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What is Z-Transform?
What is Z-Transform?
The Z-transform (ZT) is a mathematical tool which is used to convert the difference equations in time domain into the algebraic equations in z-domain.
The Z-transform is a very useful tool in the analysis of a linear shift invariant (LSI) system. An LSI discrete time system is represented by difference equations. To solve these difference equations which are in time domain, they are converted first into algebraic equations in z-domain using the Z-transform, then the algebraic equations are manipulated in z-domain and the result obtained is converted back into time domain using the inverse Z-transform.
The Z-transform may be of two types viz. unilateral (or one-sided) and bilateral (or two-sided).
Mathematically, if $\mathrm{\mathit{x\left ( n \right )}}$ is a discrete-time signal or sequence, then its bilateral or two-sided Z-transform is defined as −
$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]=X\left ( z \right )=\sum_{n=-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$
Where, z is a complex variable and it is given by,
$$\mathrm{\mathit{z\mathrm{\, =\,}r\, e^{j\, \omega }}}$$
Where, r is the radius of a circle.
Also, the unilateral or one-sided z-transform is defined as −
$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\,}X\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }x\left ( n \right )z^{-n}}}$$
The unilateral or one-sided z-transform is very useful because we mostly deal with causal sequences. Also, it is mainly suited for solving difference equations with initial conditions.
Region of Convergence (ROC) of Z-Transform
The set of points in the z-plane, for which the Z-transform of a discrete-time sequence $\mathrm{\mathit{x\left ( n \right )}}$, that is $\mathrm{\mathit{X\left ( z \right )}}$ converges is called the region of convergence (ROC) of the Z-transform $\mathrm{\mathit{X\left ( z \right )}}$.
For any given discrete-time sequence, the Z-transform may or may not converge. If there is no point in the z-plane for which the function $\mathrm{\mathit{X\left ( z \right )}}$ converges, then the sequence $\mathrm{\mathit{x\left ( n \right )}}$ is said to be having no z-transform.
Advantages and Disadvantages of Z-Transform
Following are the advantages of the Z-transform −
The Z-transform makes the analysis of a discrete-time system easier by converting the difference equations describing the system into simple linear algebraic equations.
The convolution operation in time domain is converted into multiplication in z-domain.
The Z-transform exists for the signals for which the discrete-time Fourier transform (DTFT) does not exist.
Limitations – The primary limitation of the Z-transform is that using Z-transform, the frequency domain response cannot be obtained and cannot be plotted.
Numerical Example
Find the Z-transform of the following sequence −
$$\mathrm{\mathit{y\left ( n \right )\mathrm{\, =\,}x\left ( n\mathrm{\, +\,}\mathrm{3} \right )u\left ( n \right )}}$$
Solution
The given discrete-time sequence is,
$$\mathrm{\mathit{y\left ( n \right )\mathrm{\, =\,}x\left ( n\mathrm{\, +\,}\mathrm{3} \right )u\left ( n \right )}}$$
From the definition of the Z-transform, we get,
$$\mathrm{\mathit{Z\left [ y\left ( n \right ) \right ]\mathrm{\, =\,}Y\left ( z \right )\mathrm{\, =\,}Z\left [ x\left ( n\mathrm{\, +\,}\mathrm{3} \right )u\left ( n \right ) \right ]}}$$
$$\mathrm{\mathit{Y\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left [ x\left ( n\mathrm{\, +\,}\mathrm{3} \right )u\left ( n \right ) \right ]z^{-n}\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }x\left ( n\mathrm{\, +\,}\mathrm{3} \right )z^{-n}}}$$
Let (𝑛 + 3) = 𝑚, then 𝑛 = (𝑚 − 3),
$$\mathrm{\mathit{\therefore Y\left ( z \right )\mathrm{\, =\,}\sum_{m\mathrm{\, =\,}\mathrm{3}}^{\infty }x\left ( m \right )z^{-\left ( m-\mathrm{3} \right )}\mathrm{\, =\,}z^{\mathrm{3}}\left [ \sum_{m\mathrm{\, =\,}\mathrm{3}}^{\infty }x\left ( m \right )z^{-m} \right ]}}$$
$$\mathrm{\mathit{\Rightarrow Y\left ( z \right )\mathrm{\, =\,}z^{\mathrm{3}}\left [ \sum_{m\mathrm{\, =\,}\mathrm{0}}^{\infty }x\left ( m \right )z^{-m}-x\left ( \mathrm{0} \right )-x\left ( \mathrm{1} \right )z^{\mathrm{-1}} \right ]}}$$
$$\mathrm{\mathit{\therefore Y\left ( z \right )\mathrm{\, =\,}z^{\mathrm{3}}X\left ( z \right )-z^{\mathrm{3}}x\left ( \mathrm{0} \right )-z\, x\left ( \mathrm{1} \right )}}$$
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