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Final 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,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}}$$
Final Value Theorem of Z-Transform
The final value theorem of Z-transform enables us to calculate the steady state value of a sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$, i.e., $\mathit{x}\mathrm{\left(\mathit{\infty}\right)}$ directly from its Z-transform, without the need for finding its inverse Z-transform.
Statement - If $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is a causal sequence, then the final value theorem of Z-transform states that if,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{ZT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{z}\right)}}$$
And if the Z-transform X(z) has no poles outside the unit circle, and it has no higher poles on the unit circle centred at the origin of the z-plane, then,
$$\mathrm{\mathit{x}\mathrm{\left(\mathrm{\infty }\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{n} \to \infty }\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\displaystyle \lim_{\mathit{z} \to 1 }\mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}}$$
Proof
From the definition of Z-transform of a causal sequence, we have,
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}}$$
And
$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n\mathrm{+}\mathrm{1}}\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}\:\mathrm{=}\:\mathit{z}\mathit{X}\mathrm{\left(\mathit{z}\right)}-\mathit{z}\mathit{x}\mathrm{\left(\mathrm{0}\right)}}$$
$$\mathrm{\therefore \mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n\mathrm{+}\mathrm{1}}\right)}\right]}-\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{\mathit{-n}}}-\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}}$$
$$\mathrm{\Rightarrow \mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n\mathrm{+}\mathrm{1}}\right)}\right]}-\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{z}\mathit{X}\mathrm{\left(\mathit{z}\right)}-\mathit{z}\mathit{x}\mathrm{\left(\mathrm{0}\right)}-\mathit{X}\mathrm{\left(\mathit{z}\right)}}$$
$$\mathrm{\therefore \mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}-\mathit{z}\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathrm{\left[\mathit{x}\mathrm{\left( \mathit{n}+1\right)}-\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\mathit{z^{-\mathit{n}}}}$$
$$\mathrm{\Rightarrow \mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}-\mathit{z}\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\mathrm{\left[\mathit{x}\mathrm{\left( \mathrm{1}\right)}-\mathit{x}\mathrm{\left(\mathrm{0}\right)}\right]}\mathit{z^{\mathrm{0}}}\:\mathrm{+}\:\mathrm{\left[\mathit{x}\mathrm{\left( \mathrm{2}\right)}-\mathit{x}\mathrm{\left(\mathrm{1}\right)}\right]}\mathit{z^{-\mathrm{1}}}\:\mathrm{+}\:\mathrm{\left[\mathit{x}\mathrm{\left( \mathrm{3}\right)}-\mathit{x}\mathrm{\left(\mathrm{2}\right)}\right]}\mathit{z^{-\mathrm{2}}}\:\mathrm{+}\:...}$$
Now taking limit $\mathit{z}\to 1$ on both the sides, we get,
$$\mathrm{\displaystyle \lim_{z \to 1}\mathrm{\left[ \mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}-\mathit{z}\mathit{x}\mathrm{\left(\mathrm{0}\right)}\right ]}\:\mathrm{=}\:\displaystyle \lim_{z \to 1}\mathrm{\left\{\mathrm{\left[\mathit{x}\mathrm{\left( \mathrm{1}\right)}-\mathit{x}\mathrm{\left(\mathrm{0}\right)}\right]}\mathit{z^{\mathrm{0}}}\:\mathrm{+}\:\mathrm{\left[\mathit{x}\mathrm{\left( \mathrm{2}\right)}-\mathit{x}\mathrm{\left(\mathrm{1}\right)}\right]}\mathit{z^{-\mathrm{1}}}\:\mathrm{+}\:\mathrm{\left[\mathit{x}\mathrm{\left( \mathrm{3}\right)}-\mathit{x}\mathrm{\left(\mathrm{2}\right)}\right]}\mathit{z^{-\mathrm{2}}}\:\mathrm{+}\:... \right\}}}$$
$$\mathrm{\Rightarrow \displaystyle \lim_{z \to 1}\mathrm{\left[ \mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}\right ]}-\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\mathrm{1}\right)}-\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{2}\right)}-\:\mathit{x}\mathrm{\left(\mathrm{1}\right)}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{3}\right)}-\mathit{x}\mathrm{\left(\mathrm{2}\right)}\:\mathrm{+}\:...\:\mathrm{+}\:\mathit{x}\mathrm{\left(\infty\right)}-\mathit{x}\mathrm{\left(\infty-1\right)}}$$
$$\mathrm{\Rightarrow \displaystyle \lim_{z \to 1}\mathrm{\left[ \mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}\right ]}-\mathit{x}\mathrm{\left(\mathrm{0}\right)}\:\mathrm{=}\:\mathit{x}\mathrm{\left(\infty\right)}-\mathit{x}\mathrm{\left(0\right)}}$$
$$\mathrm{\therefore \mathit{x}\mathrm{\left(\infty\right)}\:\mathrm{=}\:\displaystyle \lim_{z \to 1}\mathrm{\left[ \mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}\right ]}}$$
Numerical Example (1)
Find $\mathit{x}\mathrm{\left(\infty\right)}$ if X(z) is given by,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z^{\mathrm{2}}}}{\mathrm{\left (\mathit{z}-1 \right )}\mathrm{\left( \mathit{z-\mathrm{0.3}}\right )}}}$$
Solution
The given Z-transform of the sequence is,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z^{\mathrm{2}}}}{\mathrm{\left (\mathit{z}-1 \right )}\mathrm{\left( \mathit{z-\mathrm{0.3}}\right )}}}$$
Now, using the final value theorem for Z-transform $\mathrm{\left [ i.e,\mathit{x}\mathrm{\left(\infty\right)}\:\mathrm{=}\:\displaystyle \lim_{z \to 1}\mathrm{\left[ \mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}\right ]} \right ]}$ ,we get,
$$\mathrm{\mathit{x}\mathrm{\left(\infty\right)}\:\mathrm{=}\:\displaystyle \lim_{z \to 1}\mathrm{\left(\mathit{z-\mathrm{1}}\right)}\mathrm{\left[\frac{\mathit{z^{\mathrm{2}}}}{\mathrm{\left (\mathit{z}-1 \right )}\mathrm{\left( \mathit{z-\mathrm{0.3}}\right )}}\right]}\:\mathrm{=}\:\:\displaystyle \lim_{z \to 1}\mathrm{\left[\frac{\mathit{z^{\mathrm{2}}}}{\mathrm{\left( \mathit{z-\mathrm{0.3}}\right )}}\right]}}$$
$$\mathrm{\therefore \mathit{x}\mathrm{\left(\infty\right)}\:\mathrm{=}\:\mathrm{\left[\frac{\mathrm{1}}{\mathrm{\left( \mathrm{1-\mathrm{0.3}}\right )}}\right]}\:\mathrm{=}\:1.43}$$
Numerical Example (2)
Using the final value theorem, calculate $\mathit{x}\mathrm{\left(\infty\right)}$ if X(z) is given by,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z}+1}{\mathrm{3\left (\mathit{z}-1 \right )}\mathrm{\left( \mathit{z\mathrm{+}\mathrm{0.4}}\right )}}}$$
Solution
The given Z-transform of the sequence is,
$$\mathrm{\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z}+1}{\mathrm{3\left (\mathit{z}-1 \right )}\mathrm{\left( \mathit{z\mathrm{+}\mathrm{0.4}}\right )}}}$$
$$\mathrm{\therefore \mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\frac{\mathit{z}+1}{\mathrm{3\mathrm{\left( \mathit{z\mathrm{+}\mathrm{0.4}}\right )}}}}$$
As we can see, $\mathrm{\left ( z-1 \right )}\mathit{X}\mathrm{\left(\mathit{z}\right)}$ has no poles on or outside the unit circle. Therefore, using the final value theorem for Z-transform, we have,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{\infty}\right)}\:\mathrm{=}\:\displaystyle \lim_{z \to 1}\mathrm{\left[\frac{\mathit{z}+1}{3\mathrm{\left ( \mathit{z}+0.4\right )}} \right ]}\:\mathrm{=}\:\mathrm{\left[\frac{\mathrm{1}+1}{3\mathrm{\left(\mathrm{1}+0.4\right )}} \right ]}}$$
$$\mathrm{\therefore \mathit{x}\mathrm{\left(\mathit{\infty}\right)}\:\mathrm{=}\:\mathrm{\left[\frac{\mathrm{2}}{3\times 1.4} \right ]}\:\mathrm{=}\:0.48}$$
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