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Time Expansion 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 Expansion Property of Z-Transform
Statement – The time expansion property of Z-transform states that if
$$\mathrm{\mathit{x\left ( n \right )\overset{ZT}{\leftrightarrow}X\left ( z \right );\; \; \; \mathrm{ROC}\to \mathit{R}}} $$
Then
$$\mathrm{\mathit{x_{m}\left ( n \right )\overset{ZT}{\leftrightarrow}X\left ( z^{m} \right );\; \; \; \mathrm{ROC}\to \mathit{R}^{\mathrm{1}/m}}} $$
Where,
$$\mathrm{\mathit{x_{m}\left ( n \right )\mathrm{\, =\, }\left\{\begin{matrix} x\left ( \frac{n}{m} \right )\: ;&\mathrm{when\, \mathit{n}\,is\, an\, integral\, multiple\, of\, \mathit{m} } \ \mathrm{0}; & \mathrm{otherwise} \ \end{matrix}\right. }} $$
Also, $\mathrm{\mathit{x_{m}\left ( n \right )}}$ has $\mathrm{\mathit{\left ( m-\mathrm{1} \right )}}$ zeros inserted between successive values of the original signal.
Proof
From the definition of Z-transform, we have,
$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\, }X\left ( z \right )\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$
$$\mathrm{\mathit{\therefore Z\left [ x_{m}\left ( n \right ) \right ]\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x_{m}\left ( n \right )z^{-n}\mathrm{\, =\, }\sum_{n\mathrm{\, =\, }-\infty }^{\infty }x\left ( \frac{n}{m} \right )z^{-n}}}$$
Substituting (𝑛/𝑚) = 𝑘 in the above summation, we get,
$$\mathrm{\mathit{Z\left [ x_{m}\left ( n \right ) \right ]\mathrm{\, =\, }\sum_{k\mathrm{\, =\, }-\infty }^{\infty }x\left ( k \right )z^{-mk}}}$$
$$\mathrm{\mathit{\Rightarrow Z\left [ x_{m}\left ( n \right ) \right ]\mathrm{\, =\, }\sum_{k\mathrm{\, =\, }-\infty }^{\infty }x\left ( k \right )\left ( z^{m} \right )^{-k}\mathrm{\, =\, }X\left ( z^{m} \right )}}$$
$$\mathrm{\mathit{\therefore x_{m}\left ( n \right )\overset{ZT}{\leftrightarrow}X\left ( z^{m} \right );\; \; \; }ROC\to \mathit{R}^{\mathrm{1}/m}}$$
Numerical Example
Using time expansion property of Z-transform, find the Z-transform of the signal $\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }u\left ( \frac{n}{\mathrm{2}} \right )}}$.
Solution
The given signal is,
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }u\left ( \frac{n}{\mathrm{2}} \right )}}$$
Since, the Z-transform of the unit step sequence is given by,
$$\mathrm{\mathit{Z\left [ u\left ( n \right ) \right ]\mathrm{\, =\, }\frac{z}{z-\mathrm{1}};\; \; \mathrm{ROC}\to \left|z \right|>}1}$$
Therefore, using the time expansion property of Z-transform [i.e., $\mathrm{\mathit{x_{m}\left ( n \right )\overset{ZT}{\leftrightarrow}X\left ( z^{m} \right )}}$], we get,
$$\mathrm{\mathit{Z\left [ u\left ( \frac{n}{\mathrm{2}} \right ) \right ]\mathrm{\, =\, }\left [ \frac{z}{z-\mathrm{1}} \right ]_{z\mathrm{\, =\, }z^{\mathrm{2}}}\mathrm{\, =\, }\frac{z^{\mathrm{2}}}{\left ( Z^{\mathrm{2}}-\mathrm{1} \right )}}}$$
$$\mathrm{\mathit{\therefore u\left ( \frac{n}{\mathrm{2}} \right ) \overset{ZT}{\leftrightarrow}\frac{z^{\mathrm{2}}}{\left ( Z^{\mathrm{2}}-\mathrm{1} \right )};}\mathrm{ROC}\; \; \to \left|z \right|> \left ( \mathrm{1} \right )^{1/2}} $$
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