Signals and Systems – Long Division Method to Find Inverse Z-Transform

Signals and Systems Electronics & Electrical Digital Electronics

Inverse Z-Transform

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

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

Long Division Method to Calculate Inverse Z-Transform

If $\mathit{x}\mathrm{(\mathit{n})}$ is a two sided sequence, then its Z-transform is defined as,

$$\mathit{X}\mathrm{(z)}\:\mathrm{=}\:\displaystyle\sum\limits_{n=-\infty}^\infty \mathit{x}\mathrm{(n)}\mathit{z}^{-\mathit{n}}$$

Where, the Z-transform $\mathit{X}\mathrm{(\mathit{z})}$ 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 $\mathit{x}\mathrm{(\mathit{n})}$ is a causal sequence, then

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

i.e., $\mathit{X}\mathrm{(\mathit{z})}$ has only negative powers of z and its ROC is $|\mathit{z}|>\:\mathit{a}$.

And, if the sequence $\mathit{x}\mathrm{(\mathit{n})}$ is an anti-causal sequence, then

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

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

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

Also, the Z-transform $\mathit{X}\mathrm{(\mathit{z})}$ is a ratio of two polynomials in $\mathit{z}$ given by,

$$\mathit{X}\mathrm{(\mathit{z})}\:\mathrm{=}\:\frac{\mathit{N}\mathrm{(\mathit{z})}}{\mathit{D}\mathrm{(\mathit{z})}}\:\mathrm{=}\:\frac{\mathit{b}_\mathrm{0}\mathit{z}^{m}+\mathit{b}_\mathrm{1}\mathit{z}^{m-1}+\mathit{b}_\mathrm{2}\mathit{z}^{m-2}+\mathit{b}_\mathrm{3}\mathit{z}^{m-3}+\dotso+\mathit{b}_\mathit{m}}{\mathit{z}^{n}+\mathit{a}_\mathrm{1}\mathit{z}^{n-1}+\mathit{a}_\mathrm{2}\mathit{z}^{n-2}+\mathit{a}_\mathrm{3}\mathit{z}^{n-3}+\dotso+\mathit{a}^n}$$

Thus, by dividing the numerator of $\mathit{X}\mathrm{(\mathit{z})}$ by its denominator, we can obtain a series in z.

If the Z-transform $\mathit{X}\mathrm{(\mathit{z})}$ converges for $|\mathit{z}|>\mathit{a}$,then obtained series is given by,

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

Using this series the coefficients of $\mathit{Z}^\mathit{-n}$ can be identified if $\mathit{x}\mathrm{(\mathit{n})}$ is a causal sequence.

Similarly, if $\mathit{X}\mathrm{(\mathit{z})}$converges for $|\mathit{z}|<\:\mathit{a}$, then obtained series is given by,

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

Using this series, we can identify the coefficients of $\mathit{Z}^\mathit{-n}$ as $\mathit{x}\mathrm{(\mathit{n})}$ of an anticausal sequence.

Numerical Example

Find the inverse Z-transform of

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

Solution

The given Z-transform is,

The Z-transform is defined as,

$$\mathit{X}\mathrm{(\mathit{z})}\:\mathrm{=}\:\displaystyle\sum\limits_{n=-\infty}^\infty \mathit{x}\mathrm{(n)}\mathit{z}^{\mathit{-n}}$$ $$\Rightarrow\mathit{X}\mathrm{(\mathit{z})}\:\mathrm{=}\:\dotsm\mathit{x}\mathrm{(-3)}\mathit{z}^{\mathrm{3}}+\mathit{x}\mathrm{(-2)}\mathit{z}^{\mathrm{2}}+\mathit{x}\mathrm{(-1)}\mathit{z}+\mathit{x}\mathrm{(0)}+\mathit{x}\mathrm{(1)}\mathit{z}^{\mathrm{-1}}+\mathit{x}\mathrm{(2)}\mathit{z}^{\mathrm{-2}}+\mathit{x}\mathrm{(3)}\mathit{z}^{\mathrm{-3}}\dotsm$$

Comparing this series of $\mathit{X}\mathrm{(\mathit{z})}$ with the given series of $\mathit{X}\mathrm{(\mathit{z})}$ in the question, we get,

$$\mathit{x}\mathrm{(-3)}\:\mathrm{=}\:\mathrm{1},\mathit{x}\mathrm{(-2)}\:\mathrm{=}\:\mathrm{3},\mathit{x}\mathrm{(-1)}\:\mathrm{=}\:\mathrm{-2},\mathit{x}\mathrm{(0)}\:\mathrm{=}\:\mathrm{4},\mathit{x}\mathrm{(1)}\:\mathrm{=}\:\mathrm{-2},\mathit{x}\mathrm{(2)}\:\mathrm{=}\:\mathrm{4},\mathit{x}\mathrm{(3)}\:\mathrm{=}\:\mathrm{3},$$ $$\therefore\mathit{x}\mathrm{(\mathit{n})}\:\mathrm{=}\:\begin{Bmatrix}1,3,-2,4,-2,4,3 \ \uparrow\end{Bmatrix}$$

Manish Kumar Saini

Updated on: 07-Jan-2022

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