- Z-transform
In

mathematics andsignal processing , the**Z-transform**converts a discretetime-domain signal, which is asequence of real orcomplex number s, into a complexfrequency-domain representation.It is like a discrete equivalent of the Laplace Transform. This similarity is explored in the theory of

time scale calculus .The Z-transform and

advanced Z-transform were introduced (under the Z-transform name) byE. I. Jury in1958 in "Sampled-Data Control Systems" (John Wiley & Sons). The idea contained within the Z-transform was previously known as the "generating function method".**Definition**The Z-transform, like many other integral transforms, can be defined as either a "one-sided" or "two-sided" transform.

**Bilateral Z-transform**The "bilateral" or "two-sided" Z-transform of a discrete-time signal "x [n] " is the function "X(z)" defined as

:$X(z)\; =\; mathcal\{Z\}\{x\; [n]\; \}\; =\; sum\_\{n=-infty\}^\{infty\}\; x\; [n]\; z^\{-n\}$

where "n" is an integer and "z" is, in general, a

complex number ::$z=\; A\; e^\{jphi\}$:where "A" is the magnitude of "z", and φ is the "complex argument " (also referred to as "angle" or "phase") inradians .**Unilateral Z-transform**Alternatively, in cases where "x" ["n"] is defined only for "n" ≥ 0, the "single-sided" or "unilateral" Z-transform is defined as

:$X(z)\; =\; mathcal\{Z\}\{x\; [n]\; \}\; =\; sum\_\{n=0\}^\{infty\}\; x\; [n]\; z^\{-n\}$

In

signal processing , this definition is used when the signal is causal.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.**Geophysical Definition**In geophysics, the usual definition for the Z-transform is a polynomial in "z" as opposed to $z^\{-1\}$. This convention is used by Robinson and Treitel and by Kanasewich. The geophysical definition is

:$X(z)\; =\; mathcal\{Z\}\{x\; [n]\; \}\; =\; sum\_\{n\}\; x\; [n]\; z^\{n\}$

The two definitions are equivalent; however, the difference results in a number of changes. For example, the location of zeros and poles move from inside the unit circle, using one definition, to outside the unit circle, using the other definition (and vice versa). Thus, care is required to note which definition is being used by a particular author.

**Inverse Z-transform**The "inverse" Z-transform is

:$x\; [n]\; =\; mathcal\{Z\}^\{-1\}\; \{X(z)\; \}=\; frac\{1\}\{2\; pi\; j\}\; oint\_\{C\}\; X(z)\; z^\{n-1\}\; dz$

where $C$ is a counterclockwise closed path encircling the origin and entirely in the region of convergence (ROC). The contour or path, $C$, must encircle all of the poles of $X(z)$.

A special case of this

contour integral occurs when $C$ is the unit circle (and can be used when the ROC includes the unit circle). The inverse Z-transform simplifies to the inverse discrete-time Fourier transform::$x\; [n]\; =\; frac\{1\}\{2\; pi\}\; int\_\{-pi\}^\{+pi\}\; X(e^\{j\; omega\})\; e^\{j\; omega\; n\}\; d\; omega$ .

The 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-time Fourier transform (DTFT) (not to be confused with thediscrete Fourier transform (DFT)) is a special case of such a Z-transform obtained by restricting "z" to lie on the unit circle.**Region of convergence**The region of convergence (ROC) is the set of points in the complex plane for which the Z-transform summation converges.

:$ROC\; =\; left\{\; z\; :\; left|sum\_\{n=-infty\}^\{infty\}x\; [n]\; z^\{-n\}\; ight|\; <\; infty\; ight\}$

**Example 1 (No ROC)**Let $x\; [n]\; =\; 0.5^n$. Expanding $x\; [n]$ on the interval $(-infty,\; infty)$ it becomes

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

Looking at the sum

:$sum\_\{n=-infty\}^\{infty\}x\; [n]\; z^\{-n\}\; <\; infty\; .$

There are no such values of $z$ that satisfy this condition.

=Example 2 (causal ROC)=Let $x\; [n]\; =\; 0.5^n\; u\; [n]$ (where $u$ is the

Heaviside step function ). Expanding $x\; [n]$ on the interval $(-infty,\; infty)$ it becomes:$x\; [n]\; =\; \{...,\; 0,\; 0,\; 0,\; 1,\; 0.5,\; 0.5^2,\; 0.5^3,\; ...\}$

Looking at the sum

:$sum\_\{n=-infty\}^\{infty\}x\; [n]\; z^\{-n\}\; =\; sum\_\{n=0\}^\{infty\}0.5^nz^\{-n\}\; =\; sum\_\{n=0\}^\{infty\}left(frac\{0.5\}\{z\}\; ight)^n\; =\; frac\{1\}\{1\; -\; 0.5z^\{-1$

The last equality arises from the infinite

geometric series and the equality only holds if $left|0.5\; z^\{-1\}\; ight|\; <\; 1$ which can be rewritten in terms of $z$ as $left|z\; ight|\; >\; 0.5$. Thus, the ROC is $left|z\; ight|\; >\; 0.5$. 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 $x\; [n]\; =\; -(0.5)^n\; u\; [-n-1]$ (where $u$ is the

Heaviside step function ). Expanding $x\; [n]$ on the interval $(-infty,\; infty)$ it becomes:$x\; [n]\; =\; \{...,\; -(0.5)^\{-3\},\; -(0.5)^\{-2\},\; -(0.5)^\{-1\},\; 0,\; 0,\; 0,\; ...\}$

Looking at the sum

:$sum\_\{n=-infty\}^\{infty\}x\; [n]\; z^\{-n\}\; =\; -sum\_\{n=-infty\}^\{-1\}0.5^nz^\{-n\}\; =\; -sum\_\{n=-infty\}^\{-1\}left(frac\{z\}\{0.5\}\; ight)^\{-n\}$:$=\; -sum\_\{m=1\}^\{infty\}left(frac\{z\}\{0.5\}\; ight)^\{m\}\; =\; -frac\{0.5^\{-1\}z\}\{1\; -\; 0.5^\{-1\}z\}\; =\; frac\{z\}\{z\; -\; 0.5\}\; =\; frac\{1\}\{1\; -\; 0.5z^\{-1$

Using the infinite

geometric series , again, the equality only holds if $left|0.5^\{-1\}z\; ight|\; <\; 1$ which can be rewritten in terms of $z$ as $left|z\; ight|\; <\; 0.5$. Thus, the ROC is $left|z\; ight|\; <\; 0.5$. In this case the ROC is a disc centered at the origin and of radius 0.5.What differentiates this example from the previous example is "only" the ROC. This is intentional to demonstrate that the transform result alone is insufficient.

**Examples conclusion**Examples 2 & 3 clearly show that the Z-transform $X(z)$ of $x\; [n]$ is unique when and 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 $left|\; z\; ight|\; =\; infty$ while the anticausal system in example 3 yields an ROC that includes $left|\; z\; ight|\; =\; 0$.

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

The stability of a system can also be determined by knowing the ROC alone. If the ROC contains the unit circle (i.e., $left|\; z\; ight|\; =\; 1$) then the system is stable. In the above systems the causal system (Example 2) is stable because $left|\; z\; ight|\; >\; 0.5$ contains the unit circle.

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

* Stability

* CausalityIf you need stability then the ROC must contain the unit circle.If you need a causal system then the ROC must contain infinity and the system function will be a right-sided sequence.If you need an anticausal system then the ROC must contain the origin and the system function will be a left-sided sequence. If you need both, stability and causality, all the poles of the system function must be inside the unit circle.

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

**Properties***

**Initial value theorem**::$x\; [0]\; =lim\_\{z\; ightarrow\; infty\}X(z)$, If $x\; [n]\; ,$ causal*

**Final value theorem**:: $x\; [infty]\; =lim\_\{z\; ightarrow\; 1\}(1-z^\{-1\})X(z)$, Only if poles of $(1-z^\{-1\})X(z)$ are inside the unit circle**Table of common Z-transform pairs**Here:

* u [n] =1 for n>=0, u [n] =0 for n<0

* δ [n] = 1 for n=0, δ [n] = 0 otherwise**Relationship to Laplace**The bilateral Z-transform is simply the

two-sided Laplace transform of the ideal sampled function::$x\_\{s\}(t)\; =\; sum\_\{n=-infty\}^\{infty\}\; x(nT)\; delta(t-nT)\; =\; sum\_\{n=-infty\}^\{infty\}\; x\; [n]\; delta(t-nT)$

where $x(t)$ is the continuous-time function being sampled, $x\; [n]\; =x(nT)$ the n

^{th}sample, $T$ is the sampling period, and with the substitution: $z\; =\; e^\{sT\}$.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.The

Bilinear transform is a useful approximation for converting continuous time filters (represented in Laplace space) into discrete time filters (represented in z space), and vice versa. To do this, you can use the following substitutions in $H(s)$ or $H(z)$ :$s\; =frac\{2\}\{T\}\; frac\{z-1\}\{z+1\}$ from Laplace to z (Tustin transformation);

$z\; =frac\{2+sT\}\{2-sT\}$ from z to Laplace.

**Relationship to Fourier**The Z-transform is a generalization of the

discrete-time Fourier transform (DTFT). The DTFT can be found by evaluating the Z-transform $X(z)$ at $z=e^\{jomega\}$ or, in other words, evaluated on the unit circle. In order to determine thefrequency response of the system the Z-transform must be evaluated on the unit circle, meaning that the system's region of convergence must contain the unit circle. Otherwise, the DTFT of the system does not exist.**Linear constant-coefficient difference equation**The linear constant-coefficient difference (LCCD) equation is a representation for a linear system based on the

autoregressive moving-average equation.:$sum\_\{p=0\}^\{N\}y\; [n-p]\; alpha\_\{p\}\; =\; sum\_\{q=0\}^\{M\}x\; [n-q]\; eta\_\{q\}$

Both sides of the above equation can be divided by $alpha\_0$, if it is not zero, normalizing $alpha\_0\; =\; 1$ and the LCCD equation can be written

:$y\; [n]\; =\; sum\_\{q=0\}^\{M\}x\; [n-q]\; eta\_\{q\}\; -\; sum\_\{p=1\}^\{N\}y\; [n-p]\; alpha\_\{p\}$

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

**Transfer function**Taking the Z-transform of the above equation (using linearity and time-shifting laws) yields

:$Y(z)\; sum\_\{p=0\}^\{N\}z^\{-p\}alpha\_\{p\}\; =\; X(z)\; sum\_\{q=0\}^\{M\}z^\{-q\}eta\_\{q\}$

and rearranging results in

:$H(z)\; =\; frac\{Y(z)\}\{X(z)\}\; =\; frac\{sum\_\{q=0\}^\{M\}z^\{-q\}eta\_\{q\{sum\_\{p=0\}^\{N\}z^\{-p\}alpha\_\{p\; =\; frac\{eta\_0\; +\; z^\{-1\}\; eta\_1\; +\; z^\{-2\}\; eta\_2\; +\; cdots\; +\; z^\{-M\}\; eta\_M\}\{alpha\_0\; +\; z^\{-1\}\; alpha\_1\; +\; z^\{-2\}\; alpha\_2\; +\; cdots\; +\; z^\{-N\}\; alpha\_N\}.$

**Zeros and poles**From the

fundamental theorem of algebra thenumerator has M roots (corresponding to zeros of H) and thedenominator has N roots (corresponding to poles). Rewriting thetransfer function in terms of poles and zeros:$H(z)\; =\; frac\{(1\; -\; q\_1\; z^\{-1\})(1\; -\; q\_2\; z^\{-1\})cdots(1\; -\; q\_M\; z^\{-1\})\; \}\; \{\; (1\; -\; p\_1\; z^\{-1\})(1\; -\; p\_2\; z^\{-1\})cdots(1\; -\; p\_N\; z^\{-1\})\}$

Where $q\_k$ is the $k^\{th\}$ zero and $p\_k$ is the $k^\{th\}$ pole. The zeros and poles are commonly complex and when plotted on the complex plane (z-plane) it is called the

pole-zero plot .In simple words, zeros are the solutions to the equation obtained by setting the numerator equal to zero, while poles are the solutions to the equation obtained by setting the denominator equal to zero.

In addition, there may also exist zeros and poles at $z=0$ and $z=infty$. If we take these poles and zeros as well as multiple-order zeros and poles into consideration, the number of zeros and poles are always equal.

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 theimpulse response and the linear constant coefficient difference equation of the system.**Output response**If such a system $H(z)$ is driven by a signal $X(z)$ then the output is $Y(z)\; =\; H(z)X(z)$. By performing

partial fraction decomposition on $Y(z)$ and then taking the inverse Z-transform the output $y\; [n]$ can be found. In practice, it is often useful to fractionally decompose $frac\{Y(z)\}\{z\}$ before multiplying that quantity by $z$ to generate a form of $Y(z)$ which has terms with easily computable inverse Z-transforms.**ee also***

Advanced Z-transform

* Bilinear transform

*Finite impulse response

*Formal power series

*Laplace transform

*Laurent series

*Probability-generating function

*Zeta function regularization

*Discrete-time Fourier transform **Bibliography*** Eliahu Ibrahim Jury, "Theory and Application of the Z-Transform Method", Krieger Pub Co, 1973. ISBN 0-88275-122-0.

* Refaat El Attar, "Lecture notes on Z-Transform", Lulu Press, Morrisville NC, 2005. ISBN 1-4116-1979-X.

* Ogata, Katsuhiko, "Discrete Time Control Systems 2nd Ed", Prentice-Hall Inc, 1995, 1987. ISBN 0-13-034281-5.

* Alan V. Oppenheim and Ronald W. Schafer (1999). Discrete-Time Signal Processing, 2nd Edition, Prentice Hall Signal Processing Series. ISBN 0-13-754920-2.

**External links*** [

*http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/LaplaceZTable/LaplaceZFuncTable.html Z-Transform table of some common Laplace transforms*]

* [*http://mathworld.wolfram.com/Z-Transform.html Mathworld's entry on the Z-transform*]

* [*http://www.dsprelated.com/comp.dsp/keyword/Z_Transform.php Z-Transform threads in Comp.DSP*]

* [*http://math.fullerton.edu/mathews/c2003/ZTransformIntroMod.html Z-Transform Module by John H. Mathews*]

* [*http://www.ling.upenn.edu/courses/ling525/z.html University of Penn Z-Transform Intro*] ( [*http://dev.gentoo.org/~redhatter/misc/z-transform.pdf PDF version*] with more readable formulae)

*Wikimedia Foundation.
2010.*

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