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Inverse Z-Transform by Convolution Method
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,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}}$$
Convolution Method to Find Inverse Z-Transform
The inverse Z-transform can be calculated using the convolution theorem. In the convolution integration method, the given Z-transform X(z) is first split into $\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}$ and $\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}$ such that $\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}$.
The signals $\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}$ and $\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}$ are then obtained by taking the inverse Z-transform of $\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}$ and $\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}$ respectively. Finally, the function $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is obtained by performing the convolution of $\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}$ and $\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}$ in the time domain.
As from the definition of Z-transform of convolution of two signals, we have,
$$\mathrm{\mathit{Z}\mathrm{\left[ \mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right ]}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}}$$
Therefore, the inverse Z-transform is obtained as,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{Z}^{-\mathrm{1}}\mathrm{\left[\mathit{X}\mathrm{\left(\mathit{z}\right)}\right]}\:\mathrm{=}\:\mathit{Z}^{-\mathrm{1}}\mathrm{\left [ \mathit{Z\mathrm{\left\{\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)} \right\}}} \right ]}}$$
$$\mathrm{\therefore \mathit{Z}^{-\mathrm{1}}\mathrm{\left[\mathit{X}\mathrm{\left(\mathit{z}\right)}\right]}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\sum_{\mathit{k}=0}^{\mathit{n}}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n-k}\right)}}$$
Numerical Example
Using the convolution method, find the Z-transform of
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z^{\mathrm{2}}}}{\mathrm{\left (\mathit{z}-3 \right )}\mathrm{\left( \mathit{z-\mathrm{4}}\right )}}}$$
Solution
The given Z-transform function is,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z^{\mathrm{2}}}}{\mathrm{\left (\mathit{z}-3 \right )}\mathrm{\left( \mathit{z-\mathrm{4}}\right )}}}$$
Let,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z}}{\mathrm{\left ( \mathit{z-\mathrm{3}} \right )}}\:\frac{\mathit{z}}{\mathrm{\left ( \mathit{z-\mathrm{4}} \right )}}}$$
Taking the inverse Z-transform of $\mathit{X}_{\mathrm{1}}\mathrm{\left(\mathit{z}\right)}$ and $\mathit{X}_{\mathrm{2}}\mathrm{\left(\mathit{z}\right)}$ respectively as −
$$\mathrm{\mathit{Z}^{-\mathrm{1}}\mathrm{\left[\mathit{X_{\mathrm{1}}}\mathrm{\left(\mathit{z}\right)}\right]}\:\mathrm{=}\:\mathit{x_{\mathrm{1}}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{Z^{-\mathrm{1}}}\mathrm{\left[ \frac{\mathit{z}}{\mathrm{\left ( \mathit{z-\mathrm{3}} \right )}} \right ]}\:\mathrm{=}\:3^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$
Similarly
$$\mathrm{\mathit{Z}^{-\mathrm{1}}\mathrm{\left[\mathit{X_{\mathrm{2}}}\mathrm{\left(\mathit{z}\right)}\right]}\:\mathrm{=}\:\mathit{x_{\mathrm{2}}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{Z^{-\mathrm{1}}}\mathrm{\left[ \frac{\mathit{z}}{\mathrm{\left ( \mathit{z-\mathrm{4}} \right )}} \right ]}\:\mathrm{=}\:4^{\mathit{n}}\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$
Now, using the convolution method for finding inverse Z-transform, we have,
$$\mathrm{\mathit{Z}^{-\mathrm{1}}\mathrm{\left[\mathit{X}\mathrm{\left(\mathit{z}\right)}\right]}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{n}\right)}*\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\sum_{\mathit{k}=0}^{\mathit{n}}\mathit{x}_{\mathrm{1}}\mathrm{\left(\mathit{k}\right)}\mathit{x}_{\mathrm{2}}\mathrm{\left(\mathit{n-k}\right)}}$$
$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\sum_{\mathit{k}=0}^{\mathit{n}}3^{\mathit{k}}\mathit{u}\mathrm{\left(\mathit{k}\right)}4^{\mathit{n-k}}\mathit{u}\mathrm{\left(\mathit{n-k}\right)}\:\mathrm{=}\:\sum_{\mathit{k}=0}^{\mathit{n}}3^{\mathit{k}}\mathit{u}\mathrm{\left(\mathit{k}\right)}\:\mathrm{\left ( \frac{4^{\mathit{n}}}{4^{\mathit{k}}} \right )}\mathit{u}\mathrm{\left ( \mathit{n-k} \right )}}$$
$$\mathrm{\Rightarrow \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:4^{\mathit{n}}\sum_{\mathit{k}=0}^{\mathit{n}}\mathrm{\left( \frac{3^{\mathit{k}}}{4^{\mathit{k}}} \right )}\:\mathrm{=}\:4^{\mathit{n}}\sum_{\mathit{k}=0}^{\mathit{n}}\mathrm{\left ( \frac{3}{4} \right )}^{\mathit{k}}}$$
$$\mathrm{\Rightarrow \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:4^{\mathit{n}}\mathrm{\left [ \frac{1-\mathrm{\left ( \frac{3}{4} \right )}^{\mathit{n}+1}}{1-\mathrm{\left ( \frac{3}{4} \right )}} \right ]}\:\mathrm{=}\:4^{\mathit{n}+1}\mathrm{\left [ 1-\mathrm{\left ( \frac{3}{4} \right )}^{\mathit{n}+1} \right ]}}$$
$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:4^{\mathit{n}+1}\mathit{u}\mathrm{\left(\mathit{n}\right)}-3^{\mathit{n}+1}\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$
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