n \right )\overset{ZT}{\leftrightarrow}X\left ( z \right );\: \:}\mathrm{ROC}\mathrm{\, =\, }\mathit{R}}$$
Then,
$$\mathrm{\mathit{x\left ( -n \right )\overset{ZT}{\leftrightarrow}X\left ( \frac{\mathrm{1}}{z} \right )\mathrm{\, =\, }X\left ( z^{\mathrm{-1}} \right );\: \:}\mathrm{ROC}\mathrm{\, =\, }\frac{\mathrm{1}}{\mathit{R}}}$$
Proof
From the definition of Z-transform, we have,
$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\, }X\left ( z \right )\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$
Now, by reversing the sequence in time domain, we get,
$$\mathrm{\mathit{Z\left [ x\left ( -n \right ) \right ]\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( -n \right )z^{-n}}}$$
Substituting $\mathrm{\mathit{-n\mathrm{\, =\, }m}}$ in the above summation, we get,
$$\mathrm{\mathit{Z\left [ x\left ( -n \right ) \right ]\mathrm{\, =\, }\sum_{m\mathrm{\, =\, }\infty }^{-\infty }x\left ( m \right )z^{m}}}$$
$$\mathrm{\mathit{\Rightarrow Z\left [ x\left ( -n \right ) \right ]\mathrm{\, =\, }\sum_{m\mathrm{\, =\, }-\infty }^{-\infty }x\left ( m \right )\left ( z^{\mathrm{-1}} \right )^{m}\mathrm{\, =\, }X\left ( z^{\mathrm{-1}} \right )}}$$
$$\mathrm{\mathit{\therefore Z\left [ x\left ( -n \right ) \right ]\mathrm{\, =\, }X\left ( Z^{\mathrm{-1}} \right )\mathrm{\, =\, }Z\left ( \frac{\mathrm{1}}{z} \right )}}$$
Or it can also be represented as,
$$\mathrm{\mathit{x\left ( -n \right )\overset{ZT}{\leftrightarrow}X\left ( \frac{\mathrm{1}}{z} \right )\mathrm{\, =\, }X\left ( z^{\mathrm{-1}} \right );\; \; \mathrm{ROC}\mathrm{\, =\, }\frac{\mathrm{1}}{R} }}$$
Numerical Example
Using the time reversal property of Z-transform, find the Z-transform of the following sequence −
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }u\left ( -n \right )}}$$
Solution
Given sequence is,
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }u\left ( -n \right )}}$$
Since the Z-transform of the unit step sequence is given by,
$$\mathrm{\mathit{Z\left [ u\left ( n \right ) \right ]\mathrm{\, =\, }\frac{z}{z-\mathrm{1}};\; \; \mathrm{ROC}\to \left|z \right|> \mathrm{1}}}$$
Now, by using the time reversal property of the Z-transform [$\mathrm{\mathit{\mathrm{i.e}.,\, x\left ( -n \right )\overset{ZT}{\leftrightarrow}X\left ( \frac{\mathrm{1}}{z} \right )}}$], we get,
$$\mathrm{\mathit{Z\left [ u\left ( -n \right ) \right ]\mathrm{\, =\, }\left [ \frac{z}{z-\mathrm{1}} \right ]_{z\mathrm{\, =\, }\left ( \mathrm{1}/z \right )}}}$$
$$\mathrm{\mathit{\Rightarrow Z\left [ u\left ( -n \right ) \right ]\mathrm{\, =\, }\frac{\mathrm{1}/z}{\left ( \mathrm{1}/z \right )-\mathrm{1}}\mathrm{\, =\, }\frac{\mathrm{-1}}{z-\mathrm{1}}}}$$
$$\mathrm{\mathit{\therefore u\left ( -n \right )\overset{ZT}{\leftrightarrow}\frac{\mathrm{-1}}{z-\mathrm{1}};\: \: \: \mathrm{ROC}\to \left|z \right|<\mathrm{1}}} $$
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