![Applications of difference equations
Working Procedure:
To solve a linear difference equation with constant coefficient by Z-transform.
Step 1: Take the Z-transform on both sides of the difference equations using the formulae and the given conditions.
Step 2: Transpose all terms without 𝑈(𝑧) to the right.
Step 3: Divide by the coefficient of 𝑈(𝑧), getting 𝑈(𝑧) as a function of z.
Step 4: Express this function in terms of the Z-transform of known functions and take the inverse Z-transform on
both sides.
This gives 𝑢 𝑛 as a function of n which is the desired solution.
Example:
Using the Z-transform, solve 𝑢 𝑛+2 − 2𝑢 𝑛+1 + 𝑢 𝑛 = 3𝑛 + 5
Solution:
If 𝑍(𝑢 𝑛) = 𝑈(𝑧) then 𝑍(𝑢 𝑛+1) = 𝑧[𝑈(𝑧) − 𝑢0]
𝑍(𝑢 𝑛+2) = 𝑧2[𝑈(𝑧) − 𝑢0 − 𝑢1 𝑧−1]
𝑧
( 𝑧−1)2
+ 5
𝑧
𝑧−1
𝑍(3𝑛 + 5) = 3𝑍(𝑛) + 5𝑍(1) =3](https://image.slidesharecdn.com/z-transforms-200411063937/85/Z-transforms-2-320.jpg)
![[ since 𝑍(𝑛) =
𝑧
( 𝑧−1)2
, 𝑍(1) =
𝑧−1
𝑧
]
Given equation becomes
𝑢 𝑛+2 − 2𝑢 𝑛+1 + 𝑢 𝑛 = 3𝑛+ 5 → 𝟏
Applying Z-transform on both sides in equation 1 we get
𝑍(𝑢 𝑛+2) − 2𝑍(𝑢 𝑛+1) + 𝑍(𝑢 𝑛) = 𝑍(3𝑛 +5)
𝑧2[𝑈(𝑧) − 𝑢0 − 𝑢1 𝑧−1] − 2𝑧[𝑈(𝑧) − 𝑢0] + 𝑈(𝑧) = 3
𝑧
( 𝑧−1)2
+ 5
𝑧
𝑧−1
0𝑈(𝑧)[𝑧2 − 2𝑧 + 1] − 𝑢 [𝑧2 − 2𝑧]−𝑧2 𝑢1 𝑧
1
= 3𝑧+5𝑧( 𝑧−1)
( 𝑧−1)2
2
𝑈(𝑧)[𝑧 − 2𝑧 + 1] =
25𝑧 −2𝑧
( 𝑧−1)2
2
0 1+ 𝑢 [𝑧 − 2𝑧] + 𝑢 𝑧
2
𝑈(𝑧)(𝑧 − 1) =
25𝑧 −2𝑧
( 𝑧−1)2
2
0 1+ 𝑢 [𝑧 − 2𝑧] + 𝑢 𝑧
2 2
𝑈( 𝑧) = 5𝑧 −2𝑧
+ 𝑢0[ 𝑧 −2𝑧]
+ 𝑢1 𝑧
( 𝑧−1)4 ( 𝑧−1)2 ( 𝑧−1)2
On inversion we get
𝑢 𝑛 = 𝑍−1 [
5𝑧2 −2𝑧
(𝑧 − 1)4
] + 𝑢0 𝑍−1 [
𝑧2 − 2𝑧
(𝑧 − 1)2
] + 𝑢1 𝑍−1 [
𝑧
(𝑧 − 1)2
] ⟶ 𝟐](https://image.slidesharecdn.com/z-transforms-200411063937/85/Z-transforms-3-320.jpg)
![Note that
𝑍(1) =
𝑧
𝑧 − 1
, 𝑍(𝑛) =
𝑧
(𝑧 − 1)2
, 𝑍(𝑛2) =
𝑧2 + 𝑧
(𝑧 − 1)3
, 𝑍(𝑛3) =
𝑧3 + 4𝑧2 + 𝑧
(𝑧 − 1)4
⟶ 𝟑
Now consider
5𝑧2−2𝑧 𝑧3+4𝑧2+
𝑧 ( 𝑧−1)4 = 𝐴 [ ( 𝑧−1)4
𝑧2+𝑧 𝑧 𝑧
𝑧−1
] + 𝐵 [ ] + 𝐶 [ ] + 𝐷[ ]
( 𝑧−1)3 ( 𝑧−1)2
= 𝐴 [ ( 𝑧−1)4
] + 𝐵[
𝑧3+4𝑧2+𝑧 (𝑧2+𝑧)(𝑧−1) 𝑧( 𝑧−1)2 𝑧( 𝑧−1)3
( 𝑧−1)4 ] + 𝐶 [ ( 𝑧−1)4 ] + 𝐷[ ( 𝑧−1)4]
= 𝐴 [
𝑧3+4𝑧2+𝑧
( 𝑧−1)4
] + 𝐵[ ( 𝑧−1)4
𝑧3−𝑧2+𝑧2−𝑧) 𝑧(𝑧2−2𝑧+1)
( 𝑧−1)4
] + 𝐶 [ ] + 𝐷[
𝑧(𝑧3−3𝑧2+3𝑧−1)
( 𝑧−1)4
]
5𝑧2−2𝑧 𝑧3+4𝑧2+
𝑧 ( 𝑧−1)4 = 𝐴 [ ( 𝑧−1)4
] + 𝐵 [ 𝑧3−𝑧
] + 𝐶 [ 𝑧3−2𝑧2+𝑍
( 𝑧−1)4 ( 𝑧−1)4 ] + 𝐷[
𝑧4−3𝑧3+3𝑧2−𝑧
( 𝑧−1)4
]
Equating the like powers of z we get
𝐷 = 0 (coefficients of 𝑧4)
𝐴 + 𝐵 + 𝐶 − 3𝐷 = 0(coefficients of 𝑧3)
⇒ 𝐴 + 𝐵 + 𝐶 = 0 ⟶ 𝟒
4𝐴 + 0𝐵 − 2𝐶 + 3𝐷 = 5(coefficients of 𝑧2)
⇒ 4𝐴 − 2𝐶 = 5 ⟶ 𝟓
𝐴 − 𝐵 + 𝐶 − 𝐷 = −2(coefficients of𝑧)](https://image.slidesharecdn.com/z-transforms-200411063937/85/Z-transforms-4-320.jpg)
![⇒ 𝐴 − 𝐵 + 𝐶 = −2 ⟶ 𝟔
Solve equation 4 and 6 we get 𝐵 = 1
Put 𝐵 = 1 equation 4 we get 𝐴 + 𝐶 = −1 ⟶ 𝟕
Solve equation 3 and 2 ×equation 7 we get 𝐴 = 1
2
Put 𝐴 and 𝐵 value in equation 4 we get 𝐶 = − 3
2
2
1 3
2
∴ 𝐴 = , 𝐵 = 1, 𝐶 = − , 𝐷 = 0
2 3 2
∴ 5𝑧 −2𝑧 1 𝑧 +4𝑧 +𝑧
( 𝑧−1)4 = 2
. [ ( 𝑧−1)4
𝑧2+𝑧 3 𝑧 𝑧
𝑧−1
] + 1. [ ] − . [ ] + 0. [ ]
( 𝑧−1)3 2 ( 𝑧−1)2
∴ 𝑍−1 5𝑧2−2𝑧 1 −1[ ] = 𝑍 [
( 𝑧−1)4 2 ( 𝑧−1)4
] + 𝑍−1𝑧3+4𝑧2+𝑧 𝑧2+𝑧 3 −1 𝑧
[ ] − 𝑍 [ ]
( 𝑧−1)3 2 ( 𝑧−1)2
1 3 2 3
2 2
= 𝑛 + 𝑛 − 𝑛 by equation3
1
2
2
= 𝑛(𝑛 + 2𝑛 − 3)
1
2
( )= 𝑛 𝑛 − 1 (𝑛 + 3)
2
Consider 𝑧 −2𝑧
= 𝐴 + 𝐵
𝑧 𝑧
( 𝑧−1)2 𝑧−1 ( 𝑧−1)2
= 𝐴 𝑧(𝑧−1)
+ 𝐵 𝑧(𝑧−1)
𝑧−1 ( 𝑧−1)2](https://image.slidesharecdn.com/z-transforms-200411063937/85/Z-transforms-5-320.jpg)
![2 2
∴ 𝑧 −2𝑧
= 𝐴 𝑧 −2𝑧
+ 𝐵 𝑧
( 𝑧−1)2 𝑧−1 𝑧−1
Equating the like powers of z we get
𝐴 + 0 = 1(coefficients of 𝑧2) ∴ 𝐴 = 1
−𝐴 + 𝐵 = −2(coefficients of 𝑧)
⇒ 𝐵 = −1
2 2
∴ 𝑧 −2𝑧
= 1. 𝑧 −2𝑧
− 1. 𝑧
( 𝑧−1)2 𝑧−1 𝑧−1
2
∴ 𝑍−1 [ 𝑧 −2𝑧
] = 𝑍−1 [ 𝑧
] − 𝑍−1[
( 𝑧−1)2 𝑧−1
𝑧
( 𝑧−1)2
] = 1 − 𝑛 (by equation 3)
and 𝑍−1 [
𝑧
( 𝑧−1)2
] = 𝑛
∴ equation 2 becomes
𝑛
1
2
( )( ) ( )𝑢 = 𝑛 𝑛 − 1 𝑛 + 3 + 𝑢0 1 − 𝑛 + 𝑢1 𝑛
1
2
( )( )= 𝑛 𝑛 − 1 𝑛 + 3 + 𝑢0 + (𝑢1 − 𝑢0)𝑛
2
= 1
𝑛(𝑛 − 1)(𝑛 + 3) + 𝑐 + 𝑐 𝑛 where 𝑐 = 𝑢 , 𝑐 = 𝑢 − 𝑢0 1 0 0 1 1 0](https://image.slidesharecdn.com/z-transforms-200411063937/85/Z-transforms-6-320.jpg)
Z transforms
- 1. By A.Sujatha M.Sc.,Phil.,PGDCA., Department of Mathematics Transforms (18UMTS41) – Applications of Z-Transforms in linear difference equation
- 2. Applications of difference equations Working Procedure: To solve a linear difference equation with constant coefficient by Z-transform. Step 1: Take the Z-transform on both sides of the difference equations using the formulae and the given conditions. Step 2: Transpose all terms without 𝑈(𝑧) to the right. Step 3: Divide by the coefficient of 𝑈(𝑧), getting 𝑈(𝑧) as a function of z. Step 4: Express this function in terms of the Z-transform of known functions and take the inverse Z-transform on both sides. This gives 𝑢 𝑛 as a function of n which is the desired solution. Example: Using the Z-transform, solve 𝑢 𝑛+2 − 2𝑢 𝑛+1 + 𝑢 𝑛 = 3𝑛 + 5 Solution: If 𝑍(𝑢 𝑛) = 𝑈(𝑧) then 𝑍(𝑢 𝑛+1) = 𝑧[𝑈(𝑧) − 𝑢0] 𝑍(𝑢 𝑛+2) = 𝑧2[𝑈(𝑧) − 𝑢0 − 𝑢1 𝑧−1] 𝑧 ( 𝑧−1)2 + 5 𝑧 𝑧−1 𝑍(3𝑛 + 5) = 3𝑍(𝑛) + 5𝑍(1) =3
- 3. [ since 𝑍(𝑛) = 𝑧 ( 𝑧−1)2 , 𝑍(1) = 𝑧−1 𝑧 ] Given equation becomes 𝑢 𝑛+2 − 2𝑢 𝑛+1 + 𝑢 𝑛 = 3𝑛+ 5 → 𝟏 Applying Z-transform on both sides in equation 1 we get 𝑍(𝑢 𝑛+2) − 2𝑍(𝑢 𝑛+1) + 𝑍(𝑢 𝑛) = 𝑍(3𝑛 +5) 𝑧2[𝑈(𝑧) − 𝑢0 − 𝑢1 𝑧−1] − 2𝑧[𝑈(𝑧) − 𝑢0] + 𝑈(𝑧) = 3 𝑧 ( 𝑧−1)2 + 5 𝑧 𝑧−1 0𝑈(𝑧)[𝑧2 − 2𝑧 + 1] − 𝑢 [𝑧2 − 2𝑧]−𝑧2 𝑢1 𝑧 1 = 3𝑧+5𝑧( 𝑧−1) ( 𝑧−1)2 2 𝑈(𝑧)[𝑧 − 2𝑧 + 1] = 25𝑧 −2𝑧 ( 𝑧−1)2 2 0 1+ 𝑢 [𝑧 − 2𝑧] + 𝑢 𝑧 2 𝑈(𝑧)(𝑧 − 1) = 25𝑧 −2𝑧 ( 𝑧−1)2 2 0 1+ 𝑢 [𝑧 − 2𝑧] + 𝑢 𝑧 2 2 𝑈( 𝑧) = 5𝑧 −2𝑧 + 𝑢0[ 𝑧 −2𝑧] + 𝑢1 𝑧 ( 𝑧−1)4 ( 𝑧−1)2 ( 𝑧−1)2 On inversion we get 𝑢 𝑛 = 𝑍−1 [ 5𝑧2 −2𝑧 (𝑧 − 1)4 ] + 𝑢0 𝑍−1 [ 𝑧2 − 2𝑧 (𝑧 − 1)2 ] + 𝑢1 𝑍−1 [ 𝑧 (𝑧 − 1)2 ] ⟶ 𝟐
- 4. Note that 𝑍(1) = 𝑧 𝑧 − 1 , 𝑍(𝑛) = 𝑧 (𝑧 − 1)2 , 𝑍(𝑛2) = 𝑧2 + 𝑧 (𝑧 − 1)3 , 𝑍(𝑛3) = 𝑧3 + 4𝑧2 + 𝑧 (𝑧 − 1)4 ⟶ 𝟑 Now consider 5𝑧2−2𝑧 𝑧3+4𝑧2+ 𝑧 ( 𝑧−1)4 = 𝐴 [ ( 𝑧−1)4 𝑧2+𝑧 𝑧 𝑧 𝑧−1 ] + 𝐵 [ ] + 𝐶 [ ] + 𝐷[ ] ( 𝑧−1)3 ( 𝑧−1)2 = 𝐴 [ ( 𝑧−1)4 ] + 𝐵[ 𝑧3+4𝑧2+𝑧 (𝑧2+𝑧)(𝑧−1) 𝑧( 𝑧−1)2 𝑧( 𝑧−1)3 ( 𝑧−1)4 ] + 𝐶 [ ( 𝑧−1)4 ] + 𝐷[ ( 𝑧−1)4] = 𝐴 [ 𝑧3+4𝑧2+𝑧 ( 𝑧−1)4 ] + 𝐵[ ( 𝑧−1)4 𝑧3−𝑧2+𝑧2−𝑧) 𝑧(𝑧2−2𝑧+1) ( 𝑧−1)4 ] + 𝐶 [ ] + 𝐷[ 𝑧(𝑧3−3𝑧2+3𝑧−1) ( 𝑧−1)4 ] 5𝑧2−2𝑧 𝑧3+4𝑧2+ 𝑧 ( 𝑧−1)4 = 𝐴 [ ( 𝑧−1)4 ] + 𝐵 [ 𝑧3−𝑧 ] + 𝐶 [ 𝑧3−2𝑧2+𝑍 ( 𝑧−1)4 ( 𝑧−1)4 ] + 𝐷[ 𝑧4−3𝑧3+3𝑧2−𝑧 ( 𝑧−1)4 ] Equating the like powers of z we get 𝐷 = 0 (coefficients of 𝑧4) 𝐴 + 𝐵 + 𝐶 − 3𝐷 = 0(coefficients of 𝑧3) ⇒ 𝐴 + 𝐵 + 𝐶 = 0 ⟶ 𝟒 4𝐴 + 0𝐵 − 2𝐶 + 3𝐷 = 5(coefficients of 𝑧2) ⇒ 4𝐴 − 2𝐶 = 5 ⟶ 𝟓 𝐴 − 𝐵 + 𝐶 − 𝐷 = −2(coefficients of𝑧)
- 5. ⇒ 𝐴 − 𝐵 + 𝐶 = −2 ⟶ 𝟔 Solve equation 4 and 6 we get 𝐵 = 1 Put 𝐵 = 1 equation 4 we get 𝐴 + 𝐶 = −1 ⟶ 𝟕 Solve equation 3 and 2 ×equation 7 we get 𝐴 = 1 2 Put 𝐴 and 𝐵 value in equation 4 we get 𝐶 = − 3 2 2 1 3 2 ∴ 𝐴 = , 𝐵 = 1, 𝐶 = − , 𝐷 = 0 2 3 2 ∴ 5𝑧 −2𝑧 1 𝑧 +4𝑧 +𝑧 ( 𝑧−1)4 = 2 . [ ( 𝑧−1)4 𝑧2+𝑧 3 𝑧 𝑧 𝑧−1 ] + 1. [ ] − . [ ] + 0. [ ] ( 𝑧−1)3 2 ( 𝑧−1)2 ∴ 𝑍−1 5𝑧2−2𝑧 1 −1[ ] = 𝑍 [ ( 𝑧−1)4 2 ( 𝑧−1)4 ] + 𝑍−1𝑧3+4𝑧2+𝑧 𝑧2+𝑧 3 −1 𝑧 [ ] − 𝑍 [ ] ( 𝑧−1)3 2 ( 𝑧−1)2 1 3 2 3 2 2 = 𝑛 + 𝑛 − 𝑛 by equation3 1 2 2 = 𝑛(𝑛 + 2𝑛 − 3) 1 2 ( )= 𝑛 𝑛 − 1 (𝑛 + 3) 2 Consider 𝑧 −2𝑧 = 𝐴 + 𝐵 𝑧 𝑧 ( 𝑧−1)2 𝑧−1 ( 𝑧−1)2 = 𝐴 𝑧(𝑧−1) + 𝐵 𝑧(𝑧−1) 𝑧−1 ( 𝑧−1)2
- 6. 2 2 ∴ 𝑧 −2𝑧 = 𝐴 𝑧 −2𝑧 + 𝐵 𝑧 ( 𝑧−1)2 𝑧−1 𝑧−1 Equating the like powers of z we get 𝐴 + 0 = 1(coefficients of 𝑧2) ∴ 𝐴 = 1 −𝐴 + 𝐵 = −2(coefficients of 𝑧) ⇒ 𝐵 = −1 2 2 ∴ 𝑧 −2𝑧 = 1. 𝑧 −2𝑧 − 1. 𝑧 ( 𝑧−1)2 𝑧−1 𝑧−1 2 ∴ 𝑍−1 [ 𝑧 −2𝑧 ] = 𝑍−1 [ 𝑧 ] − 𝑍−1[ ( 𝑧−1)2 𝑧−1 𝑧 ( 𝑧−1)2 ] = 1 − 𝑛 (by equation 3) and 𝑍−1 [ 𝑧 ( 𝑧−1)2 ] = 𝑛 ∴ equation 2 becomes 𝑛 1 2 ( )( ) ( )𝑢 = 𝑛 𝑛 − 1 𝑛 + 3 + 𝑢0 1 − 𝑛 + 𝑢1 𝑛 1 2 ( )( )= 𝑛 𝑛 − 1 𝑛 + 3 + 𝑢0 + (𝑢1 − 𝑢0)𝑛 2 = 1 𝑛(𝑛 − 1)(𝑛 + 3) + 𝑐 + 𝑐 𝑛 where 𝑐 = 𝑢 , 𝑐 = 𝑢 − 𝑢0 1 0 0 1 1 0