Final Value Theorem of Z-Transform

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Z-Transform

The Z-transform is a mathematical tool that is used to convert the difference equations in the discrete-time domain into algebraic equations in the z-domain. Mathematically, if x(n) is a discrete-time function, then its Z-transform is defined as,

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

Final Value Theorem of Z-Transform

The final value theorem of the Z-transform enables us to calculate the steady-state value of a sequence x(n), i.e., $\mathrm{x(\infty)}$, directly from its Z-transform, without the need for finding its inverse Z-transform.

Statement

If x(n) is a causal sequence, the final value theorem of Z-transform states that if:

$$\mathrm{x(n) \:\overset{ZT}\longleftrightarrow\: X(z)}$$

And if the Z-transform X(z) has no poles outside the unit circle, and it has no higher poles on the unit circle centered at the origin of the z-plane, then:

$$\mathrm{x(\infty) \:=\: \lim_{n \to \infty}\: x(n) \:=\: \lim_{z \to 1}\: (z \:-\: 1)\: X(z)}$$

Proof

From the definition of the Z-transform of a causal sequence, we have,

$$\mathrm{Z[x(n)] \:=\: X(z) \:=\: \sum_{n=0}^{\infty}\: x(n) z^{-n}}$$

And:

$$\mathrm{Z[x(n\:+\:1)] \:=\: \sum_{n=0}^{\infty}\: x(n) z^{-n} \:=\: z X(z) \:-\: z x(0)}$$

$$\mathrm{\therefore\:Z[x(n\:+\:1)] \:-\: Z[x(n)] \:=\: \sum_{n=0}^{\infty}\: x(n) z^{-n} \:-\: \sum_{n=0}^{\infty}\:x(n) z^{-n}}$$

$$\mathrm{\Rightarrow\:Z[x(n\:+\:1)] \:-\: Z[x(n)] \:=\: zX(z) \:-\: zx(0) \:-\: X(z)}$$

$$\mathrm{\therefore\: (z\:-\:1)X(z) \:-\: zx(0) \:=\: \sum_{n=0}^{\infty}\: \left[ x(n\:+\:1) \:-\: x(n) \right] z^{-n}}$$

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

Now, taking the limit as $\mathrm{z \to 1}$ on both sides, we get

$$\mathrm{\lim_{z \to 1} \left[ (z\:-\:1)X(z) \:-\: zx(0) \right] \:=\: \lim_{z \to 1} \:\left\{ \left[ x(1) \:-\: x(0) \right] z^0 \:+\: \left[ x(2) \:-\: x(1) \right] z^{-1} \:+\: \left[ x(3) \:-\: x(2) \right] z^{-2} \:+\: \dots \right\}}$$

$$\mathrm{\lim_{z \to 1}\: \left[ (z \:-\: 1) X(z) \:-\: z x(0) \right] \:=\: x(1) \:-\: x(0) \:+\: x(2) \:-\: x(1) \:+\: x(3) \:-\: x(2) \:+\: \dots \:+\: x(\infty) \:-\: x(\infty \:-\: 1)}$$

$$\mathrm{\Rightarrow\:\lim_{z \to 1}\: \left[ (z \:-\: 1) X(z) \right] \:-\: x(0) \:=\: x(\infty) \:-\: x(0)}$$

$$\mathrm{\therefore\:x(\infty) \:=\: \lim_{z \to 1}\: \left[ (z \:-\: 1) X(z) \right]}$$

Numerical Examples

Example 1

Find $\mathrm{x(\infty)}$ if $\mathrm{X(z)}$ is given by,

$$\mathrm{X(z) \:=\: \frac{z^2}{(z \:-\: 1)(z \:-\: 0.3)}}$$

Solution

The given Z-transform of the sequence is,

$$\mathrm{X(z) \:=\: \frac{z^2}{(z \:-\: 1)(z \:-\: 0.3)}}$$

Now, using the final value theorem for Z-transform $\mathrm{\left[\text{i.e, }x(\infty)\:=\:\lim_{z\to 1}\:[(z\:-\:1)X(z)]\right]}$, we get,

$$\mathrm{x(\infty) \:=\: \lim_{z \to 1} \:(z \:-\: 1) \left[ \frac{z^2}{(z \:-\: 1)(z \:-\: 0.3)} \right]\:=\:\lim_{z\to 1}\:\left[\frac{z^2}{(z\:-\:0.3)} \right]}$$

$$\mathrm{\therefore\:x(\infty)\:=\:\left[\frac{1}{(1\:-\:0.3)} \right]\:=\:1.43}$$

Example 2

Using the final value theorem, calculate $\mathrm{x(\infty)}$ if $\mathrm{X(z)}$ is given by:

$$\mathrm{X(z) \:=\: \frac{z \:+\: 1}{3(z \:-\: 1)(z \:+\: 0.4)}}$$

Solution

The given Z-transform of the sequence is,

$$\mathrm{X(z) \:=\: \frac{z \:+\: 1}{3(z \:-\: 1)(z \:+\: 0.4)}}$$

$$\mathrm{\therefore\: (z\:-\:1)X(z) \:=\: z \:+\: \frac{z\:+\:1}{3(z \:+\: 0.4)}}$$

As we can see, $\mathrm{(z \:-\: 1) X(z)}$ has no poles on or outside the unit circle. Therefore, using the final value theorem for Z-transform, we have,

$$\mathrm{x(\infty) \:=\: \lim_{z \to 1} \left[ \frac{z \:+\:1}{3(z \:+\: 0.4)} \right] \:=\: \left[ \frac{1 \:+\:1}{3(1 \:+\: 0.4)} \right]}$$

$$\mathrm{\therefore\:x(\infty) \:=\: \left[ \frac{2}{3 \:\times\: 1.4} \right] \:=\: 0.48}$$