Long Division Method to Find Inverse Z-Transform

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Inverse Z-Transform

The inverse Z-transform is defined as the process of finding the time domain signal x(n) from its Z-transform X(z). The inverse Z-transform is denoted as:

$$\mathrm{x(n) \:=\: Z^{-1}[X(z)]}$$

Long Division Method to Calculate Inverse Z-Transform

If x(n) is a two sided sequence, then its Z-transform is defined as,

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

Where, the Z-transform X(z) has both positive powers of z as well as negative powers of z. Using the long division method, a two sided sequence cannot be obtained. Therefore, if the sequence x(n) is a causal sequence, then

$$\mathrm{X(z) \:=\: \sum_{n=0}^{\infty}\: x(n) z^{-n} \:=\: x(0) \:+\: x(1) z^{-1} \:+\: x(2) z^{-2} \:+\: x(3) z^{-3} \:+\: \dotso}$$

i.e., X(z) has only negative powers of z and its ROC is |z| > a.

And, if the sequence x(n) is an anti-causal sequence, then

$$\mathrm{X(z) \:=\: \sum_{n=-\infty}^{0}\: x(n) z^{-n} \:=\: \dotso \:+\: x(-3) z^3 \:+\: x(-2) z^2 \:+\: x(-1) z \:+\: x(0)}$$

That is, for anti-causal sequence, the X(z) has only positive powers of z and its ROC is |z| < a .

As the determination of the inverse Z-transform of X(z) is only the determination of sequence x(n), i.e., if x(n) is causal then x(0), x(1), x(2), ... or if x(n) is anti-causal,then x(0), x(−1), x(−2), ....

Also, the Z-transform X(z) is a ratio of two polynomials in z given by,

$$\mathrm{X(z) \:=\: \frac{N(z)}{D(z)} \:=\: \frac{b_0 z^m \:+\: b_1 z^{m-1} \:+\: b_2 z^{m-2} \:+\: b_3 z^{m-3} \:+\: \dotso \:+\: b_m z^n}{a_1 z^{n-1} \:+\: a_2 z^{n-2} \:+\: a_3 z^{n-3} \:+\: \dotso \:+\: a_n}}$$

Thus, by dividing the numerator of X(z) by its denominator, we can obtain a series in z.

If the Z-transform X(z) converges for |z| > a ,then obtained series is given by,

$$\mathrm{X(z) \:=\: x(0) \:+\: x(1) z^{-1} \:+\: x(2) z^{-2} \:+\: x(3) z^{-3} \:+\:\dotso}$$

Using this series the coefficients of Z^−n can be identified if x(n) is a causal sequence.

Similarly, if X(z) converges for |z| < a, then obtained series is given by,

$$\mathrm{X(z) \:=\: x(0) \:+\: x(-1) z^1 \:+\: x(-2) z^2 \:+\: x(-3) z^3 \:+\: \dotso}$$

Using this series, we can identify the coefficients of Z^−n as x(n) of an anticausal sequence.

Numerical Example

Find the inverse Z-transform of

$$\mathrm{X(z) \:=\: z^3 \:+\: 3z^2 - 2z + 4 - 2z^{-1} + 4z^{-2} + 3z^{-3}}$$

Solution

The given Z-transform is,

$$\mathrm{X(z) \:=\: z^3 \:+\: 3z^2 - 2z + 4 - 2z^{-1} + 4z^{-2} + 3z^{-3}}$$

The Z-transform is defined as,

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

$$\mathrm{\Rightarrow\: X(z) \:=\: \dotso \:+\: x(-3) z^3 \:+\: x(-2) z^2 \:+\: x(-1) z \:+\: x(0) \:+\: x(1) z^{-1} \:+\: x(2) z^{-2} \:+\: x(3) z^{-3} \:+\: \dotso}$$

Comparing this series of X(z) with the given series of X(z) in the question, we get,

$$\mathrm{x(-3) \:=\: 1,\: x(-2) \: = \: 3, \: x(-1) \: = \: -2, \: x(0) \: = \: 4, \: x(1) \: = \: -2, \: x(2) \: = \: 4, \: x(3) \: = \: 3}$$

$$\mathrm{\therefore\: x(n) \:=\: \{1,\: 3,\: -2,\: 4,\: -2,\: 4,\: 3,\: \,\: \uparrow\}}$$