JUST THE MATHS SLIDES NUMBER 16.8 Z-TRANSFORMS 1 (Definition and - PDF document

“JUST THE MATHS” SLIDES NUMBER 16.8 Z-TRANSFORMS 1 (Definition and rules) by A.J.Hobson 16.8.1 Introduction 16.8.2 Standard Z-Transform definition and results 16.8.3 Properties of Z-Transforms

UNIT 16.8 - Z TRANSFORMS 1 DEFINITION AND RULES 16.8.1 INTRODUCTION Linear Difference Equations We consider “linear difference equations with constant coefficients”. DEFINITION 1 A first-order linear difference equation with constant coefficients has the general form, a 1 u n +1 + a 0 u n = f ( n ); a 0 , a 1 are constants; n is a positive integer; f ( n ) is a given function of n (possibly zero); u n is the general term of an infinite sequence of numbers, { u n } ≡ u 0 , u 1 , u 2 , u 3 , . . . 1

DEFINITION 2 A second order linear difference equation with constant coefficients has the general form a 2 u n +2 + a 1 u n +1 + a 0 u n = f ( n ); a 0 , a 1 , a 2 are constants; n is an integer; f ( n ) is a given function of n (possibly zero); u n is the general term of an infinite sequence of numbers, { u n } ≡ u 0 , u 1 , u 2 , u 3 , . . . Notes: (i) We shall assume that u n = 0 whenever n < 0. (ii) “Boundary conditions” will be given as follows: The value of u 0 for a first-order equation; The values of u 0 and u 1 for a second-order equation. 2

ILLUSTRATION Certain simple difference equations may be solved by very elementary methods. For example, to solve u n +1 − ( n + 1) u n = 0 , subject to the boundary condition that u 0 = 1, we may rewrite the difference equation as u n +1 = ( n + 1) u n . By using this formula repeatedly, we obtain u 1 = u 0 = 1 , u 2 = 2 u 1 = 2 , u 3 = 3 u 2 = 3 × 2 , u 4 = 4 u 3 = 4 × 3 × 2 , . . . . Hence, u n = n ! 3

16.8.2 STANDARD Z-TRANSFORM DEFINITION AND RESULTS THE DEFINITION OF A Z-TRANSFORM (WITH EXAMPLES) The Z-Transform of the sequence of numbers { u n } ≡ u 0 , u 1 , u 2 , u 3 , . . . . is defined by the formula, ∞ r =0 u r z − r , Z { u n } = � provided that the series converges (allowing for z to be a complex number if necessary). EXAMPLES 1. Determine the Z-Transform of the sequence, { u n } ≡ { a n } , where a is a non-zero constant. Solution ∞ Z { a n } = r =0 a r z − r . � That is, z + a 2 z 2 + a 3 1 Z { a n } = 1 + a z z 3 + . . . . = = z − a, 1 − a z by properties of infinite geometric series. 4

2. Determine the Z-Transform of the sequence, { u n } = { n } . Solution ∞ r =0 rz − r . Z { n } = � That is, Z { n } = 1 z + 2 z 2 + 3 z 3 + 4 z 4 + . . . . or  1 z + 1 z 2 + 1  1 z 2 + 1 z 3 + 1      +  + z 3 + . . . . z 4 + . . . .      1 z 3 + 1    , z 4 + . . . .   giving 1 1 1 1   1 z 2 z 3 z z Z { n } = + + + . . . . =   ,    1 − 1 1 − 1 1 − 1 1 − 1 1 − 1    z z z z z by properties of infinite geometric series. Thus, z z Z { n } = (1 − z ) 2 = ( z − 1) 2 5

A SHORT TABLE OF Z-TRANSFORMS { u n } Z { u n } Region z { 1 } | z | > 1 z − 1 z { a n } ( a constant) | z | > | a | z − a z { n } | z | > 1 ( z − 1) 2 e − nT z | z | > e − T � � ( T constant) z − e − T z sin T sin nT ( T constant) z 2 − 2 z cos T +1 | z | > 1 z ( z − cos T ) cos nT ( T constant) z 2 − 2 z cos T +1 | z | > 1 1 for n = 0 1 All z 0 for n > 0 (Unit Pulse sequence) 1 0 for n = 0 | z | > | a | z − a { a n − 1 } for n > 0 6

16.8.3 PROPERTIES OF Z-TRANSFORMS (a) Linearity If { u n } and { v n } are sequences of numbers, while A and B are constants, then Z { Au n + Bv n } ≡ A.Z { u n } + B.Z { v n } . Proof: The left-hand side is equivalent to ∞ ∞ ∞ r =0 ( Au r + Bv r ) z − r ≡ A r =0 u r z − r + B r =0 v r z − r . � � � This is equivalent to the right-hand side. EXAMPLE 5 z 3 z Z { 5 . 2 n − 3 n } = z − 2 − ( z − 1) 2 . 7

(b) The First Shifting Theorem Z { u n − 1 } ≡ 1 z.Z { u n } , where { u n − 1 } denotes the sequence whose first term, cor- responding to n = 0, is taken as zero and whose subse- quent terms, corresponding to n = 1 , 2 , 3 , 4 , . . . . , are the terms u 0 , u 1 , u 3 , u 4 , . . . of the original sequence. Proof: The left-hand side is equivalent to r =0 u r − 1 z − r ≡ u 0 z + u 1 z 2 + u 2 z 3 + u 3 ∞ z 4 + . . . ., � since u n = 0 whenever n < 0. Thus, Z { u n − 1 } ≡ 1  u 0 + u 1 z + u 2 z 2 + u 3    . z 3 + . . . . z. This is equivalent to the right-hand side. 8

Note: A more general form of the First Shifting Theorem states that Z { u n − k } ≡ 1 z k .Z { u n } , where { u n − k } denotes the sequence whose first k terms, corresponding to n = 0 , 1 , 2 , . . . . , k − 1 , are taken as zero and whose subsequent terms, corresponding to n = k, k +1 , k +2 , . . . . are the terms u 0 , u 1 , u 2 , . . . . of the original sequence. ILLUSTRATION Given that { u n } ≡ { 4 n } , we may say that Z { u n − 2 } ≡ 1 z 2 .Z { u n } ≡ 1 1 z z − 4 ≡ z 2 . z ( z − 4) . Note: { u n − 2 } has terms 0 , 0 , 1 , 4 , 4 2 , 4 3 , . . . and, by applying the definition of a Z-Transform directly, we would obtain z 3 + 4 2 z 4 + 4 3 Z { u n − 2 } = 1 z 2 + 4 z 5 . . . . 9

Hence , Z { u n } ≡ 1 1 1 ≡ z 2 . z ( z − 4) , 1 − 4 z by properties of infinite geometric series (c) The Second Shifting Theorem Z { u n +1 } ≡ z.Z { u n } − z.u 0 Proof: The left-hand side is equivalent to r =0 u r +1 z − r ≡ u 1 + u 2 z + u 3 z 2 + u 4 ∞ z 4 + . . . . � That is,  u 0 + u 1 z + u 2 z 2 + u 3 z 3 + u 4    − z.u 0 z. z 4 + . . . . This is equivalent to the right-hand side Note: Z { u n +2 } ≡ z.Z { u n +1 }− z.u 1 ≡ z 2 .Z { u n }− z 2 .u 0 − z.u 1 10

Z-TRANSFORMS (AN ENGINEERING INTRODUCTION) The mathematics of certain engineering problems leads to what are know as “difference equations” For example, consider the following electrical “ladder network” . R 1 R 1 R 1 R 1 R 1 R 1 ❜ R 2 R 2 R 2 R 2 R 2 R 2 R 2 R 2 ❜ ✻ ✻ ✻ ✻ ✻ ✻ i 0 i 1 i n i n +1 i n +2 i N It may be shown that R 1 i n +1 + R 2 ( i n +1 − i n ) + R 2 ( i n +1 − i n +2 ) = 0 where 0 ≤ n ≤ N − 2. The question which arises is “how can we determine a formula for i n in terms of n ” ? 11