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

It can be considered as a discrete-time equivalent of the Laplace transform. This similarity is explored in the theory of time scale calculus.

History

The basic idea now known as the Z-transform was known to Laplace, and re-introduced in 1947 by W. Hurewicz as a tractable way to solve linear, constant-coefficient difference equations.[1] It was later dubbed "the z-transform" by Ragazzini and Zadeh in the sampled-data control group at Columbia University in 1952.[2][3]

The modified or advanced Z-transform was later developed and popularized by E. I. Jury.[4][5]

The idea contained within the Z-transform is also known in mathematical literature as the method of generating functions which can be traced back as early as 1730 when it was introduced by de Moivre in conjunction with probability theory.[6] From a mathematical view the Z-transform can also be viewed as a Laurent series where one views the sequence of numbers under consideration as the (Laurent) expansion of an analytic function.
Definition

The Z-transform, like many 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 formal power series 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:

$$z = A e^{j\phi} = A(\cos{\phi}+j\sin{\phi})\,$$

where A is the magnitude of z, j is the imaginary unit, and \phi is the complex argument (also referred to as angle or phase) in radians.
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 can be used to evaluate the Z-transform of the unit impulse response of a discrete-time causal system.

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 power series in z as opposed to $$z^{-1}$$. This convention is used by Robinson and Treitel and by Kanasewich.[citation needed] 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. 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 which is always guaranteed when X(z)\ is stable, i.e. all the poles are within 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 the discrete 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.

$$ROC = \left\{ z : \left|\sum_{n=-\infty}^{\infty}x[n]z^{-n}\right| < \infty \right\}$$

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

$$\sum_{n=-\infty}^{\infty}x[n]z^{-n} \rightarrow \infty\ .$$

Therefore, there are no values of z\ that satisfy this condition.

Example 2 (causal ROC)
ROC shown in blue, the unit circle as a dotted grey circle (appears reddish to the eye) and the circle $$\left|z\right| = 0.5$$ is shown as a dashed black circle

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}\right)^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}\right| < 1\$$ which can be rewritten in terms of z\ as \left|z\right| > 0.5\ . Thus, the ROC is \left|z\right| > 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)
ROC shown in blue, the unit circle as a dotted grey circle and the circle $$\left|z\right| = 0.5$$ is shown as a dashed black circle

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, 0, ...\}.\$$

Looking at the sum

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

Using the infinite geometric series, again, the equality only holds if $$\left|0.5^{-1}z\right| < 1\$$ which can be rewritten in terms of $$z\$$ as $$\left|z\right| < 0.5\$$ . Thus, the ROC is $$\left|z\right| < 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 $$\left| z \right| = \infty\$$ while the anticausal system in example 3 yields an ROC that includes \left| z \right| = 0\ .
ROC shown as a blue ring $$0.5 < \left| z \right| < 0.75\$$

In systems with multiple poles it is possible to have an ROC that includes neither $$\left| z \right| = \infty\ nor \left| z \right| = 0\$$. The ROC creates a circular band. For example, $$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 \right| < 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., $$\left| z \right| = 1\$$ ) then the system is stable. In the above systems the causal system (Example 2) is stable because $$\left| z \right| > 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
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 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

Time domain Z-domain Proof ROC
Notation $$x[n]={\mathcal {Z}}^{-1}\{X(z)\}$$ $$X(z)={\mathcal {Z}}\{x[n]\}$$
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 $${\displaystyle x_{K}[n]={\begin{cases}x[r],&n=Kr\\0,&n\notin K\mathbb {Z} \end{cases}}}$$

with $${\displaystyle K\mathbb {Z} :=\{Kr:r\in \mathbb {Z} \}}$$

$$X(z^{K})$$ {\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}}
Decimation $${\displaystyle x[Kn]}$$ $${\frac {1}{K}}\sum _{p=0}^{K-1}X\left(z^{\tfrac {1}{K}}\cdot e^{-i{\tfrac {2\pi }{K}}p}\right)$$ ohio-state.edu  or  ee.ic.ac.uk
Time shifting $${\displaystyle x[n-k]}$$ $$z^{-k}X(z)$$ {\begin{aligned}Z\{x[n-k]\}&=\sum _{n=0}^{\infty }x[n-k]z^{-n}\\&=\sum _{j=-k}^{\infty }x[j]z^{-(j+k)}&&j=n-k\\&=\sum _{j=-k}^{\infty }x[j]z^{-j}z^{-k}\\&=z^{-k}\sum _{j=-k}^{\infty }x[j]z^{-j}\\&=z^{-k}\sum _{j=0}^{\infty }x[j]z^{-j}&&x[\beta ]=0,\beta <0\\&=z^{-k}X(z)\end{aligned}} ROC, except z = 0 if k > 0 and z = ∞ if k < 0
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}}
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}}
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 $$\operatorname {Im} \{x[n]\}$$ $${\tfrac {1}{2j}}\left[X(z)-X^{*}(z^{*})\right]$$
Differentiation $$nx[n]$$ $$-z{\frac {dX(z)}{dz}}$$ {\begin{aligned}{\mathcal {Z}}\{nx(n)\}&=\sum _{n=-\infty }^{\infty }nx(n)z^{-n}\\&=z\sum _{n=-\infty }^{\infty }nx(n)z^{-n-1}\\&=-z\sum _{n=-\infty }^{\infty }x(n)(-nz^{-n-1})\\&=-z\sum _{n=-\infty }^{\infty }x(n){\frac {d}{dz}}(z^{-n})\\&=-z{\frac {dX(z)}{dz}}\end{aligned}}

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 $$r_{x_{1},x_{2}}=x_{1}^{*}[-n]*x_{2}[n]$$ $$R_{x_{1},x_{2}}(z)=X_{1}^{*}({\tfrac {1}{z^{*}}})X_{2}(z)$$
First difference $$x[n]-x[n-1]$$ $$(1-z^{-1})X(z)$$ Contains the intersection of ROC of X1(z) and z ≠ 0
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 +x[-\infty ])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 $$x_{1}[n]x_{2}[n]$$ $${\frac {1}{j2\pi }}\oint _{C}X_{1}(v)X_{2}({\tfrac {z}{v}})v^{-1}\mathrm {d} v$$ -

Parseval's relation $$\sum_{n=-\infty}^{\infty} x_1[n]x^*_2[n]\ \frac{1}{j2\pi}\oint_C X_1(v)X^*_2(\frac{1}{v^*})v^{-1}\mathrm{d}v \$$

Initial value theorem

$$x[0]=\lim_{z\rightarrow \infty}X(z) \ , If x[n]\,$$ causal

Final value theorem

$$x[\infty]=\lim_{z\rightarrow 1}(z-1)X(z) \$$ , Only if poles of (z-1)X(z) \ are inside the unit circle

Table of common Z-transform pairs

Here:

u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \end{cases}
\delta[n] = \begin{cases} 1, & n = 0 \\ 0, & n \ne 0 \end{cases}

Signal, x[n] Z-transform, X(z) ROC
1 $$\delta[n] \, 1\, \mbox{all }z\,$$
2 $$\delta[n-n_0] \, z^{-n_0} \, z \neq 0\,$$
3 $$u[n] \, \frac{1}{1-z^{-1} } |z| > 1\,$$
4 $$\, e^{-\alpha n} u[n] 1 \over 1-e^{-\alpha }z^{-1} |z| > |e^{-\alpha}| \,$$
5 $$- u[-n-1] \, \frac{1}{1 - z^{-1}} |z| < 1\,$$
6 $$n u[n] \, \frac{z^{-1}}{( 1-z^{-1} )^2} |z| > 1\,$$
7 $$- n u[-n-1] \, \frac{z^{-1} }{ (1 - z^{-1})^2 } |z| < 1 \,$$
8 $$n^2 u[n] \, \frac{ z^{-1} (1 + z^{-1} )}{(1 - z^{-1})^3} |z| > 1\,$$
9 $$- n^2 u[-n - 1] \, \frac{ z^{-1} (1 + z^{-1} )}{(1 - z^{-1})^3} |z| < 1\,$$
10 $$n^3 u[n] \, \frac{z^{-1} (1 + 4 z^{-1} + z^{-2} )}{(1-z^{-1})^4} |z| > 1\,$$
11 $$- n^3 u[-n -1] \, \frac{z^{-1} (1 + 4 z^{-1} + z^{-2} )}{(1-z^{-1})^4} |z| < 1\,$$
12 $$a^n u[n] \, \frac{1}{1-a z^{-1}} |z| > |a|\,$$
13 $$-a^n u[-n-1] \, \frac{1}{1-a z^{-1}} |z| < |a|\,$$
14 $$n a^n u[n] \, \frac{az^{-1} }{ (1-a z^{-1})^2 } |z| > |a|\,$$
15 $$-n a^n u[-n-1] \, \frac{az^{-1} }{ (1-a z^{-1})^2 } |z| < |a|\,$$
16 $$n^2 a^n u[n] \, \frac{a z^{-1} (1 + a z^{-1}) }{(1-a z^{-1})^3} |z| > |a|\,$$
17 $$-( n^2 a^n u[-n -1] \, \frac{a z^{-1} (1 + a z^{-1}) }{(1-a z^{-1})^3} |z| < |a|\,$$
18 $$\cos(\omega_0 n) u[n] \, \frac{ 1-z^{-1} \cos(\omega_0) }{ 1-2z^{-1}\cos(\omega_0)+ z^{-2} } |z| >1\,$$
19 $$\sin(\omega_0 n) u[n] \, \frac{ z^{-1} \sin(\omega_0) }{ 1-2z^{-1}\cos(\omega_0)+ z^{-2} } |z| >1\,$$
20 $$a^n \cos(\omega_0 n) u[n] \, \frac{ 1-a z^{-1} \cos( \omega_0) }{ 1-2az^{-1}\cos(\omega_0)+ a^2 z^{-2} } |z| > |a|\,$$
21 $$a^n \sin(\omega_0 n) u[n] \, \frac{ az^{-1} \sin(\omega_0) }{ 1-2az^{-1}\cos(\omega_0)+ a^2 z^{-2} } |z| > |a|\,$$
Relationship to Laplace transform

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):

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

from Laplace to z (Tustin transformation), or

$$\, z =\frac{2+sT}{2-sT} \quad$$

from z to Laplace. Through the bilinear transformation, the complex s-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 j \Omega axis of the s-plane onto the unit circle in the z-plane. As such, the Fourier transform (which is the Laplace transform evaluated on the j \Omega axis) becomes the discrete-time Fourier transform. This assumes that the Fourier transform exists; i.e., that the j \Omega axis is in the region of convergence of the Laplace transform.
Relationship to Fourier transform

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^{j\omega}\ (where \omega is the normalized frequency) 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, 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]\beta_{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]\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

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}\$$

and rearranging results in

$$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

From the fundamental theorem of algebra the numerator has M roots (corresponding to zeros of H) and the denominator has N roots (corresponding to poles). Rewriting the transfer 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 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 the impulse 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.

Star transform
Bilinear transform
Finite impulse response
Formal power series
Laplace transform
Laurent series
Probability-generating function
Zeta function regularization
Discrete-time Fourier transform
Difference equation (recurrence relation)
Discrete convolution

References

^ E. R. Kanasewich (1981). Time sequence analysis in geophysics (3rd ed.). University of Alberta. pp. 185–186. ISBN 978-0-88864-074-1.
^ J. R. Ragazzini and L. A. Zadeh (1952). "The analysis of sampled-data systems". Trans. Am. Inst. Elec. Eng. 71 (II): 225–234.
^ 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.