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, x(n) if is a discrete time function, then its Z-transform is defined as,

$$\mathrm{Z[x(n)]\:=\:X(z)\:=\:\sum_{n=-\infty}^{\infty}\:x(n)z^{-n}}$$

Conjugation Property of Z-Transform

Statement

The conjugation property of Z-transform states that if

$$\mathrm{x(n)\:\overset{ZT}\longleftrightarrow\:X(z);\:\:\text{ROC }\:=\:R}$$

Then,

$$\mathrm{x^*(n)\:\overset{ZT}\longleftrightarrow\:X^*(z^*);\:\:\text{ROC }\:=\:R}$$

Proof

From the definition of Z-transform, we have,

$$\mathrm{Z[x(n)]\:=\:X(z)\:=\:\sum_{n=-\infty}^{\infty}\:x(n)z^{-n}}$$

$$\mathrm{\therefore\:Z[x^*(n)]\:=\:\sum_{n=-\infty}^{\infty}\:x^*(n)z^{-n}}$$

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

$$\mathrm{\therefore\:Z[x^*(n)]\:=\:X^*(z^*)}$$

This can also be represented as,

$$\mathrm{x^*(n)\:\overset{ZT}\longleftrightarrow\:X^*(z^*)}$$

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

Accumulation Property of Z-Transform

Statement

The accumulation property of Z-transform states that if

$$\mathrm{x(n)\:\overset{ZT}\longleftrightarrow\:X(z)}$$

Then,

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

Proof

From the definition of Z-transform, we have,

$$\mathrm{Z[x(n)] \:=\: X(z) \:=\: \sum_{n=-\infty}^{\infty}\: x(n)\: z^{-n}}$$

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

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

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

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

By interchanging the order of summations, we get,

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

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

Also, it can be represented as,

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

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