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Initial Value Theorem of Z-Transform
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,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}}$$
Initial Value Theorem of Z-Transform
The initial value theorem enables us to calculate the initial value of a signal $\mathit{x}\mathrm{\left(\mathit{n}\right)}$, i.e., $\mathit{x}\mathrm{\left(\mathrm{0}\right)}$ directly from its Z-transform X(z) without the need for finding the inverse Z-transform of X(z).
Statement - The initial value theorem of Z-transform states that if
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{z}\right)}}$$
Where, $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is a causal sequence. Then,
$$\mathrm{\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{n} \to 0}\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{z} \to \infty }\mathit{X}\mathrm{\left(\mathit{z}\right)}}$$
Proof
From the definition of Z-transform of a causal sequence, we have,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}}$$
$$\mathrm{\Rightarrow \mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{1}\right)}\mathit{z^{-\mathrm{1}}}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{2}\right)}\mathit{z^{-\mathrm{2}}}\:\mathrm{+}\:...}$$
$$\mathrm{\Rightarrow \mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{+}\:\frac{\mathit{x}\mathrm{\left(\mathrm{1}\right)}}{\mathit{z}}\:\mathrm{+}\:\frac{\mathit{x}\mathrm{\left(\mathrm{2}\right)}}{\mathit{z^{\mathrm{2}}}}\:\mathrm{+}\:\frac{\mathit{x}\mathrm{\left(\mathrm{3}\right)}}{\mathit{z^{\mathrm{3}}}}\:\mathrm{+}\:...}$$
Now, by taking the limit $\mathit{z}\to \infty$ on the both sides, we get,
$$\mathrm{\displaystyle \lim_{z \to \infty }\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\displaystyle \lim_{z \to \infty}\mathrm{\left[\:\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{+}\:\frac{\mathit{x}\mathrm{\left(\mathrm{1}\right)}}{\mathit{z}}\:\mathrm{+}\:\frac{\mathit{x}\mathrm{\left(\mathrm{2}\right)}}{\mathit{z^{\mathrm{2}}}}\:\mathrm{+}\:\frac{\mathit{x}\mathrm{\left(\mathrm{3}\right)}}{\mathit{z^{\mathrm{3}}}}\:\mathrm{+}\:... \right]}}$$
$$\mathrm{\Rightarrow \displaystyle \lim_{z \to \infty }\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{+}\:\mathrm{0}\:\:\mathrm{+}\:\mathrm{0}\:\:\mathrm{+}\:\mathrm{0}\:\mathrm{+}\:...\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathrm{0}\right)}}$$
$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{n} \to 0}\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{z} \to \infty }\mathit{X}\mathrm{\left(\mathit{z}\right)}}$$
Therefore, it gives the initial value of the function $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ directly from its Z-transform.
Numerical Example (1)
Using the initial value theorem for Z-transform, find the initial value of signal $\mathit{x}\mathrm{\left(\mathit{n}\right)}$, i.e., $\mathit{x}\mathrm{\left(\mathrm{0}\right)}$if X(z) is given by,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z^{\mathrm{2}}}\mathrm{+}\:\mathit{z}\:\mathrm{+}\:1}{\mathrm{\left( \mathit{z}+2\right)}\mathrm{\left( \mathit{z}+1\right)}}}$$
Solution
The given Z-transform of the signal is,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z^{\mathrm{2}}}\mathrm{+}\:\mathit{z}\:\mathrm{+}\:1}{\mathrm{\left( \mathit{z}+2\right)}\mathrm{\left( \mathit{z}+1\right)}}}$$
$$\mathrm{\Rightarrow\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{1+\mathrm{\left( \frac{1}{\mathit{z}}\right)}+\mathrm{\left( \frac{1}{\mathit{z^{2}}}\right )}}{\mathrm{\left( 1+\frac{2}{\mathit{z}}\right)}\mathrm{\left( 1+\frac{1}{\mathit{z}}\right )}}}$$
Now, using the initial value theorem of Z-transform $\mathrm{\left[ i.e,\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{n} \to 0}\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{z} \to \infty }\mathit{X}\mathrm{\left(\mathit{z}\right)} \right]}$ .we get,
$$\mathrm{\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{z} \to \infty}\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{z} \to \infty}\mathrm{\left[ \frac{1+\mathrm{\left( \frac{1}{\mathit{z}}\right)}+\mathrm{\left ( \frac{1}{\mathit{z}^{2}}\right )}}{\mathrm{\left( 1+\frac{2}{\mathit{z}}\right)}\mathrm{\left ( 1+\frac{1}{\mathit{z}} \right )}}\right]}}$$
$$\mathrm{\Rightarrow\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\frac{1\:\mathrm{+}\:0\:\mathrm{+}\:0}{\mathrm{\left(1+0 \right)}\mathrm{\left( 1+0\right)}}\:\mathrm{=}\:1}$$
Hence, the initial value of the given function is $\mathit{x}\mathrm{\left(\mathrm{0}\right)}$ = 1.
Numerical Example (2)
Find the initial value $\mathit{x}\mathrm{\left(\mathrm{0}\right)}$ of a sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$, if X(z) is given by,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z}\:\mathrm{+}\:\mathrm{4}}{\mathrm{\left( \mathit{z}+2\right)}\mathrm{\left( \mathit{z}+1\right)}}}$$
Solution
The given Z-transform of the sequence is,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z}\:\mathrm{+}\:\mathrm{4}}{\mathrm{\left( \mathit{z}+2\right)}\mathrm{\left( \mathit{z}+1\right)}}}$$
$$\mathrm{\Rightarrow\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{z\mathrm{\left( 1+\frac{4}{\mathit{z}}\right )}}{\mathrm{\mathit{z^{\mathrm{2}}}\left( 1+\frac{2}{\mathit{z}}\right)}\mathrm{\left( 1+\frac{1}{\mathit{z}}\right )}}\:\mathrm{=}\:\frac{\mathrm{\left( 1+\frac{4}{\mathit{z}}\right )}}{\mathrm{\mathit{z}\left( 1+\frac{2}{\mathit{z}}\right)}\mathrm{\left( 1+\frac{1}{\mathit{z}}\right )}}}$$
Using the initial value theorem of Z-transform, we get,
$$\mathrm{\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{z} \to \infty}\mathrm{\left[ \frac{\mathrm{\left( 1+\frac{4}{\mathit{z}}\right )}}{\mathrm{\mathit{z}\left( 1+\frac{2}{\mathit{z}}\right)}\mathrm{\left( 1+\frac{1}{\mathit{z}}\right )}}\right ]}\:\mathrm{=}\:0}$$
Therefore, the initial of the given sequence is equal to $\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:0 $
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