Z Transform: Formation of difference equations

Formation of difference equations:

 

Def. Difference equations:

A difference equation is a relation between the differences of an unknown function at one or more general values of the argument.

Thus

 Δy_{(n + 1)} + y_(n) = 2              ……..(1)

and Δy_{(n + 1)} + Δ²y_{(n − 1)} = 1       …….(2)

are difference equations.

 

Def: Order of a difference equation

The order of a difference equation is the difference between the largest and the smallest arguments occurring in the difference equation divided by the unit of increment.

 

Def: Solution of a difference equation:

The solution of a difference equation is an expression for y(n) which satisfies the given difference equation.

 

Def: The general solution of a difference equation:

The general solution of a difference equation is that in which the number of arbitrary constants is equal to the order of the difference equation.

 

Def: The particular solution of a difference equation:

A particular solution is that the solution which is obtained from the general solution by giving particular values to the constants.

 

Example Problems - Formation of difference equations :

 

1. Form the difference equation corresponding to the family of curves y = ax + bx²

Solution:

У_х = ax + bx²    …… (1)

y_x+1 = a (x + 1) + b (x + 1)²  ………(2)

y_x+2 = a(x + 2) + b(x + 2)² ………(3)

Eliminating a and b from (1), (2) and (3), we get

 y_х[(x+1)(x+1)² ‒ (x+2)( x+1)²] − y_x+1[x(x + 2)² − x²(x + 2)] + y_x+2[x (x + 1)² ‒ x²(x + 1)] = 0

y_x(x + 1)(x+2)(x+2‒x‒1] ‒ y_x+1 [x (x + 2)] [x + 2‒x] + y_x+2x(x + 1) [x+1‒x] = 0

Hint: Expand the determinant through

 y_x(x +1)(x + 2) ‒ y_x+12x(x+2) + y_x+2x(x + 1) = 0

(x² + 3x + 2)y_x − 2 (x² + 2x) + y_x+1 + (x² + x)y_x+2 = 0

 

2. From y_n = a2ⁿ + b(‒2)ⁿ, derive a difference equation by eliminating the arbitrary constants.

Solution:

Given: y_n = a2ⁿ + b(−2)ⁿ

 y_n+1 = a2ⁿ⁺¹ + b(‒2)ⁿ⁺¹ = 2a2ⁿ ‒ 2b(‒2)ⁿ   ... (2)

y_n+2 = a2ⁿ⁺² + b(‒2)ⁿ⁺²

 = a(2ⁿ)4 + b(‒2)ⁿ(‒2)² = 4a2ⁿ + 4b(‒2)ⁿ   ... (2)

Eliminating a(2)ⁿ & b(‒2)ⁿ from (1), (2) and (3), we get

y_n(8+8) ‒ y_n+1(4‒4) + y_n+2(‒2‒2) = 0

16y_n ‒ (0) y_n+1 ‒ 4y_n+2 = 0

 y_n+2 ‒ 4y_n = 0 which is the desired difference equation.

 

3. Derive the difference equation from

 y_n = (A + Bn) (‒3)ⁿ.

Solution:

Given :

 y_n =(A + Bn) (‒3)ⁿ = A (‒3)ⁿ + Вn (‒3)ⁿ         ..... (1)

  y_n+1 =[A + B (n + 1)] (−3)ⁿ⁺¹

 = A (‒3)ⁿ(‒3) + B(n + 1)(‒3)ⁿ(‒3)

 = ‒3A (‒3)ⁿ ‒ 3B(n + 1)(‒3)ⁿ      ……. (2)

 y_{n + 2} = [A + B(n + 2)] (−3)ⁿ⁺²

= A(−3)ⁿ⁺² + B(n + 2)(−3)ⁿ⁺²

= 9A (−3)ⁿ + 9B(n + 2) (−3)ⁿ

Eliminating A(‒3)ⁿ and B(‒3)ⁿ from (1), (2) & (3), we get

 y_n [−27 (n + 2) + 27 (n + 1)] − y_n+1 [9 (n + 2) ‒ 9n] + y_n+2 [‒3 (n + 1) + 3n] = 0

 ‒27y_n ‒ 18y_n+1 ‒ 3y_n+2 = 0

 y_n+2 + 6y_n+1 + 9y_n = 0

 

4. Derive the difference equation from

u_n = A2ⁿ + Bn

Solution:

u_n = A2ⁿ + Bn     …..(1)

u_n+1 = A2ⁿ⁺¹ + B(n+1)

= A2ⁿ + (n+1)B     ……..(2)

u_n+2 = A2ⁿ⁺² + B(n+2)

= A2ⁿ + (n+2)B     ……..(2)

Eliminating A2ⁿ and B from (1), (2) and (3), we get

 u_n [2(n + 2) ‒ 4(n+1)] ‒ u_n+1[n+2‒4n] + u_n+2[n+1‒2n] = 0

u_n[‒2n] ‒ u_n+1[‒3n+2] + u_n+2[‒n+1] = 0

 (1‒n)u_n+2 + (3n‒2)u_n+1 ‒ 2nu_n = 0

 

5. Derive the difference equation from

 y_n = (A + Bn) 2ⁿ

Solution:

Given:

 y_n = (A + Bn)2ⁿ

 y_n = A2ⁿ + Вn2ⁿ   …… (1)

 y_{n +1} = A2ⁿ⁺¹ + В(n+1)2ⁿ⁺¹

= 2A2ⁿ + 2В(n+1)2ⁿ   …… (2)

y_{n +2} = A2ⁿ⁺² + В(n+1)2ⁿ⁺²

= 4A2ⁿ + 4В(n+2)2ⁿ   …… (3)

Eliminating A2ⁿ and B2ⁿ from (1), (2) and (3), we get

 y_n [8 (n+2)‒8 (n + 1)] ‒ y_n+1[4 (n + 2) ‒ 4n] + y_n+2[2(n + 1) − 2n] = 0

y_n [8] ‒ y_n+1[8] + y_n+2[2] = 0

y_n+2  ‒ 4y_n+1 + 4y_n = 0

Note: Formulae

Δy_r = y_{r +1} ‒ y_r

Δ²y_r = Δy_{r +1} ‒ Δy_r

Δ³y_r = Δ²y_{r +1} ‒ Δ²y_r

In general, Δ^py_r = Δ^p-1y_{r +1} ‒ Δ^p-1yr

 

6. Write the difference equation Δ³y_x + Δ²y_x + Δy_x + y_x = 0 in the subscript notation:

Solution:

Δ³y_x = Δ²y_x+1 ‒ Δ²y_x

Δ²y_x = Δy_x+1 ‒ Δy_x

Δy_x = y_x+1 ‒ y_x

Δy_x+1 = y_x+2 ‒ y_x+1

Δ²y_x+1 = Δy_x+2 ‒ Δy_x+1

Δy_x+2 = y_x+3 ‒ y_x+2

Given:

 Δ³y_x + Δ²y_x + Δy_x + y_x = 0

 (Δ²y_x+1 ‒ Δ²y_x) + Δ²y_x + Δу_x + y_x = 0

 Δ²y_{x +1} + Δy_x + y_x = 0

Δy_x+2 ‒ Δy_x+1 + y_x+1 ‒ y_x + y_x = 0

 y_x+3 ‒ y_x+2 ‒ y_x+2 + y_x+1 + y_x+1 = 0

 y_x+3 ‒ 2y_x+2 ‒ 2y_x+1 = 0

 

7. Find the difference equation satisfied by y = ax² ‒ bx.

Solution:

Given:

 y = ax² ‒ bx

i.e., y_x = ax² ‒ bx     .. (1)

y_x+1 = a(x + 1)² − b (x + 1)       .. (2)

y_x+2 = a(x + 2)² − b(x + 2)       .. (3)

Eliminating a and b from (1), (2) and (3), we get

 y_x [ − (x + 1)² (x + 2) + (x + 2)² (x + 1)] − y_x+1[−x²(x + 2) + x(x + 2)²] + y_x+2[ ‒x²(x + 1) + x (x + 1)²] = 0

 y_x(x + 1)(x+2) [−x ‒1+x+2] ‒ y_x+1x(x + 2) [−x + x + 2] + y_x+2x(x + 1) [‒x+x+1] = 0

⇒ y_x(x + 1)(x + 2) ‒ y_x+1x(x+2)² + y_x+2(x+1) = 0

⇒ (x + 1)(x+2)y_x ‒ 2x (x+2)y_x+1 + x(x+1)y_x+2 = 0

 

8.  Form the difference equation generated by y_x = ax + b2^x

Solution: Given: y_x = ax + b2^x  …….. (1)

 y_x+1 = a (x+1) + b2^x+1

= (x+1)a + 2b2^x  …….. (1)

y_x+2 = a(x+2) + b2^x+2

= (x + 2)a + 4b2^x …….. (1)

Eliminating a and b2^k from (1), (2) and (3), we get

 y_x[4(x+1)‒2(x+2)] ‒ y_x+1[4x‒(x+2)] + y_x+2[2x − (x + 1)] = 0

 y_x[2x] ‒ y_x+1[3x‒2] + y_x+2[x‒1] = 0

 (x‒1)y_x+2 + (‒3x + 2)y_x+1 + 2xy_x = 0

 

9. Form the difference equation generated by y_x = a2^x + b3^x + c.

Solution: Given:

y_x = a2^x + b3^x + c    …..(1)

y_x+1 = a2^x+1 + b3^x+1 + c = 2a2^x + 3b3^x + c   …..(2)

y_x+2 = a2^x+2 + b3^x+2 + c = 4a2^x + 9b3^x + c   …..(3)

y_x+3 = 8a2^x +27b3^x + c   …..(4)

Eliminating a 2^x, b3^x, c from (1), (2), (3) and (4), we get

 (y_x+1 ‒ y_x) [78‒56] ‒ 1 [26 (y_x+2 ‒ y_x) ‒ 8(y_x+3 ‒ y_x)] + 2[7(y_x+2 ‒ y_x) ‒ 3(y_x+3 ‒ y_x)] = 0

 22y_x+1 ‒ 22y_x ‒ 26y_x+2 + 26y_x + 8y_x+3 ‒ 8y_x + 14y_x+2 ‒ 14y_x ‒ 6y_x+3 + 6y_x = 0

2y_x+3 ‒ 12y_x+2 + 22y_x+1 ‒ 12y_x = 0

i.e., y_x+3 ‒ 6y_x+2 + 11y_x+1 ‒ 6y_x = 0

 

10. Form the difference equation from y_n = a + b3ⁿ.

Solution:

 Given

 y_n = a + b3ⁿ    …(1)

 y_n+1 = a + b3ⁿ⁺¹ = a + (b)(3)3ⁿ   …(2)

 y_n+2 = a + b3ⁿ⁺² = a + (b)(9)3ⁿ   …(3)

 Eliminating a and b3ⁿ from (1), (2) and (3) we get

 y_n[9 − 3] − y_n+1 [9 ‒ 1] + y_n+2 [3−1]  = 0

 6y_n ‒ 8y_n+1 + 2y_n+2 = 0

 y_n+2 ‒ 4y_n+1 + 3y_n = 0

 

11. Form the difference equation from u_n = a2ⁿ⁺¹_.

Solution:

Given: u_n = a2ⁿ⁺¹           ... (1)

 u_n+1 = a 2^(n+1)+1 = a 2ⁿ⁺¹2         ... (2)

Eliminating a2ⁿ⁺¹ from (1) & (2) we get

i.e., 2u_n ‒ u_n+1 = 0

i.e., u_n+1 ‒ 2u_n = 0