# Conjugation and Accumulation Properties 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, $\mathrm{\mathit{x\left ( n \right )}}$ if is a discrete time function, then its Z-transform is defined as,

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\,=\,}X\left ( z \right )\mathrm{\,=\,}\sum_{n\mathrm{\,=\,}-\infty }^{\infty }x\left ( n \right )z^{-n} }}$$

## Conjugation Property of Z-Transform

Statement – The conjugation property of Z-transform states that if

$$\mathrm{\mathit{x\left ( n \right )\overset{ZT}{\leftrightarrow}X\left ( z \right );\; \; \mathrm{ROC}\mathrm{\,=\,}\mathit{R} }}$$

Then,

$$\mathrm{\mathit{x^{*}\left ( n \right )\overset{ZT}{\leftrightarrow}X^{*}\left ( z^{*} \right );\; \; \mathrm{ROC}\mathrm{\,=\,}\mathit{R}}}$$

### Proof

From the definition of Z-transform, we have,

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\,=\,}X\left ( z \right )\mathrm{\,=\,}\sum_{n\mathrm{\,=\,}-\infty }^{\infty }x\left ( n \right )z^{-n} }}$$

$$\mathrm{\mathit{\therefore Z\left [ x^{\ast }\left ( n \right ) \right ]\mathrm{\,=\,}\sum_{n\mathrm{\,=\,}-\infty }^{\infty }x^{\ast }\left ( n \right )z^{-n}}}$$

$$\mathrm{\mathit{\Rightarrow Z\left [ x^{\ast }\left ( n \right ) \right ]\mathrm{\,=\,}\left [ \sum_{n\mathrm{\,=\,}-\infty }^{\infty }x\left ( n \right )\left ( z^{\ast } \right )^{-n} \right ]^{\ast }\mathrm{\,=\,}\left [ X\left ( z^{\ast } \right ) \right ]^{\ast }}}$$

$$\mathrm{\mathit{\therefore Z\left [ x^{\ast }\left ( n \right ) \right ]\mathrm{\,=\,}X^{\ast }\left ( z^{\ast } \right )}}$$

This can also be represented as,

$$\mathrm{\mathit{x^{\ast }\left ( n \right )\overset{ZT}{\leftrightarrow}X^{\ast }\left ( z^{\ast } \right )}}$$

Hence, it proves the conjugation property of Z-transform.

## Accumulation Property of Z-Transform

Statement – The accumulation property of Z-transform states that if

$$\mathrm{\mathit{x\left ( n \right )\overset{ZT}{\leftrightarrow}X\left ( z \right )}}$$

Then,

$$\mathrm{\mathit{\sum_{p\mathrm{\,=\,}-\infty }^{n}x\left ( p \right )\overset{ZT}{\leftrightarrow}\left ( \frac{z}{z-\mathrm{1}} \right )X\left ( z \right )}}$$

### Proof

From the definition of Z-transform, we have,

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\,=\,}X\left ( z \right )\mathrm{\,=\,}\sum_{n\mathrm{\,=\,}-\infty }^{\infty }x\left ( n \right )z^{-n} }}$$

$$\mathrm{\mathit{\therefore Z\left [\sum_{p\mathrm{\,=\,}-\infty }^{n} x\left ( p \right ) \right ]\mathrm{\,=\,}\sum_{n\mathrm{\,=\,}-\infty }^{\infty }\left [ \sum_{p\mathrm{\,=\,}-\infty }^{n }x\left ( p \right ) \right ]z^{-n} }}$$

Substituting $\mathrm{\mathit{\left ( n-p \right )\mathrm{\,=\,}m\:\: \mathrm{or}\: \: n\mathrm{\,=\,}\left ( p\mathrm{\,+\,}m \right )\: \: \mathrm{or}\: \: p\mathrm{\,=\,}\left ( n-m \right )}}$ in the RHS of the above equation, we get,

$$\mathrm{\mathit{Z\left [\sum_{p\mathrm{\,=\,}-\infty }^{n} x\left ( p \right ) \right ]\mathrm{\,=\,}\sum_{p\mathrm{\,=\,}-\infty-m }^{p\mathrm{\,=\,}\infty-m }\left [ \sum_{m\mathrm{\,=\,}n-\left ( -\infty \right ) }^{m\mathrm{\,=\,}n-n }x\left ( p \right ) \right ]z^{-\left ( p\mathrm{\,+\,}m \right )}}}$$

$$\mathrm{\mathit{\Rightarrow Z\left [\sum_{p\mathrm{\,=\,}-\infty }^{n} x\left ( p \right ) \right ]\mathrm{\,=\,}\sum_{p\mathrm{\,=\,}-\infty }^{\infty}\sum_{m\mathrm{\,=\,}\infty }^{\mathrm{0}}x\left ( p \right )z^{-p}z^{-m}}}$$

By interchanging the order of summations, we get,

$$\mathrm{\mathit{\Rightarrow Z\left [\sum_{p\mathrm{\,=\,}-\infty }^{n} x\left ( p \right ) \right ]\mathrm{\,=\,}\sum_{m\mathrm{\,=\,}\mathrm{0} }^{\infty}z^{-m}\sum_{p\mathrm{\,=\,}-\infty }^{\infty }x\left ( p \right )z^{-p}\mathrm{\,=\,}\left (\sum_{m\mathrm{\,=\,}\mathrm{0} }^{\infty}z^{-m} \right )X\left ( z \right ) }}$$

$$\mathrm{\mathit{\therefore Z\left [\sum_{p\mathrm{\,=\,}-\infty }^{n} x\left ( p \right ) \right ]\mathrm{\,=\,}\left ( \frac{\mathrm{1}}{\mathrm{1}-z^{\mathrm{-1}}} \right )X\left ( z \right )\mathrm{\,=\,}\left ( \frac{z}{z-\mathrm{1}} \right )X\left ( z \right )}}$$

Also, it can be represented as,

$$\mathrm{\mathit{\sum_{p\mathrm{\,=\,}-\infty }^{n} x\left ( p \right )\overset{ZT}{\leftrightarrow}\left ( \frac{z}{z-\mathrm{1}} \right )X\left ( z \right )}}$$

Thus, it proves the accumulation property of Z-transform.