Differentiation in z-Domain Property of Z-Transform

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

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

Differentiation in z-Domain Property of Z-Transform

Statement

The differentiation in z-domain property of Z-transform states that the multiplication by n in time domain corresponds to the differentiation in z-domain. This property is also called the multiplication by n property of Z-transform. Therefore, if

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

Then, according to the differentiation in z-domain property,

$$\mathrm{nx(n) \:\overset{ZT}\longleftrightarrow\: -z\: \frac{d}{dz} \:X(z);\: \text{ ROC } \:=\: R}$$

Proof

From the definition of Z-transform, we have,

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

Differentiating the above equation on both sides with respect to z, we get,

$$\mathrm{\frac{d}{dz}\: X(z) \:=\: \frac{d}{dz}\: \left[ \sum_{n=-\infty}^{\infty}\: x(n) z^{-n} \right]}$$

$$\mathrm{\frac{d}{dz}\: X(z) \:= \:\sum_{n=-\infty}^{\infty}\: x(n)\: \frac{d(z^{-n})}{dz}}$$

$$\mathrm{\frac{d}{dz}\: X(z) \:=\: \sum_{n=-\infty}^{\infty}\: x(n) \:(-n) z^{-n\:-\:1}}$$

$$\mathrm{\frac{d}{dz}\: X(z) \:=\: \sum_{n=-\infty}^{\infty}\: n x(n) \:z^{-n}\: (-z^{-1})}$$

$$\mathrm{\frac{d}{dz}\: X(z) \:=\: (-z^{-1}) \sum_{n=-\infty}^{\infty}\: n x(n) z^{-n} \:=\: (-z^{-1}) Z[nx(n)]}$$

$$\mathrm{\therefore\: Z[nx(n)] \:=\: -z \frac{d}{dz}\: X(z)}$$

Also, it can be represented as,

$$\mathrm{nx(n) \:\overset{ZT}\longleftrightarrow\: -z \:\frac{d}{dz}\: X(z)}$$

Similarly, if the signal is multiplied by nk in time domain, then

$$\mathrm{n^kx(n) \:\overset{ZT}\longleftrightarrow\: (-1)^k\: z^k \:\frac{d^k}{dz^k}\: X(z)}$$

Numerical Example

Find the Z-transform of the signal $\mathrm{x(n) \:=\: n^2 u(n)}$.

Solution

The given signal is,

$$\mathrm{x(n) \:=\: n^2 u(n)}$$

Since the Z-transform of the unit step sequence is given by,

$$\mathrm{Z[u(n)] \:=\: \frac{z}{z\:-\:1}}$$

Now, using the multiplication by n property of Z-transform $\mathrm{\left[\text{i.e. },\: nx(n) \:\overset{ZT}\longleftrightarrow\: -z\: \frac{d}{dz} \:X(z)\right]}$, we get,

$$\mathrm{Z[nu(n)] \:=\: -z\frac{d}{dz}\{Z[u(n)]\}\:=\: -z \frac{d}{dz} \left(\frac{z}{z\:-\:1} \right)}$$

$$\mathrm{\Rightarrow\:Z[nu(n)] \:=\: -z \left[ \frac{(z\:-\:1)(1) \:-\: z(1)}{(z\:-\:1)^2} \right]}$$

$$\mathrm{\therefore\:Z[nu(n)] \:=\: \frac{z}{(z\:-\:1)^2}}$$

Aging, using the multiplication by n property of Z-transform, we get,

$$\mathrm{Z[n^2 u(n)] \:=\: -z \frac{d}{dz} \left( \frac{z}{(z\:-\:1)^2} \right)}$$

$$\mathrm{\Rightarrow\:Z[n^2 u(n)] \:=\: -z \left[ \frac{(z\:-\:1)^2(1) \:-\: z(2)(z\:-\:1)}{(z\:-\:1)^4} \right]}$$

$$\mathrm{Z[n^2 u(n)] \:=\: - \frac{z \left[ (z\:-\:1)^2 \:-\: 2z(z\:-\:1) \right]}{(z\:-\:1)^4}}$$

$$\mathrm{Z[n^2 u(n)] \:=\: \frac{z (z\:+\:1)(z\:-\:1)^3}{(z\:-\:1)^4}}$$

$$\mathrm{Z[n^2 u(n)] \:=\: \frac{z (z\:+\:1)}{(z\:-\:)}}$$