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Z-transform

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From Wikipedia, the free encyclopedia

Linear transform from the time domain to the frequency domain

Not to be confused withFisher z-transformation.

Inmathematics andsignal processing, theZ-transform converts adiscrete-time signal, which is asequence ofreal orcomplex numbers, into a complex valuedfrequency-domain (thez-domain orz-plane) representation.^[1]^[2]^[3]

It can be considered a discrete-time counterpart of theLaplace transform (thes-domain ors-plane).^[4] This similarity is explored in the theory oftime-scale calculus.

While thecontinuous-time Fourier transform is evaluated on the s-domain's vertical axis (the imaginary axis), thediscrete-time Fourier transform is evaluated along the z-domain'sunit circle. The s-domain's lefthalf-plane maps to the area inside the z-domain's unit circle, while the s-domain's right half-plane maps to the area outside of the z-domain's unit circle.

In signal processing, one of the means of designingdigital filters is to take analog designs, subject them to abilinear transform which maps them from the s-domain to the z-domain, and then produce the digital filter by inspection, manipulation, or numerical approximation. Such methods tend not to be accurate except in the vicinity of the complex unity, i.e. at low frequencies.

History

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The basic idea now known as the Z-transform was known toLaplace, and it was re-introduced in 1947 byW. Hurewicz^[5]^[6] and others as a way to treat sampled-data control systems used with radar. It gives a tractable way to solve linear, constant-coefficientdifference equations. It was later dubbed "the z-transform" byRagazzini andZadeh in the sampled-data control group at Columbia University in 1952.^[7]^[8]

The modified oradvanced Z-transform was later developed and popularized byE. I. Jury.^[9]^[10]

The idea contained within the Z-transform is also known in mathematical literature as the method ofgenerating functions which can be traced back as early as 1730 when it was introduced byde Moivre in conjunction with probability theory.^[11] From a mathematical view the Z-transform can also be viewed as aLaurent series where one views the sequence of numbers under consideration as the (Laurent) expansion of an analytic function.

Definition

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The Z-transform can be defined as either aone-sided ortwo-sided transform. (Just as we have theone-sided Laplace transform and thetwo-sided Laplace transform.)^[12]

Bilateral Z-transform

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Thebilateral ortwo-sided Z-transform of a discrete-time signal $x[n]$ is theformal power series $X(z)$ defined as:

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

where $n {\displaystyle n}$ is an integer and $z {\displaystyle z}$ is, in general, acomplex number. Inpolar form, $z {\displaystyle z}$ may be written as:

z=Ae^{i\phi }=A\cdot (\cos {\phi }+i\sin {\phi })

where $A {\displaystyle A}$ is the magnitude of $z {\displaystyle z}$ , $i {\displaystyle i}$ is theimaginary unit, and $\phi$ is thecomplex argument (also referred to asangle orphase) inradians.

Unilateral Z-transform

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Alternatively, in cases where $x[n]$ is defined only for $n\geq 0$ , thesingle-sided orunilateral Z-transform is defined as:

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

Insignal processing, this definition can be used to evaluate the Z-transform of theunit impulse response of a discrete-timecausal system.

An important example of the unilateral Z-transform is theprobability-generating function, where the component $x[n]$ is the probability that a discrete random variable takes the valuen. The properties of Z-transforms (listed in§ Properties) have useful interpretations in the context of probability theory.

Inverse Z-transform

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Theinverse Z-transform is:

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

where $C {\displaystyle C}$ is a counterclockwise closed path encircling the origin and entirely in theregion of convergence (ROC). In the case where the ROC is causal (seeExample 2), this means the path $C {\displaystyle C}$ must encircle all of the poles of $X(z)$ .

A special case of thiscontour integral occurs when $C {\displaystyle C}$ is the unit circle. This contour can be used when the ROC includes the unit circle, which is always guaranteed when $X(z)$ is stable, that is, when all the poles are inside the unit circle. With this contour, the inverse Z-transform simplifies to theinverse discrete-time Fourier transform, orFourier series, of the periodic values of the Z-transform around the unit circle:

$x[n]={\frac {1}{2\pi }}\int _{-\pi }^{\pi }X(e^{i\omega })e^{i\omega n}d\omega .$

The Z-transform with a finite range of $n {\displaystyle n}$ and a finite number of uniformly spaced $z {\displaystyle z}$ values can be computed efficiently viaBluestein's FFT algorithm. Thediscrete-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 $z {\displaystyle z}$ to lie on the unit circle.

The following three methods are often used for the evaluation of the inverse -transform,

Direct evaluation by contour integration

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This method involves applying theCauchy Residue Theorem to evaluate the inverse Z-transform. By integrating around a closed contour in the complex plane, the residues at the poles of the Z-transform function inside the ROC are summed. This technique is particularly useful when working with functions expressed in terms of complex variables.

Expansion into a series of terms in the variablesz andz⁻¹

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In this method, the Z-transform is expanded into a power series. This approach is useful when the Z-transform function is rational, allowing for the approximation of the inverse by expanding into a series and determining the signal coefficients term by term.

Partial-fraction expansion and table lookup

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This technique decomposes the Z-transform into a sum of simpler fractions, each corresponding to known Z-transform pairs. The inverse Z-transform is then determined by looking up each term in a standard table of Z-transform pairs. This method is widely used for its efficiency and simplicity, especially when the original function can be easily broken down into recognizable components.

Example

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^[13]

A) Determine the inverse Z-transform of the following by series expansion method, $X(z)={\frac {1}{1-1.5z^{-1}+0.5z^{-2}}}$

Solution:

Case 1:

ROC: $\left\vert Z\right\vert >1$

Since the ROC is the exterior of a circle, $x(n)$ is causal (signal existing forn ≥ 0). $X(z)={1 \over 1-{3 \over 2}z^{-1}+{1 \over 2}z^{-2}}=1+{{3 \over 2}z^{-1}}+{{7 \over 4}z^{-2}}+{{15 \over 8}z^{-3}}+{{31 \over 16}z^{-4}}+....$ thus, ${\begin{aligned}x(n)&=\left\{1,{\frac {3}{2}},{\frac {7}{4}},{\frac {15}{8}},{\frac {31}{16}}\ldots \right\}\\&\qquad \!\uparrow \\\end{aligned}}$ (arrow indicates term atx(0) = 1).

Note that in each step of long division process we eliminate lowest power term of $z^{-1}$ .

Case 2:

ROC: $\left\vert Z\right\vert <0.5$

Since the ROC is the interior of a circle, $x(n)$ is anticausal (signal existing forn < 0).

By performing long division we get $X(z)={\frac {1}{1-{\frac {3}{2}}z^{-1}+{\frac {1}{2}}z^{-2}}}=2z^{2}+6z^{3}+14z^{4}+30z^{5}+\ldots$

{\begin{aligned}x(n)&=\{30,14,6,2,0,0\}\\&\qquad \qquad \qquad \quad \ \ \,\uparrow \\\end{aligned}}

(arrow indicates term atx(0) = 0).

Note that in each step of long division process we eliminate lowest power term of $z {\displaystyle z}$ .

Note:

When the signal is causal, we get positive powers of $z {\displaystyle z}$ and when the signal is anticausal, we get negative powers of $z {\displaystyle z}$ .
$z^{k}$ indicates term at $x(-k)$ and $z^{-k}$ indicates term at $x(k)$ .

B) Determine the inverse Z-transform of the following by series expansion method,

Eliminating negative powers if $z {\displaystyle z}$ and dividing by $z {\displaystyle z}$ , ${\frac {X(z)}{z}}={\frac {z^{2}}{z(z^{2}-1.5z+0.5)}}={\frac {z}{z^{2}-1.5z+0.5}}$

By partial fraction expansion, ${\begin{aligned}{\frac {X(z)}{z}}&={\frac {z}{(z-1)(z-0.5)}}={\frac {A_{1}}{z-0.5}}+{\frac {A_{2}}{z-1}}\\[4pt]&A_{1}=\left.{\frac {(z-0.5)X(z)}{z}}\right\vert _{z=0.5}={\frac {0.5}{(0.5-1)}}=-1\\[4pt]&A_{2}=\left.{\frac {(z-1)X(z)}{z}}\right\vert _{z=1}={\frac {1}{1-0.5}}={2}\\[4pt]{\frac {X(z)}{z}}&={\frac {2}{z-1}}-{\frac {1}{z-0.5}}\end{aligned}}$

Case 1:

ROC: $\left\vert Z\right\vert >1$

Both the terms are causal, hence $x(n)$ is causal.

${\begin{aligned}x(n)&=2{(1)^{n}}u(n)-1{(0.5)^{n}}u(n)\\&=(2-0.5^{n})u(n)\\\end{aligned}}$

Case 2:

ROC: $\left\vert Z\right\vert <0.5$

Both the terms are anticausal, hence $x(n)$ is anticausal.

${\begin{aligned}x(n)&=-2{(1)^{n}}u(-n-1)-(-1{(0.5)^{n}}u(-n-1))\\&=(0.5^{n}-2)u(-n-1)\\\end{aligned}}$

Case 3:

ROC: $0.5<\left\vert Z\right\vert <1$

One of the terms is causal (p=0.5 provides the causal part) and other is anticausal (p=1 provides the anticausal part), hence $x(n)$ is both sided.

${\begin{aligned}x(n)&=-2{(1)^{n}}u(-n-1)-1{(0.5)^{n}}u(n)\\&=-2u(-n-1)-0.5^{n}u(n)\\\end{aligned}}$

Region of convergence

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Theregion of convergence (ROC) is the set of points in the complex plane for which the Z-transform summationabsolutely converges:

\mathrm {ROC} =\left\{z:\sum _{n=-\infty }^{\infty }\left|x[n]z^{-n}\right|<\infty \right\}

Example 1 (no ROC)

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Let $x[n]=(0.5)^{n}\ .$ Expanding $x[n]$ on the interval $(-\infty ,\infty )$ it becomes

x[n]=\left\{\dots ,(0.5)^{-3},(0.5)^{-2},(0.5)^{-1},1,(0.5),(0.5)^{2},(0.5)^{3},\dots \right\}=\left\{\dots ,2^{3},2^{2},2,1,(0.5),(0.5)^{2},(0.5)^{3},\dots \right\}.

Looking at the sum

\sum _{n=-\infty }^{\infty }x[n]z^{-n}\to \infty .

Therefore, there are no values of $z {\displaystyle z}$ that satisfy this condition.

Example 2 (causal ROC)

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ROC (blue), |z| = 0.5 (dashed black circle), and the unit circle (dotted grey circle).

Let $x[n]=(0.5)^{n}\,u[n]$ (where $u {\displaystyle u}$ is theHeaviside step function). Expanding $x[n]$ on the interval $(-\infty ,\infty )$ it becomes

x[n]=\left\{\dots ,0,0,0,1,(0.5),(0.5)^{2},(0.5)^{3},\dots \right\}.

Looking at the sum

\sum _{n=-\infty }^{\infty }x[n]z^{-n}=\sum _{n=0}^{\infty }(0.5)^{n}z^{-n}=\sum _{n=0}^{\infty }\left({\frac {0.5}{z}}\right)^{n}={\frac {1}{1-(0.5)z^{-1}}}.

The last equality arises from the infinitegeometric series and the equality only holds if $|(0.5)z^{-1}|<1,$ which can be rewritten in terms of $z {\displaystyle z}$ as $|z|>(0.5).$ Thus, the ROC is $|z|>(0.5).$ 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)

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Let $x[n]=-(0.5)^{n}\,u[-n-1]$ (where $u {\displaystyle u}$ is theHeaviside step function). Expanding $x[n]$ on the interval $(-\infty ,\infty )$ it becomes

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

Looking at the sum

{\begin{aligned}\sum _{n=-\infty }^{\infty }x[n]\,z^{-n}&=-\sum _{n=-\infty }^{-1}(0.5)^{n}\,z^{-n}\\&=-\sum _{m=1}^{\infty }\left({\frac {z}{0.5}}\right)^{m}\\&=-{\frac {(0.5)^{-1}z}{1-(0.5)^{-1}z}}\\&=-{\frac {1}{(.5)z^{-1}-1}}\\&={\frac {1}{1-(0.5)z^{-1}}}\\\end{aligned}}

and using the infinitegeometric series again, the equality only holds if $|(0.5)^{-1}z|<1$ which can be rewritten in terms of $z {\displaystyle z}$ as $|z|<(0.5).$ Thus, the ROC is $|z|<(0.5).$ 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 isonly the ROC. This is intentional to demonstrate that the transform result alone is insufficient.

Examples conclusion

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Examples 2 and 3 clearly show that the Z-transform $X(z)$ of $x[n]$ is unique when and only when specifying the ROC. Creating thepole–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 willnever contain poles.

In example 2, the causal system yields a ROC that includes $|z|=\infty$ while the anticausal system in example 3 yields an ROC that includes $|z|=0.$

ROC shown as a blue ring 0.5 < |z| < 0.75

In systems with multiple poles it is possible to have a ROC that includes neither $|z|=\infty$ nor $|z|=0.$ The ROC creates a circular band. For example,

x[n]=(0.5)^{n}\,u[n]-(0.75)^{n}\,u[-n-1]

has poles at 0.5 and 0.75. The ROC will be 0.5 < |z| < 0.75, which includes neither the origin nor infinity. Such a system is called a mixed-causality system as it contains a causal term $(0.5)^{n}\,u[n]$ and an anticausal term $-(0.75)^{n}\,u[-n-1].$

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

Let us assume we are provided a Z-transform of a system without a ROC (i.e., an ambiguous $x[n]$ ). We can determine a unique $x[n]$ provided we desire the following:

Stability
Causality

For stability the ROC must contain the unit circle. If we need a causal system then the ROC must contain infinity and the system function will be a right-sided sequence. If we need an anticausal system then the ROC must contain the origin and the system function will be a left-sided sequence. If we 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

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**Properties of the z-transform**
Property	Time domain	Z-domain	Proof	ROC
Definition of Z-transform	$x[n]$	$X(z)$	$X(z)={\mathcal {Z}}\{x[n]\}$ (definition of the z-transform) $x[n]={\mathcal {Z}}^{-1}\{X(z)\}$ (definition of the inverse z-transform)	$r_{2}<\|z\|<r_{1}$
Linearity	$a_{1}x_{1}[n]+a_{2}x_{2}[n]$	$a_{1}X_{1}(z)+a_{2}X_{2}(z)$	${\begin{aligned}X(z)&=\sum _{n=-\infty }^{\infty }(a_{1}x_{1}[n]+a_{2}x_{2}[n])z^{-n}\\&=a_{1}\sum _{n=-\infty }^{\infty }x_{1}[n]\,z^{-n}+a_{2}\sum _{n=-\infty }^{\infty }x_{2}[n]\,z^{-n}\\&=a_{1}X_{1}(z)+a_{2}X_{2}(z)\end{aligned}}$	Contains ROC₁ ∩ ROC₂
Time expansion	$x_{K}[n]={\begin{cases}x[r],&n=Kr\\0,&n\notin K\mathbb {Z} \end{cases}}$ with $K\mathbb {Z} :=\{Kr:r\in \mathbb {Z} \}$	$X(z^{K})$	${\begin{aligned}X_{K}(z)&=\sum _{n=-\infty }^{\infty }x_{K}[n]z^{-n}\\&=\sum _{r=-\infty }^{\infty }x[r]z^{-rK}\\&=\sum _{r=-\infty }^{\infty }x[r](z^{K})^{-r}\\&=X(z^{K})\end{aligned}}$	$R^{\frac {1}{K}}$
Decimation	$x[Kn]$	${\frac {1}{K}}\sum _{p=0}^{K-1}X\left(z^{\tfrac {1}{K}}\cdot e^{-i{\tfrac {2\pi }{K}}p}\right)$	ohio-state.edu oree.ic.ac.uk
Time delay	$x[n-k]$ with $k>0$ and $x:x[n]=0\ \forall \,n<0$	$z^{-k}X(z)$	${\begin{aligned}{\mathcal {Z}}\{x[n-k]\}&=\sum _{n=0}^{\infty }x[n-k]z^{-n}\\&=\sum _{m=-k}^{\infty }x[m]z^{-(m+k)}&&m=n-k\\&=\sum _{m=-k}^{\infty }x[m]z^{-m}z^{-k}\\&=z^{-k}\sum _{m=-k}^{\infty }x[m]z^{-m}\\&=z^{-k}\sum _{m=0}^{\infty }x[m]z^{-m}&&x[\beta ]=0,\forall \beta <0\\&=z^{-k}X(z)\end{aligned}}$	ROC, except $z{=}0$ if $k>0$ and $z{=}\infty$ if $k<0$
Time advance	$x[n+k]$ with $k>0$	Bilateral Z-transform: $z^{k}X(z)$ Unilateral Z-transform:^[14] $z^{k}\,X(z)-z^{k}\sum _{n=0}^{k-1}x[n]\,z^{-n}$
First difference backward	$x[n]-x[n-1]$ with $x[n]{=}0$ for $n<0$	$(1-z^{-1})\,X(z)$		Contains the intersection of ROC of $X_{1}(z)$ and $z\neq 0$
First difference forward	$x[n+1]-x[n]$	$(z-1)\,X(z)-z\,x[0]$
Time reversal	$x[-n]$	$X(z^{-1})$	${\begin{aligned}{\mathcal {Z}}\{x(-n)\}&=\sum _{n=-\infty }^{\infty }x[-n]z^{-n}\\&=\sum _{m=-\infty }^{\infty }x[m]z^{m}\\&=\sum _{m=-\infty }^{\infty }x[m]{(z^{-1})}^{-m}\\&=X(z^{-1})\\\end{aligned}}$	${\tfrac {1}{r_{1}}}<\|z\|<{\tfrac {1}{r_{2}}}$
Scaling in the z-domain	$a^{n}x[n]$	$X(a^{-1}z)$	${\begin{aligned}{\mathcal {Z}}\left\{a^{n}x[n]\right\}&=\sum _{n=-\infty }^{\infty }a^{n}x[n]z^{-n}\\&=\sum _{n=-\infty }^{\infty }x[n](a^{-1}z)^{-n}\\&=X(a^{-1}z)\end{aligned}}$	$\|a\|r_{2}<\|z\|<\|a\|r_{1}$
Complex conjugation	$x^{*}[n]$	$X^{}(z^{})$	${\begin{aligned}{\mathcal {Z}}\{x^{}(n)\}&=\sum _{n=-\infty }^{\infty }x^{}[n]z^{-n}\\&=\sum _{n=-\infty }^{\infty }\left[x[n](z^{})^{-n}\right]^{}\\&=\left[\sum _{n=-\infty }^{\infty }x[n](z^{})^{-n}\right]^{}\\&=X^{}(z^{})\end{aligned}}$
Real part	$\operatorname {Re} \{x[n]\}$	${\tfrac {1}{2}}\left[X(z)+X^{}(z^{})\right]$
Imaginary part	$\operatorname {Im} \{x[n]\}$	${\tfrac {1}{2i}}\left[X(z)-X^{}(z^{})\right]$
Differentiation in the z-domain	$n\,x[n]$	$-z{\frac {dX(z)}{dz}}$	${\begin{aligned}{\mathcal {Z}}\{n\,x(n)\}&=\sum _{n=-\infty }^{\infty }n\,x[n]z^{-n}\\&=z\sum _{n=-\infty }^{\infty }n\,x[n]z^{-n-1}\\&=-z\sum _{n=-\infty }^{\infty }x[n](-n\,z^{-n-1})\\&=-z\sum _{n=-\infty }^{\infty }x[n]{\frac {d}{dz}}(z^{-n})\\&=-z{\frac {dX(z)}{dz}}\end{aligned}}$	ROC, if $X(z)$ is rational; ROC possibly excluding the boundary, if $X(z)$ is irrational^[15]
Convolution	$x_{1}[n]*x_{2}[n]$	$X_{1}(z)\,X_{2}(z)$	${\begin{aligned}{\mathcal {Z}}\{x_{1}(n)*x_{2}(n)\}&={\mathcal {Z}}\left\{\sum _{l=-\infty }^{\infty }x_{1}[l]x_{2}[n-l]\right\}\\&=\sum _{n=-\infty }^{\infty }\left[\sum _{l=-\infty }^{\infty }x_{1}[l]x_{2}[n-l]\right]z^{-n}\\&=\sum _{l=-\infty }^{\infty }x_{1}[l]\left[\sum _{n=-\infty }^{\infty }x_{2}[n-l]z^{-n}\right]\\&=\left[\sum _{l=-\infty }^{\infty }x_{1}(l)z^{-l}\right]\!\!\left[\sum _{n=-\infty }^{\infty }x_{2}[n]z^{-n}\right]\\&=X_{1}(z)X_{2}(z)\end{aligned}}$	Contains ROC₁ ∩ ROC₂
Cross-correlation	$r_{x_{1},x_{2}}=x_{1}^{}[-n]x_{2}[n]$	$R_{x_{1},x_{2}}(z)=X_{1}^{}({\tfrac {1}{z^{}}})X_{2}(z)$		Contains the intersection of ROC of $X_{1}({\tfrac {1}{z^{*}}})$ and $X_{2}(z)$
Accumulation	$\sum _{k=-\infty }^{n}x[k]$	${\frac {1}{1-z^{-1}}}X(z)$	${\begin{aligned}\sum _{n=-\infty }^{\infty }\sum _{k=-\infty }^{n}x[k]z^{-n}&=\sum _{n=-\infty }^{\infty }(x[n]+\cdots )z^{-n}\\&=X(z)\left(1+z^{-1}+z^{-2}+\cdots \right)\\&=X(z)\sum _{j=0}^{\infty }z^{-j}\\&=X(z){\frac {1}{1-z^{-1}}}\end{aligned}}$
Multiplication	$x_{1}[n]\,x_{2}[n]$	${\frac {1}{2\pi i}}\oint _{C}X_{1}(v)X_{2}({\tfrac {z}{v}})v^{-1}\mathrm {d} v$		-

Parseval's theorem

\sum _{n=-\infty }^{\infty }x_{1}[n]x_{2}^{*}[n]\quad =\quad {\frac {1}{2\pi i}}\oint _{C}X_{1}(v)X_{2}^{*}({\tfrac {1}{v^{*}}})v^{-1}\mathrm {d} v

Initial value theorem: If $x[n]$ is causal, then

x[0]=\lim _{z\to \infty }X(z).

Final value theorem: If the poles of $(z-1)X(z)$ are inside the unit circle, then

x[\infty ]=\lim _{z\to 1}(z-1)X(z).

Table of common Z-transform pairs

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Here:

u:n\mapsto u[n]={\begin{cases}1,&n\geq 0\\0,&n<0\end{cases}}

is theunit (or Heaviside) step function and

\delta :n\mapsto \delta [n]={\begin{cases}1,&n=0\\0,&n\neq 0\end{cases}}

is thediscrete-time unit impulse function (cf.Dirac delta function, which is a continuous-time version). The two functions are chosen together so that the unit step function is the accumulation (running total) of the unit impulse function.

	Signal, $x[n]$	Z-transform, $X(z)$	ROC
1	$\delta [n]$	1	allz
2	$\delta [n-n_{0}]$	$z^{-n_{0}}$	$z\neq 0$
3	$u[n]\,$	${\frac {1}{1-z^{-1}}}$	$\|z\|>1$
4	$-u[-n-1]$	${\frac {1}{1-z^{-1}}}$	$\|z\|<1$
5	$nu[n]$	${\frac {z^{-1}}{(1-z^{-1})^{2}}}$	$\|z\|>1$
6	$-nu[-n-1]\,$	${\frac {z^{-1}}{(1-z^{-1})^{2}}}$	$\|z\|<1$
7	$n^{2}u[n]$	${\frac {z^{-1}(1+z^{-1})}{(1-z^{-1})^{3}}}$	$\|z\|>1\,$
8	$-n^{2}u[-n-1]\,$	${\frac {z^{-1}(1+z^{-1})}{(1-z^{-1})^{3}}}$	$\|z\|<1\,$
9	$n^{3}u[n]$	${\frac {z^{-1}(1+4z^{-1}+z^{-2})}{(1-z^{-1})^{4}}}$	$\|z\|>1\,$
10	$-n^{3}u[-n-1]$	${\frac {z^{-1}(1+4z^{-1}+z^{-2})}{(1-z^{-1})^{4}}}$	$\|z\|<1\,$
11	$a^{n}u[n]$	${\frac {1}{1-az^{-1}}}$	$\|z\|>\|a\|$
12	$-a^{n}u[-n-1]$	${\frac {1}{1-az^{-1}}}$	$\|z\|<\|a\|$
13	$na^{n}u[n]$	${\frac {az^{-1}}{(1-az^{-1})^{2}}}$	$\|z\|>\|a\|$
14	$-na^{n}u[-n-1]$	${\frac {az^{-1}}{(1-az^{-1})^{2}}}$	$\|z\|<\|a\|$
15	$n^{2}a^{n}u[n]$	${\frac {az^{-1}(1+az^{-1})}{(1-az^{-1})^{3}}}$	$\|z\|>\|a\|$
16	$-n^{2}a^{n}u[-n-1]$	${\frac {az^{-1}(1+az^{-1})}{(1-az^{-1})^{3}}}$	$\|z\|<\|a\|$
17	$\left({\begin{array}{c}n+m-1\\m-1\end{array}}\right)a^{n}u[n]$ ^[14]	${\frac {1}{(1-az^{-1})^{m}}}$ , for positive integer $m {\displaystyle m}$ ^[15]	$\|z\|>\|a\|$
18	$(-1)^{m}\left({\begin{array}{c}-n-1\\m-1\end{array}}\right)a^{n}u[-n-m]$	${\frac {1}{(1-az^{-1})^{m}}}$ , for positive integer $m {\displaystyle m}$ ^[15]	$\|z\|<\|a\|$
19	$\cos(\omega _{0}n)u[n]$	${\frac {1-z^{-1}\cos(\omega _{0})}{1-2z^{-1}\cos(\omega _{0})+z^{-2}}}$	$\|z\|>1$
20	$\sin(\omega _{0}n)u[n]$	${\frac {z^{-1}\sin(\omega _{0})}{1-2z^{-1}\cos(\omega _{0})+z^{-2}}}$	$\|z\|>1$
21	$a^{n}\cos(\omega _{0}n)u[n]$	${\frac {1-az^{-1}\cos(\omega _{0})}{1-2az^{-1}\cos(\omega _{0})+a^{2}z^{-2}}}$	$\|z\|>\|a\|$
22	$a^{n}\sin(\omega _{0}n)u[n]$	${\frac {az^{-1}\sin(\omega _{0})}{1-2az^{-1}\cos(\omega _{0})+a^{2}z^{-2}}}$	$\|z\|>\|a\|$

Relationship to Fourier series and Fourier transform

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Further information:Discrete-time Fourier transform § Relationship to the Z-transform

For values of $z {\displaystyle z}$ in the region $|z|{=}1$ , known as theunit circle, we can express the transform as a function of a single real variable $\omega$ by defining $z{=}e^{i\omega }.$ And the bi-lateral transform reduces to aFourier series:

$\sum _{n=-\infty }^{\infty }x[n]\ z^{-n}=\sum _{n=-\infty }^{\infty }x[n]\ e^{-i\omega n},$

Eq.1

which is also known as thediscrete-time Fourier transform (DTFT) of the $x[n]$ sequence. This $2\pi$ -periodic function is theperiodic summation of aFourier transform, which makes it a widely used analysis tool. To understand this, let $X(f)$ be the Fourier transform of any function, $x(t)$ , whose samples at some interval $T {\displaystyle T}$ equal the $x[n]$ sequence. Then the DTFT of the $x[n]$ sequence can be written as follows.

$\underbrace {\sum _{n=-\infty }^{\infty }\overbrace {x(nT)} ^{x[n]}\ e^{-2i\pi fnT}} _{\text{DTFT}}={\frac {1}{T}}\sum _{k=-\infty }^{\infty }X(f-k/T),$

Eq.2

where $T {\displaystyle T}$ has units of seconds, $f {\displaystyle f}$ has units ofhertz. Comparison of the two series reveals that $\omega {=}2\pi fT$ is anormalized frequency with unit ofradian per sample. The value $\omega {=}2\pi$ corresponds to ${\textstyle f{=}{\frac {1}{T}}}$ . And now, with the substitution ${\textstyle f{=}{\frac {\omega }{2\pi T}},}$ Eq.1 can be expressed in terms of $X({\tfrac {\omega -2\pi k}{2\pi T}})$ (a Fourier transform):

$\sum _{n=-\infty }^{\infty }x[n]\ e^{-i\omega n}={\frac {1}{T}}\sum _{k=-\infty }^{\infty }\underbrace {X\left({\tfrac {\omega }{2\pi T}}-{\tfrac {k}{T}}\right)} _{X\left({\frac {\omega -2\pi k}{2\pi T}}\right)}.$

Eq.3

As parameterT changes, the individual terms ofEq.2 move farther apart or closer together along thef-axis. InEq.3 however, the centers remain 2π apart, while their widths expand or contract. When sequence $x(nT)$ represents theimpulse response of anLTI system, these functions are also known as itsfrequency response. When the $x(nT)$ sequence is periodic, its DTFT is divergent at one or more harmonic frequencies, and zero at all other frequencies. This is often represented by the use of amplitude-variantDirac delta functions at the harmonic frequencies. Due to periodicity, there are only a finite number of unique amplitudes, which are readily computed by the much simplerdiscrete Fourier transform (DFT). (SeeDiscrete-time Fourier transform § Periodic data.)

Relationship to Laplace transform

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Further information:Laplace transform § Z-transform

Bilinear transform

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Main article:Bilinear transform

Thebilinear transform can be used to convert continuous-time filters (represented in the Laplace domain) into discrete-time filters (represented in the Z-domain), and vice versa. The following substitution is used:

s={\frac {2}{T}}{\frac {(z-1)}{(z+1)}}

to convert some function $H(s)$ in the Laplace domain to a function $H(z)$ in the Z-domain (Tustin transformation), or

z=e^{sT}\approx {\frac {1+sT/2}{1-sT/2}}

from the Z-domain to the Laplace domain. Through the bilinear transformation, the complexs-plane (of the Laplace transform) is mapped to the complex z-plane (of the z-transform). While this mapping is (necessarily) nonlinear, it is useful in that it maps the entire $i\omega$ axis of thes-plane onto theunit circle in the z-plane. As such, the Fourier transform (which is the Laplace transform evaluated on the $i\omega$ axis) becomes the discrete-time Fourier transform. This assumes that the Fourier transform exists; i.e., that the $i\omega$ axis is in the region of convergence of the Laplace transform.

Starred transform

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Main article:Starred transform

Given a one-sided Z-transform $X(z)$ of a time-sampled function, the correspondingstarred transform produces a Laplace transform and restores the dependence on $T {\displaystyle T}$ (the sampling parameter):

{\bigg .}X^{*}(s)=X(z){\bigg |}_{\displaystyle z=e^{sT}}

The inverse Laplace transform is a mathematical abstraction known as animpulse-sampled function.

Linear constant-coefficient difference equation

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The linear constant-coefficient difference (LCCD) equation is a representation for a linear system based on theautoregressive moving-average equation:

\sum _{p=0}^{N}y[n-p]\alpha _{p}=\sum _{q=0}^{M}x[n-q]\beta _{q}.

Both sides of the above equation can be divided by $\alpha _{0}$ if it is not zero. By normalizing with $\alpha _{0}{=}1,$ the LCCD equation can be written

y[n]=\sum _{q=0}^{M}x[n-q]\beta _{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

[edit]

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}\beta _{q}

where $X(z)$ and $Y(z)$ are the z-transform of $x[n]$ and $y[n],$ respectively. (Notation conventions typically use capitalized letters to refer to the z-transform of a signal denoted by a corresponding lower case letter, similar to the convention used for notating Laplace transforms.)

Rearranging results in the system'stransfer function:

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}+\cdots +z^{-M}\beta _{M}}{\alpha _{0}+z^{-1}\alpha _{1}+z^{-2}\alpha _{2}+\cdots +z^{-N}\alpha _{N}}}.

Zeros and poles

[edit]

From thefundamental theorem of algebra thenumerator has $M {\displaystyle M}$ roots (corresponding to zeros of $H {\displaystyle H}$ ) and thedenominator has $N {\displaystyle N}$ roots (corresponding to poles). Rewriting thetransfer function in terms ofzeros and poles

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 $k {\displaystyle k}$ th zero and $p_{k}$ is the $k {\displaystyle k}$ th pole. The zeros and poles are commonly complex and when plotted on the complex plane (z-plane) it is called thepole–zero plot.

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

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If such a system $H(z)$ is driven by a signal $X(z)$ then the output is $Y(z)=H(z)X(z).$ By performingpartial 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 $\textstyle {\frac {Y(z)}{z}}$ before multiplying that quantity by $z {\displaystyle z}$ to generate a form of $Y(z)$ which has terms with easily computable inverse Z-transforms.

References

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^Mandal, Jyotsna Kumar (2020). "Z-Transform-Based Reversible Encoding".Reversible Steganography and Authentication via Transform Encoding. Studies in Computational Intelligence. Vol. 901. Singapore: Springer Singapore. pp. 157–195.doi:10.1007/978-981-15-4397-5_7.ISBN 978-981-15-4396-8.ISSN 1860-949X.S2CID 226413693.Z is a complex variable. Z-transform converts the discrete spatial domain signal into complex frequency domain representation. Z-transform is derived from the Laplace transform.
^Lynn, Paul A. (1986). "The Laplace Transform and thez-transform".Electronic Signals and Systems. London: Macmillan Education UK. pp. 225–272.doi:10.1007/978-1-349-18461-3_6.ISBN 978-0-333-39164-8.Laplace Transform and thez-transform are closely related to the Fourier Transform.z-transform is especially suitable for dealing with discrete signals and systems. It offers a more compact and convenient notation than the discrete-time Fourier Transform.
^Jury, Eliahu Ibrahim (1964).Theory and application of the z-transform method. New York: John Wiley & Sons. pp. XIII, 330 s.
^Palani, S. (2021-08-26). "Thez-Transform Analysis of Discrete Time Signals and Systems".Signals and Systems. Cham: Springer International Publishing. pp. 921–1055.doi:10.1007/978-3-030-75742-7_9.ISBN 978-3-030-75741-0.S2CID 238692483.z-transform is the discrete counterpart of Laplace transform.z-transform converts difference equations of discrete time systems to algebraic equations which simplifies the discrete time system analysis. Laplace transform andz-transform are common except that Laplace transform deals with continuous time signals and systems.
^E. R. Kanasewich (1981).Time Sequence Analysis in Geophysics. University of Alberta. pp. 186, 249.ISBN 978-0-88864-074-1.
^E. R. Kanasewich (1981).Time sequence analysis in geophysics (3rd ed.). University of Alberta. pp. 185–186.ISBN 978-0-88864-074-1.
^Ragazzini, J. R.; Zadeh, L. A. (1952). "The analysis of sampled-data systems".Transactions of the American Institute of Electrical Engineers, Part II: Applications and Industry.71 (5):225–234.doi:10.1109/TAI.1952.6371274.S2CID 51674188.
^Cornelius T. Leondes (1996).Digital control systems implementation and computational techniques. Academic Press. p. 123.ISBN 978-0-12-012779-5.
^Eliahu Ibrahim Jury (1958).Sampled-Data Control Systems. John Wiley & Sons.
^Eliahu Ibrahim Jury (1973).Theory and Application of the Z-Transform Method. Krieger Pub Co.ISBN 0-88275-122-0.
^Eliahu Ibrahim Jury (1964).Theory and Application of the Z-Transform Method. John Wiley & Sons. p. 1.
^Jackson, Leland B. (1996). "The z Transform".Digital Filters and Signal Processing. Boston, MA: Springer US. pp. 29–54.doi:10.1007/978-1-4757-2458-5_3.ISBN 978-1-4419-5153-3.z transform is to discrete-time systems what the Laplace transform is to continuous-time systems.z is a complex variable. This is sometimes referred to as the two-sidedz transform, with the one-sided z transform being the same except for a summation fromn = 0 to infinity. The primary use of the one sided transform ... is for causal sequences, in which case the two transforms are the same anyway. We will not, therefore, make this distinction and will refer to ... as simply the z transform ofx(n).
^Proakis, John; Manolakis, Dimitris.Digital Signal Processing Principles, Algorithms and Applications (3rd ed.). PRENTICE-HALL INTERNATIONAL, INC.
^Bolzern, Paolo; Scattolini, Riccardo; Schiavoni, Nicola (2015).Fondamenti di Controlli Automatici (in Italian). MC Graw Hill Education.ISBN 978-88-386-6882-1.
^^a ^b ^cA. R. Forouzan (2016). "Region of convergence of derivative of Z transform".Electronics Letters.52 (8):617–619.Bibcode:2016ElL....52..617F.doi:10.1049/el.2016.0189.S2CID 124802942.

External links

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"Z-transform".Encyclopedia of Mathematics.EMS Press. 2001 [1994].
Merrikh-Bayat, Farshad (2014). "Two Methods for Numerical Inversion of the Z-Transform".arXiv:1409.1727 [math.NA].
Z-Transform table of some common Laplace transforms
Mathworld's entry on the Z-transform Archived 2013-01-30 at theWayback Machine
Z-Transform threads in Comp.DSP Archived 2012-06-15 at theWayback Machine
A graphic of the relationship between Laplace transform s-plane to Z-plane of the Z transform
A video-based explanation of the Z-Transform for engineers
What is the z-Transform?

v t e Digital signal processing
Theory	Detection theory Discrete signal Estimation theory Nyquist–Shannon sampling theorem
Sub-fields	Audio signal processing Digital image processing Speech processing Statistical signal processing
Techniques	Z-transform Advanced z-transform Matched Z-transform method Bilinear transform Constant-Q transform Discrete cosine transform (DCT) Discrete Fourier transform (DFT) Discrete-time Fourier transform (DTFT) Impulse invariance Integral transform Laplace transform Post's inversion formula Starred transform Zak transform
Sampling	Aliasing Anti-aliasing filter Downsampling Nyquist rate /frequency Oversampling Quantization Sampling rate Undersampling Upsampling

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[8]ページ先頭

Movatterモバイル変換

Z-transform

History

Definition

Bilateral Z-transform

Unilateral Z-transform

Inverse Z-transform

Direct evaluation by contour integration

Expansion into a series of terms in the variablesz andz⁻¹

Partial-fraction expansion and table lookup

Example

Region of convergence

Example 1 (no ROC)

Example 2 (causal ROC)

Example 3 (anticausal ROC)

Examples conclusion

Properties

Table of common Z-transform pairs

Relationship to Fourier series and Fourier transform

Relationship to Laplace transform

Bilinear transform

Starred transform

Linear constant-coefficient difference equation

Transfer function

Zeros and poles

Output response

See also

References

Further reading

External links

Movatterモバイル変換

History

Definition

Bilateral Z-transform

Unilateral Z-transform

Inverse Z-transform

Direct evaluation by contour integration

Expansion into a series of terms in the variablesz andz−1

Partial-fraction expansion and table lookup

Example

Region of convergence

Example 1 (no ROC)

Example 2 (causal ROC)

Example 3 (anticausal ROC)

Examples conclusion

Properties

Table of common Z-transform pairs

Relationship to Fourier series and Fourier transform

Relationship to Laplace transform

Bilinear transform

Starred transform

Linear constant-coefficient difference equation

Transfer function

Zeros and poles

Output response

See also

References

Further reading

External links

Expansion into a series of terms in the variablesz andz⁻¹