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Signals and Systems β What is Inverse Z-Transform?
The Inverse Z-Transform
The inverse Z-transform is defined as the process of finding the time domain signal $\mathrm{\mathit{x\left ( n \right )}}$ from its Z-transform $\mathrm{\mathit{X\left ( z \right )}}$. The inverse Z-transform is denoted as −
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\,}Z^{-\mathrm{1}}\left [ X\left ( z \right ) \right ]}}$$
Since the Z-transform is defined as,
$$\mathrm{\mathit{X\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}-\infty }^{\infty }x\left ( n \right )z^{-n}\; \; \; \cdot \cdot \cdot \left ( \mathrm{1} \right )}}$$
Where, z is a complex variable and is given by,
$$\mathrm{\mathit{z\mathrm{\, =\,}r\, e^{j\, \omega }}}$$
Where, r is the radius of a circle in z-plane. Hence, on substituting the value of z in eq. (1), we get,
$$\mathrm{\mathit{X\left ( z \right )\mathrm{\, =\,}X\left ( r\, e^{j\, \omega } \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}-\infty }^{\infty }\left [ x\left ( n \right )r^{-n} \right ]e^{-j\, \omega \, n}\; \; \; \cdot \cdot \cdot \left ( \mathrm{2} \right )}}$$
The equation (2) is the discrete-time Fourier transform (DTFT) of the signal $\mathrm{\mathit{\left [ x\left ( n \right )r^{-n} \right ]}}$ Therefore, the inverse DTFT of the function $\mathrm{\mathit{X\left ( r\, e^{j\, \omega } \right )}}$ must be $\mathrm{\mathit{\left [ x\left ( n \right )r^{-n} \right ]}}$. Hence, we can write,
$$\mathrm{\mathit{F^{-\mathrm{1}}\left [ X\left ( r\, e^{j\, \omega } \right ) \right ]\mathrm{\, =\,}x\left ( n \right )r^{-n}\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\pi }^{\pi }X\left ( re^{j\, \omega } \right )e^{j\, \omega \, n}d\omega}}$$
$$\mathrm{\mathit{\Rightarrow x\left ( n \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}\pi }\int_{-\pi }^{\pi }X\left ( re^{j\, \omega } \right )\left ( re^{j\, \omega } \right )^{n}\: d\omega\; \; \; \cdot \cdot \cdot \left ( \mathrm{3} \right )}}$$
$$\mathrm{\mathit{\because z\mathrm{\, =\,}re^{j\, \omega }}}$$
Then,
$$\mathrm{\mathit{\frac{dz}{d\omega }\mathrm{\, =\,}jre^{j\, \omega }\; \; or\: \: d\omega \mathrm{\, =\,}\frac{dz}{jre^{j\, \omega }}}}$$
Substituting the values of z and dω in eq. (3), we have,
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint_{c}X\left ( z \right )z^{\left ( n-1 \right )}\, dz\; \; \; \cdot \cdot \cdot \left ( \mathrm{4} \right ) }}$$
The integration given in the eq. (4) represents the integration around the circle of radius $\mathrm{\mathit{\left|z \right|\mathrm{\, =\,}r}}$ in the counter clockwise direction. This is the direct method of finding inverse Z-transform. The direct method is quite tedious. Hence, indirect methods are used for finding the inverse Z-transform.
Methods to Find the Inverse Z-Transform
Generally, there are following four methods which are used for finding the inverse Z-transform −
Long Division Method or Power Series Method – The long division method is simple and the advantage of this method is that it is more general and can be applied to any problem. But, the disadvantage of this method is that it does not give the solution in the closed form. Also, it can be used only if the region of convergence (ROC) of the given Z-transform $\mathrm{\mathit{X\left ( z \right )}}$ is either of the form of |$\mathrm{\mathit{z}}$| > π or of the form of |$\mathrm{\mathit{z}}$| < π.
Partial Fraction Expansion Method – In this method, the proper fraction $\mathrm{\mathit{X\left ( z \right )}}$⁄$\mathrm{\mathit{z}}$ is written in terms of partial fractions and inverse Ztransform of each partial fraction is found by the standard Z-transform pairs and then, all of them are added.
Residue Method or Complex Inversion Integral Method – In the residue method, the inverse Z-transform is obtained by using the following equation −
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint_{c}X\left ( z \right )z^{\left ( n-1 \right )}\, dz }}$$
Convolution Integral Method – The convolution integral method uses the convolution property of the Z-transform and it can be used when the given Z-transform $\mathrm{\mathit{X\left ( z \right )}}$ can be written as the product of two functions.
Numerical Example
Determine the inverse Z-transform −
$$\mathrm{\mathit{X\left (z \right )\mathrm{\, =\,}\frac{\mathrm{4}z^{-\mathrm{1}}}{\left [ \mathrm{1-\left ( \frac{1}{3} \right )}z^{-\mathrm{1}} \right ]^{\mathrm{2}}};\; \; ROC\to \left|z \right|>\left ( \mathrm{\frac{1}{3}} \right )}}$$
Solution
The inverse Z-transform can be obtained by using the equation −
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}\pi j}\oint_{c}X\left ( z \right )z^{\left ( n-1 \right )}\, dz }}$$
This equation can be evaluated by finding the sum of all residues of the poles that are inside the circle c in the z-plane in the ROC. Thus, the above eqn. can also be written as
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\,}\sum }Residues\: of\: \mathit{X\left ( z \right )z^{n-\mathrm{1}}}\: at\: poles\: inside\: c}$$
$$\mathrm{\Rightarrow \mathit{x\left ( n \right )\mathrm{\, =\,}\sum_{i}\left ( z-z_{i} \right )X\left ( z \right )z^{n-\mathrm{1}}|_{z\mathrm{\, =\,}z_{i}}}}$$
Now, if there is a pole of multiplicity k, then the residue at that pole is given by,
$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left ( k-\mathrm{1} \right )\mathrm{\, !\,}}\frac{d^{k-\mathrm{1}}}{dz^{k-\mathrm{1}}}\left [ \left ( z-z_{i} \right )^{k}X\left ( z \right )z^{n-\mathrm{1}} \right ]_{z\mathrm{\, =\,}z_{i}}}}$$
The given Z-transform is,
$$\mathrm{\mathit{X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{4}z^{-\mathrm{1}}}{\left [ \mathrm{1-\left ( \frac{1}{3} \right )} z^{\mathrm{-1}}\right ]^{\mathrm{2}}}\mathrm{\, =\,}\frac{\mathrm{4}z}{\left [ \mathrm{\mathit{z}-\left ( \frac{1}{3} \right )}\right ]^{\mathrm{2}}}}}$$
As the given Z-transform $\mathrm{\mathit{X\left ( z \right )}}$ has a pole of order 2 at $\mathrm{\mathit{z}}$ = (1/3).
$$\mathrm{\mathit{\therefore x\left ( n \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\left ( \mathrm{2}-\mathrm{1} \right )\mathrm{\, !\,}}\frac{d^{\left ( \mathrm{2-1} \right )}}{dz^{\left ( \mathrm{2-1} \right )}}\left [ \left ( z-\mathrm{\frac{1}{3}} \right )^{\mathrm{2}}\frac{\mathrm{4}z\: z^{n-\mathrm{1}}}{\left [ z-\left ( \mathrm{\frac{1}{3}} \right ) \right ]^{\mathrm{2}}} \right ]_{z\mathrm{\, =\,}\mathrm{\left ( 1/3 \right )}}}}$$
$$\mathrm{\mathit{\Rightarrow x\left ( n \right )\mathrm{\, =\,}\frac{d}{dz}\left [ \left ( z-\mathrm{\frac{1}{3}} \right )^{\mathrm{2}}\frac{\mathrm{4}z^{n}}{\left [ z-\mathrm{\left ( \frac{1}{3} \right )} \right ]^{\mathrm{2}}} \right ]_{z\mathrm{\, =\,}\left ( \mathrm{1/3} \right )}\; \; \mathrm{\, =\,}\left [ \mathrm{4}nz^{n-\mathrm{1}} \right ]_{z\mathrm{\, =\,}\left ( \mathrm{1/3} \right )}}}$$
$$\mathrm{\mathit{\therefore x\left ( n \right )\mathrm{\, =\,}\mathrm{4}n\left ( \mathrm{\frac{1}{3}} \right )^{n-\mathrm{1}}u\left ( n \right )}}$$
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