Solution of difference equations using Z-transform

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.

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

Solve :

1. Solve yn+1 -2yn = 0 given y0 = 3

Solution: Given: yn+1 -2yn = 0

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

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

Solution: Given: yn+2 – 4yn = 0

3. Solve yn+2 - 4yn = 0

Solution: Given

4. Using Z-transform, solve

un+2 + 3un+1 + 2un = 0 given u0 = 1, u1 = 2.

Solution: Given:

5. Solve the difference equation

y (n + 3) − 3y (n + 1) + 2y(n) = 0 given that y(0) 4, y(1) = 0 and y(2) = 8. 

Solution: 

6. Solve Yn + 2 - 2 cos α Yn +1 + Yn = 0 given that y0 = 1, y1 = cos α.

Solution: 

7. Solve the difference equation

y (k + 2) - 4y (k + 1) + 4y (k) = 0 where y(0) = 1, y (1) = 0.

Solution :

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

Solution:

Let 3z + 7 = A (z − 1) + B (z + 4)

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

Solution: 

Let 1 = A (z − 2) + B (z − 1)

10. Using Z-transform method solve yn+2 + Yn = 2 given that y0 = = y1 = 0.

Solution: 

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

Solution: 

12. Using the Z-transform, solve

un+2 + 4un+1 + 3un = 2n with u0 = 0, u1 = 1

Solution: 

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

Solution: 

14. yn + 2 + 4yn +1 + 3yn = 3n with y0 = 0, y1 = 1

Solution: 

15. Solve yn+2 + Yn = n2n.

Solution: 

equating z2 on both sides, we get

16. Using Z-transform, solve yn+2 + 4yn+1 − 5yn = 24n - 8 given that y0 = 3 and yı = -5.

Solution: 

put z = -5, we get

-375 + 25 + 95 + 39 = A (-6)3

-216 = -216 A

A = 1

put z = 1, we get

3 + 1 - 19 + 39 = D(6)

24 = D(6)

D = 4

equating z3 on both sides, we get

3 = A+ B

3 = 1 + B

B = 2

put z = 0, we get

0 + 0 + 0 + 39 = -A + 5B - 5C + 5D

39 = −1 + 10 – 5C + 20

39 = 29 - 5C 

5C = 29 - 39

5C = -10

C = -2

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

Solution :

18. Find the response of the system :

yn + 2 - 5yn+1 + 6yn = un, with y0 = 0, y1 = 1 and un = 1 for n = 0, 1, 2, by Z-transform method. 

Solution: Given: yn +2 - 5yn +1 + 6yn = un

Taking Z-transform of both sides of the given equation we get

19. Solve the simultaneous difference equations xn+1 = 5xn+7,

yn+1 = xn + 2yn given that x0 = 0, and y0 = 1.

Solution: 

20. Solve the system using Z-transform

xn+1 = 7xn + 10yn

yn+1 = xn + 4yn given that x0 = 3, y0 = 2.

Solution: 

z = 2 is a simple pole

z = 9 is a simple pole

z = 2 is a simple pole

z = 9 is a simple pole

EXERCISE - 5.1

I. Find the Z-transform of the following sequences.

EXERCISE - 5.2

1. Find the inverse Z-transform of the following sequence by any method.

EXERCISE - 5.3

Solve the following equations, using Z-transform.

Miscellaneous Problems

Find the bilateral Z - transform of the following :

1. Find

Solution: 

2. Find Z{f(n)} if_

Solution :

3. Find Z [f(n)] if f(n) =

Solution :

4. Find the Z-transform of

Solution : 

5. Find the Z-transform of the sequence x(n) = {1, 2, 3, 4, 0, 6, 7}

Solution : 

As x (n) has values from n = 0 to n = 6

X (z) has finite values except at z = 0.

At z = 0, X (z) is infinite.

Hence, X (z) is convergent for all values of z, except z = 0

⸫ The region of convergence (ROC) is the entire z plane except z = 0.

6. Find Z [et sin 2 t]

Sol. 

7. If then find ƒ (0), using initial value theorem.

Solution : By initial value theorem