About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculum development project

About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculum development project undertaken by a consortium of five English universities led by Loughborough University, funded by the Higher Education Funding Council for England under the Fund for the Development of Teaching and Learning for the period October 2002 September 2005. HELM aims to enhance the mathematical education of engineering undergraduates through a range of flexible learning resources in the form of Workbooks and web-delivered interactive segments. HELM supports two CAA regimes: an integrated web-delivered implementation and a CD-based version. HELM learning resources have been produced primarily by teams of writers at six universities: Hull, Loughborough, Manchester, Newcastle, Reading, Sunderland. HELM gratefully acknowledges the valuable support of colleagues at the following universities and colleges involved in the critical reading, trialling, enhancement and revision of the learning materials: Aston, Bournemouth & Poole College, Cambridge, City, Glamorgan, Glasgow, Glasgow Caledonian, Glenrothes Institute of Applied Technology, Harper Adams University College, Hertfordshire, Leicester, Liverpool, London Metropolitan, Moray College, Northumbria, Nottingham, Nottingham Trent, Oxford Brookes, Plymouth, Portsmouth, Queens Belfast, Robert Gordon, Royal Forest of Dean College, Salford, Sligo Institute of Technology, Southampton, Southampton Institute, Surrey, Teesside, Ulster, University of Wales Institute Cardiff, West Kingsway College (London), West Notts College. HELM Contacts: Post: HELM, Mathematics Education Centre, Loughborough University, Loughborough, LE11 3TU. Email: helm@lboro.ac.uk Web: http://helm.lboro.ac.uk HELM Workbooks List 1 Basic Algebra 26 Functions of a Complex Variable 2 Basic Functions 27 Multiple Integration 3 Equations, Inequalities & Partial Fractions 28 Differential Vector Calculus 4 Trigonometry 29 Integral Vector Calculus 5 Functions and Modelling 30 Introduction to Numerical Methods 6 Exponential and Logarithmic Functions 31 Numerical Methods of Approximation 7 Matrices 32 Numerical Initial Value Problems 8 Matrix Solution of Equations 33 Numerical Boundary Value Problems 9 Vectors 34 Modelling Motion 10 Complex Numbers 35 Sets and Probability 11 Differentiation 36 Descriptive Statistics 12 Applications of Differentiation 37 Discrete Probability Distributions 13 Integration 38 Continuous Probability Distributions 14 Applications of Integration 1 39 The Normal Distribution 15 Applications of Integration 2 40 Sampling Distributions and Estimation 16 Sequences and Series 41 Hypothesis Testing 17 Conics and Polar Coordinates 42 Goodness of Fit and Contingency Tables 18 Functions of Several Variables 43 Regression and Correlation 19 Differential Equations 44 Analysis of Variance 20 Laplace Transforms 45 Non-parametric Statistics 21 -Transforms 46 Reliability and Quality Control 22 Eigenvalues and Eigenvectors 47 Mathematics and Physics Miscellany 23 Fourier Series 48 Engineering Case Studies 24 Fourier Transforms 49 Student s Guide 25 Partial Differential Equations 50 Tutor s Guide Copyright Loughborough University, 2006

Contents 21 -Transforms 21.1 The -Transform 2 21.2 Basics of -Transform Theory 12 21.3 -Transforms and Difference Equations 36 21.4 Engineering Applications of -Transforms 64 21.5 Sampled Functions 85 Learning outcomes In this Workbook you will learn about the properties and applications of the -transform, a major mathematical tool for the analysis and design of discrete systems including digital control systems.

Basics of -Transform Theory 21.2 Introduction In this Section, which is absolutely fundamental, we define what is meant by the -transform of a sequence. We then obtain the -transform of some important sequences and discuss useful properties of the transform. Most of the results obtained are tabulated at the end of the Section. The -transform is the major mathematical tool for analysis in such areas as digital control and digital signal processing. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... understand sigma (Σ) notation for summations be familiar with geometric series and the binomial theorem have studied basic complex number theory including complex exponentials define the -transform of a sequence obtain the -transform of simple sequences from the definition or from basic properties of the -transform 12 HELM (2008): Workbook 21: -Transforms

1. The -transform If you have studied the Laplace transform either in a Mathematics course for Engineers and Scientists or have applied it in, for example, an analog control course you may recall that 1. the Laplace transform definition involves an integral 2. applying the Laplace transform to certain ordinary differential equations turns them into simpler (algebraic) equations 3. use of the Laplace transform gives rise to the basic concept of the transfer function of a continuous (or analog) system. The -transform plays a similar role for discrete systems, i.e. ones where sequences are involved, to that played by the Laplace transform for systems where the basic variable t is continuous. Specifically: 1. the -transform definition involves a summation 2. the -transform converts certain difference equations to algebraic equations 3. use of the -transform gives rise to the concept of the transfer function of discrete (or digital) systems. Key Point 1 Definition: For a sequence {y n } the -transform denoted by Y () is given by the infinite series Y () =y 0 + y 1 1 + y 2 2 +...= y n n (1) n=0 Notes: 1. The -transform only involves the terms y n, n =0, 1, 2,...of the sequence. Terms y 1,y 2,... whether ero or non-ero, are not involved. 2. The infinite series in (1) must converge for Y () to be defined as a precise function of. We shall discuss this point further with specific examples shortly. 3. The precise significance of the quantity (strictly the variable ) need not concern us except to note that it is complex and, unlike n, is continuous. Key Point 2 We use the notation Z{y n } = Y () to mean that the -transform of the sequence {y n } is Y (). HELM (2008): Section 21.2: Basics of -Transform Theory 13

-Transforms and Difference Equations 21.3 Introduction In this we apply -transforms to the solution of certain types of difference equation. We shall see that this is done by turning the difference equation into an ordinary algebraic equation. We investigate both first and second order difference equations. A key aspect in this process in the inversion of the -transform. As well as demonstrating the use of partial fractions for this purpose we show an alternative, often easier, method using what are known as residues. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... have studied carefully Section 21.2 be familiar with simple partial fractions invert -transforms using partial fractions or residues where appropriate solve constant coefficient linear difference equations using -transforms 36 HELM (2008): Workbook 21: -Transforms

Engineering Applications of -Transforms 21.4 Introduction In this Section we shall apply the basic theory of -transforms to help us to obtain the response or output sequence for a discrete system. This will involve the concept of the transfer function and we shall also show how to obtain the transfer functions of series and feedback systems. We will also discuss an alternative technique for output calculations using convolution. Finally we shall discuss the initial and final value theorems of -transforms which are important in digital control. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... be familiar with basic -transforms, particularly the shift properties obtain transfer functions for discrete systems including series and feedback combinations state the link between the convolution summation of two sequences and the product of their -transforms 64 HELM (2008): Workbook 21: -Transforms

1. Applications of -transforms Transfer (or system) function Consider a first order linear constant coefficient difference equation y n + ay n 1 = bx n n =0, 1, 2,... (1) where {x n } is a given sequence. Assume an initial condition y 1 is given. Task Take the -transform of (1), insert the initial condition and obtain Y () in terms of X(). Using the right shift theorem Y ()+a( 1 Y ()+y 1 )=bx() where X() is the -transform of the given or input sequence {x n } and Y () is the -transform of the response or output sequence {y n }. Solving for Y () so Y ()(1 + a 1 )=bx() ay 1 Y () = bx() 1+a ay 1 1 1+a 1 (2) The form of (2) shows us clearly that Y () is made up of two components, Y 1 () and Y 2 () say, where (i) Y 1 () = bx() which depends on the input X() 1+a 1 (ii) Y 2 () = ay 1 1+a which depends on the initial condition y 1. 1 HELM (2008): Section 21.4: Engineering Applications of -Transforms 65

Sampled Functions 21.5 Introduction A sequence can be obtained by sampling a continuous function or signal and in this Section we show first of all how to extend our knowledge of -transforms so as to be able to deal with sampled signals. We then show how the -transform of a sampled signal is related to the Laplace transform of the unsampled version of the signal. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... HELM (2008): Section 21.5: Sampled Functions possess an outline knowledge of Laplace transforms and of -transforms take the -transform of a sequence obtained by sampling state the relation between the -transform of a sequence obtained by sampling and the Laplace transform of the underlying continuous signal 85

1. Sampling theory If a continuous-time signal f(t) is sampled at terms t =0,T,2T,...nT,... then a sequence of values {f(0), f(t ), f(2t ),...f(nt ),...} is obtained. The quantity T is called the sample interval or sample period. f(t) - - - - - - T 2T nt t Figure 18 In the previous Sections of this Workbook we have used the simpler notation {f n } to denote a sequence. If the sequence has actually arisen by sampling then f n is just a convenient notation for the sample value f(nt ). Most of our previous results for -transforms of sequences hold with only minor changes for sampled signals. So consider a continuous signal f(t); its -transform is the -transform of the sequence of sample values i.e. Z{f(t)} = Z{f(nT )} = f(nt ) n n=0 We shall briefly obtain -transforms of common sampled signals utiliing results obtained earlier. You may assume that all signals are sampled at 0,T,2T,...nT,... Unit step function 1 t 0 u(t) = 0 t<0 Since the sampled values here are a sequence of 1 s, Z{u(t)} = Z{u n } = = 1 1 1 1 > 1 where {u n } = {1, 1, 1,...} is the unit step sequence. 86 HELM (2008): Workbook 21: -Transforms

Ramp function r(t) = t t 0 0 t<0 The sample values here are {r(nt )} = {0, T,2T,...} The ramp sequence {r n } = {0, 1, 2,...} has -transform Hence Z{r(nT )} = T ( 1) 2 since {r(nt )} = T {r n }. ( 1) 2. Task Obtain the -transform of the exponential signal e αt t 0 f(t) = 0 t<0. [Hint: use the -transform of the geometric sequence {a n }.] The sample values of the exponential are {1, e αt,e α2t,...,e αnt,...} i.e. f(nt )=e αnt =(e αt ) n. But Z{a n } = a Z{(e αt ) n } = e = 1 αt 1 e αt 1 HELM (2008): Section 21.5: Sampled Functions 87

Sampled sinusoids Earlier in this Workbook we obtained the -transform of the sequence {cos ωn} i.e. Z{cos ωn} = 2 cos ω 2 2 cos ω +1 Hence, since sampling the continuous sinusoid f(t) = cos ωt yields the sequence {cos nωt } we have, simply replacing ω by ωt in the -transform: Z{cos ωt} = Z{cos nωt } = 2 cos ωt 2 2 cos ωt +1 Task Obtain the -transform of the sampled version of the sine wave f(t) = sin ωt. Z{sin ωn} = sin ω 2 2 cos ω +1 Z{sin ωt} = Z{sin nωt } = sin ωt 2 2 cos ωt +1 88 HELM (2008): Workbook 21: -Transforms

Shift theorems These are similar to those discussed earlier in this Workbook but for sampled signals the shifts are by integer multiples of the sample period T. For example a simple right shift, or delay, of a sampled signal by one sample period is shown in the following figure: f(nt ) t f(nt T ) T 2T 3T t T 2T 3T 4T Figure 19 The right shift properties of -transforms can be written down immediately. (Look back at the shift properties in Section 21.2 subsection 5, if necessary:) If y(t) has -transform Y () which, as we have seen, really means that its sample values {y(nt )} give Y (), then for y(t) shifted to the right by one sample interval the -transform becomes Z{y(t T )} = y( T )+ 1 Y () The proof is very similar to that used for sequences earlier which gave the result: Z{y n 1 } = y 1 + 1 Y () Task Using the result Z{y n 2 } = y 2 + y 1 1 + 2 Y () write down the result for Z{y(t 2T )} Z{y(t 2T )} = y( 2T )+y( T ) 1 + 2 Y () These results can of course be generalised to obtain Z{y(t mt )} where m is any positive integer. In particular, for causal or one-sided signals y(t) (i.e. signals which are ero for t<0): Z{y(t mt )} = m Y () Note carefully here that the power of is still m not mt. HELM (2008): Section 21.5: Sampled Functions 89

Examples: For the unit step function we saw that: Z{u(t)} = 1 = 1 1 1 Hence from the shift properties above we have immediately, since u(t) is certainly causal, and so on. Z{u(t T )} = 1 1 = Z{u(t 3T )} = 3 1 = 1 1 1 3 1 1 u(t T ) T 2T 3T t u(t 3T ) T 2T 3T 4T 5T t Figure 20 2. -transforms and Laplace transforms In this Workbook we have developed the theory and some applications of the -transform from first principles. We mentioned much earlier that the -transform plays essentially the same role for discrete systems that the Laplace transform does for continuous systems. We now explore the precise link between these two transforms. A brief knowledge of Laplace transform will be assumed. At first sight it is not obvious that there is a connection. The -transform is a summation defined, for a sampled signal f n f(nt ), as F () = f(nt ) n n=0 while the Laplace transform written symbolically as L{f(t)} is an integral, defined for a continuous time function f(t), t 0 as F (s) = 0 f(t)e st dt. 90 HELM (2008): Workbook 21: -Transforms

Thus, for example, if f(t) = e αt (continuous time exponential) L{f(t)} = F (s) = 1 s + α which has a (simple) pole at s = α = s 1 say. As we have seen, sampling f(t) gives the sequence {f(nt )} = {e αnt } with -transform F () = 1 1 e αt = 1 e. αt The -transform has a pole when = 1 where 1 = e αt = e s 1T [Note the abuse of notations in writing both F (s) and F () here since in fact these are different functions.] Task The continuous time function f(t) =te αt has Laplace transform F (s) = 1 (s + α) 2 Firstly write down the pole of this function and its order: 1 F (s) = has its pole at s = s (s + α) 2 1 = α. The pole is second order. Now obtain the -transform F () of the sampled version of f(t), locate the pole(s) of F () and state the order: HELM (2008): Section 21.5: Sampled Functions 91

Consider f(nt )=nt e αnt =(nt )(e αt ) n The ramp sequence {nt } has -transform f(nt ) has -transform F () = TeαT (e αt 1) 2 = T ( 1) 2 Te αt ( e αt ) 2 (see Key Point 8) This has a (second order) pole when = 1 = e αt = e s 1T. We have seen in both the above examples a close link between the pole s 1 of the Laplace transform of f(t) and the pole 1 of the -transform of the sampled version of f(t) i.e. 1 = e s 1T where T is the sample interval. Multiple poles lead to similar results i.e. if F (s) has poles s 1,s 2,... then F () has poles 1, 2,... where i = e s it. The relation (1) between the poles is, in fact, an example of a more general relation between the values of s and as we shall now investigate. (1) Key Point 19 The unit impulse function δ(t) can be defined informally as follows: 1 ɛ P ɛ (t) ɛ t Figure 21 The rectangular pulse P (t) of width ε and height 1 shown in Figure 21 encloses unit area and has ε Laplace transform P ε (s) = ε 0 1 ε e st = 1 εs (1 e εs ) (2) As ε becomes smaller P ε (t) becomes taller and narrower but still encloses unit area. The unit impulse function δ(t) (sometimes called the Dirac delta function) can be defined as 92 HELM (2008): Workbook 21: -Transforms

δ(t) = lim ε 0 P ε (t) The Laplace transform, say (s), of δ(t) can be obtained correspondingly by letting 0 in (2), i.e. 1 (s) = lim ε 0 εs (1 e εs ) (εs) 2 1 (1 εs + = lim 2! ε 0 εs...) (Using the Maclaurin seies expansion of e εs ) (εs) 2 εs + (εs)3 = lim 2! 3! ε 0 εs = 1 +... i.e. Lδ(t) =1 (3) Task A shifted unit impulse δ(t nt ) is defined as lim ε 0 P ε (t nt ) as illustrated below. P ɛ (t nt ) 1 ɛ nt nt + ɛ t Obtain the Laplace transform of this rectangular pulse and, by letting ε 0, obtain the Laplace transform of δ(t nt ). HELM (2008): Section 21.5: Sampled Functions 93

L{P ε (t nt )} = nt +ε nt 1 ε e st dt = 1 εs = 1 εs e st nt +ε nt e snt s(nt e +ε) = 1 εs e snt (1 e sε ) e snt as ε 0 Hence L{δ(t nt )} = e snt which reduces to the result (3) L{δ(t)} =1 when n =0 (4) These results (3) and (4) can be compared with the results Z{δ n } =1 Z{δ n m } = m for discrete impulses of height 1. Now consider a continuous function f(t). Suppose, as usual, that this function is sampled at t = nt for n =0, 1, 2,... f(t) T 2T 3T 4T - - - - - - t Figure 22 This sampled equivalent of f(t), say f (t) can be defined as a sequence of equidistant impulses, the strength of each impulse being the sample value f(nt )i.e. f (t) = f(nt )δ(t nt ) n=0 This function is a continuous-time signal i.e. is defined for all t. Using (4) it has a Laplace transform F (s) = f(nt )e snt (5) n=0 If, in this sum (5) we replace e st by we obtain the -transform of the sequence {f(nt )} of samples: f(nt ) n n=0 94 HELM (2008): Workbook 21: -Transforms

The Laplace transform F (s) = Key Point 20 f(nt )e snt n=0 of a sampled function is equivalent to the -transform F () of the sequence {f(nt )} of sample values with = e st. Table 2: -transforms of some sampled signals This table can be compared with the table of the -transforms of sequences on the following page. f(t) f(nt ) F () Radius of convergence t 0 n =0, 1, 2,... R 1 1 t nt 1 ( 1) 2 1 1 t 2 (nt ) 2 T 2 ( +1) ( 1) 3 1 e αt e αnt e αt e αt sin ωt sin nωt sin ωt 2 2 cos ωt +1 1 cos ωt cos nωt ( cos ωt) 2 2 cos ωt +1 1 te αt nt e αnt Te αt ( e αt ) 2 e αt e αt sin ωt e αt cos ωt e αnt sin ωnt e αnt cos ωnt e αt 1 sin ωt 1 2e αt 1 cos ωt + e 2aT 2 e αt 1 e αt 1 cos ωt 1 2e αt 1 cos ωt + e 2aT 2 e αt Note: R is such that the closed forms of F () (those listed in the above table) are valid for >R. HELM (2008): Section 21.5: Sampled Functions 95

Table of -transforms f n F () Name δ n 1 unit impulse δ n m u n a n e αn sinh αn cosh αn sin ωn cos ωn e αn sin ωn m 1 a e α sinh α 2 2 cosh α +1 2 cosh α 2 2 cosh α +1 sin ω 2 2 cos ω +1 2 cos ω 2 2 cos ω +1 e α sin ω 2 2e α cos ω + e 2α unit step sequence geometric sequence e αn cos ωn n 2 e α cos ω 2 2e α cos ω + e 2α ( 1) 2 ramp sequence n 2 ( +1) ( 1) 3 n 3 ( 2 +4 +1) ( 1) 4 a n f n F a nf n df d 96 HELM (2008): Workbook 21: -Transforms

Clearly, from (2), if y 1 =0(ero initial condition) then Y () =Y 1 () and hence the term ero-state response is sometimes used for Y 1 (). Similarly if {x n } and hence X() =0(ero input) Y () =Y 2 () and hence the term ero-input response can be used for Y 2 (). In engineering the difference equation (1) is regarded as modelling a system or more specifically a linear discrete time-invariant system. The terms linear and time-invariant arise because the difference equation (1) is linear and has constant coefficients i.e. the coefficients do not involve the index n. The term discrete is used because sequences of numbers, not continuous quantities, are involved. As noted above, the given sequence {x n } is considered to be the input sequence and {y n }, the solution to (1), is regarded as the output sequence. {x n } input (stimulus) system Figure 8 {y n } output (response) A more precise block diagram representation of a system can be easily drawn since only two operations are involved: 1. Multiplying the terms of a sequence by a constant. 2. Shifting to the right, or delaying, the terms of the sequence. A system which consists of a single multiplier is denoted as shown by a triangular symbol: {x n } {y n } A y n = Ax n Figure 9 As we have seen earlier in this workbook a system which consists of only a single delay unit is represented symbolically as follows {x n } {y n } 1 y n = x n 1 Figure 10 The system represented by the difference equation (1) consists of two multipliers and one delay unit. Because (1) can be written y n = bx n ay n 1 a symbolic representation of (1) is as shown in Figure 11. 66 HELM (2008): Workbook 21: -Transforms

{x n } {y n } + b + 1 a Figure 11 The circle symbol denotes an adder or summation unit whose output is the sum of the two (or more) sequences that are input to it. We will now concentrate upon the ero state response of the system i.e. we will assume that the initial condition y 1 is ero. Thus, using (2), so Y () = Y () X() = bx() 1+a 1 b 1+a 1 The quantity Y (), the ratio of the output -transform to the input -transform, is called the X() transfer function of the discrete system. It is often denoted by H(). (3) Key Point 16 The transfer function H() of a discrete system is defined by H() = Y () X() when the initial conditions are ero. = -transform of output sequence -transform of input sequence HELM (2008): Section 21.4: Engineering Applications of -Transforms 67

Task (a) Write down the transfer function H() of the system represented by (1) (i) using negative powers of (ii) using positive powers of. (b) Write down the inverse -transform of H(). (a) From (3) (i) H() = (ii) H() = b 1+a 1 b + a (b) Referring to the Table of -transforms at the end of the Workbook: {h n } = b( a) n n =0, 1, 2,... We can represent any discrete system as follows {x n } {y n } H() X() Y () Figure 12 From the definition of the transfer function it follows that Y () =X()H() (at ero initial conditions). The corresponding relation between {y n }, {x n } and the inverse -transform {h n } of the transfer function will be discussed later; it is called a convolution summation. The significance of {h n } is readily obtained. 1 n =0 Suppose {x n } = 0 n =1, 2, 3,... i.e. {x n } is the unit impulse sequence that is normally denoted by δ n. Hence, in this case, X() =Z{δ n } =1 so Y () =H() and {y n } = {h n } In words: {h n } is the response or output of a system where the input is the unit impulse sequence {δ n }. Hence {h n } is called the unit impulse response of the system. 68 HELM (2008): Workbook 21: -Transforms

Key Point 17 For a linear, time invariant discrete system, the unit impulse response and the system transfer function are a -transform pair: H() =Z{h n } {h n } = Z 1 {H()} It follows from the previous Task that for the first order system (1) H() = b 1+a 1 = b + a is the transfer function and {h n } = {b( a) n } is the unit impulse response sequence. Task Write down the transfer function of (a) a single multiplier unit (b) a single delay unit. (a) {y n } = {A x n } if the multiplying factor is A using the linearity property of -transform so Y () =AX() H() = Y () X() = A (b) {y n } = {x n 1 } is the required transfer function. so Y () = 1 X() (remembering that initial conditions are ero) H() = 1 is the transfer function of the single delay unit. HELM (2008): Section 21.4: Engineering Applications of -Transforms 69

Task Obtain the transfer function of the system. y n + a 1 y n 1 = b 0 x n + b 1 x n 1 n =0, 1, 2,... where {x n } is a known sequence with x n =0for n = 1, 2,... [Remember that the transfer function is only defined at ero initial condition i.e. assume y 1 =0also.] Taking -transforms Y ()+a 1 1 Y () = b 0 X()+b 1 1 X() Y ()(1 + a 1 1 ) = (b 0 + b 1 1 )X() so the transfer function is H() = Y () X() = b 0 + b 1 1 1+a 1 1 = b 0 + b 1 + a 1 Second order systems Consider the system whose difference equation is y n + a 1 y n 1 + a 2 y n 2 = bx n n =0, 1, 2,... (4) where the input sequence x n =0, n = 1, 2,... In exactly the same way as for first order systems it is easy to show that the system response has a -transform with two components. 70 HELM (2008): Workbook 21: -Transforms

Task Take the -transform of (4), assuming given initial values y 1, y 2. Show that Y () has two components. Obtain the transfer function of the system (4). From (4) Y () = Y ()+a 1 ( 1 Y ()+y 1 )+a 2 ( 2 Y ()+ 1 y 1 + y 2 )=bx() Y ()(1 + a 1 1 + a 2 2 )+a 1 y 1 + a 2 1 y 1 + a 2 y 2 = bx() bx() 1+a 1 1 + a 2 (a 1y 1 + a 2 1 y 1 + a 2 y 2 ) = Y 2 1+a 1 1 + a 2 2 1 ()+Y 2 () say. At ero initial conditions, Y () =Y 1 () so the transfer function is H() = b 1+a 1 1 + a 2 2 = b 2 2 + a 1 + a 2. Example Obtain (i) the unit impulse response (ii) the unit step response of the system specified by the second order difference equation y n 3 4 y n 1 + 1 8 y n 2 = x n (5) Note that both these responses refer to the case of ero initial conditions. Hence it is convenient to first obtain the transfer function H() of the system and then use the relation Y () =X()H() in each case. We write down the transfer function of (5), using positive powers of. Taking the -transform of (5) at ero initial conditions we obtain Y () 3 4 1 Y ()+ 1 8 2 Y () = X() Y () 1 34 1 + 18 2 = X() H() = Y () X() = 2 2 3 4 + 1 8 = 2 ( 1 2 )( 1 4 ) HELM (2008): Section 21.4: Engineering Applications of -Transforms 71

We now complete the problem for inputs (i) x n = δ n (ii) x n = u n, the unit step sequence, using partial fractions. H() = 2 = 1 2 1 4 2 1 2 1 4 (i) With x n = δ n so X() =1the response is, as we saw earlier, Y () =H() so y n = h n n n 1 1 where h n = Z 1 H() =2 n =0, 1, 2,... 2 4 (ii) The -transform of the unit step is so the unit step response has -transform 1 Y () = 2 1 2 1 4 = 2 1 2 + 1 3 1 4 ( 1) + 8 3 1 Hence, taking inverse -transforms, the unit step response of the system is n 1 y n =( 2) + 1 n 1 2 3 + 8 n =0, 1, 2,... 4 3 Notice carefully the form of this unit step response - the first two terms decrease as n increases and are called transients. Thus y n 8 3 as n and the term 8 3 is referred to as the steady state part of the unit step response. Combinations of systems The concept of transfer function enables us to readily analyse combinations of discrete systems. Series combination Suppose we have two systems S 1 and S 2 with transfer functions H 1 (), H 2 () in series with each other. i.e. the output from S 1 is the input to S 2. {x n } {y S 1 (n)} = {x 2 (n)} {y n } 1 S 2 H 1 () H 2 () X() Y 1 () = X 2 () Y () Figure 13 72 HELM (2008): Workbook 21: -Transforms

Clearly, at ero initial conditions, Y 1 () = H 1 ()X() Y () = H 2 ()X 2 () = H 2 ()Y 1 () Y () = H 2 ()H 1 ()X() so the ratio of the final output transform to the input transform is Y () X() = H 2() H 1 () i.e. the series system shown above is equivalent to a single system with transfer function H 2 () H 1 () {x n } {y n } H 1 ()H 2 () X() Y () Figure 14 (6) Task Obtain (a) the transfer function (b) the governing difference equation of the system obtained by connecting two first order systems S 1 and S 2 in series. The governing equations are: S 1 : S 2 : y n ay n 1 = bx n y n cy n 1 = dx n (a) Begin by finding the transfer function of S 1 and S 2 and then use (6): HELM (2008): Section 21.4: Engineering Applications of -Transforms 73

S 1 : Y () a 1 Y () =bx() so H 1 () = S 2 : H 2 () = d 1 c 1 so the series arrangement has transfer function b 1 a 1 H() = = bd (1 a 1 )(1 c 1 ) bd 1 (a + c) 1 + ac 2 If X() and Y () are the input and output transforms for the series arrangement, then Y () =H() X() = bdx() 1 (a + c) 1 + ac 2 (b) By transfering the denominator from the right-hand side to the left-hand side and taking inverse -transforms obtain the required difference equation of the series arrangement: We have Y ()(1 (a + c) 1 + ac 2 )=bdx() Y () (a + c) 1 Y ()+ac 2 Y () =bdx() from which, using the right shift theorem, y n (a + c)y n 1 + acy n 2 = bd x n. which is the required difference equation. You can see that the two first order systems in series have an equivalent second order system. 74 HELM (2008): Workbook 21: -Transforms

Feedback combination {x n } X() + + {w n } W () H 1 () Y () 1 H 2 () Figure 15 For the above negative feedback arrangement of two discrete systems with transfer functions H 1 (), H 2 () we have, at ero initial conditions, Y () =W ()H 1 () where W () =X() H 2 ()Y () Task Eliminate W () and hence obtain the transfer function of the feedback system. so Y () = (X() H 2 ()Y ())H 1 () = X()H 1 () H 2 ()H 1 ()Y () Y ()(1 + H 2 ()H 1 ()) = X()H 1 () Y () X() = H 1 () 1+H 2 ()H 1 () This is the required transfer function of the negative feedback system. HELM (2008): Section 21.4: Engineering Applications of -Transforms 75

2. Convolution and -transforms Consider a discrete system with transfer function H() {x n } {y n } H() X() Y () Figure 16 We know, from the definition of the transfer function that at ero initial conditions Y () =X()H() We now investigate the corresponding relation between the input sequence {x n } and the output sequence {y n }. We have seen earlier that the system itself can be characterised by its unit impulse response {h n } which is the inverse -transform of H(). We are thus seeking the inverse -transform of the product X()H(). We emphasie immediately that this is not given by the product {x n }{h n }, a point we also made much earlier in the workbook. We go back to basic definitions of the -transform: Y () =y 0 + y 1 1 + y 2 2 + y 3 3 +... X() =x 0 + x 1 1 + x 2 2 + x 3 3 +... H() =h 0 + h 1 1 + h 2 2 + h 3 3 +... Hence, multiplying X() by H() we obtain, collecting the terms according to the powers of 1 : x 0 h 0 +(x 0 h 1 + x 1 h 0 ) 1 +(x 0 h 2 + x 1 h 1 + x 2 h 0 ) 2 +... (7) Task Write out the terms in 3 in the product X()H() and, looking at the emerging pattern, deduce the coefficient of n. 76 HELM (2008): Workbook 21: -Transforms

(x 0 h 3 + x 1 h 2 + x 2 h 1 + x 3 h 0 ) 3 which suggests that the coefficient of n is x 0 h n + x 1 h n 1 + x 2 h n 2 +...+ x n 1 h 1 + x n h 0 Hence, comparing corresponding terms in Y () and X()H() 0 : y 0 = x 0 h 0 1 : y 1 = x 0 h 1 + x 1 h 0 2 : y 2 = x 0 h 2 + x 1 h 1 + x 2 h 0 3 : y 3 = x 0 h 3 + x 1 h 2 + x 2 h 1 + x 3 h 0 (8).. n : y n = x 0 h n + x 1 h n 1 + x 2 h n 2 +...+ x n 1 h 1 + x n h 0 (9) = = n x k h n k k=0 n h k x n k k=0 (Can you see why (10b) also follows from (9)?) (10a) (10b) The sequence {y n } whose n th term is given by (9) and (10) is said to be the convolution (or more precisely the convolution summation) of the sequences {x n } and {h n }, The convolution of two sequences is usually denoted by an asterisk symbol ( ). We have shown therefore that Z 1 {X()H()} = {x n } {h n } = {h n } {x n } where the general term of {x n } {h n } is in (10a) and that of {h n } {x n } is in (10b). In words: the output sequence {y n } from a linear time invariant system is given by the convolution of the input sequence with the unit impulse response sequence of the system. This result only holds if initial conditions are ero. HELM (2008): Section 21.4: Engineering Applications of -Transforms 77

Key Point 18 {x n } {y n } H() X() Y () Figure 17 We have, at ero initial conditions Y () =X()H() (definition of transfer function) {y n } = {x n } {h n } (convolution summation) where y n is given in general by (9) and (10) with the first four terms written out explicitly in (8). Although we have developed the convolution summation in the context of linear systems the proof given actually applies to any sequences i.e. for arbitrary causal sequences say {v n }{w n } with - transforms V () and W () respectively: Z 1 {V ()W ()} = {v n } {w n } or, equivalently, Z({v n } {w n })=V ()W (). Indeed it is simple to prove this second result from the definition of the -transform for any causal sequences {v n } = {v 0,v 1,v 2,...} and {w n } = {w 0,w 1,w 2,...} n Thus since the general term of {v n } {w n } is v k w n k we have Z({v n } {w n })= n v k w n k n=0 k=0 or, since w n k =0if k>n, Z({v n } {w n })= v k w n k n n=0 k=0 Putting m = n k or n = m + k we obtain Z({v n } {w n })= v k w m (m+k) Finally, Z({v n } {w n })= m=0 k=0 w m m m=0 which completes the proof. k=0 n k=0 v k k = W ()V () (Why is the lower limit m =0correct?) 78 HELM (2008): Workbook 21: -Transforms

Example 2 Calculate the convolution {y n } of the sequences {v n } = {a n } {w n } = {b n } a = b (i) directly (ii) using -transforms. Solution (i) We have from (10) n n y n = v k w n k = a k b n k k=0 k=0 n a k = b n b k=0 a a 2 a n = b 1+ n + +... b b b The bracketed sum involves n +1terms of a geometric series of common ratio a b. a n+1 1 y n = b n b 1 a b = (bn+1 a n+1 ) (b a) (ii) The -transforms are V () = a W () = b so 2 y n = Z 1 { ( a)( b) } = bn+1 a n+1 (b a) using partial fractions or residues HELM (2008): Section 21.4: Engineering Applications of -Transforms 79

Task Obtain by two methods the convolution of the causal sequence {2 n } = {1, 2, 2 2, 2 3,...} with itself. (a) By direct use of (10) if {y n } = {2 n } {2 n } n n y n = 2 k 2 n k =2 n 1=(n + 1)2 n k=0 (b) Using -transforms: Z{2 n } = 2 2 k=0 so {y n } = Z 1 { ( 2) } 2 We will find this using the residue method. Y () n 1 has a second order pole at =2. n+1 y n = Res ( 2), 2 2 d = d n+1 =(n + 1)2 n 2 80 HELM (2008): Workbook 21: -Transforms

3. Initial and final value theorems of -transforms These results are important in, for example, Digital Control Theory where we are sometimes particularly interested in the initial and ultimate behaviour of systems. Initial value theorem. If f n is a sequence with -transform F () then the initial value f 0 is given by f 0 = lim F () (provided, of course, that this limit exists). This result follows, at least informally, from the definition of the -transform: F () =f 0 + f 1 1 + f 2 2 +... from which, taking limits as the required result is obtained. Task Obtain the -transform of f(n) =1 a n, 0 <a<1 Verify the initial value theorem for the -transform pair you obtain. Using standard -transforms we obtain Z{f n } = F () = = 1 a 1 1 1 1 1 a 1 hence, as : F () 1 1=0 Similarly, as n 0 f n 1 1=0 so the initial value theorem is verified for this case. HELM (2008): Section 21.4: Engineering Applications of -Transforms 81

Final value theorem Suppose again that {f n } is a sequence with -transform F (). We further assume that all the poles of F () lie inside the unit circle in the plane (i.e. have magnitude less than 1) apart possibly from a first order pole at =1. The final value of f n i.e. lim n f n is then given by Proof: Recalling the left shift property we have Z{f n+1 } = F() f 0 Z{f n+1 f n } = lim k k (f n+1 f n ) n = F() f 0 F () n=0 or, alternatively, dividing through by on both sides: (1 1 )F () f 0 = lim k k (f n+1 f n ) (n+1) n=0 Hence (1 1 )F () =f 0 +(f 1 f 0 ) 1 +(f 2 f 1 ) 2 +... or as 1 lim f n = lim(1 1 )F () n 1 lim 1 1 )F () = f 0 +(f 1 f 0 )+(f 2 f 1 )+... = lim f k k Example Again consider the sequence f n =1 a n 0 <a<1 and its -transform F () = 1 a = 1 1 1 1 1 a 1 Clearly as n then f n 1. Considering the right-hand side (1 1 )F () =1 (1 1 ) 1 0=1 as 1. 1 a 1 Note carefully that F () = 1 a has a pole at a (0 <a<1) and a simple pole at =1. The final value theorem does not hold for -transform poles outside the unit circle e.g. f n =2 n F () = 2 Clearly f n as n whereas 1 (1 1 )F () = ( 2) 0 as 1 82 HELM (2008): Workbook 21: -Transforms

1. A low pass digital filter is characterised by Exercises y n =0.1x n +0.9y n 1 Two such filters are connected in series. Deduce the transfer function and governing difference equation for the overall system. Obtain the response of the series system to (i) a unit step and (ii) a unit alternating input. Discuss your results. 2. The two systems y n = x n 0.7x n 1 +0.4y n 1 y n =0.9x n 1 0.7y n 1 are connected in series. Find the difference equation governing the overall system. 3. A system S 1 is governed by the difference equation y n =6x n 1 +5y n 1 It is desired to stabilise S 1 by using a feedback configuration. The system S 2 in the feedback loop is characterised by y n = αx n 1 + βy n 1 Show that the feedback system S 3 has an overall transfer function H 3 () = H 1 () 1+H 1 ()H 2 () and determine values for the parameters α and β if H 3 () is to have a second order pole at =0.5. Show briefly why the feedback systems S 3 stabilies the original system. 4. Use -transforms to find the sum of squares of all integers from 1 to n: y n = n k=1 k 2 [Hint: y n y n 1 = n 2 ] 5. Evaluate each of the following convolution summations (i) directly (ii) using -transforms: (a) a n b n a = b (b) a n a n (c) δ n 3 δ n 5 1 n =0, 1, 2, 3 (d) x n x n where x n = 0 n =4, 5, 6, 7... HELM (2008): Section 21.4: Engineering Applications of -Transforms 83

s 1. Step response: y n =1 (0.99)(0.9) n 0.09n(0.9) n Alternating response: y n = 1 361 ( 1)n + 2.61 361 (0.9)n + 1.71 361 n(0.9)n 2. y n +0.3y n 1 0.28y n 2 =0.9x n 1 0.63x n 2 3. α =3.375 β = 4 4. n k 2 = k=1 5. (a) (2n + 1)(n +1)n 6 1 (a b) (an+1 b n+1 ) (b) (n +1)a n (c) δ n 8 (d) {1, 2, 3, 4, 3, 2, 1} 84 HELM (2008): Workbook 21: -Transforms

1. Solution of difference equations using -transforms Using -transforms, in particular the shift theorems discussed at the end of the previous Section, provides a useful method of solving certain types of difference equation. In particular linear constant coefficient difference equations are amenable to the -transform technique although certain other types can also be tackled. In fact all the difference equations that we looked at in Section 21.1 were linear: y n+1 = y n + d y n+1 = Ay n y n+2 = y n+1 + y n (1st order) (1st order) (2nd order) Other examples of linear difference equations are y n+2 +4y n+1 3y n = n 2 y n+1 + y n = n 3 n (2nd order) (1st order) The key point is that for a difference equation to be classified as linear the terms of the sequence {y n } arise only to power 1 or, more precisely, the highest subscript term is obtainable as a linear combination of the lower ones. All the examples cited above are consequently linear. Note carefully that the term n 2 in our fourth example does not imply non-linearity since linearity is determined by the y n terms. Examples of non-linear difference equations are y n+1 = y n +1 y 2 n+1 +2y n = 3 y n+1 y n = n cos(y n+1 ) = y n We shall not consider the problem of solving non-linear difference equations. The five linear equations listed above also have constant coefficients; for example: y n+2 +4y n+1 3y n = n 2 has the constant coefficients 1, 4, 3. The (linear) difference equation ny n+2 y n+1 + y n =0 has one variable coefficient vi n and so is not classified as a constant coefficient difference equation. Solution of first order linear constant coefficient difference equations Consider the first order difference equation y n+1 3y n =4 n =0, 1, 2,... The equation could be solved in a step-by-step or recursive manner, provided that y 0 is known because y 1 =4+3y 0 y 2 =4+3y 1 y 3 =4+3y 2 and so on. This process will certainly produce the terms of the solution sequence {y n } but the general term y n may not be obvious. HELM (2008): Section 21.3: -Transforms and Difference Equations 37

So consider y n+1 3y n =4 n =0, 1, 2,... (1) with initial condition y 0 =1. We multiply both sides of (1) by n and sum each side over all positive integer values of n and ero. We obtain (y n+1 3y n ) n = 4 n or n=0 y n+1 n 3 n=0 n=0 y n n =4 n=0 n (2) n=0 The three terms in (2) are clearly recognisable as -transforms. 4 The right-hand side is the -transform of the constant sequence {4, 4,...} which is 1. If Y () = y n n denotes the -transform of the sequence {y n } that we are seeking then n=0 y n+1 n = Y() y 0 (by the left shift theorem). n=0 Consequently (2) can be written Y() y 0 3 Y () = 4 1 Equation (3) is the -transform of the original difference equation (1). The intervening steps have been included here for explanation purposes but we shall omit them in future. The important point is that (3) is no longer a difference equation. It is an algebraic equation where the unknown, Y (), is the -transform of the solution sequence {y n }. We now insert the initial condition y 0 =1and solve (3) for Y (): (3) ( 3)Y () = ( 3)Y () = 4 ( 1) 4 1 + = 2 +3 1 so Y () = 2 +3 ( 1)( 3) The final step consists of obtaining the sequence {y n } of which (4) is the -transform. As it stands (4) is not recogniable as any of the standard transforms that we have obtained. Consequently, one method of inverting (4) is to use a partial fraction expansion. (We assume that you are familiar with simple partial fractions. See 3.6) (4) 38 HELM (2008): Workbook 21: -Transforms

Thus Y () = ( +3) ( 1)( 3) = 2 1 + 3 3 (in partial fractions) so Y () = 2 1 + 3 3 Now, taking inverse -transforms, the general term y n is, using the linearity property, y n = 2Z 1 { 1 } +3Z 1 { 3 } The symbolic notation Z 1 is common and is short for the inverse -transform of. Task Using standard -transforms write down y n explicitly, where y n = 2Z 1 { 1 } +3Z 1 { 3 } y n = 2+3 3 n = 2+3 n+1 n =0, 1, 2,... (5) Checking the solution: From this solution (5) y n = 2+3 n+1 we easily obtain y 0 = 2+3=1 y 1 = 2+3 2 =7 y 2 = 2+3 3 = 25 y 3 = 2+3 4 = 79 (as given) etc. HELM (2008): Section 21.3: -Transforms and Difference Equations 39

These agree with those obtained by recursive solution of the given problem (1): y n+1 3y n =4 y 0 =1 which yields y 1 =4+3y 0 =7 y 2 =4+3y 1 = 25 y 3 =4+3y 2 = 79 etc. More conclusively we can put the solution (5) back into the left-hand side of the difference equation (1). If y n = 2+3 n+1 then 3y n = 6+3 n+2 and y n+1 = 2+3 n+2 So, on the left-hand side of (1), y n+1 3y n = 2+3 n+2 ( 6+3 n+2 ) which does indeed equal 4, the given right-hand side, and so the solution has been verified. Key Point 13 To solve a linear constant coefficient difference equation, three steps are involved: 1. Replace each term in the difference equation by its -transform and insert the initial condition(s). 2. Solve the resulting algebraic equation. (Thus gives the -transform Y () of the solution sequence.) 3. Find the inverse -transform of Y (). The third step is usually the most difficult. We will consider the problem of finding inverse - transforms more fully later. 40 HELM (2008): Workbook 21: -Transforms

Task Solve the difference equation y n+1 y n = d n =0, 1, 2,... y 0 = a (6) where a and d are constants. (The solution will give the n th term of an arithmetic sequence with a constant difference d and initial term a.) Start by replacing each term of (6) by its -transform: If Y () =Z{y n } we obtain the algebraic equation Y() y 0 Y () = d ( 1) Note that the right-hand side transform is that of a constant sequence {d, d,...}. Note also the use of the left shift theorem. Now insert the initial condition y 0 = a and then solve for Y (): HELM (2008): Section 21.3: -Transforms and Difference Equations 41

( 1)Y () = d ( 1) + a Y () = d ( 1) 2 + a 1 Finally take the inverse -transform of the right-hand side. [Hint: Recall the -transform of the ramp sequence {n}.] We have y n = d Z 1 { ( 1) } + a 2 Z 1 { 1 } y n = dn + a n =0, 1, 2,... (7) using the known -transforms of the ramp and unit step sequences. Equation (7) may well be a familiar result to you an arithmetic sequence whose eroth term is y 0 = a has general term y n = a + nd. i.e. {y n } = {a, a + d,... a + nd,...} This solution is of course readily obtained by direct recursive solution of (6) without need for - transforms. In this case the general term (a + nd) is readily seen from the form of the recursive solution: (Make sure you really do see it). N.B. If the term a is labelled as the first term (rather than the eroth) then y 1 = a, y 2 = a + d, y 3 a +2d, so in this case the n th term is y n = a +(n 1)d rather than (7). 42 HELM (2008): Workbook 21: -Transforms

Use of the right shift theorem in solving difference equations The problem just solved was given by (6), i.e. y n+1 y n = d with y 0 = a n =0, 1, 2,... We obtained the solution y n = a + nd n =0, 1, 2,... Now consider the problem y n y n 1 = d n =0, 1, 2,... (8) with y 1 = a. The only difference between the two problems is that the initial condition in (8) is given at n = 1 rather than at n =0. Writing out the first few terms should make this clear. (6) (8) y 1 y 0 = d y 0 y 1 = d y 2 y 1 = d y 1 y 0 = d. y n+1 y n = d y 0 = a. y n y n 1 = d y 1 = a The solution to (8) must therefore be the same as for (6) but with every term in the solution (7) of (6) shifted 1 unit to the left. Thus the solution to (8) is expected to be y n = a +(n +1)d n = 1, 0, 1, 2,... (replacing n by (n +1)in the solution (7)). Task Use the right shift theorem of -transforms to solve (8) with the initial condition y 1 = a. (a) Begin by taking the -transform of (8), inserting the initial condition and solving for Y (): HELM (2008): Section 21.3: -Transforms and Difference Equations 43

We have, for the -transform of (8) Y () ( 1 Y ()+y 1 ) = Y ()(1 1 ) a = 1 Y () = Y () = d 1 d 1 d ( 1) + a d 2 ( 1) + a 2 1 [Note that here d means d ] (9) The second term of Y () has the inverse -transform {a u n } = {a, a, a,...}. The first term is less straightforward. However, we have already reasoned that the other term in y n here should be (n +1)d. (b) Show that the -transform of (n +1)d is step: d 2. Use the standard transform of the ramp and ( 1) 2 We have Z{(n +1)d} = dz{n} + dz{1} by the linearity property Z{(n +1)d} = d ( 1) + d 2 1 = 1+ 1 d ( 1) 2 = d 2 ( 1) 2 as expected. 44 HELM (2008): Workbook 21: -Transforms

(c) Finally, state y n : Returning to (9) the inverse -transform is y n =(n +1)d + au n i.e. y n = a +(n +1)d n = 1, 0, 1, 2,... as we expected. Task Earlier in this Section (pages 37-39) we solved y n+1 3y n =4 n =0, 1, 2,... with y 0 =1. Now solve y n 3y n 1 =4 n =0, 1, 2,... with y 1 =1. (10) Begin by obtaining the -transform of y n : We have, taking the -transform of (10), Y () 3( 1 Y ()+1)= 4 1 (using the right shift property and inserting the initial condition.) Y () 3 1 Y () = 3+ 4 1 ( 3) Y () = 3+ 4 1 so Y () = 3 3 + 4 2 ( 1)( 3) HELM (2008): Section 21.3: -Transforms and Difference Equations 45

Write the second term as 4 and obtain the partial fraction expansion of the ( 1)( 3) bracketed term. Then complete the -transform inversion. ( 1)( 3) = 1 3 2 1 + 2 3 We now have so Y () = 3 3 2 1 + 6 3 y n =3 3 n 2+6 3 n = 2+9 3 n = 2+3 n+2 (11) Compare this solution (11) to that of the previous problem (5) on page 39: Solution (11) is just the solution sequence (5) moved 1 unit to the left. We anticipated this since the difference equation (10) and associated initial condition is the same as the difference equation (1) but shifted one unit to the left. 46 HELM (2008): Workbook 21: -Transforms

2. Second order difference equations You will learn in this section about solving second order linear constant coefficient difference equations. In this case two initial conditions are required, typically either y 0 and y 1 or y 1 and y 2. In the first case we use the left shift property of the -transform, in the second case we use the right shift property. The same three basic steps are involved as in the first order case. Task By solving y n+2 = y n+1 + y n (12) y 0 = y 1 =1 obtain the general term y n of the Fibonacci sequence. Begin by taking the -transform of (12), using the left shift property. Then insert the initial conditions and solve the resulting algebraic equation for Y (), the -transform of {y n }: 2 Y () 2 y 0 y 1 = Y () y 0 + Y () (taking -transforms ) so 2 Y () 2 = Y () + Y () ( 2 1)Y () = 2 Y () = 2 2 1 (inserting initial conditions) (solving for Y ()). HELM (2008): Section 21.3: -Transforms and Difference Equations 47

Now solve the quadratic equation 2 1=0 and hence factorie the denominator of Y (): 2 1=0 = 1 ± 1+4 2 so if a = 1+ 5, b = 1 5 2 2 Y () = 2 ( a)( b) = 1 ± 5 2 This form for Y () often arises in solving second order difference equations. fractions and find y n, leaving a and b as general at this stage: Write it in partial Y () = ( a)( b) where A = a a b and B = b b a Hence, taking inverse -transforms y n = Aa n + Bb n = = A a + B ( b) in partial fractions 1 (a b) (an+1 b n+1 ) (13) 48 HELM (2008): Workbook 21: -Transforms

Now complete the Fibonacci problem: With a = 1+ 5 2 we obtain, using (13) y n = 1 1+ 5 5 2 b = 1 5 2 n+1 so a b = 5 1 n 5 n =2, 3, 4,... 2 for the n th term of the Fibonacci sequence. With an appropriate computational aid you could (i) check that this formula does indeed give the familiar sequence {1, 1, 2, 3, 5, 8, 13,...} and (ii) obtain, for example, y 50 and y 100. The inverse -transform of Y () = 2 ( a)( b) Key Point 14 a = b is y n = 1 (a b) (an+1 b n+1 ) HELM (2008): Section 21.3: -Transforms and Difference Equations 49

Task Use the right shift property of -transforms to solve the second order difference equation y n 7y n 1 +10y n 2 =0 with y 1 = 16 and y 2 =5. [Hint: the steps involved are the same as in the previous Task] Y () 7( 1 Y () + 16) + 10( 2 Y ()+16 1 +5)=0 Y ()(1 7 1 +10 2 ) 112 + 160 1 +50=0 2 7 +10 Y () 2 = 62 160 1 Y () = 62 2 2 7 +10 160 2 7 +10 (62 160) = ( 2)( 5) = 12 2 + 50 5 in partial fractions so y n = 12 2 n +50 5 n n =0, 1, 2,... We now give an Example where a quadratic equation with repeated solutions arises. 50 HELM (2008): Workbook 21: -Transforms

Example 1 (a) Obtain the -transform of {f n } = {na n }. (b) Solve y n 6y n 1 +9y n 2 =0 n =0, 1, 2,... y 1 =1 y 2 =0 [Hint: use the result from (a) at the inversion stage.] Solution (a) Z{n} = Z{na n } = ( 1) 2 property Z{f n a n } = F a /a (/a 1) 2 = a where we have used the ( a) 2 (b) Taking the -transform of the difference equation and inserting the initial conditions: Y () 6( 1 Y () + 1) + 9( 2 Y ()+ 1 )=0 Y ()(1 6 1 +9 2 )=6 9 1 Y ()( 2 6 +9)=6 2 9 Y () = 62 9 6 9 ( 3) = = 2 ( 3) 2 6 3 + 9 ( 3) 2 from which, using the result (a) on the second term, y n =6 3 n +3n 3 n = (6 + 3n)3 n We shall re-do this inversion by an alternative method shortly. in partial fractions Task Solve the difference equation y n+2 + y n =0 with y 0,y 1 arbitrary. (14) Start by obtaining Y () using the left shift theorem: HELM (2008): Section 21.3: -Transforms and Difference Equations 51

2 Y () 2 y 0 y 1 + Y () = 0 ( 2 +1)Y () = 2 y 0 + y 1 Y () = 2 2 +1 y 0 + 2 +1 y 1 To find the inverse -transforms recall the results for Z{cos ωn} and Z{sin ωn} from Key Point 6 (page 21) and some of the particular cases discussed in Section 21.2. Hence find y n here: Taking Z{cos ωn} and Z{sin ωn} with ω = π 2 nπ Z cos 2 nπ Z sin 2 = = 2 2 +1 2 +1 Hence y n = y 0 Z 1 { 2 2 +1 } + y 1Z 1 nπ { 2 +1 } = y 0 cos 2 nπ + y 1 sin 2 Those of you who are familiar with differential equations may know that d 2 y dt + y =0 y(0) = y 0, y (0) = y 2 0 (16) has solutions y 1 =cost and y 2 = sin t and a general solution y = c 1 cos t + c 2 sin t (17) where c 1 = y 0 and c 2 = y 0. This differential equation is a model for simple harmonic oscillations. The difference equation (14) and its solution (15) are the discrete counterparts of (16) and (17). (15) 52 HELM (2008): Workbook 21: -Transforms

3. Inversion of -transforms using residues This method has its basis in a branch of mathematics called complex integration. You may recall that the quantity of -transforms is a complex quantity, more specifically a complex variable. However, it is not necessary to delve deeply into the theory of complex variables in order to obtain simple inverse -transforms using what are called residues. In many cases inversion using residues is easier than using partial fractions. Hence reading on is strongly advised. Pole of a function of a complex variable If G() is a function of the complex variable and if G() = G 1() ( 0 ) k where G 1 ( 0 ) is non-ero and finite then G() is said to have a pole of order k at = 0. For example if G() = 6( 2) ( 3)( 4) 2 then G() has the following 3 poles. (i) pole of order 1 at =0 (ii) pole of order 1 at =3 (iii) pole of order 2 at =4. (Poles of order 1 are sometimes known as simple poles.) Note that when =2, G() =0. Hence =2is said to be a ero of G(). (It is the only ero in this case). Task Write down the poles and eros of G() = 3( +4) 2 (2 + 1)(3 9) State the order of each pole. (18) G() has a ero when = 4. G() has first order poles at = 1/2, =3. G() has a second order pole at =0. HELM (2008): Section 21.3: -Transforms and Difference Equations 53

Residue at a pole The residue of a complex function G() at a first order pole 0 is Res (G(), 0 )=[G()( 0 )] 0 (19) The residue at a second order pole 0 is d Res (G(), 0 )= d (G()( 0) 2 ) (20) 0 You need not worry about how these results are obtained or their full mathematical significance. (Any textbook on Complex Variable Theory could be consulted by interested readers.) Example Consider again the function (18) in the previous guided exercise. G() = = 3( +4) 2 (2 + 1)(3 9) ( +4) 2 2 + 1 2 ( 3) The second form is the more convenient for the residue formulae to be used. Using (19) at the two first order poles: Res G(), 1 2 = = Res [G(), 3] = G() 1 2 1 2 ( +4) = 18 2 2 ( 3) 1 2 ( +4) 2 2 + 1 2 Using (20) at the second order pole d Res (G(), 0) = d (G()( 0)2 ) The differentiation has to be carried out before the substitution of =0of course. Res (G(), 0) = = 1 2 0 3 5 = 1 9 d ( +4) d 2 + 1 ( 3) 2 d +4 d 2 5 2 3 2 0 0 54 HELM (2008): Workbook 21: -Transforms

Task Carry out the differentiation shown on the last line of the previous page, then substitute =0and hence obtain the required residue. Differentiating by the quotient rule then substituting =0gives Res (G(), 0) = 17 9 If G() has a k th order pole at = 0 Key Point 15 Residue at a Pole of Order k i.e. G() = G 1() G ( 0 ) k 1 ( 0 ) = 0and finite 1 d k 1 Res (G(), 0 )= (k 1)! d (G() ( 0) k ) (21) k 1 0 This formula reduces to (19) and (20) when k =1and 2 respectively. HELM (2008): Section 21.3: -Transforms and Difference Equations 55

Inverse -transform formula Recall that, by definition, the -transform of a sequence {f n } is F () =f 0 + f 1 1 + f 2 2 +...f n n +... If we multiply both sides by n 1 where n is a positive integer we obtain F () n 1 = f 0 n 1 + f 1 n 2 + f 2 n 3 +...f n 1 + f n+1 2 +... Using again a result from complex integration it can be shown from this expression that the general term f n is given by f n = sum of residues of F () n 1 at its poles (22) The poles of F () n 1 will be those of F () with possibly additional poles at the origin. To illustrate the residue method of inversion we shall re-do some of the earlier examples that were done using partial fractions. Example: so Y () = 2 ( a)( b) n+1 a = b Y () n 1 = = G(), say. ( a)( b) G() has first order poles at = a, = b so using (19). Res (G(), a) = Res (G(), b) = n+1 b n+1 a a b = an+1 a b = bn+1 b a = bn+1 a b We need simply add these residues to obtain the required inverse -transform f n = as before. 1 (a b) (an+1 b n+1 ) 56 HELM (2008): Workbook 21: -Transforms

Task Obtain, using (22), the inverse -transform of Y () = 62 9 ( 3) 2 Firstly, obtain the pole(s) of G() =Y () n 1 and deduce the order: G() =Y () n 1 = 6n+1 9 n ( 3) 2 whose only pole is one of second order at =3. Now calculate the residue of G() at =3using (20) and hence write down the required inverse -transform y n : HELM (2008): Section 21.3: -Transforms and Difference Equations 57

Res (G(), 3) = d d (6n+1 9 n ) 3 = 6(n +1) n 9n n 1 3 = 6(n + 1)3 n 9n3 n 1 = 6 3 n +3n3 n This is the same as was found by partial fractions, but there is considerably less labour by the residue method. In the above examples all the poles of the various functions G() were real. situation but the residue method will cope with complex poles. Example We showed earlier that 2 2 +1 and cos nπ 2 formed a -transform pair. This is the easiest We will now obtain y n if Y () = Using residues with, from (22), 2 2 +1 using residues. G() = n+1 2 +1 = n+1 ( i)( + i) where i2 = 1. we see that G() has first order poles at the complex conjugate points ± i. Using (19) n+1 Res (G(), i)= = in+1 Res (G(), i) = ( i)n+1 + i 2i ( 2i) i (Note the complex conjugate residues at the complex conjugate poles.) Hence Z 1 { 2 2 +1 } = 1 i n+1 ( i) n+1 2i But i = e iπ/2 and i = e iπ/2, so the inverse -transform is 1 e i(n+1)π/2 e i(n+1)π/2 = sin(n +1) π nπ 2i 2 =cos 2 as expected. 58 HELM (2008): Workbook 21: -Transforms

Task Show, using residues, that Z 1 nπ { 2 +1 } = sin 2 Using (22): G() = n 1 2 +1 = n 2 +1 = n ( + i)( i) Res (G(), i) = in 2i Res (G(), i) = ( i)n 2i Z 1 { 2 +1 } = 1 2i (in ( i) n ) = 1 2i (einπ/2 e inπ/2 ) nπ = sin 2 HELM (2008): Section 21.3: -Transforms and Difference Equations 59

4. An application of difference equations currents in a ladder network The application we will consider is that of finding the electric currents in each loop of the ladder resistance network shown, which consists of (N + 1) loops. The currents form a sequence {i 0,i 1,...i N } V i o i 1 i n i n+1 i N Figure 7 All the resistors have the same resistance R so loops 1 to N are identical. The ero th loop contains an applied voltage V. In this ero th loop, Kirchhoff s voltage law gives V = Ri 0 + R(i 0 i 1 ) from which i 1 =2i 0 V R (23) Similarly, applying the Kirchhoff law to the (n + 1) th loop where there is no voltage source and 3 resistors 0=Ri n+1 + R(i n+1 i n+2 )+R(i n+1 i n ) from which i n+2 3i n+1 + i n =0 n =0, 1, 2,...(N 2) (24) (24) is the basic difference equation that has to be solved. Task Using the left shift theorems obtain the -transform of equation (24). Denote by I() the -transform of {i n }. Simplify the algebraic equation you obtain. 60 HELM (2008): Workbook 21: -Transforms

We obtain Simplifying 2 I() 2 i 0 i 1 3(I() i 0 )+I() =0 ( 2 3 +1)I() = 2 i 0 + i 1 3i 0 (25) If we now eliminate i 1 using (23), the right-hand side of (25) becomes 2 i 0 + 2i 0 V 3i 0 = 2 i 0 i 0 V R R = i 0 2 V i 0 R Hence from (25) I() = 1+ V i 0 2 i 0 R 2 3 +1 Our final task is to find the inverse -transform of (26). (26) Task Look at the table of -transforms on page 35 (or at the back of the Workbook) and suggest what sequences are likely to arise by inverting I() as given in (26). The most likely candidates are hyperbolic sequences because both {cosh αn} and {sinh αn} have -transforms with denominator 2 2 cosh α +1 which is of the same form as the denominator of (26), remembering that cosh α 1. (Why are the trigonometric sequences {cos ωn} and {sin ωn} not plausible here?) To proceed, we introduce a quantity α such that α is the positive solution of 2 cosh α =3from which (using cosh 2 α sinh 2 α 1) we get HELM (2008): Section 21.3: -Transforms and Difference Equations 61

sinh α = 9 4 1= Hence (26) can be written I() =i 0 5 2 2 1+ V i 0 R 2 2 cosh α +1 (27) To further progress, bearing in mind the -transforms of {cosh αn} and {sinh αn}, we must subtract and add cosh α to the numerator of (27), where cosh α = 3 2. 2 cosh α + 3 I() = i 0 2 1+ V i 0 R 2 2 cosh α +1 3 = i 0 ( 2 cosh α) 2 2 cosh α +1 + 2 1 V i 0 R 2 2 cosh α +1 The first term in the square bracket is the -transform of {cosh αn}. The second term is 1 2 V i 0 R 2 2 cosh α +1 = 1 2 V 2 5 5 i 0 R 2 2 2 cosh α +1 which has inverse -transform 1 2 V 2 5 sinh αn i 0 R Hence we have for the loop currents i0 i n = i 0 cosh(αn)+ 2 V 2 5 sinh(αn) R n =0, 1,...N (27) where cosh α = 3 2 determines the value of α. Finally, by Kirchhoff s law applied to the rightmost loop 3i N = i N 1 from which, with (27), we could determine the value of i 0. 62 HELM (2008): Workbook 21: -Transforms

Exercises 1. Deduce the inverse -transform of each of the following functions: 2 2 3 (a) 2 3 4 (b) 22 + ( 1) 2 2 2 (c) 2 2 2 +2 (d) 32 +5 4 2. Use -transforms to solve each of the following difference equations: (a) y n+1 3y n =4 n y 0 =0 (b) y n 3y n 1 =6 y 1 =4 (c) y n 2y n 1 = n y 1 =0 (d) y n+1 5y n =5 n+1 y 0 =0 (e) y n+1 +3y n =4δ n 2 y 0 =2 (f) y n 7y n 1 +10y n 2 =0 y 1 = 16, y 2 =5 (g) y n 6y n 1 +9y n 2 =0 y 1 =1, y 2 =0 s 1 (a) ( 1) n +4 n (b) 2+3n (c) cos(nπ/3) (d) 3δ n 2 +5δ n 4 2 (a) y n =4 n 3 n (b) y n = 21 3 n 3 (c) y n =2 2 n 2 n (d) y n = n5 n (e) y n =2 ( 3) n +4 ( 3) n 3 u n 2 (f) y n = 12 2 n +50 5 n (g) y n = (6 + 3n)3 n HELM (2008): Section 21.3: -Transforms and Difference Equations 63

Less strictly one might write Zy n = Y (). Some texts use the notation y n Y () to denote that (the sequence) y n and (the function) Y () form a -transform pair. We shall also call {y n } the inverse -transform of Y () and write symbolically {y n } = Z 1 Y (). 2. Commonly used -transforms Unit impulse sequence (delta sequence) This is a simple but important sequence denoted by δ n and defined as 1 n =0 δ n = 0 n = ±1, ±2,... The significance of the term unit impulse is obvious from this definition. By the definition (1) of the -transform Z{δ n } = 1+0 1 +0 2 +... = 1 If the single non-ero value is other than at n =0the calculation of the -transform is equally simple. For example, 1 n =3 δ n 3 = 0 otherwise From (1) we obtain Z{δ n 3 } = 0+0 1 +0 2 + 3 +0 4 +... = 3 Task Write down the definition of δ n m where m is any positive integer and obtain its -transform. 1 n = m δ n m = 0 otherwise Z{δ n m } = m 14 HELM (2008): Workbook 21: -Transforms

Key Point 3 Z{δ n m } = m m =0, 1, 2,... Unit step sequence As we saw earlier in this Workbook the unit step sequence is 1 n =0, 1, 2,... u n = 0 n = 1, 2, 3,... Then, by the definition (1) Z{u n } =1+1 1 +1 2 +... The infinite series here is a geometric series (with a constant ratio 1 between successive terms). Hence the sum of the first N terms is S N = 1+ 1 +...+ (N 1) = 1 N 1 1 As N S N 1 1 1 provided 1 < 1 Hence, in what is called the closed form of this -transform we have the result given in the following Key Point: Z{u n } = Key Point 4 1 1 = 1 1 U() say, 1 < 1 The restriction that this result is only valid if 1 < 1 or, equivalently > 1 means that the position of the complex quantity must lie outside the circle centre origin and of unit radius in an Argand diagram. This restriction is not too significant in elementary applications of the -transform. HELM (2008): Section 21.2: Basics of -Transform Theory 15

The geometric sequence {a n } Task For any arbitrary constant a obtain the -transform of the causal sequence 0 n = 1, 2, 3,... f n = a n n =0, 1, 2, 3,... We have, by the definition in Key Point 1, F () =Z{f n } =1+a 1 + a 2 2 +... which is a geometric series with common ratio a 1. Hence, provided a 1 < 1, the closed form of the -transform is 1 F () = 1 a = 1 a. The -transform of this sequence {a n }, which is itself a geometric sequence is summaried in Key Point 5. Z{a n } = Key Point 5 1 1 a = 1 a > a. Notice that if a =1we recover the result for the -transform of the unit step sequence. 16 HELM (2008): Workbook 21: -Transforms

Task Use Key Point 5 to write down the -transform of the following causal sequences (a) 2 n (b) ( 1) n, the unit alternating sequence (c) e n (d) e αn where α is a constant. (a) Using a =2 Z{2 n 1 } = 1 2 = 1 2 > 2 (b) Using a = 1 Z{( 1) n 1 } = 1+ = 1 +1 > 1 (c) Using a = e 1 Z{e n } = e 1 >e 1 (d) Using a = e α Z{e αn } = e α > e α The basic -transforms obtained have all been straightforwardly found from the definition in Key Point 1. To obtain further useful results we need a knowledge of some of the properties of -transforms. HELM (2008): Section 21.2: Basics of -Transform Theory 17

3. Linearity property and applications Linearity property This simple property states that if {v n } and {w n } have -transforms V () and W () respectively then Z{av n + bw n } = av ()+bw () for any constants a and b. (In particular if a = b =1this property tells us that adding sequences corresponds to adding their -transforms). The proof of the linearity property is straightforward using obvious properties of the summation operation. By the -transform definition: Z{av n + bw n } = (av n + bw n ) n = n=0 (av n n + bw n n ) n=0 = a v n n + b w n n n=0 = av ()+bv () n=0 We can now use the linearity property and the exponential sequence {e αn } to obtain the -transforms of hyperbolic and of trigonometric sequences relatively easily. For example, sinh n = en e n 2 Hence, by the linearity property, Z{sinh n} = 1 2 Z{en } 1 2 Z{e n } = 1 2 e e 1 e 1 ( e) = 2 2 (e + e 1 ) +1 = e e 1 2 2 (2 cosh 1) +1 = sinh 1 2 2 cosh 1 + 1 Using αn instead of n in this calculation, where α is a constant, we obtain Z{sinh αn} = sinh α 2 2 cosh α +1 18 HELM (2008): Workbook 21: -Transforms

Task Using cosh αn eαn + e αn 2 {1, cosh α, cosh 2α,...} obtain the -transform of the sequence {cosh αn} = We have, by linearity, Z{cosh αn} = 1 2 Z{eαn } + 1 2 Z{e αn } = 1 2 e + 1 α e α = 2 (e α + e α ) 2 2 2 cosh α +1 = 2 cosh α 2 2 cosh α +1 Trigonometric sequences If we use the result Z{a n } = a > a with, respectively, a = e iω and a = e iω where ω is a constant and i denotes 1 we obtain Z{e iωn } = Z{e iωn } = e +iω e iω Hence, recalling from complex number theory that cos x = eix + e ix 2 we can state, using the linearity property, that HELM (2008): Section 21.2: Basics of -Transform Theory 19

Z{cos ωn} = 1 2 Z{eiωn } + 1 2 Z{e iωn } = 1 2 e + 1 iω e iω = 2 (e iω + e iω ) 2 2 (e iω + e iω ) +1 = 2 cos ω 2 2 cos ω +1 (Note the similarity of the algebra here to that arising in the corresponding hyperbolic case. Note also the similarity of the results for Z{cosh αn} and Z{cos ωn}.) Task By a similar procedure to that used above for Z{cos ωn} obtain Z{sin ωn}. 20 HELM (2008): Workbook 21: -Transforms

We have Z{sin ωn} = 1 2i Z{eiωn } 1 2i Z{e iωn } Z{sin ωn} = 2i = 2i = 1 e 1 iω e iω e iω + e iω 2 2 cos ω +1 sin ω 2 2 cos ω +1 (Don t miss the i factor here!) Key Point 6 Z{cos ωn} = Z{sin ωn} = 2 cos ω 2 2 cos ω +1 sin ω 2 2 cos ω +1 Notice the same denominator in the two results in Key Point 6. Key Point 7 Z{cosh αn} = Z{sinh αn} = 2 cosh α 2 2 cosh α +1 sinh α 2 2 cosh α +1 Again notice the denominators in Key Point 7. Compare these results with those for the two trigonometric sequences in Key Point 6. HELM (2008): Section 21.2: Basics of -Transform Theory 21

Task Use Key Points 6 and 7 to write down the -transforms of (a) sin n (b) {cos 3n} (c) {sinh 2n} (d) {cosh n} 2 (a) Z sin n 2 = (b) Z{cos 3n} = (c) Z{sinh 2n} = (d) Z{cosh n} = sin 1 2 2 2 cos 1 2 +1 2 cos 3 2 2 cos 3 + 1 sinh 2 2 2 cosh 2 + 1 2 cosh 1 2 2 cosh 1 + 1 22 HELM (2008): Workbook 21: -Transforms

Task Use the results for Z{cos ωn} and Z{sin ωn} in Key Point 6 to obtain the - transforms of nπ nπ (a) {cos(nπ)} (b) sin (c) cos 2 2 Write out the first few terms of each sequence. (a) With ω = π Z{cos nπ} = 2 cos π 2 2 cos π +1 = {cos nπ} = {1, 1, 1, 1,...} = {( 1) n } 2 + 2 +2 +1 = +1 We have re-derived the -transform of the unit alternating sequence. (See Task on page 17). (b) With ω = π 2 Z sin nπ sin π 2 = 2 2 2 cos = π 2 +1 2 +1 where sin nπ = {0, 1, 0, 1, 0,...} 2 (c) With ω = π Z cos nπ = 2 cos π 2 = 2 2 2 2 +1 2 +1 where cos nπ = {1, 0, 1, 0, 1,...} 2 (These three results can also be readily obtained from the definition of the -transform. Try!) HELM (2008): Section 21.2: Basics of -Transform Theory 23

4. Further -transform properties We showed earlier that the results Z{v n + w n } = V ()+W () and similarly Z{v n w n } = V () W () follow from the linearity property. You should be clear that there is no comparable result for the product of two sequences. Z{v n w n } is not equal to V ()W () For two specific products of sequences however we can derive useful results. Multiplication of a sequence by a n Suppose f n is an arbitrary sequence with -transform F (). Consider the sequence {v n } where v n = a n f n i.e. {v 0,v 1,v 2,...} = {f 0, af 1,a 2 f 2,...} By the -transform definition Z{v n } = v 0 + v 1 1 + v 2 2 +... = = But F () = = f 0 + af 1 1 + a 2 f 2 2 +... a n f n n n=0 f n a n=0 f n n n=0 n Thus we have shown that Z{a n f n } = F a Key Point 8 Z{a n f n } = F a That is, multiplying a sequence {f n } by the sequence {a n } does not change the form of the - transform F (). We merely replace by in that transform. a 24 HELM (2008): Workbook 21: -Transforms

For example, using Key Point 6 we have Z{cos n} = 2 cos 1 2 2 cos 1 + 1 So, replacing by 1 =2, 2 Z n 1 cos n = (2)2 (2)cos1 2 (2) 2 4 cos 1 + 1 Task Using Key Point 8, write down the -transform of the sequence {v n } where v n = e 2n sin 3n We have, Z{sin 3n} = sin 3 2 2 cos 3 + 1 so with a = e 2 we replace by e +2 to obtain Z{v n } = Z{e 2n sin 3n} = = e 2 sin 3 (e 2 ) 2 2e 2 cos 3 + 1 e 2 sin 3 2 2e 2 cos 3 + e 4 HELM (2008): Section 21.2: Basics of -Transform Theory 25

Task Using the property just discussed write down the -transform of the sequence {w n } where w n = e αn cos ωn We have, Z{cos ωn} = 2 cos ω 2 2 cos ω +1 So replacing by e α we obtain Z{w n } = Z{e αn cos ωn} = = (e α ) 2 e α cos ω (e α ) 2 2e α cos ω +1 2 e α cos ω 2 2e α cos ω + e 2α Key Point 9 Z{e αn cos ωn} = Z{e αn sin ωn} = 2 e α cos ω 2 2e α cos ω + e 2α e α sin ω 2 2e α cos ω + e 2α Note the same denominator in each case. 26 HELM (2008): Workbook 21: -Transforms

Multiplication of a sequence by n An important sequence whose -transform we have not yet obtained is the unit ramp sequence {r n }: 0 n = 1, 2, 3,... r n = n n =0, 1, 2,... 3 2 1 r n 0 1 2 3 n Figure 5 Figure 5 clearly suggests the nomenclature ramp. We shall attempt to use the -transform of {r n } from the definition: Z{r n } =0+1 1 +2 2 +3 3 +... This is not a geometric series but we can write 1 +2 2 +3 3 = 1 (1 + 2 1 +3 2 +...) = 1 (1 1 ) 2 1 < 1 where we have used the binomial theorem ( 16.3). Hence Z{r n } = Z{n} = = 1 2 1 1 > 1 ( 1) 2 Key Point 10 The -transform of the unit ramp sequence is Z{r n } = = R() (say) ( 1) 2 Recall now that the unit step sequence has -transform Z{u n } = the subject of the next Task. ( 1) = U() (say) which is HELM (2008): Section 21.2: Basics of -Transform Theory 27

Task Obtain the derivative of U()= ( 1) with respect to. We have, using the quotient rule of differentiation: du d = d ( 1)1 ()(1) = d 1 ( 1) 2 = 1 ( 1) 2 We also know that R() = ( 1) =( ) 1 = du 2 ( 1) 2 d Also, if we compare the sequences u n = {0, 0, 1, 1, 1, 1,...} r n = {0, 0, 0, 1, 2, 3,...} we see that r n = nu n, (4) so from (3) and (4) we conclude that Z{n u n } = du d Now let us consider the problem more generally. Let f n be an arbitrary sequence with -transform F (): F () =f 0 + f 1 1 + f 2 2 + f 3 3 +...= f n n 28 HELM (2008): Workbook 21: -Transforms n=0 (3)

We differentiate both sides with respect to the variable, doing this term-by-term on the right-hand side. Thus df d = f 1 2 2f 2 3 3f 3 4...= ( n)f n n 1 n=1 = 1 (f 1 1 +2f 2 2 +3f 3 3 +...)= 1 nf n n n=1 But the bracketed term is the -transform of the sequence {n f n } = {0,f 1, 2f 2, 3f 3,...} Thus if F () =Z{f n } we have shown that df d = 1 Z{n f n } or Z{n f n } = df d We have already (equations (3) and (4) above) demonstrated this result for the case f n = u n. Key Point 11 If Z{f n } = F () then Z{n f n } = df d Task By differentiating the -transform R() of the unit ramp sequence obtain the - transform of the causal sequence {n 2 }. HELM (2008): Section 21.2: Basics of -Transform Theory 29

We have so Z{n} = ( 1) 2 Z{n 2 } = Z{n.n} = d d ( 1) 2 By the quotient rule d d ( 1) 2 Multiplying by we obtain Z{n 2 } = = ( 1)2 ()(2)( 1) ( 1) 4 = 1 2 ( 1) 3 = 1 ( 1) 3 + 2 (1 + ) = ( 1) 3 ( 1) 3 Clearly this process can be continued to obtain the transforms of {n 3 }, {n 4 },... etc. 5. Shifting properties of the -transform In this subsection we consider perhaps the most important properties of the -transform. properties relate the -transform Y () of a sequence {y n } to the -transforms of (i) right shifted or delayed sequences {y n 1 }{y n 2 } etc. (ii) left shifted or advanced sequences {y n+1 }, {y n+2 } etc. These The results obtained, formally called shift theorems, are vital in enabling us to solve certain types of difference equation and are also invaluable in the analysis of digital systems of various types. Right shift theorems Let {v n } = {y n 1 } i.e. the terms of the sequence {v n } are the same as those of {y n } but shifted one place to the right. The -transforms are, by definition, Y () = y 0 + y 1 1 + y 2 2 + y j 3 +... V () = v 0 + v 1 1 + v 2 2 + v 3 3 +... = y 1 + y 0 1 + y 1 2 + y 2 3 +... = y 1 + 1 (y 0 + y 1 1 + y 2 2 +...) i.e. V () =Z{y n 1 } = y 1 + 1 Y () 30 HELM (2008): Workbook 21: -Transforms

Task Obtain the -transform of the sequence {w n } = {y n 2 } using the method illustrated above. The -transform of {w n } is W () =w 0 + w 1 1 + w 2 2 + w 3 3 +... or, since w n = y n 2, W () =y 2 + y 1 1 + y 0 2 + y 1 3 +... = y 2 + y 1 1 + 2 (y 0 + y 1 1 +...) i.e. W () =Z{y n 2 } = y 2 + y 1 1 + 2 Y () Clearly, we could proceed in a similar way to obtains a general result for Z{y n m } where m is any positive integer. The result is Z{y n m } = y m + y m+1 1 +...+ y 1 m+1 + m Y () For the particular case of causal sequences (where y 1 = y 2 =...=0) these results are particularly simple: Z{y n 1 } = 1 Y () Z{y n 2 } = 2 Y () (causal systems only) Z{y n m } = m Y () You may recall from earlier in this Workbook that in a digital system we represented the right shift operation symbolically in the following way: {y n } {y n 1 } {y n 2 } 1 1 Figure 6 The significance of the 1 factor inside the rectangles should now be clearer. If we replace the input and output sequences by their -transforms: Z{y n } = Y () Z{y n 1 } = 1 Y () it is evident that in the -transform domain the shift becomes a multiplication by the factor 1. N.B. This discussion applies strictly only to causal sequences. HELM (2008): Section 21.2: Basics of -Transform Theory 31

Notational point: A causal sequence is sometimes written as y n u n where u n is the unit step sequence 0 n = 1, 2,... u n = 1 n =0, 1, 2,... The right shift theorem is then written, for a causal sequence, Z{y n m u n m } = m Y () Examples Recall that the -transform of the causal sequence {a n } is theorems that (i) Z{a n 1 } = Z{0, 1, a, a 2,...} = 1 a = 1 a (ii) Z{a n 2 } = Z{0, 0, 1, a, a 2,...} = 1 a = 1 ( a). It follows, from the right shift a Task Write the following sequence f n as a difference of two unit step sequences. Hence obtain its -transform. f n 1 0 1 2 3 4 5 6 7 n 32 HELM (2008): Workbook 21: -Transforms

Since {u n } = and {u n 5 } = it follows that f n = u n u n 5 1 n =0, 1, 2,... 0 n = 1, 2,... 1 n =5, 6, 7,... 0 otherwise Hence F () = 1 5 1 = 4 1 Left shift theorems Recall that the sequences {y n+1 }, {y n+2 }...denote the sequences obtained by shifting the sequence {y n } by 1, 2,...units to the left respectively. Thus, since Y () =Z{y n } = y 0 + y 1 1 + y 2 2 +... then Z{y n+1 } = y 1 + y 2 1 + y 3 2 +... = y 1 + (y 2 2 + y 3 3 +...) The term in brackets is the -transform of the unshifted sequence {y n } apart from its first two terms: thus Z{y n+1 } = y 1 + (Y () y 0 y 1 1 ) Z{y n+1 } = Y () y 0 Task Obtain the -transform of the sequence {y n+2 } using the method illustrated above. HELM (2008): Section 21.2: Basics of -Transform Theory 33

Z{y n+2 } = y 2 + y 3 1 + y 4 2 +... = y 2 + 2 (y 3 3 + y 4 4 +...) = y 2 + 2 (Y () y 0 y 1 1 y 2 2 ) Z{y n+2 } = 2 Y () 2 y 0 y 1 These left shift theorems have simple forms in special cases: if y 0 =0 Z{y n+1 } = Y() if y 0 = y 1 =0 Z{y n+2 } = 2 Y () if y 0 = y 1 =...y m 1 =0 Z{y n+m } = m Y () Key Point 12 The right shift theorems or delay theorems are: Z{y n 1 } = y 1 + 1 Y () Z{y n 2 } = y 2 + y 1 1 + 2 Y ().... Z{y n m } = y m + y m+1 1 +...+ y 1 m+1 + m Y () The left shift theorems or advance theorems are: Z{y n+1 } = Y () y 0 Z{y n+2 } = 2 Y () 2 y 0 y 1.. Z{y n m } = m Y () m y 0 m 1 y 1... y m 1 Note carefully the occurrence of positive powers of in the left shift theorems and of negative powers of in the right shift theorems. 34 HELM (2008): Workbook 21: -Transforms

Table 1: -transforms f n F () Name δ n 1 unit impulse δ n m u n a n e αn sinh αn cosh αn sin ωn cos ωn e αn sin ωn e αn cos ωn m 1 a e α sinh α 2 2 cosh α +1 2 cosh α 2 2 cosh α +1 sin ω 2 2 cos ω +1 2 cos ω 2 2 cos ω +1 e α sin ω 2 2e α cos ω + e 2α 2 e α cos ω 2 2e α cos ω + e 2α unit step sequence geometric sequence n ( 1) 2 ramp sequence n 2 ( +1) ( 1) 3 n 3 ( 2 +4 +1) ( 1) 4 a n f n nf n F a df d This table has been copied to the back of this Workbook (page 96) for convenience. HELM (2008): Section 21.2: Basics of -Transform Theory 35

The -Transform 21.1 Introduction The -transform is the major mathematical tool for analysis in such topics as digital control and digital signal processing. In this introductory Section we lay the foundations of the subject by briefly discussing sequences, shifting of sequences and difference equations. Readers familiar with these topics can proceed directly to Section 21.2 where -transforms are first introduced. Prerequisites Before starting this Section you should... Learning Outcomes On completion you should be able to... have competence with algebra explain what is meant by a sequence and by a difference equation distinguish between first and second order difference equations shift sequences to the left or right 2 HELM (2008): Workbook 21: -Transforms

1. Preliminaries: Sequences and Difference Equations Sequences A sequence is a set of numbers formed according to some definite rule. For example the sequence {1, 4, 9, 16, 25,...} (1) is formed by the squares of the positive integers. If we write y 1 =1, y 2 =4, y 3 =9,... then the general or n th term of the sequence (1) is y n = n 2. The notations y(n) and y[n] are also used sometimes to denote the general term. The notation {y n } is used as an abbreviation for a whole sequence. An alternative way of considering a sequence is to view it as being obtained by sampling a continuous function. In the above example the sequence of squares can be regarded as being obtained from the function y(t) =t 2 by sampling the function at t =1, 2, 3,... as shown in Figure 1. 9 y = t 2 4 1 1 2 3 t Figure 1 The notation y(n), as opposed to y n, for the general term of a sequence emphasies this sampling aspect. Task Find the general term of the sequence {2, 4, 8, 16, 32,...}. The terms of the sequence are the integer powers of 2: y 1 =2=2 1 y 2 =4=2 2 y 3 =8=2 3... so y n =2 n. HELM (2008): Section 21.1: The -Transform 3

Here the sequence {2 n } are the sample values of the continuous function y(t) =2 t at t =1, 2, 3,... An alternative way of defining a sequence is as follows: (i) give the first term y 1 of the sequence (ii) give the rule for obtaining the (n +1) th term from the n th. A simple example is y n+1 = y n + d y 1 = a where a and d are constants. It is straightforward to obtain an expression for y n in terms of n as follows: y 2 = y 1 + d = a + d y 3 = y 2 + d = a + d + d = a +2d y 4 = y 3 + d = a +3d. y n = a +(n 1)d (2) This sequence characterised by a constant difference between successive terms y n+1 y n = d n =1, 2, 3,... is called an arithmetic sequence. Task Calculate the n th term of the arithmetic sequence defined by y n+1 y n =2 y 1 =9. Write out the first 4 terms of this sequence explicitly. Suggest why an arithmetic sequence is also known as a linear sequence. 4 HELM (2008): Workbook 21: -Transforms

We have, using (2), y n =9+(n 1)2 or y n =2n +7 so y 1 =9(as given), y 2 = 11, y 3 = 13, y 4 = 15,... A graph of y n against n would be just a set of points but all lie on the straight line y =2x +7, hence the term linear sequence. y n 13 11 9 y(x) = 2x + 7 1 2 3 n Nomenclature The equation y n+1 y n = d (3) is called a difference equation or recurrence equation or more specifically a first order, constant coefficient, linear, difference equation. The sequence whose n th term is y n = a +(n 1)d (4) is the solution of (3) for the initial condition y 1 = a. The coefficients in (3) are the numbers preceding the terms y n+1 and y n so are 1 and 1 respectively. The classification first order for the difference equation (3) follows because the difference between the highest and lowest subscripts is n +1 n =1. Now consider again the sequence Clearly {y n } = {2 n } y n+1 y n =2 n+1 2 n =2 n so the difference here is dependent on n i.e. is not constant. Hence the sequence {2 n } = {2, 4, 8,...} is not an arithmetic sequence. HELM (2008): Section 21.1: The -Transform 5

Task For the sequence {y n } =2 n calculate y n+1 2y n. Hence write down a difference equation and initial condition for which {2 n } is the solution. y n+1 2y n =2 n+1 2 2 n =2 n+1 2 n+1 =0 Hence y n =2 n is the solution of the homogeneous difference equation y n+1 2y n =0 with initial condition y 1 =2. (5) The term homogeneous refers to the fact that the right-hand side of the difference equation (5) is ero. More generally it follows that y n+1 Ay n =0 y 1 = A has solution sequence {y n } with general term y n = A n A second order difference equation Second order difference equations are characterised, as you would expect, by a difference of 2 between the highest and lowest subscripts. A famous example of a constant coefficient second order difference equation is y n+2 = y n+1 + y n or y n+2 y n+1 y n =0 (6) The solution {y n } of (6) is a sequence where any term is the sum of the two preceding ones. 6 HELM (2008): Workbook 21: -Transforms

Task What additional information is needed if (6) is to be solved? Two initial conditions, the values of y 1 and y 2 must be specified so we can calculate and so on. y 3 = y 2 + y 1 y 4 = y 3 + y 2 Task Find the first 6 terms of the solution sequence of (6) for each of the following sets of initial conditions (a) y 1 =1 y 2 =3 (b) y 1 =1 y 2 =1 (a) {1, 3, 4, 7, 11, 18...} (b) {1, 1, 2, 3, 5, 8,...} (7) The sequence (7) is a very famous one; it is known as the Fibonacci Sequence. It follows that the solution sequence of the difference equation (6) y n+1 = y n+1 + y n with initial conditions y 1 = y 2 =1is the Fibonacci sequence. What is not so obvious is what is the general term y n of this sequence. One way of obtaining y n in this case, and for many other linear constant coefficient difference equations, is via a technique involving Z transforms which we shall introduce shortly. HELM (2008): Section 21.1: The -Transform 7

Shifting of sequences Right Shift Recall the sequence {y n } = {n 2 } or, writing out the first few terms explicitly, {y n } = {1, 4, 9, 16, 25,...} The sequence {v n } = {0, 1, 4, 9, 16, 25,...} contains the same numbers as y n but they are all shifted one place to the right. The general term of this shifted sequence is v n =(n 1) 2 n =1, 2, 3,... Similarly the sequence {w n } = {0, 0, 1, 4, 9, 16, 25,...} has general term (n 2) 2 n =2, 3,... w n = 0 n =1 Task For the sequence {y n } = {2 n } = {2, 4, 8, 16,...} write out explicitly the first 6 terms and the general terms of the sequences v n and w n obtained respectively by shifting the terms of {y n } (a) one place to the right (b) three places the the right. (a) (b) 2 n 1 n =2, 3, 4,... {v n } = {0, 2, 4, 8, 16, 32...} v n = 0 n =1 {w n } = {0, 0, 0, 2, 4, 8...} w n = 2 n 3 n =4, 5, 6,... 0 n =1, 2, 3 8 HELM (2008): Workbook 21: -Transforms

The operation of shifting the terms of a sequence is an important one in digital signal processing and digital control. We shall have more to say about this later. For the moment we just note that in a digital system a right shift can be produced by delay unit denoted symbolically as follows: {y n } 1 {y n 1 } Figure 2 A shift of 2 units to the right could be produced by 2 such delay units in series: {y n } {y n 1 } {y n 2 } 1 1 Figure 3 (The significance of writing 1 will emerge later when we have studied transforms.) Left Shift Suppose we again consider the sequence of squares {y n } = {1, 4, 9, 16, 25,...} with y n = n 2. Shifting all the numbers one place to the left (or advancing the sequence) means that the sequence {v n } generated has terms v 0 = y 1 =1 v 1 = y 2 =4 v 2 = y 3 =9... and so has general term v n =(n +1) 2 n =0, 1, 2,... = y n+1 Notice here the appearance of the ero subscript for the first time. Shifting the terms of {v n } one place to the left or equivalently the terms of {y n } two places to the left generates a sequence {w n } where w 1 = v 0 = y 1 =1 w 0 = v 1 = y 2 =4 and so on. The general term is w n =(n +2) 2 n = 1, 0, 1, 2,... = y n+2 HELM (2008): Section 21.1: The -Transform 9

Task If {y n } = {1, 1, 2, 3, 5,...} n =1, 2, 3,... is the Fibonacci sequence, write out the terms of the sequences {y n+1 }, {y n+2 }. y n+1 = {1, 1, 2, 3, 5,...} where y 0 =1(arrowed), y 1 =1,y 2 =2,... y n+2 = {1, 1, 2, 3, 5,...} where y 1 =1, y 0 =1(arrowed), y 1 =2,y 2 =3,... It should be clear from this discussion of left shifted sequences that the simpler idea of a sequence beginning at n =1and containing only terms y 1,y 2,... has to be modified. We should instead think of a sequence as two-sided i.e. {y n } defined for all integer values of n and ero. In writing out the middle terms of a two sided sequence it is convenient to show by an arrow the term y 0. For example the sequence {y n } = {n 2 } {...9, 4, 1, 0, 1, 4, 9,...} n =0, ±1, ±2,... could be written A sequence which is ero for negative integers n is sometimes called a causal sequence. For example the sequence, denoted by {u n }, 0 n = 1, 2, 3,... u n = 1 n =0, 1, 2, 3,... is causal. Figure 4 makes it clear why {u n } is called the unit step sequence. 1 u n 3 2 1 0 1 2 n Figure 4 The curly bracket notation for the unit step sequence with the n =0term arrowed is {u n } = {...,0, 0, 0, 1, 1, 1,...} 10 HELM (2008): Workbook 21: -Transforms

Task Draw graphs of the sequences {u n 1 }, {u n 2 }, {u n+1 } where {u n } is the unit step sequence. u n 1 1 3 2 1 0 1 2 3 n u n 2 1 3 2 1 0 1 2 3 n u n+1 1 3 2 1 0 1 2 n HELM (2008): Section 21.1: The -Transform 11

Index for Workbook 21 Causal sequences 10 Convolution summation 68, 76-80 Currents in a ladder network 60 Delta sequence 14 Difference equation 5, 37 - first order 5, 37 - homogeneous 6 - second order 6, 47 Engineering applications 60, 64-84 Feedback convolution 75 Fibonacci sequence 7 Final value theorem 82 Geometric sequence 16 Homogeneous difference equation 6 Initial value theorem 81 Inverse transform formula 56 Inverse -transform 14 Inversion by residues 53-59 Ladder network 60 Laplace transforms 90 Left shift of a sequence 9, 89 Left shift theorem 33-34 Linearity property 18-19 Multiplication by -a n 24 -n 27 Pole of a function 53 Residue at a pole 54, 55 Right shift of a sequence 8, 89 Right shift theorem 30, 34 Sampled functions 85-95 Sampling 3, 85-95 Second order system 70 Sequences 3 Series combination of systems 72 Shifting of sequences 8, 30 Signal sampling 85-95 Steady state 72 Systems 64-84 Tables of -transforms 35, 95, 96 Transfer function 65, 67 Transients 72 Trigonometric sequences 19 Unit impulse function 92 Unit impulse sequence 14 Unit ramp sequence 27, 29, 87 Unit step function 86 Unit step sequence 10, 15 Zero input response 66 Zero state response 66 -transform 13 EXERCISES 63, 83