Initial Value Theorem of Z-Transform

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Z-Transform

The Z-transform is a mathematical tool used to convert 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:

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

Initial Value Theorem of Z-Transform

The initial value theorem enables us to calculate the initial value of a signal x(n), i.e., x(0), 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

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

Where x(n) is a causal sequence, then,

$$\mathrm{x(0) \:=\: \lim_{n \to 0}\: x(n) \:=\: \lim_{z \to \infty}\: X(z)}$$

Proof

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

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

$$\mathrm{\Rightarrow\:Z[x(n)]\:=\:X(z) \:=\: x(0) \:+\: x(1) z^{-1} \:+\: x(2) z^{-2} \:+\: \dots}$$

$$\mathrm{\Rightarrow\:Z[x(n)]\:=\:X(z)\:=\:x(0)\:+\:\frac{x(1)}{z}\:+\:\frac{x(2)}{z^2}\:+\:\frac{x(3)}{z^3}\:+\:\dotso}$$

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

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

$$\mathrm{\Rightarrow\:\lim_{z \to \infty} X(z) \:=\: x(0) \:+\: 0 \:+\: 0 \:+\: 0 \:+\: \dotso \:=\: x(0)}$$

$$\mathrm{\therefore\:x(0) \:=\: \lim_{n \to 0} x(n)\:=\:\lim_{z\to \infty}\:X(z)}$$

Therefore, it gives the initial value of the function x(n) directly from its Z-transform.

Numerical Examples

Example 1

Using the initial value theorem for Z-transform, find the initial value of the signal x(n), i.e., x(0), if X(z) is given by,

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

Solution

The given Z-transform of the signal is,

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

$$\mathrm{\Rightarrow\:X(z) \:=\: \frac{1 \:+\: \left(\frac{1}{z}\right) \:+\: \left(\frac{1}{z^2}\right)}{\left(1 \:+\: \frac{2}{z}\right)\left(1 \:+\: \frac{1}{z}\right)}}$$

Now, using the initial value theorem of Z-transform $\mathrm{\left[\text{i.e., }\:x(0)\:=\:\lim_{n\to 0}\:x(n) \:=\: \lim_{z\to \infty}\:X(z)\right]}$. we get,

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

$$\mathrm{\Rightarrow\:x(0) \:=\: \frac{1 \:+\: 0 \:+\: 0}{(1\:+\:0)(1\:+\:0)} \:=\: 1}$$

Hence, the initial value of the given function is x(0) = 1.

Example 2

Find the initial value x(0) of a sequence x(n), if X(z) is given by,

$$\mathrm{X(z) \:=\: \frac{z \:+\: 4}{(z \:+\: 2)(z \:+\: 1)}}$$

Solution

The given Z-transform of the sequence is,

$$\mathrm{X(z) \:=\: \frac{z \:+\: 4}{(z \:+\: 2)(z \:+\: 1)}}$$

$$\mathrm{\Rightarrow\: X(z) \:=\: \frac{z\left(1 \:+\: \frac{4}{z}\right)}{z^2\left(1 \:+\: \frac{2}{z}\right)\left(1 \:+\: \frac{1}{z}\right)} \:=\: \frac{\left(1 \:+\: \frac{4}{z}\right)}{z\left(1 \:+\: \frac{2}{z}\right)\left(1 \:+\: \frac{1}{z}\right)}}$$

Using the initial value theorem of Z-transform, we get,

$$\mathrm{x(0) \:=\: \lim_{z \to \infty} \left[ \frac{\left(1 \:+\: \frac{4}{z}\right)}{z\left(1 \:+\: \frac{2}{z}\right)\left(1 \:+\: \frac{1}{z}\right)} \right]\:=\:0}$$

Therefore, the initial value of the given sequence is x(0) = 0.