Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2014, Article ID 905791, 9 pages http://dx.doi.org/10.1155/2014/905791

Research Article Transformation Matrix for Time Discretization Based on Tustin’s Method

Yiming Jiang, Xiaodong Hu, and Sen Wu

College of Precision Instrument & Optoelectronics Engineering, Tianjin University, Tianjin 300072, China

Correspondence should be addressed to Xiaodong Hu; [email protected]

Received 16 April 2014; Revised 27 June 2014; Accepted 10 July 2014; Published 18 August 2014

Academic Editor: Ivanka Stamova

Copyright © 2014 Yiming Jiang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper studies rules in transformation of through time discretization. A method of using transformation matrix to realize bilinear transform (also known as Tustin’s method) is presented. This method can be described as the conversion between the coefficients of transfer functions, which are expressed as transform by certain matrix. For a polynomial of degree n,the corresponding transformation matrix of order n exists and is unique. Furthermore, the transformation matrix can be decomposed into an upper triangular matrix multiplied with another lower triangular matrix. And both have obvious regularity. The proposed method can achieve rapid bilinear transform used in automatic design of . The result of numerical simulation verifies the correctness of the theoretical results. Moreover, it also can be extended to other similar problems. Example in the last throws light on this point.

1. Introduction to enrich the method and to make it easier to understand. Of course they are all simplified. In this paper we proposed a new method of using trans- formation matrix to realize bilinear transform. We describe 1.1. About Tustin’s Method. Tustin’s method is also called the the principles and characteristics of the method in Section 2. bilinear transform. As a convenient tool for time discretiza- Also we have examinations and simulations on the method tion, it is widely used in designing digital filters to obtain in Section 3. expected discrete time system with desired response char- In Section 2, Section 2.1 starts with exploration of how to acteristics. The transform, as a method to convert 𝑆 domain express derivative of a function with discrete points under the function to 𝑍 domain function, is a first-order approximation method of bilinear transform. By studying the rules of that of the 𝑆-to-𝑍 mapping. Under the , it can conversion, Section 2.2 reveals regularity in the arithmetic be regarded as a discrete-time approximation [1]: of bilinear transform operated on transfer function, so that ∞ it can be replaced by systematic mathematical derivations. 𝑒𝑇𝑠/2 ∑ (1/𝑛!) ⋅ (𝑇𝑠/2)𝑛 1 + 𝑇𝑠/2 𝑧=𝑒𝑇𝑠 = = 𝑛=0 ≈ . Finally, the transformation matrix to realize Tustin transform −𝑇𝑠/2 ∞ 𝑛 (1) 𝑒 ∑𝑛=0 (1/𝑛!) ⋅ (−𝑇𝑠/2) 1 − 𝑇𝑠/2 is obtained by recursion. Also in the process, some inner rules of the Tustin transform can be revealed. Equivalently, In Section 3, Section 3.1 provides a simple application 2 1−𝑧−1 to verify the mathematical correctness of the method and 𝑠= ⋅ , (2) 𝑇 1+𝑧−1 uses the obtained transformation matrix as an example. Section 3.2 explains how it works when the parameters of where 𝑇 is the step size of the trape- functions are uncertain. Section 3.3 shows an example about zoidal rule used in the bilinear transform derivation. itsexpandedapplicationasitcanbeusedtoconstruct Compared with 𝑍 transform based on the pulse response transformation matrixes in other situations. The purpose is method, the Tustin transform eliminates the effect 2 Mathematical Problems in Engineering generated by overlapping of frequency spectrum, relying less than or equal to 𝑛). If the sampling step size is 𝑇,then on the linear transformation from 𝑆 domain to 𝑍 domain 𝑥𝑛−𝑖 = 𝑥(𝑛𝑇 − 𝑖𝑇), 𝑖=0,1,2,...,𝑚. as a one-to-one mapping. Consequently, it is suitable for If (𝑥𝑛 −𝑥𝑛−1)/𝑇 represents first-order derivative with the design of high-pass and band-stop filters. Furthermore, discrete values 𝑥𝑛, 𝑥𝑛−1 of two adjacent points, such recursion 𝑆 2 because of the simple algebraic relationship between and can be obtained as (𝑥𝑛 −2𝑥𝑛−1 +𝑥𝑛−2)/𝑇 represents the 𝑍 , the transfer function of digital filter can be obtained second-order derivative with three, (𝑥𝑛 −3𝑥𝑛−1 +3𝑥𝑛−2 − 𝑆 𝑍 3 conveniently by directing the -to- mapping into the cor- 𝑥𝑛−3)/𝑇 represents the third-order derivative with four, and responding transfer function of analog filter [1–3]. so on. Here we define 𝑖 𝑑 𝑥⃗ = ∑(−1)𝑗𝐶𝑗𝑥 , 1.2. Problems to Be Solved. Because the bilinear transform 𝑖 𝑛 𝑖 𝑛−𝑗 (3) 𝑗=0 is a kind of fractional linear transformation, the amount of calculation increases sharply with the increase of the where 𝑑𝑖 and 𝑥𝑛⃗ are (𝑚 + 1)-dimensional row and column order of system. The problem of complex calculation appears 𝑗 vectors and (𝑚 + 1) is equal the number of samples. Let 𝐶𝑖 = when operating bilinear transform for high-order system. 𝑖!/((𝑖 − 𝑗)!𝑗!) (𝑖, 𝑗 are nonnegative integers, and 𝑖≥𝑗): Particularly, in many situations the parameters of system are 0 1 𝑖 𝑖 changeable; thus then traditional method fails to work. For 𝑑𝑖 =[𝐶𝑖 −𝐶𝑖 ⋅⋅⋅ (−1) 𝐶𝑖 0 ⋅⋅⋅ 0], these reasons, programs to realize automatic operation are (4) 𝑇 required. An efficient way is to express the transform with 𝑥𝑛⃗ =[𝑥𝑛 𝑥𝑛−1 ⋅⋅⋅ 𝑥𝑛−𝑚] . matrix. 𝑖 Many scholars have proposed methods to obtain the Thus 𝑑𝑖𝑥𝑛⃗ /𝑇 is the expression of the 𝑖th order derivative 𝑖 𝑖 matrix for bilinear transform, of which the literatures [4– 𝑑 𝑥(𝑡)/𝑑𝑡 for continuous time signal 𝑥 with (𝑖 + 1) values. In 6] were to derive coefficient matrix of different order under addition, each sampling point can be expressed by the linear the discussion of relationships between the matrix elements, combination of 𝑑𝑖𝑥𝑛⃗ (𝑖=0,1,2,...)as below: while [7, 8] used method of induction to derive coefficient 𝑥 =𝑥, matrix of order n from coefficient matrix of order zero. Other 𝑛 𝑛 relative reference as [9–14] also studied quick algorithms for 𝑥 =𝑥 −(𝑥 −𝑥 ), different forms of linear 𝑆-to-𝑍 transform. 𝑛−1 𝑛 𝑛 𝑛−1 The purpose of this paper is to introduce an easier way to 𝑥𝑛−2 =𝑥𝑛 −2(𝑥𝑛 −𝑥𝑛−1)+(𝑥𝑛 −2𝑥𝑛−1 +𝑥𝑛−2), obtain the transformation matrix. The method applies when (5) the order of transfer function is large, using computer as a 𝑥𝑛−3 =𝑥𝑛 −3(𝑥𝑛 −𝑥𝑛−1)+3(𝑥𝑛 −2𝑥𝑛−1 +𝑥𝑛−2) direct operation, or when the system function is uncertain, using the method to get the dynamic differential equation. −(𝑥𝑛 −3𝑥𝑛−1 +3𝑥𝑛−2 −𝑥𝑛−3) ⋅⋅⋅ .

2. Derivations That is, Bilinear transform is a bridge between 𝑆 domain and 𝑍 𝑚 𝑖 𝑖 𝑖 𝑗 𝑗 domain. Laplace transform can change a time domain func- 𝑥𝑛−𝑚 = ∑(−1) 𝐶𝑚 (∑(−1) 𝐶𝑖 𝑥𝑛−𝑗) tion with t into a frequency domain function with 𝑠,andithas 𝑖=0 𝑗=0 (6) a special quality that it can deform differential equations into 𝑚 algebraic equations to simplify mathematical operations. 𝑍 𝑖 𝑖 = ∑(−1) 𝐶𝑚 (𝑑𝑖𝑥𝑛⃗ ). transform is similar to Laplace transform, but it is used to deal 𝑖=0 with discrete series of 𝑡. It can change a time domain function with 𝑡 into 𝑍 domain. The result of every mapping from 𝑆 To facilitate the expression, let domain to 𝑍 domain is replacement of continuous function 𝐶0 𝐶0 𝐶0 ⋅⋅⋅ 𝐶0 by discrete points. Accordingly, when processing a continu- [ 0 1 2 𝑚 ] [ 1 1 1 ] ousfunction,theresultofmappingisexpressionofrelations [ −𝐶1 −𝐶2 ⋅⋅⋅ −𝐶𝑚 ] [ ] between numerical values, integration, and differentiation by [ ] [ 2 2 ] =𝐷 . discrete points. Because transfer functions can be converted [ 𝐶2 ⋅⋅⋅ 𝐶𝑚 ] 𝑚 (7) [ ] to differential equations, expressions of differential forms by [ ] [ . ] discrete points should be studied. d . 𝑚 𝑚 [ (−1) 𝐶𝑚]

2.1. Definition for Derivatives with Discrete Points. In order 𝐷𝑚 is an upper triangular matrix of (𝑚 + 1) × (𝑚 +1) and to discuss distinctions between deformations to differential satisfies the following properties: forms before and after 𝑆-to-𝑍 mapping by bilinear transform, −1 𝐷 =(𝐷 ) =[𝑑𝑇 𝑑𝑇 ⋅⋅⋅ 𝑑𝑇 ], such relations below will be necessary to deduct. 𝑚 𝑚 0 1 𝑚 (8) Suppose {𝑥𝑛,𝑥𝑛−1,𝑥𝑛−2,...,𝑥𝑛−𝑚} are (𝑚 + 1) values at 𝑥⃗𝑇 =[𝑑 𝑥⃗ 𝑑 𝑥⃗ ⋅⋅⋅ 𝑑 𝑥⃗ ]𝐷 . sampling points of discrete digital signal 𝑥(𝑡) (𝑚 is an integer 𝑛 0 𝑛 1 𝑛 𝑚 𝑛 𝑚 (9) Mathematical Problems in Engineering 3

Extensionally, if 𝑥𝑛−𝑖⃗ is defined as an (𝑚 + 1)-dimensional where, 𝑋(𝑠) and 𝑌(𝑠) are polynomials of 𝑠,and𝑏𝑝 =0̸ .The 𝑇 column vector [𝑥𝑛−𝑖 𝑥𝑛−𝑖−1 ⋅⋅⋅ 𝑥𝑛−𝑚+1 𝑥𝑛−𝑚 0⋅⋅⋅0] by general method of using bilinear transformation is substitut- 𝑆 𝑍 𝑍 moving forward the elements of 𝑥𝑛⃗ for 𝑖 steps, we get another ing the -to- mapping (2) into the transfer function. The 𝐻(𝑧) property (10)as domain transfer function canbeshownasfollows: −1 −(𝑝−1) −𝑝 𝛼0 +𝛼1𝑧 +⋅⋅⋅+𝛼𝑝−1𝑧 +𝛼𝑝𝑧 𝐻 (𝑧) = . (14) [𝑑𝑚−𝑖𝑥𝑛⃗ 𝑑𝑚−𝑖𝑥𝑛−1⃗ ⋅⋅⋅ 𝑑𝑚−𝑖𝑥𝑛−𝑖⃗ ] −1 −(𝑝−1) −𝑝 𝛽0 +𝛽1𝑧 +⋅⋅⋅+𝛽𝑝−1𝑧 +𝛽𝑝𝑧 (10) ⃗ ⃗ ⃗ =[𝑑𝑚−𝑖𝑥𝑛 𝑑𝑚−𝑖+1𝑥𝑛 ⋅⋅⋅ 𝑑𝑚𝑥𝑛]𝐷𝑖. According to the definition of 𝑍 transform, (14)isequal to the difference equation as follows: 𝑚=3 𝑖=2 For example, when , , 𝛽0𝑦𝑛 +𝛽1𝑦𝑛−1 +⋅⋅⋅+𝛽𝑝𝑦𝑛−𝑝 (15) 1111 =𝛼0𝑥𝑛 +𝛼1𝑥𝑛−1 +⋅⋅⋅+𝛼𝑝𝑥𝑛−𝑝. 11 1 [ −1 −2 −3] 𝐷 = [ ] ,𝐷= [ −1 −2] , 𝑎 𝑎 ⋅⋅⋅ 𝑎 𝑇 ⃗ 𝑏 𝑏 ⋅⋅⋅ 𝑏 𝑇 3 [ 13] 2 Here define 𝑎=[⃗ 0 1 𝑝] , 𝑏=[0 1 𝑝] , [ 1 ] 𝛼 𝛼 ⋅⋅⋅ 𝛼 𝑇 ⃗ 𝛽 𝛽 ⋅⋅⋅ 𝛽 𝑇 [ −1] 𝛼=[⃗ 0 1 𝑝] ,and𝛽=[ 0 1 𝑝] .Combined with the relation between 𝑋 and 𝑌 obtained by the transfer 𝑇 𝑥𝑛⃗ =[𝑥𝑛 𝑥𝑛−1 𝑥𝑛−2 𝑥𝑛−3] function 𝐻(𝑠),equationof𝑋 and 𝑌 can be obtained. The 𝑎⃗ 𝑏⃗ 11 1 1 following are deduced on the relationship between and , 𝛼⃗ 𝛽⃗ [ −1 −2 −3] ,and . =[𝑑 𝑥⃗𝑇 𝑑 𝑥⃗𝑇 𝑑 𝑥⃗𝑇 𝑑 𝑥⃗𝑇] [ ] , 𝑇 ⃗ 𝑇 0 𝑛 1 𝑛 2 𝑛 3 𝑛 [ 13] (11) Consider the simple case when 𝑎=[⃗ 10] , 𝑏=[01] , 𝐻(𝑠) = 𝑌(𝑠)/𝑋(𝑠) =1/𝑠 [ −1] and ;then

𝑇 𝑇 𝑇 𝑠𝑌 (𝑠) −𝑋(𝑠) =0. (16) [𝑑1𝑥𝑛⃗ 𝑑1𝑥𝑛−1⃗ 𝑑1𝑥𝑛−2⃗ ] According to the Laplace transform, (16)isequalto(17): =[(𝑥 −𝑥 )(𝑥 −𝑥 )(𝑥 −𝑥 )] 𝑛 𝑛−1 𝑛−1 𝑛−2 𝑛−2 𝑛−3 𝑑𝑦 (𝑡) =𝑥(𝑡) . (17) 11 1 𝑑𝑡 =[𝑑 𝑥⃗𝑇 𝑑 𝑥⃗𝑇 𝑑 𝑥⃗𝑇] [ −1 −2] , 1 𝑛 2 𝑛 3 𝑛 Take input sequence {𝑥𝑛,𝑥𝑛−1} from signal 𝑥 and output 1 [ ] sequence {𝑦𝑛,𝑦𝑛−1} fromsignal,andthestepsizeis𝑇. Meanwhile, the average values are 𝑥 = (1/2)(𝑥𝑛 +𝑥𝑛−1) and 𝑦 = (1/2)(𝑦 +𝑦 ) {𝐶0/2, 𝐶1/2} where, 𝑑0 =[1000], 𝑑1 =[1−100], 𝑛 𝑛−1 weighed by 1 1 . 𝑦 𝑑 𝑦⃗ /𝑇 𝑑2 =[1−210], 𝑑3 =[1−33−1],and Replace the first-order derivative of by 1 𝑛 ; accord- 𝑇 𝑇 𝑇 𝑇 ing to (17), 𝑑1𝑦𝑛⃗ /𝑇 equals 𝑥,as 𝑥𝑛⃗ =[𝑥𝑛 𝑥𝑛−1 𝑥𝑛−2 𝑥𝑛−3] , 𝑥𝑛−1⃗ =[𝑥𝑛−1 𝑥𝑛−2 𝑥𝑛−3 0] , 𝑇 𝑇 𝑦𝑛 −𝑦𝑛−1 1 and 𝑥𝑛−2⃗ =[𝑥𝑛−2 𝑥𝑛−3 00] . =𝑥= (𝑥 +𝑥 ) 𝑇 2 𝑛 𝑛−1 (18)

2.2. Principles and Steps to Obtain Transformation Matrix. which can be expressed as With the definitions above, the transformation matrix for 𝑑 𝑦⃗ 1 1 1 1 𝑛 = (𝑑 𝑥⃗ +𝑑 𝑥⃗ ) = ∑ (𝐶𝑖 𝑑 𝑥⃗ ) . bilinear transform will be presented and the relations 𝑇 2 0 𝑛 0 𝑛−1 2 1 0 𝑛−𝑖 (19) between coefficients before and after the transform willbe 𝑖=0 deducted. Thus it can be concluded that when taking input sequence 𝑑 𝑥⃗ /𝑇𝑖 Since the digital signal is not continuous, 𝑖 𝑛 is used {𝑥𝑛,𝑥𝑛−1} and output sequence {𝑦𝑛,𝑦𝑛−1} into related opera- 𝑖 𝑥 as the th order derivative of noncontinuous signal .Fora tions, as a kind of approximation, (1/2)(𝑑0𝑥𝑛⃗ +𝑑0𝑥𝑛−1⃗ ) takes general transfer function, the position of 𝑥 (𝑑0𝑥𝑛⃗ =𝑥𝑛 and 𝑑0𝑥𝑛−1⃗ =𝑥𝑛−1), and 𝑑1𝑦𝑛⃗ /𝑇 takes the position of the first-order derivative of y. 𝑌 (𝑠) 𝑎 +𝑎𝑠+⋅⋅⋅+𝑎 𝑠𝑝−1 +𝑎 𝑠𝑝 The mapping can be shown as below 𝐻 (𝑠) = = 0 1 𝑝−1 𝑝 𝑝−1 𝑝 (12) 𝑖 1−𝑖 𝑗 ⃗ 𝑋 (𝑠) 𝑏0 +𝑏1𝑠+⋅⋅⋅+𝑏𝑝−1𝑠 +𝑏𝑝𝑠 𝑑 𝑘 (𝑡) ∑ (𝐶 𝑑𝑖𝑘𝑛−𝑗) 󳨀→ 𝑗=0 1−𝑖 , (𝑖=0,1) (20) 𝑑𝑡𝑖 21−𝑖𝑇𝑖

is equal to the differential equation as follows: Take input sequence {𝑥𝑛,𝑥𝑛−1,𝑥𝑛−2} and output sequence {𝑦𝑛,𝑦𝑛−1,𝑦𝑛−2} sequentially. According to (17), 𝑑𝑦 (𝑡) 𝑑𝑝𝑦 (𝑡) 𝑦𝑛 −𝑦𝑛−1 1 𝑏0𝑦 (𝑡) +𝑏1 +⋅⋅⋅+𝑏𝑝 = (𝑥 +𝑥 ) , 𝑑𝑡 𝑑𝑡𝑝 𝑇 2 𝑛 𝑛−1 (21) (13) 𝑑𝑥 (𝑡) 𝑑𝑝𝑥 (𝑡) 𝑦 −𝑦 1 =𝑎𝑥 (𝑡) +𝑎 +⋅⋅⋅+𝑎 , 𝑛−1 𝑛−2 = (𝑥 +𝑥 ). 0 1 𝑑𝑡 𝑝 𝑑𝑡𝑝 𝑇 2 𝑛−1 𝑛−2 (22) 4 Mathematical Problems in Engineering

−1 Subtract (22)from(21) and then multiply by (𝑇) ;weget With conclusions above, taking input sequence {𝑥𝑛,𝑥𝑛−1,𝑥𝑛−2} and output sequence {𝑦𝑛,𝑦𝑛−1,𝑦𝑛−2},(30) 𝑦 −2𝑦 +𝑦 1 (𝑥 −𝑥 ) 𝑛 𝑛−1 𝑛−2 = 𝑛 𝑛−2 transforms into the following: 𝑇2 2 𝑇 (23) 𝑦 +2𝑦 +𝑦 𝑑 𝑦⃗ +𝑑 𝑦⃗ 𝑑 𝑦⃗ 1 (𝑥𝑛 −𝑥𝑛−1)+(𝑥𝑛−1 −𝑥𝑛−2) 𝑛 𝑛−1 𝑛−2 1 𝑛 1 𝑛−1 2 𝑛 = 𝑏0 ( )+𝑏1 ( )+𝑏2 2 𝑇 4 2𝑇 𝑇2 𝑥 +2𝑥 +𝑥 as = 𝑛 𝑛−1 𝑛−2 . 4 𝑑 𝑦⃗ 1 (𝑑 𝑥⃗ +𝑑 𝑥⃗ ) 1 1 2 𝑛 = 1 𝑛 1 𝑛−1 = ∑ (𝐶𝑖 𝑑 𝑥⃗ ). (31) 2 1 1 𝑛−𝑖 (24) 𝑇 2 𝑇 2𝑇 𝑖=0

The coefficients of {𝑥𝑛,𝑥𝑛−1,𝑥𝑛−2} and {𝑦𝑛,𝑦𝑛−1,𝑦𝑛−2} in 𝑑 𝑦⃗ /𝑇2 Because 2 𝑛 is the expression of second-order (31) are the same as coefficients in the difference equation (1/2)((𝑑 𝑥⃗ +𝑑 𝑥⃗ )/𝑇) derivative, 1 𝑛 1 𝑛−1 is the average first-order after bilinear transform. {𝐶0/2, 𝐶1/2} derivative weighed by 1 1 .Similarto(17), (24)can By this method, take input sequence {𝑥𝑛,𝑥𝑛−1,...,𝑥𝑛−𝑚} be expressed by the differential equation as follows: and output sequence {𝑦𝑛,𝑦𝑛−1,...,𝑦𝑛−𝑚} with (𝑚 + 1) points, and the mapping can be expressed as 𝑑2𝑦 (𝑡) 𝑑𝑥 (𝑡) = . (25) 𝑑𝑡2 𝑑𝑡 𝑖 𝑚−𝑖 𝑗 ⃗ 𝑑 𝑘 (𝑡) ∑ (𝐶 𝑑𝑖𝑘𝑛−𝑗) 󳨀→ 𝑗=0 𝑚−𝑖 , (𝑖=0,1,2,...,𝑚) . (32) After the Laplace transform, (26) can be obtained which 𝑑𝑡𝑖 2𝑚−𝑖𝑇𝑖 canbededucedby(16):

2 𝑠 𝑌 (𝑠) −𝑠𝑋(𝑠) =0. (26) Such rules can be used to deal with a general transfer function (12), for (13) can be expressed as Add (22)to(21)andweget 𝑝 𝑝−1 𝑦 −𝑦 1 𝑏 𝑏 𝑏𝑝 𝑛 𝑛−2 0 ∑ (𝐶𝑖 𝑑 𝑦⃗ )+ 1 ∑ (𝐶𝑖 𝑑 𝑦⃗ )+⋅⋅⋅+ 𝑑 𝑦⃗ = (𝑥𝑛 +2𝑥𝑛−1 +𝑥𝑛−2) (27) 𝑖 𝑝 0 𝑛−𝑖 𝑖−1 𝑝−1 1 𝑛−𝑖 𝑖 𝑝 𝑛 𝑇 2 2 𝑖=0 2 𝑇 𝑖=0 𝑇 as 𝑎 𝑝 𝑎 𝑝−1 = 0 ∑ (𝐶𝑖 𝑑 𝑥⃗ )+ 1 ∑ (𝐶𝑖 𝑑 𝑥⃗ ) 1 (𝑑1𝑦𝑛⃗ +𝑑1𝑦𝑛−1⃗ ) 1 𝑖 𝑝 0 𝑛−𝑖 𝑖−1 𝑝−1 1 𝑛−𝑖 = (𝑑 𝑥⃗ +2𝑑 𝑥⃗ +𝑑 𝑥⃗ ) 2 𝑖=0 2 𝑇 𝑖=0 2 𝑇 4 0 𝑛 0 𝑛−1 0 𝑛−2 𝑎𝑝 2 (28) +⋅⋅⋅+ 𝑑 𝑥⃗ . 1 𝑖 𝑇𝑖 𝑝 𝑛 = ∑ (𝐶2𝑑0𝑥𝑛−𝑖⃗ ). 4 𝑖=0 (33) ⃗ ⃗ Because (1/2)((𝑑1𝑦𝑛 +𝑑1𝑦𝑛−1)/𝑇) is the expression 𝑘 ⃗ ⃗ As a conclusion, for a discrete digital signal whose of first-order derivative 𝑑𝑦(𝑡)/𝑑𝑡, (1/4)(𝑑0𝑥𝑛 +2𝑑0𝑥𝑛−1 + 𝑇 𝑚 𝑑 𝑥⃗ ) {𝑥 ,𝑥 ,𝑥 } interval is , in order to get its th order derivative, at least 0 𝑛−2 is the average value of 𝑛 𝑛−1 𝑛−2 weighed by (𝑚 + 1) values are needed to take into operations. If the mth {𝐶0/4, 𝐶1/4, 𝐶2/4} 2 2 2 .Equation(28) can be expressed by the order derivative of 𝑘 is expressed by (𝑚 + 𝑙) values, such differential equation as16 ( ). mapping can be shown as follows: Thus it can be concluded that, when taking input sequence {𝑥𝑛,𝑥𝑛−1,𝑥𝑛−2} and output sequence {𝑦𝑛,𝑦𝑛−1,𝑦𝑛−2} (1/4)(𝑑 𝑥⃗ +2𝑑 𝑥⃗ +𝑑 𝑥⃗ ) 𝑑𝑚𝑘 (𝑡) 𝑙−1 into related operations, 0 𝑛 0 𝑛−1 0 𝑛−2 takes 󳨀→ 𝑇 −𝑚 ⋅2−(𝑙−1)∑ (𝐶𝑖 (𝑑 𝑘⃗ )) . 𝑚 𝑙−1 𝑚 𝑛−𝑖 (34) the position of 𝑥, (1/2)((𝑑1𝑦𝑛⃗ +𝑑1𝑦𝑛−1⃗ )/𝑇) takes the position 𝑑𝑡 2 𝑖=0 of the first-order derivative of 𝑦,and𝑑2𝑦𝑛⃗ /𝑇 takes the 𝑦 position of the second-order derivative of .Themappingcan 𝑙 be shown as follows: As changes, the mapping shown by (34) satisfies the distribution rule of binomial coefficients [4, 5], shown in 𝑖 2−𝑖 𝑗 ⃗ 𝑑 𝑘 (𝑡) ∑ (𝐶 𝑑𝑖𝑘𝑛−𝑗) Figure 1. 󳨀→ 𝑗=0 2−𝑖 , (𝑖 = 0, 1,) 2 . (29) 𝑑𝑡𝑖 22−𝑖𝑇𝑖 The tree structure in Figure 1 canbeexpressedby Figure 2,inwhich𝑘𝑖𝑗 is the element in 𝑖th row and 𝑇 𝑇 𝑗 𝑎=[⃗ 100] 𝑏=[⃗ 𝑏 𝑏 𝑏 ] th column. Meanwhile, (34) can also be expressed as Now consider when , 0 1 2 ,and 𝑚 𝑚 −𝑚 −(𝑙−1) 𝑙 ⃗ 2 𝑑 𝑘(𝑡)/𝑑𝑡 →𝑇 ⋅2 ∑𝑗=1(𝑘𝑙𝑗 (𝑑𝑚𝑘𝑛−𝑗+1)). 𝐻(𝑠) = 1/(𝑏0 +𝑏1𝑠+𝑏2𝑠 ). The corresponding differential equation is For example, according to that theory, while taking {𝑘𝑛,𝑘𝑛−1,...,𝑘𝑛−𝑚} from signal 𝑘, according to (34), 𝑑𝑦 (𝑡) 𝑑2𝑦 (𝑡) (𝑇)−𝑚𝑑 𝑘⃗ 𝑚 𝑘 𝑏 𝑦 (𝑡) +𝑏 +𝑏 =𝑥(𝑡) . 𝑚 𝑛 represents the th order derivative of ; 0 1 2 2 (30) −𝑖 −(𝑚−𝑖) 𝑖 𝑗 ⃗ 𝑑𝑡 𝑑𝑡 2 (𝑇) ∑𝑗=0(𝐶𝑖 𝑑𝑚−𝑖𝑘𝑛−𝑗) represents the (𝑚 − 𝑖)th order Mathematical Problems in Engineering 5

→ → → → → dm k n dm k n−1 dm k n−2 dm k n−3 dm k n−4 ··· conclusions above, the right side of (12) can be transformed ··· ··· ··· ··· ··· ··· by such mapping: 𝑑𝑥 (𝑡) 𝑑𝑝𝑥 (𝑡) 2𝑝 (𝑎 𝑥 (𝑡) +𝑎 +⋅⋅⋅+𝑎 ) 1 4641 l=5 0 1 𝑑𝑡 𝑝 𝑑𝑥𝑝

𝑝 (𝑝−𝑖) 𝑖 (37) 1 331 l=4 2 𝑗 󳨀→ ∑ (𝑎𝑖( ) ∑ (𝐶𝑖 𝑑𝑝−𝑖𝑥𝑛−𝑗⃗ )) . 𝑖=0 𝑇 𝑗=0 1 2 1 l=3 According to (34), the equation can be shown as follows:

11 l=2 𝑝 2 (𝑝−𝑖) 𝑖 ∑ (𝑎 ( ) ∑ (𝐶𝑗𝑑 𝑥⃗ )) 𝑖 𝑇 𝑖 𝑝−𝑖 𝑛−𝑗 𝑖=0 𝑗=0 (38) 1 l=1

=[𝑑0𝑥𝑛⃗ 𝑑1𝑥𝑛⃗ ⋅⋅⋅ 𝑑𝑝𝑥𝑛⃗ ]𝐻𝑝𝑎,⃗ Figure 1: Using (𝑚 + 𝑙) values to express 𝑚th order derivative of 𝑘. where 𝐻𝑝 is lower triangular matrix of (𝑝 + 1) × (𝑝; +1) 𝐻𝑝 is defined as k ij 2 2 (𝑝−𝑖) 2 𝑝 𝐻 =[ℎ⃗ ( ) ℎ⃗ ⋅⋅⋅ ( ) ℎ⃗ ⋅⋅⋅ ( ) ℎ⃗ ]. 𝑝 𝑝 𝑇 𝑝−1 𝑇 𝑝−𝑖 𝑇 0 kij =k(i−1)(j−1) +k(i−1)j (39) ⃗ ⃗ ⃗ ℎ0, ℎ1,...,ℎ𝑝 are (𝑝 + 1)-dimensional column vectors, and k(i−1)(j−1) k(i−1)j

Figure 2: Relations between elements of Figure 1 [4]. 𝑂 0 1 𝑖 𝑇 ℎ𝑖 =[ ][𝐶𝑖 𝐶𝑖 ⋅⋅⋅ 𝐶𝑖] , (40) 𝐷𝑖

𝑂 where 𝑖=0,1,2,...,𝑝, [ 𝐷 ] is partitioned matrix of (𝑝 + 1) × 𝑘 𝑖≤𝑚 𝑖 derivative of ( ). And according to (10), such equation (𝑖 + 1) and 𝑂 is zero matrix of (𝑝 − 𝑖) × (𝑖 +1). exists as follows: According to (9), the right side of (15)transformsas

𝑖 follows: ∑ (𝐶𝑗𝑑 𝑘⃗ ) 𝑖 𝑚−𝑖 𝑛−𝑗 𝛼0𝑥𝑛 +𝛼1𝑥𝑛−1 +⋅⋅⋅+𝛼𝑝𝑥𝑛−𝑝 𝑗=0 𝑇 0 = 𝑥𝑛⃗ 𝛼⃗ (41) 𝐶𝑖 [ ] (35) [ ] =[𝑑 𝑥⃗ 𝑑 𝑥⃗ ⋅⋅⋅ 𝑑 𝑥⃗ ]𝐷 𝛼.⃗ [ 1] 0 𝑛 1 𝑛 𝑝 𝑛 𝑝 ⃗ ⃗ ⃗ 𝐶𝑖 =[𝑑𝑚−𝑖𝑘𝑛 𝑑𝑚−𝑖+1𝑘𝑛 ⋅⋅⋅ 𝑑𝑚𝑘𝑛]𝐷𝑖 [ ] . [ . ] 𝐷 𝛼⃗ [ . ] Obviously 𝑝 is the coefficient vector of 𝑖 [𝑑0𝑥𝑛⃗ 𝑑1𝑥𝑛⃗ ⋅⋅⋅ 𝑑𝑝𝑥𝑛⃗ ] [𝐶𝑖 ] .Since(13)equals(15)and(38) equals (41), consequently, 𝑚=3 𝑖=2 For example, when , , 𝐷𝑝𝛼=𝐻⃗ 𝑝𝑎.⃗ (42)

2 1 Inthesameway, 𝑗 ⃗ ⃗ ⃗ ⃗ [ ] ∑ (𝐶 𝑑1𝑘𝑛−𝑗)=[𝑑 𝑘 𝑑 𝑘 𝑑 𝑘 ] 2 2 1 𝑛 1 𝑛−1 1 𝑛−2 𝐷 𝛽=𝐻⃗ 𝑏.⃗ 𝑗=0 [1] 𝑝 𝑝 (43) 1 −1 According to (8), (𝐷𝑝) =𝐷𝑝, the coefficient of (15)can ⃗ ⃗ ⃗ [2] =[𝑑1𝑘𝑛 𝑑2𝑘𝑛 𝑑3𝑘𝑛]𝐷2 (36) be expressed as follows: [1] ⃗ ⃗ 𝛼=𝐷⃗ 𝐻 𝑎,⃗ 𝛽=𝐷 𝐻 𝑏, (44) 4 𝑝 𝑝 𝑝 𝑝 =[ ⃗ ⃗ ⃗ ] [−4] . 𝑑1𝑘𝑛 𝑑2𝑘𝑛 𝑑3𝑘𝑛 𝑎⃗ 𝑏⃗ 1 where and express the coefficients of the numerator [ ] and denominator polynomials of 𝐻(𝑠). The transformation matrix for bilinear transformation is obtained by multiplying 𝑎⃗ Back to the deduction on the relationship between and 𝐷𝑝 and 𝐻𝑝, 𝐷𝑝,and𝐻𝑝 exist and are unique when 𝑝 is set to ⃗ ⃗ 𝑝 𝑏,and𝛼⃗ and 𝛽, first multiply both sides by 2 ; referring to the afixedvalue. 6 Mathematical Problems in Engineering

3. Examples where

3.1. An Example of Digital Filter Design. For example, the 11 1 11 design of a 4-order Butterworth low-pass filter with cut-off 𝐷 =1, 𝐷 =[ ], 𝐷 = [ −1 −2] , 0 1 −1 2 frequency 𝜔𝑐 = 0.5𝜋. [ 1 ] Set 𝑇=1s; then the cut-off frequency of the correspond- ing analog filter Ω𝑐 can be obtained [4, 5]: 1111 (51) [ ] [ −1 −2 −3] 𝐷3 = . 2 𝜔𝑐𝑇 [ 13] Ω𝑐 = tan =2rad/s . (45) 𝑇 2 [ −1]

According to 4-order Butterworth normalized system 𝐷 𝐻 𝑎⃗ 𝑏⃗ 𝛼⃗ 𝛽⃗ function [2], Take 4 and 4, and into (47); and can be obtained as 1 𝐻 (𝑢) = . 𝑇 𝑇 𝑛 1 + 2.613𝑢 + 3.414𝑢2 + 2.613𝑢3 +𝑢4 (46) 𝛼⃗ = 16[14641] =[16 64 96 64 16] , (52) ⃗ 𝑇 Take 𝑢 = 𝑠/Ω𝑐 into 𝐻𝑛(𝑢);thus 𝛽=[170.24 0 82.752 0 3.008] .

16 𝐻 (𝑠) = Thus, 16 + 20.904𝑠 + 13.656𝑠2 + 5.226𝑠3 +𝑠4 [1𝑠𝑠2 𝑠3 𝑠4] 𝑎⃗ (47) 𝐻 (𝑧) = . [1𝑠𝑠2 𝑠3 𝑠4] 𝑏⃗ [1𝑧−1 𝑧−2 𝑧−3 𝑧−4] 𝛼⃗ = [1𝑧−1 𝑧−2 𝑧−3 𝑧−4] 𝛽⃗ 𝑇 ⃗ So 𝑎=[⃗ 160000] and 𝑏= 𝑇 16 + 64𝑧−1 + 96𝑧−2 + 64𝑧−3 + 16𝑧−4 [16 20.904 13.656 5.226 1] ; according to (41), = 170.24 + 82.752𝑧−2 + 3.008𝑧−4 𝛼=𝐷⃗ 𝐻 𝑎,⃗ 𝛽⃗ =𝐷𝐻 𝑏⃗, (48) 0.0940 + 0.376𝑧−1 + 0.564𝑧−2 + 0.376𝑧−3 + 0.0940𝑧−4 4 4 4 4 = . 1 + 0.486𝑧−2 + 0.0177𝑧−4 where, according to (7), (39), and 𝑇=1, (53)

11111 𝐻(𝑧) could be expressed by difference equation: [ ] [ −1 −2 −3 −4] 𝐷 = [ 136] , 4 [ ] 𝑦 + 0.486𝑦 + 0.0177𝑦 [ −1 −4] 𝑛 𝑛−2 𝑛−4 1 [ ] = 0.0940𝑥𝑛 + 0.376𝑥𝑛−1 + 0.564𝑥𝑛−2 + 0.376𝑥𝑛−3 (54) (49) 16 + 0.0940𝑥 . [ ] 𝑛−4 [−32 16 ] [ ] 𝐻4 = [ 24 −24 16 ] . [ −8 12 −16 16 ] The amplitude- of Butterworth digital 1−24−816 low-pass filter obtained by Tustin transform is shown in [ ] Figure 3.

For 𝐻4, each column can be calculated by such rules: 3.2. An Example of Computing 𝑍-Domain Transfer Function. 16 1 Reference [15] mentioned design of switch capacitor circuit 16 1 𝑆 [−32] [4] by Tustin transform. The -domain transfer function of first- [ ] [ ] [−24] [3] [ 24 ] =𝐷 [6] , [ ] =2⋅𝐷 [ ] , order and second-order are as follows: [ ] 4 [ ] [ 12 ] 3 [3] [ −8 ] [4] −2 1 𝐴 +𝐴 𝑠 1 1 [ ] [ ] 0 1 [ ] [ ] 𝐻1 (𝑠) = (±) , (50) 𝐵0 +𝑠 16 1 (55) 2 16 3 1 2 [ ] [ ] 𝐴0 +𝐴1𝑠+𝐴2𝑠 −16 =2 ⋅𝐷2 2 ,[]=2 ⋅𝐷1 [ ], 𝐻 (𝑠) = (±) , −8 1 2 2 [ 4 ] [1] 𝐵0 +𝐵1𝑠+𝑠 16 = 24 ⋅𝐷 , 0 where 𝐴0, 𝐴1, 𝐴2, 𝐵0,and𝐵1 are real. Mathematical Problems in Engineering 7

1 1 0.8 0.8 0.6 0.6 0.4 0.4 amplitude amplitude Normalized 0.2 Normalized 0.2

0 0.5 1 1.5 2 2.5 3 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Frequency (Hz) Frequency (𝜔/𝜋)

(a) The amplitude-frequency response of normalized analog low-pass (b) The amplitude-frequency response of digital low-pass filter filter Figure 3: Simulation of amplitude frequency response.

2𝐵 4 In order to get 𝑍-domain transfer function, operate 𝐻1(𝑠) 𝐵 + 1 + [ 0 2 ] by Tustin transformation as [ 𝑇 𝑇 ] [ ] [ 8 ] = [ 2𝐵0 − ] , [ 𝑇2 ] [ ] 2𝐴 2𝐵 4 𝐴 + 1 𝐵 − 1 + 20 [ 0 ] [ 0 2 ] 11 𝐴0 [ 𝑇 ] 𝑇 𝑇 𝛼⃗1 =𝐷1𝐻1𝑎1⃗ =[ ][ 2 ][ ]=[ ] , 0−1 −1 𝐴1 2𝐴 (58) 𝑇 𝐴 − 1 [ 0 ] 𝑇 𝐻2 (𝑧) 20 2 2𝐴 4𝐴 8𝐴 𝐵0 + 1 2 2 −1 11[ ] 𝐵 [ 𝑇] = (±) (𝐴0 + + +(2𝐴0 − )𝑧 𝛽⃗ =𝐷𝐻 𝑏⃗ =[ ] [ ] [ 0]=[ ] , 𝑇 𝑇2 𝑇2 1 1 1 1 0−1 2 1 [ 2 ] −1 𝐵 − 2𝐴 4𝐴 [ 𝑇] [ 0 𝑇] +(𝐴 − 1 + 2 )𝑧−2) 0 2 (56) 𝑇 𝑇 (59) 2𝐵 4 8 𝐴 +2𝐴 /𝑇 + (𝐴 −2𝐴 /𝑇)−1 𝑧 ×(𝐵 + 1 + +(2𝐵 − )𝑧−1 𝐻 (𝑧) = (±) 0 1 0 1 ; 0 𝑇 𝑇2 0 𝑇2 1 −1 (57) 𝐵0 +2/𝑇+(𝐵0 − 2/𝑇) 𝑧 2𝐵 4 −1 +(𝐵 − 1 + )𝑧−2) . 0 𝑇 𝑇2

then operate 𝐻2(𝑠) by Tustin transformation as Thus, 𝐻1(𝑧) and 𝐻2(𝑧) are the 𝑍-domain transfer func- tions.

𝛼⃗2 =𝐷2𝐻2𝑎2⃗ 3.3. An Example of Extensional Application. Reference [9] presented a matrix method for a biquadratic transformation. 400 By the algorithm recommended in this paper, we get the [ 4 ] 11 1 [−4 0 ] 𝐴0 matrix with little deformation. [ ] [ ] [ ] 𝑛 = 0−1−2 [ 𝑇 ] 𝐴1 For an th order discrete-time polynomial 00 1 [ 2 4 ] 𝐴 [ ] 1− [ 2] 𝑛 2 𝑖 𝑛 ⃗ [ 𝑇 𝑇 ] 𝐷 (𝑧) = ∑𝑑𝑖𝑧 =[1 𝑧 ⋅⋅⋅ 𝑧 ] 𝑑, (60) 2𝐴 4𝐴 𝑖=0 𝐴 + 1 + 2 [ 0 𝑇 𝑇2 ] 𝑑⃗ 𝑛 𝑑=⃗ [ ] where is an -dimensional column vector, [ ] [𝑑 𝑑 ⋅⋅⋅ 𝑑 ]𝑇 [ 8𝐴2 ] 0 1 𝑛 . = [ 2𝐴0 − ] , [ 𝑇2 ] Take the transformation form (61)into𝐷(𝑧): [ ] 2 2𝐴1 4𝐴2 𝑠 +𝑠+1 𝐴0 − + 𝑧= ; (61) [ 𝑇 𝑇2 ] 𝑠2 −𝑠+1 ⃗ ⃗ 𝐷(𝑧) 𝛽2 =𝐷2𝐻2𝑏2 then transformed into a continuous-time polynomial as 2𝑛 400 𝑖 2𝑛 [ 4 ] 𝐶 (𝑠) = ∑𝑐𝑖𝑠 = [1 𝑠 ⋅⋅⋅ 𝑠 ] 𝑐,⃗ (62) 11 1 [ ] 𝐵0 [ ] [−4 0 ] [ ] 𝑖=0 = 0−1−2 [ 𝑇 ] 𝐵1 00 1 [ 2 4 ] 1 where 𝑐⃗ is an 2𝑛-dimensional column vector, [ ] 1− [ ] 𝑇 [ 𝑇 𝑇2 ] 𝑐=[⃗ 𝑐0 𝑐1 ⋅⋅⋅ 𝑐2𝑛] . 8 Mathematical Problems in Engineering

The relationship between coefficients in(60)and(62)can Finally, 𝑄𝑛 canbeexpressedas be expressed as 𝑄𝑛 =𝐺𝑛𝐷𝑛𝐻𝑛 (70) ⃗ 𝑐=𝑄⃗ 𝑛𝑑, (63) while 𝑛=2: where 𝑄𝑛 is the (2𝑛+1)×(𝑛+1)transformation matrix. 𝑄𝑛 100 canbeobtainedbythismethod. [ ] [0−10] 11 1 Equal to (61), the transformation form can be deformed [ ] [ ] 𝐺2 = [201] ,𝐷2 = 0−1−2 , as [ ] 0−10 [00 1] 2 𝑠2 +𝑠+1 1+(𝑠/(𝑠 +1)) 𝑢=−(𝑠/(𝑠2+1)) 1−𝑢 [100] (71) 𝑧= = 󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀→ 𝑧= . 2 2 𝑠 −𝑠+1 1−(𝑠/(𝑠 +1)) 1+𝑢 400 (64) [ ] 𝐻2 = −420 (𝑇=2) ; 1−11 Take (64)into𝐷(𝑧),then𝐷(𝑧) transforms into 𝐹(𝑢) as [ ] follows: thus 𝑄2 can be obtained by (70), which is the same as shown 𝐷 (𝑧) =[1 𝑧 ⋅⋅⋅𝑛 𝑧 ] 𝑑⃗ in [8]: (65) 𝑧=(1−𝑢)/(1+𝑢) 111 󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀→𝐹(𝑢) =[1𝑢⋅⋅⋅𝑢𝑛] 𝑓.⃗ [ ] [−2 0 2] 𝑄 =𝐺𝐷 𝐻 = [ 313] . ⃗ ⃗ 2 2 2 2 [ ] (72) According to (44), the relation between 𝑑 and 𝑓 can be [−2 0 2] 𝑇=2 expressed as (66); here consider the step as : [ 111] 𝑓=𝐷⃗ 𝐻 𝑑.⃗ (66) 𝑛 𝑛 4. Conclusions Last, take the equation between 𝑢 and 𝑠 into 𝐶(𝑠),as follows: The method of using corresponding matrix for Tustin trans- form representation reflects the inherent properties; on the 𝐹 (𝑢) =[1 𝑢 ⋅⋅⋅𝑛 𝑢 ] 𝑓⃗ other hand, it allows computer programs to deal with the 𝑛 (67) transfer functions of order automatically. In this paper, the 𝑢=−(𝑠/(𝑠2+1)) 𝐷 𝐻 󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀󳨀→𝐶(𝑠) =[1 𝑠 ⋅⋅⋅2𝑛 𝑠 ] 𝑐.⃗ transformation matrix is obtained by multiplying 𝑛 and 𝑛, which dramatically reduces the complexity of the program without any tabulation. Furthermore, this method can also be 𝑓⃗ Similar to the deduction of (44), the relation between modifiedproperlytosuitsimilarproblemsaboutcoefficients 𝑐⃗ and canbeexpressedas after linear transform. ⃗ 𝑐=𝐺⃗ 𝑛𝑓, (68) Conflict of Interests where 𝐺𝑛 is a (2𝑛 + 1) × (𝑛 +1) matrix: The authors declare that there is no conflict of interests 𝐶0 00⋅⋅⋅0 regarding the publication of this paper. [ 𝑛 ] [ 0 . ] [ 0−𝐶 0⋅⋅⋅ . ] [ 𝑛−1 . ] [ 𝐶1 0𝐶0 ⋅⋅⋅ 0 ] References [ 𝑛 𝑛−2 ] [ . . . ] 𝐺𝑛 = [ . . . ⋅⋅⋅ (−1)𝑛𝐶0] [1] Z. Y. Wu, Digital , China Higher Education [ . . . 0] [𝐶𝑛−1 0𝐶𝑛−2 ⋅⋅⋅ 0 ] Press, 2004. [ 𝑛 𝑛−2 ] [ ] [2] Z. Y. Song, “Design of digital filter using bilinear Z transform,” [ 𝑛−1 . ] 0−𝐶𝑛−1 0⋅⋅⋅ . China Computer & Network,vol.4,no.1,pp.33–66,1978. 𝑛 [ 𝐶𝑛 00⋅⋅⋅0] [3] V. Havu and J. Malinen, “The Cayley transform as a time discretization scheme,” NumericalFunctionalAnalysisandOpti- 𝑖 𝑛 =[𝑔𝑛⃗ −𝑔𝑛−1⃗ ⋅⋅⋅ (−1) 𝑔𝑛−𝑖⃗ ⋅⋅⋅ (−1) 𝑔0⃗ ] mization,vol.28,no.7-8,pp.825–851,2007. [4] Q.Y.Shen,X.Y.Miao,andA.L.Song,“Atransformationmatrix 𝑇 ⏟⏟⏟⏟⏟⏟⏟⋅⋅⋅ 0𝐶𝑘−1 0𝐶𝑘 0𝐶𝑘+1 0⋅⋅⋅⏟⏟⏟⏟⏟⏟⏟ algorithm for computing bilinear transformation,” Information 𝑔2𝑘⃗ =[ 2𝑘 2𝑘 2𝑘 ] 𝑛−3 𝑛−3 and Control,vol.11,no.4,pp.25–30,1982. [5] B. Pˇsenicka,ˇ F. Garc´ıa-Ugalde, and A. Herrera-Camacho, “The 𝑘−1 𝑘 𝑘+1 𝑘+2 𝑇 ⏟⏟⏟⏟⏟⏟⏟⋅⋅⋅ 𝐶 0𝐶 0𝐶 0𝐶 ⏟⏟⏟⏟⏟⏟⏟⋅⋅⋅ bilinear Z transform by Pascal matrix and its application in the 𝑔2𝑘+1⃗ =[ 2𝑘+1 2𝑘+1 2𝑘+1 2𝑘+1 ] 𝑛−3 𝑛−3 design of digital filters,” IEEE Signal Processing Letters,vol.9,no. 𝑘=0,1,2,.... 11,pp.368–370,2002. [6]S.Chivapreecha,S.Sriyapong,S.Junnapiya,andK.Dejhan, (69) “Bilinear s-z with frequency transformation using pascal matrix Mathematical Problems in Engineering 9

operation,” in Proceedings of the International Symposium on Communications and Information Technologies (ISCIT ’05),pp. 764–767, October 2005. [7] G.L.LiuandP.K.Li,“Asimplerecursivemethodforcomputing bilinear,” Journal of North China Institute of Electric Power,vol. 14, no. 1, pp. 75–82, 1988. [8] M. Jiang and D. Z. Cao, “A simple algorithm for bilinear trans- forming,” Journal of Southeast University,vol.21,no.3,pp.124– 127, 1991. [9] D. Kim and Y. Choo, “Computing biquadratic transformation matrix,” Journal of Institute of Control, Robotics and Systems,vol. 15,no.2,pp.128–131,2009. [10] C. Zhu and Y. Zou, “Improved recursive algorithm for frac- tional-order system solution based on PSE and Tustin trans- form,” Systems Engineering and Electronics, vol. 31, no. 11, pp. 2736–2741, 2009. [11] G. Zhang, X. Chen, and T. Chen, “Digital redesign via the generalised bilinear transformation,” International Journal of Control,vol.82,no.4,pp.741–754,2009. [12] S. Yan, N. Shiratori, and L. Xu, “Simple state-space formulations of 2-D frequency transformation and double bilinear transfor- mation,” Multidimensional Systems and Signal Processing,vol.21, no.1,pp.3–23,2010. [13] S. Yan, N. Shiratori, H. Shieh, and L. Xu, “A general state-space representation of n-variable bilinear transformation,” Signal Processing,vol.91,no.2,pp.185–190,2011. [14] P. Karimaghaee and N. Noroozi, “Frequency weighted discrete- time controller order reduction using bilinear transformation,” Journal of Electrical Engineering,vol.62,no.1,pp.44–48,2011. [15] X. J. Yang, “Design of switched-capacitor one-order and two- order sections using bilinear transformation,” Acta Electronica Sinica,vol.21,no.4,pp.31–40,1982. Advances in Advances in Journal of Journal of Operations Research Decision Sciences Applied Mathematics Algebra Probability and Statistics Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

The Scientific International Journal of World Journal Differential Equations Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

Submit your manuscripts at http://www.hindawi.com

International Journal of Advances in Combinatorics Mathematical Physics Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

Journal of Journal of Mathematical Problems Abstract and Discrete Dynamics in Complex Analysis Mathematics in Engineering Applied Analysis Nature and Society Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

International Journal of Journal of Mathematics and Mathematical Discrete Mathematics Sciences

Journal of International Journal of Journal of Function Spaces Stochastic Analysis Optimization Hindawi Publishing Corporation Hindawi Publishing Corporation Volume 2014 Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014