JUST THE MATHS SLIDES NUMBER 16.10 Z-TRANSFORMS 3 (Solution of - PDF document

“JUST THE MATHS” SLIDES NUMBER 16.10 Z-TRANSFORMS 3 (Solution of linear difference equations) by A.J.Hobson 16.10.1 First order linear difference equations 16.10.2 Second order linear difference equations

UNIT 16.10 - Z TRANSFORMS 3 THE SOLUTION OF LINEAR DIFFERENCE EQUATIONS Linear Difference Equations may be solved by constructing the Z-Transform of both sides of the equa- tion. 16.10.1 FIRST ORDER LINEAR DIFFERENCE EQUATIONS EXAMPLES 1. Solve the linear difference equation, u n +1 − 2 u n = (3) − n , given that u 0 = 2 / 5. Solution Using the second shifting theorem, Z { u n +1 } = z.Z { u n } − z. 2 5 . Taking the Z-Transform of the difference equation, z.Z { u n } − 2 z 5 .z − 2 Z { u n } = . z − 1 3 1

On rearrangement, Z { u n } = 2 z z 5 . z − 2 + z − 1 � � ( z − 2) 3 − 3 3   ≡ 2 z 5 5 5 . z − 2 + z. +     z − 1   z − 2   3 z − 2 − 3 z z ≡ 5 . . z − 1 3 Taking the inverse Z-Transform of this function of z ,  (2) n − 3   5(3) − n   { u n } ≡  .     2. Solve the linear difference equation, u n +1 + u n = f ( n ) , given that  1 when n = 0;  f ( n ) ≡  0 when n > 0. and u 0 = 5. Solution Using the second shifting theorem, Z { u n +1 } = z.Z { u n } − z. 5 2

Taking the Z-Transform of the difference equation, z.Z { u n } − 5 z + Z { u n } = 1 . On rearrangement, 1 5 z Z { u n } = z + 1 + z + 1 . Hence, 5 when n = 0;   { u n } =  ( − 1) n − 1 + 5( − 1) n ≡ 4( − 1) n when n > 0.   16.10.2 SECOND ORDER LINEAR DIFFERENCE EQUATIONS EXAMPLES 1. Solve the linear difference equation, u n +2 = u n +1 + u n , given that u 0 = 0 and u 1 = 1. Solution Using the second shifting theorem, Z { u n +1 } = z.Z { u n − z. 0 ≡ z.Z { u n } . 3

and Z { u n +2 } = z 2 Z { u n } − z. 1 ≡ z 2 Z { u n } − z. Taking the Z-Transform of the difference equation, z 2 .Z { u n } − z = z.Z { u n } + Z { u n } . On rearrangement, z Z { u n } = z 2 − z − 1 . This may be written z Z { u n } = ( z − α )( z − β ) . From the quadratic formula, √ α, β = 1 ± 5 . 2 Using partial fractions, 1  z z  Z { u n } = z − α −  .     α − β z − β  Taking the inverse Z-Transform of this function of z gives  1  α − β [( α ) n − ( β ) n ]     { u n } ≡  .        4

2. Solve the linear difference equation, u n +2 − 7 u n +1 + 10 u n = 16 n, given that u 0 = 6 and u 1 = 2. Solution Using the second shifting theorem, Z { u n +1 } = z.Z { u n } − 6 z and Z { u n +2 } = z 2 .Z { u n } − 6 z 2 − 2 z. Taking the Z-Transform of the difference equation, 16 z z 2 .Z { u n }− 6 z 2 − 2 z − 7 [ z.Z { u n } − 6 z ]+10 Z { u n } = ( z − 1) 2 . On rearrangement, 16 z Z { u n } [ z 2 − 7 z + 10] − 6 z 2 + 40 z = ( z − 1) 2 . Hence, 6 z 2 − 40 z 16 z Z { u n } = ( z − 1) 2 ( z − 5)( z − 2) + ( z − 5)( z − 2) . 5

Using partial fractions, 4 3 4 5   Z { u n } = z. z − 2 − z − 5 + ( z − 1) 2 +  .     z − 1  The solution to the difference equation is therefore { u n } ≡ { 4(2) n − 3(5) n + 4 n + 5 } . 3. Solve the linear difference equation, u n +2 + 2 u n = 0 , √ given that u 0 = 1 and u 1 = 2. Solution Using the second shifting theorem, √ Z { u n +2 } = z 2 Z { u n } − z 2 − z 2 . Taking the Z-Transform of the difference equation, √ z 2 Z { u n } − z 2 − z 2 + 2 Z { u n } = 0 . On rearrangement, √ √ √ Z { u n } = z 2 + z 2 ≡ z.z + 2 z + 2 √ √ z 2 + 2 ≡ z. 2) . z 2 + 2 ( z + j 2)( z − j 6

Using partial fractions, √ √   2(1 + j ) 2(1 − j ) √ √ √ √ Z { u n } = z 2) +       j 2 2( z − j − j 2 2( z + j 2)   or  (1 − j ) (1 + j )  √ √ Z { u n } ≡ z. 2) +  .     2( z − j 2( z + j 2)  Hence, √ √ 1 2) n + 1   2) n   { u n } ≡ 2(1 − j )( j 2(1 + j )( − j       √ 1   2) n [(1 − j )( j ) n + (1 + j )( − j ) n ]   ≡ 2(       √ � √ √ 1  � 2 e − j π 4 .e j nπ 2 e j π 4 .e − j nπ 2) n 2 +   ≡ 2(   2     √ 1    e j (2 n − 1) π + e − j (2 n − 1) π   2) n +1   ≡ 2(   4 4      √  1 2) n +1 . 2 cos (2 n − 1) π      ≡ 2(   4       √  2) n +1 cos (2 n − 1) π      ≡  (  .   4     7