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, a1un+1 + a0un = f(n); a0, a1 are constants; n is a positive integer; f(n) is a given function of n (possibly zero); un is the general term of an infinite sequence of numbers, {un} ≡ u0, u1, u2, u3, . . .

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DEFINITION 2 A second order linear difference equation with constant coefficients has the general form a2un+2 + a1un+1 + a0un = f(n); a0, a1, a2 are constants; n is an integer; f(n) is a given function of n (possibly zero); un is the general term of an infinite sequence of numbers, {un} ≡ u0, u1, u2 , u3, . . . Notes: (i) We shall assume that un = 0 whenever n < 0. (ii) “Boundary conditions” will be given as follows: The value of u0 for a first-order equation; The values of u0 and u1 for a second-order equation.

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ILLUSTRATION Certain simple difference equations may be solved by very elementary methods. For example, to solve un+1 − (n + 1)un = 0, subject to the boundary condition that u0 = 1, we may rewrite the difference equation as un+1 = (n + 1)un. By using this formula repeatedly, we obtain u1 = u0 = 1, u2 = 2u1 = 2, u3 = 3u2 = 3 × 2, u4 = 4u3 = 4 × 3 × 2, . . . . Hence, un = n!

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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 {un} ≡ u0, u1, u2, u3, . . . . is defined by the formula, Z{un} =

∞

• r=0 urz−r,

provided that the series converges (allowing for z to be a complex number if necessary). EXAMPLES

• 1. Determine the Z-Transform of the sequence,

{un} ≡ {an}, where a is a non-zero constant. Solution Z{an} =

∞

• r=0 arz−r.

That is, Z{an} = 1 + a z + a2 z2 + a3 z3 + . . . . = 1 1 − a

z

= z z − a, by properties of infinite geometric series.

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• 2. Determine the Z-Transform of the sequence,

{un} = {n}. Solution Z{n} =

∞

• r=0 rz−r.

That is, Z{n} = 1 z + 2 z2 + 3 z3 + 4 z4 + . . . .

• r

  1

z + 1 z2 + 1 z3 + . . . .

   +    1

z2 + 1 z3 + 1 z4 + . . . .

   +    1

z3 + 1 z4 + . . . .

   ,

giving Z{n} =

1 z

1 − 1

z

+

1 z2

1 − 1

z

+

1 z3

1 − 1

z

+ . . . . = 1 1 − 1

z

    

1 z

1 − 1

z

     ,

by properties of infinite geometric series. Thus, Z{n} = z (1 − z)2 = z (z − 1)2

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A SHORT TABLE OF Z-TRANSFORMS {un} Z{un} Region {1}

z z−1

|z| > 1 {an} (a constant)

z z−a

|z| > |a| {n}

z (z−1)2

|z| > 1

• e−nT
• (T constant)

z z−e−T

|z| > e−T sin nT (T constant)

z sin T z2−2z cos T+1 |z| > 1

cos nT (T constant)

z(z−cos T) z2−2z cos T+1 |z| > 1

1 for n = 0 1 All z 0 for n > 0 (Unit Pulse sequence) 0 for n = 0

1 z−a

|z| > |a| {an−1} for n > 0

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16.8.3 PROPERTIES OF Z-TRANSFORMS (a) Linearity If {un} and {vn} are sequences of numbers, while A and B are constants, then Z{Aun + Bvn} ≡ A.Z{un} + B.Z{vn}. Proof: The left-hand side is equivalent to

∞

• r=0 (Aur + Bvr)z−r ≡ A

∞

• r=0 urz−r + B

∞

• r=0 vrz−r.

This is equivalent to the right-hand side. EXAMPLE Z{5.2n − 3n} = 5z z − 2 − 3z (z − 1)2.

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(b) The First Shifting Theorem Z{un−1} ≡ 1 z.Z{un}, where {un−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 u0, u1, u3, u4, . . . of the original sequence. Proof: The left-hand side is equivalent to

∞

• r=0 ur−1z−r ≡ u0

z + u1 z2 + u2 z3 + u3 z4 + . . . ., since un = 0 whenever n < 0. Thus, Z{un−1} ≡ 1 z.

 u0 + u1

z + u2 z2 + u3 z3 + . . . .

  .

This is equivalent to the right-hand side.

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Note: A more general form of the First Shifting Theorem states that Z{un−k} ≡ 1 zk.Z{un}, where {un−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 u0, u1, u2, . . . .

• f the original sequence.

ILLUSTRATION Given that {un} ≡ {4n}, we may say that Z{un−2} ≡ 1 z2.Z{un} ≡ 1 z2. z z − 4 ≡ 1 z(z − 4). Note: {un−2} has terms 0, 0, 1, 4, 42, 43, . . . and, by applying the definition of a Z-Transform directly, we would obtain Z{un−2} = 1 z2 + 4 z3 + 42 z4 + 43 z5 . . . .

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Hence, Z{un} ≡ 1 z2. 1 1 − 4

z

≡ 1 z(z − 4), by properties of infinite geometric series (c) The Second Shifting Theorem Z{un+1} ≡ z.Z{un} − z.u0 Proof: The left-hand side is equivalent to

∞

• r=0 ur+1z−r ≡ u1 + u2

z + u3 z2 + u4 z4 + . . . . That is, z.

 u0 + u1

z + u2 z2 + u3 z3 + u4 z4 + . . . .

  − z.u0

This is equivalent to the right-hand side Note: Z{un+2} ≡ z.Z{un+1}−z.u1 ≡ z2.Z{un}−z2.u0−z.u1

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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”.

❜ ❜

R2 R1 i0

✻

R2 R1 i1

✻

R2 R2 R1 in

✻

R2 R1 in+1

✻

R2 R1 in+2

✻

R2 R2 R1 iN

✻

It may be shown that R1in+1 + R2(in+1 − in) + R2(in+1 − in+2) = 0 where 0 ≤ n ≤ N − 2. The question which arises is “how can we determine a formula for in in terms of n” ?

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