Application: Solution of difference equations using Z transform

Application: Solution of difference equations using Z‒transform.

We know that Laplace Transforms are very useful to solve linear differential equations

The Z‒transforms are useful to solve linear difference equations.

Formula:

(1) Z[y_n]  = Y(z)

(2) Z [y_n+1] = zY(z) ‒ zy (0)

(3) Z [y_n+2] = z²Y(z) ‒ z²y(0) ‒ zy(1)

(4) Z [y_n+3] = z³Y(z) ‒ z²y(0) ‒ z²y(1) ‒ zy(2)

(5) Z [y_n‒1] = z^‒1Y(z)

Standard formulae:

Problems based on Solution of the difference equations using Z‒transform

Solve:

1. y_n+1 ‒ 2y_n = 0 given y₀ = 3

2. y_n+2 ‒ 4y_n = 0 given y₀ = 0, y₁ = 2

3. y_n+2 ‒ 4y_n = 0

4. u_n+2 + 3u_n+1 + 2u_n = 0 given u₀ = 1, u₁ = 2

5. y (n + 3) ‒ 3y (n + 1) + 2y (n) = 0 given y(0) = 4, y(1) = 0, y (2) = 8

6. y_n+2 ‒ 2cos a y_n+1 + y_n = 0 given y₀ = 1, y₁ = cosa

7. y(k + 2) − 4y(k + 1) + 4y(k) = 0 given y(0) = 1, y(1) = 0

8. y(n) + 3y(n‒1) ‒ 4y(n‒2) = 0 given y(0) = 3, y(1) = ‒2, n ≥ 2

9. x(n + 1) − 2x(n) = 1, given x(0) = 0

10. y_n+2 + y_n = 2 given y₀=y₁ = 0

11. y_n+2 + 6у_n+1 + 9y_n = 2ⁿ given y₀ = y₁ = 0

12. u_n+2 + 4u_n+1 + 3u_n = 2ⁿ given u₀ = 0, u₁ = 1

13. u_n+2 ‒ 5u_n+1 + 6u_n = 4ⁿ given u₀ = 0, u₁ = 1

14. y_n+2 + 4y_n+1 + 3y_n = 3ⁿ given y₀ = 0, y₁ = 1

15. y_n+2 + y_n = n2ⁿ

16. y_n+2 + 4y_n+1 ‒ 5y_n = 24n ‒ 8 given y₀ = 3, y₁ = ‒5

17. y(n) − y(n − 1) = u(n) + u(n − 1) given u(n) = n, u(n − 1) = n – 1

18. y_n+2 − 5y_n+1 + 6y_n = u_n, y₀ = 0, y₁ = 1, u_n = 1

19. x_n+1 = 5x_n+7; y_n+1 = x_n + 2y_n,  x₀ = 0, y₀ = 1

20. x_n+1 = 7x_n + 10y_n, y_n+1 = x_n + 4y_n, x₀ = 3, y₀ = 2

1. Solve y_n+1 ‒ 2y_n = 0 given y₀ = 3

Solution:

Given: y_n+1 ‒ 2y_n = 0

Taking Z‒transform on both sides of the difference equation, we get

 Z[y_n+1] ‒ 2Z [y_n] = Z(0)

2. Using Z-transform, solve y_n+2 ‒ 4y_n = 0 given that y₀ = 0, y₁ = 2

Solution:

3. Solve y_n+2 ‒ 4y_n = 0

Solution:

4. Using Z-transform, solve u_n+2 + 3u_n+1 + 2u_n = 0 given u₀ = 1, u₁ = 2

Solution:

5. Solve the difference equation y (n + 3) ‒ 3y (n + 1) + 2y (n) = 0 given y(0) = 4, y(1) = 0, y (2) = 8

Solution:

6. Solve y_n+2 ‒ 2cos a y_n+1 + y_n = 0 given that y₀ = 1, y₁ = cosa

Solution:

7. Solve the difference equation y(k + 2) − 4y(k + 1) + 4y(k) = 0 given y(0) = 1, y(1) = 0

Solution:

8. Using Z-transform, solve y(n) + 3y(n‒1) ‒ 4y(n‒2) = 0 given y(0) = 3, y(1) = ‒2, n ≥ 2.

Solution:

9. Solve x(n + 1) − 2x(n) = 1, given x(0) = 0

Solution:

10. Using Z-transform, method solve y_n+2 + y_n = 2 given y₀=y₁ = 0

Solution:

11. Solve y_n+2 + 6у_n+1 + 9y_n = 2ⁿ given y₀ = y₁ = 0

Solution:

12. Using Z-transform, solve u_n+2 + 4u_n+1 + 3u_n = 2ⁿ given u₀ = 0, u₁ = 1

Solution:

13. Using Z-transform, solve u_n+2 ‒ 5u_n+1 + 6u_n = 4ⁿ given that u₀ = 0, u₁ = 1

Solution:

14. Solve y_n+2 + 4y_n+1 + 3y_n = 3ⁿ given y₀ = 0, y₁ = 1

Solution:

15. Solve y_n+2 + y_n = n2ⁿ

Solution:

16. Using Z-transform, solve y_n+2 + 4y_n+1 ‒ 5y_n = 24n ‒ 8 given that y₀ = 3, y₁ = ‒5

Solution:

17. Solve y(n) − y(n − 1) = u(n) + u(n − 1) given u(n) = n, u(n − 1) = n – 1

Solution:

18. Find the response of the system: y_n+2 − 5y_n+1 + 6y_n = u_n, y₀ = 0, y₁ = 1, u_n = 1 for n=0,1,2,… by Z-transform method.

Solution:

19. Solve the simultaneous difference equation x_n+1 = 5x_n+7; y_n+1 = x_n + 2y_n given that x₀ = 0, y₀ = 1.

Solution:

20. Solve the system using Z-transform x_n+1 = 7x_n + 10y_n, y_n+1 = x_n + 4y_n given that x₀ = 3, y₀ = 2

Solution: