“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
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