Time Expansion Property of Z-Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

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,

$$\mathrm{\mathit{Z\left [ u\left ( \frac{n}{\mathrm{2}} \right ) \right ]\mathrm{\, =\, }\left [ \frac{z}{z-\mathrm{1}} \right ]_{z\mathrm{\, =\, }z^{\mathrm{2}}}\mathrm{\, =\, }\frac{z^{\mathrm{2}}}{\left ( Z^{\mathrm{2}}-\mathrm{1} \right )}}}$$

$$\mathrm{\mathit{\therefore u\left ( \frac{n}{\mathrm{2}} \right ) \overset{ZT}{\leftrightarrow}\frac{z^{\mathrm{2}}}{\left ( Z^{\mathrm{2}}-\mathrm{1} \right )};}\mathrm{ROC}\; \; \to \left|z \right|> \left ( \mathrm{1} \right )^{1/2}} $$

raja
Updated on 31-Jan-2022 10:41:10

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