# Z-transform

In mathematics and signal processing, the Z-transform converts a discrete time domain signal, which is a sequence of real numbers, into a complex frequency domain representation.

The Z-transform and advanced Z-transform were introduced (under the Z-transform name) by E. I. Jury in 1958 in Sampled-Data Control Systems (John Wiley & Sons). The idea contained within the Z-transform was previously known as the "generating function method".

Z-transform is a placeholder name, akin to calling the Laplace transform the "s-transform". More accurate would be "Laurent transform", because it is based on the Laurent series.

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## Definition

The Z-transform of a discrete-time signal x[n] is the function X(z) defined by

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

where n is an integer and z is a complex circular number of the form [itex] z=r \ e^{j\omega} [itex].

Sometimes we are only interested in the values of the signal x[n] for non-negative values of n. If such is the case, the Z-transform is defined as

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

The former, with limits from negative to positive infinity, is known as the bilateral or doubly infinite Z-transform. The latter, with limits from 0 to [itex] + \infty \ [itex], is sometimes called the unilateral Z-transform. In signal processing, the latter definition is used when the signal is causal in nature.

An important example of the unilateral Z-transform is the probability-generating function, where the component x[n] is the probability that a discrete random variable takes the value n, and the function X(z) is usually written as X(s), in terms of s = z−1. The properties of Z-transforms (below) have useful interpretations in the context of probability theory.

## Region of convergence

The region of convergence (ROC) is where the Z-transform of a signal has a finite sum for a region in the complex plane.

[itex]ROC = \{z : \sum_{n=-\infty}^{\infty}x[n]z^{-n} < \infty\}\ [itex]

### Example 1 (No ROC)

Let [itex]x[n] = 0.5^n\ [itex]. Expanding [itex]x[n]\ [itex] on [itex](-\infty, \infty)\ [itex] it becomes

[itex]x[n] = \{..., 0.5^{-3}, 0.5^{-2}, 0.5^{-1}, 1, 0.5, 0.5^2, 0.5^3, ...\} = \{..., 2^3, 2^2, 2, 1, 0.5, 0.5^2, 0.5^3, ...\}\ [itex]

Looking at the sum

[itex]\sum_{n=-\infty}^{\infty}x[n]z^{-n} < \infty\ [itex]

There are no such values of [itex]z\ [itex] that satisfy this condition.

### Example 2 (causal ROC)

Let [itex]x[n] = 0.5^n u[n]\ [itex]. Expanding [itex]x[n]\ [itex] on [itex](-\infty, \infty)\ [itex] it becomes

[itex]x[n] = \{..., 0, 0, 0, 1, 0.5, 0.5^2, 0.5^3, ...\}\ [itex]

Looking at the sum

[itex]\sum_{n=-\infty}^{\infty}x[n]z^{-n} = \sum_{n=0}^{\infty}0.5^nz^{-n} = \sum_{n=0}^{\infty}\left(\frac{0.5}{z}\right)^n = \frac{1}{1 - 0.5z^{-1}}\ [itex]

The last equality arises from the infinite geometric series and the equality only holds if [itex]\left|0.5 z^{-1}\right| < 1\ [itex] which can be rewritten in terms of [itex]z\ [itex] as [itex]\left|z\right| > 0.5\ [itex]. Thus, the ROC is [itex]\left|z\right| > 0.5\ [itex]. In this case the ROC is the complex plane with a disc of radius 0.5 at the origin "punched out".

### Example 3 (anticausal ROC)

Let [itex]x[n] = -(0.5)^n u[-n-1]\ [itex] (where [itex]u[itex] is the Heaviside step function). Expanding [itex]x[n]\ [itex] on [itex](-\infty, \infty)\ [itex] it becomes

[itex]x[n] = \{..., -(0.5)^{-3}, -(0.5)^{-2}, -(0.5)^{-1}, 0, 0, 0, ...\}\ [itex]

Looking at the sum

[itex]\sum_{n=-\infty}^{\infty}x[n]z^{-n} = -\sum_{n=-\infty}^{-1}0.5^nz^{-n} = -\sum_{n=-\infty}^{-1}\left(\frac{z}{0.5}\right)^{-n}\ [itex]
[itex]= -\sum_{m=1}^{\infty}\left(\frac{z}{0.5}\right)^{m} = -\frac{0.5^{-1}z}{1 - 0.5^{-1}z} = \frac{z}{z - 0.5} = \frac{1}{1 - 0.5z^{-1}}\ [itex]

Using the infinite geometric series, again, the equality only holds if [itex]\left|0.5^{-1}z\right| < 1\ [itex] which can be rewritten in terms of [itex]z\ [itex] as [itex]\left|z\right| < 0.5\ [itex]. Thus, the ROC is [itex]\left|z\right| < 0.5\ [itex]. In this case the ROC is a disc centered at the origin and of radius 0.5.

### Examples conclusion

Examples 2 & 3 clearly show that the Z-transform [itex]X(z)\ [itex] of [itex]x[n]\ [itex] is unique only when specifying the ROC. Creating the pole-zero plot for the causal and anticausal case show that the ROC for either case does not include the pole that is at 0.5. This extends to cases with multiple poles: the ROC will never contain poles.

In example 2, the causal system yields an ROC that includes [itex]\left| z \right| = \infty\ [itex] while the anticausal system in example 3 yields an ROC that includes [itex]\left| z \right| = 0\ [itex].

In systems with multiple poles it is possible to have an ROC that includes neither [itex]\left| z \right| = \infty\ [itex] nor [itex]\left| z \right| = 0\ [itex]. The ROC creates a circular band. For example, [itex]x[n] = 0.5^nu[n] - 0.75^nu[-n-1]\ [itex] has poles at 0.5 and 0.75. The ROC will be [itex]0.5 < \left| z \right| < 0.75\ [itex], which includes neither the origin nor infinity. Such a system is called a mixed-causality system as it contains a causal term [itex]0.5^nu[n]\ [itex] and an anticausal term [itex]-(0.75)^nu[-n-1]\ [itex].

The stability of a system can also be determined by knowing the ROC alone. If the ROC contains the unit circle (i.e., [itex]\left| z \right| = 1\ [itex]) then the system is stable. In the above systems the causal system is stable because [itex]\left| z \right| > 0.5\ [itex] contains the unit circle.

If you are provided a Z-transform of a system without an ROC (i.e., an ambiguous [itex]x[n]\ [itex]) you can determine a unique [itex]x[n]\ [itex] provided you desire the following:

• Stability
• Causality

If you need stability then the ROC must contain the unit circle. If you need a causal system then the ROC must contain infinity. If you need an anticausal system then the ROC must contain the origin.

The unique [itex]x[n]\ [itex] can then be found.

## Properties

• Linearity. The Z-transform of the linear combination of two signals is the linear combination of the individual Z-transforms.
[itex]Z(a_1 x_1[n] + a_2 x_2[n]) = a_1 Z(x_1[n]) + a_2 Z(x_2[n]) \ [itex]
• Shift. Time-shifting the signal by a distance of k to the right results in multiplying the Z-transform by z−k.
[itex]Z(x[n-k]) = z^{-k}Z(x[n]) \ [itex]
• Convolution. The Z-transform of the convolution of two sequences is the product of the individual Z-transforms.
[itex]Z(\{x[n]\} \bigodot \! \! \! \! \! \! \star \ \ \{y[n]\}) = Z(\{x[n]\}) Z(\{y[n]\}) \ [itex]
• Differentiation.
[itex]Z(\{nx[n]\}) = \ -z \frac{dZ(\{x[n]\})}{dz} \ [itex]

The inverse Z-transform can be computed as follows:

[itex]x[n]=\frac{1}{2\pi i}\oint_CX(z)z^{n-1}\,dz\ [itex]

where [itex]C \ [itex] is any closed curve around the origin and lying in the region of convergence of [itex]X(z) \ [itex].

The (unilateral) Z-transform is to discrete time domain signals what the Laplace transform is to continuous time domain signals.

Z-transform with a finite range of n and a finite number of uniformly-spaced z values can be computed efficiently via Bluestein's FFT algorithm. The discrete Fourier transform is a special case of such a Z-transform obtained by restricting z to lie on the unit circle.

## Relationship to Laplace

The bilateral Z-transform is simply the two-sided Laplace transform of the ideal sampled function

[itex] x(t) \sum_{n=-\infty}^{\infty} \delta(t-nT) = \sum_{n=-\infty}^{\infty} x[n] \delta(t-nT) \ [itex]

where [itex] x(t) \ [itex] is the continuous-time function being sampled, [itex] x[n]=x(nT) \ [itex] the nth sample, [itex] T \ [itex] is the sampling period, and with the substitution: [itex] z = e^{sT} \ [itex].

Likewise the unilateral Z-transform is simply the one-sided Laplace transform of the ideal sampled function. Both assume that the sampled function is zero for all negative time indices.

## Relationship to Fourier

The Z-transform is a more generalization of the discrete fourier transform (DFT). The DFT can be found by evaluating the Z-transform [itex]X(z)\ [itex] at [itex]z=e^{j\omega}\ [itex] or, in other words, evaluated on the unit circle. In order to determine the frequency response of the system the Z-transform must be evaluated on the unit circle.

## LCCD equation

Linear constant coefficient difference (LCCD) equation is a representation for a linear system based on the ARMA equation.

[itex]\sum_{p=0}^{N}y[n-p]\alpha_{p} = \sum_{q=0}^{M}x[n-q]\beta_{q}\ [itex]

Both sides of the above equation can be divided by [itex]\alpha_0 \ [itex], if it is not zero, normalizing [itex]\alpha_0 = 1\ [itex] and the LCCD equation can be written

[itex]y[n] = \sum_{q=0}^{M}x[n-q]\beta_{q} - \sum_{p=1}^{N}y[n-p]\alpha_{p}\ [itex]

This form of the LCCD equation is favorable to make it more explicit that the "current" output [itex]y[{n}]\ [itex] is a function of past outputs [itex]y[{n-p}]\ [itex], current input [itex]x[{n}]\ [itex], and previous inputs [itex]x[{n-q}]\ [itex].

### Transfer function

Taking the Z-transform of the equation yields

[itex]Y(z) \sum_{p=0}^{N}z^{-p}\alpha_{p} = X(z) \sum_{q=0}^{M}z^{-q}\beta_{q}\ [itex]

and rearranging results in

[itex]H(z) = \frac{Y(z)}{X(z)} = \frac{\sum_{q=0}^{M}z^{-q}\beta_{q}}{\sum_{p=0}^{N}z^{-p}\alpha_{p}} = \frac{\beta_0 + z^{-1} \beta_1 + z^{-2} \beta_2 + ... + z^{-M} \beta_M}{\alpha_0 + z^{-1} \alpha_1 + z^{-2} \alpha_2 + ... + z^{-N} \alpha_N}\ [itex]

### Zeros and poles

From the fundamental theorem of algebra the numerator has M roots (called zeros) and the denominator has N roots (called poles). Rewriting the transfer function in terms of poles

[itex]H(z) = \frac{(1 + q_1 z^{-1})(1 + q_2 z^{-1})...(1 + q_M z^{-1})}{(1 + p_1 z^{-1})(1 + p_2 z^{-1})...(1 + p_N z^{-1})}\ [itex]

Where [itex]q_k\ [itex] is the [itex]k^{th}\ [itex] zero and [itex]p_k\ [itex] is the [itex]k^{th}\ [itex] pole. The zeros and poles are commonly complex and when plotted on the complex plane it is called the pole-zero plot.

By factoring the denominator, partial fraction decomposition can be used, which can then be transformed back to the time domain. Doing so would result in the impulse response of the system.

### Output response

If such a system [itex]H(z)\ [itex] is driven by a signal [itex]X(z)\ [itex] then the output is [itex]Y(z) = H(z)X(z)\ [itex]. By performing partial fraction decomposition on [itex]Y(z)\ [itex] and then taking the inverse Z-transform the output [itex]y[n]\ [itex] can be found.

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