Differentiation in z-Domain 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 $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is a discrete time function, then its Z-transform is defined as,

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

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 zdomain. This property is also called the multiplication by n property of Ztransform. Therefore, if

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

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

$$\mathrm{\mathit{n}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{-z}\frac{\mathit{d}}{\mathit{dz}}\mathit{X}\mathrm{\left(\mathit{z}\right)};\:\:\mathrm{ROC}\:\mathrm{=}\:\mathit{R}}$$

Proof

From the definition of Z-transform, we have,

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

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

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

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

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

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

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

Also, it can be represented as,

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

Similarly, if the signal is multiplied by $\mathit{n}^{\mathit{k}}$ in time domain, then

$$\mathrm{\mathit{n}^{\mathit{k}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathrm{\left ( -1 \right )}^{\mathit{k}}\mathit{z}^{\mathit{k}}\frac{\mathit{d}^{\mathit{k}}}{\mathit{dz}^{\mathit{k}}}\mathit{X}\mathrm{\left(\mathit{z}\right)}}$$

Numerical Example

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

Solution

The given signal is,

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

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

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

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

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

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

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

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

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

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

$$\mathrm{\Rightarrow\mathit{Z}\mathrm{\left[\mathit{n}^{\mathrm{2}}\mathit{u}\mathrm{\left(\mathit{n}\right)}\right ]}\:\mathrm{=}\: \mathit{-Z}\mathrm{\left [ \frac{\mathit{z-\mathrm{1}-\mathrm{2}\mathit{z}}}{\mathrm{\left ( \mathit{z-\mathrm{1}} \right )}^{\mathrm{3}}}\right ]}\:\mathrm{=}\:\mathit{-Z}\mathrm{\left [ \frac{\mathit{-z-\mathrm{1}}}{\mathrm{\left ( \mathit{z-\mathrm{1}} \right )}^{\mathrm{3}}}\right ]}}$$

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

raja
Updated on 24-Jan-2022 08:48:55

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