5 Z-Transform and difference equations

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1 5 Z-Transform and difference equations Z-transform - Elementary properties - Inverse Z-transform - Convolution theorem - Formation of difference equations - Solution of difference equations using Z-transform. 5. Z-Transform 5.. Two sided or bilateral Z x (n) = n n x (n) n = X () where, x(n) is a sequence of numbers with n (all integers) is a complex number One sided or unilateral Z x (n) = n n=0 x (n) n = X () where, n = 0,,... and is a complex number Discrete values of t = nt Z x (t) = where t = nt, n = 0,,... n n=0 T = sampling period is a complex number. Note :, ], ( ) are same. Formulae : x (t) n = n n=0 = a function of = X () x (nt ) n 275

2 276 Unit V : Z-Transform and difference equations(z.t.&d.e.). + x + x = ( x), x < 2. x + x 2... = ( + x), x < x + 3x = ( x) 2, x < 4. 2x + 3x 2... = ( + x) 2, x < 5. x + x2 2 + x = log ( x) = log ( x) = log 6. x0 0! + x! + x2 2! +... = ex ] ( x) 5..4 Problems under Z - transforms Find the values of the following: I set: constants (Use Unilateral Z-transform formula). Z (), > 2. Z (2) 3. Z ( 3) 2 ( ( 3 II set: power n (Use Unilateral Z-transform formula) 4. Z (a n ) 5. Z (3 n ) 6. Z ( 3) n ] 7. Z (e an ) 8. Z ( e an) ), > ), >, > a a 3, > 3, > e a, > ea e a, > e a III set : n (or) n (or) n! 9. Z (n) (Use Unilateral Z-transform formula) ( ) 2, >

3 MA635 Transforms and Partial Differential Equations by K A Niranjan Kumar 277 ( ) ( ) 0. Z log n ( ) ( ). Z log n + ( ) 2. Z ( )] log n ( ) 3. Z e / n! ( ) ( ) 4. Z e / (n + )! 5. Z (n + 2) ( ) IV set : Modulus (Use Bilateral Z-transform formula) ) 6. Z (a n a a + a ( ( ) ) n 5 ( + 5) 7. Z 5 (5 ) V set : θ (Use Unilateral Z-transform formula) ( cos θ) 8. Z (cos nθ) 2 2 cos θ + sin θ 9. Z (sin nθ) 2 2 cos θ + ( 20. Z cos nπ ) ( 2. Z sin nπ ) Z (r n ( r cos θ) cos nθ) 2 2r cos θ + r Z (r n r sin θ sin nθ) 2 2r cos θ + r 2 VI set : Partial Fractions (Use Unilateral Z-transform formula)

4 278 Unit V : Z-Transform and difference equations(z.t.&d.e.) ( 24. Z n (n + ) ( n Z n (n + ) ( 26. Z (n + ) (n + 2) ) ( ) log ) log ) ( 2) log ( 2 + 3) + VII set : t (Use Unilateral Z-transform formula for t ) 27. Z (t) T ( ) Z ( e at) e at 29. Z ( e at) e at 30. Z ( e at+2) e 2 e at sin ωt 3. Z (sin ωt) 2 2 cos ωt Properties of Z-transforms: Property : Linearity property If Z x (n)] = X (), Z y (n)] = Y (), then Z ax (n) + by (n)] = az x (n)] + bz y (n)] = ax () + by () Property 2 : Change of scale If Z x (n)] = X (), Z y (n)] = Y (), then Z a n x (n)] = Z x (n)] a = X ()] a Property 3 : Differentiation in Z - domain If Z x (n)] = X (), Z y (n)] = Y (), then ( = X a) Z nx (n)] = d d Z x (n)] = d d X () Property 4 : First shifting propertyfrequency shifting] If Z x (n)] = X (), Z y (n)] = Y (), then

5 MA635 Transforms and Partial Differential Equations by K A Niranjan Kumar 279 (i)z e at x (t) ] = Z x (t)] e at = X ()] e at (ii)z e at x (t) ] = Z x (t)] e at = X ()] e at Property 5 : Second shifting propertyuse in solving difference equation] If Z x (n)] = X (), Z y (n)] = Y (), then (i)z x (n + )] = Z x (n)] x (0) = X () x (0)] (ii)z x (n + 2)] = 2 Z x (n)] 2 x (0) x () = 2 X () x (0) x () ] (iii)z x (n + 3)] = 3 Z x (n)] 3 x (0) 2 x () x (2) = 3 X () x (0) x () x (2) ] 2 (iv)z x (n k)] = k Z x (n)] = k X () 5.2. Problems under Properties VIII set... In general 32. Z ( n 2) + ( ) Z (ne an e a ) ( e a ) Z ( n 3) ( ) 35. Z (n + ) (n + 2)] =Z ( n 2 + 3n + 2 ) ( + ) ( ) ( ) ( ) 4 IX set ( ) a n 36. Z e ( ) a n! ( ) 2 n 37. Z ( ) n + 2 log 2 X set e at 38. Z ( e at t ) T (e at ) 2

6 280 Unit V : Z-Transform and difference equations(z.t.&d.e.) 39. Z ( ( t 2 e t) T 2eT e T + ) 40. Z ( e iat) (e T ) 3 e iat e iat Initial value theorem If Z f (t)] = F (), then f (0) = lim F () Final value theorem If Z f (t)] = F (), then lim f (t) = lim ( ) F () t Problems under initial and final value theorems Use initial value theorem and final value theorem, find f (0), lim t f (t) 4. F () = 2 e T 2, F () = ( cos at ) 2 2 cos at +, Inverse Z-transforms If Z x (n)] = X (), then Z X ()] = x (n) 5.3. Methods to solve inverse Z-transforms (i) Partial fraction method (iv) Expansion (ii) Cauchy residue method (v) Power series (iii) Convolution method (vi) Long division

7 MA635 Transforms and Partial Differential Equations by K A Niranjan Kumar Partial fraction method If Z X ()], is given: Step (i) : Consider X () and then apply partial fraction. Step (ii) : After partial fraction method, take to to R.H.S., then apply Z on both sides Cauchy s Residue theorem : Z X ()] = x (n) = sum of the residues If Z X ()], is given: Step (i) : Consider X (). Then multiply by n both sides. i.e., X () n = R.H.S. Step (ii) : Simplify the R.H.S. and then equate the denominator = 0. Then, we will get poles. According to the poles we will get residues. Definition Residue of the simple pole (pole of order ) Res =a X () = lim a ( a) R.H.S.] Residue of the double pole (pole of order 2) = Res ( =a twice Residue of the pole of order m ) X () d = lim ( a) 2 R.H.S.] a d = ( =a Res ) X () d m = lim a m times d ( m a)m R.H.S.] Problems under Z :by partial fraction method (or/and ) by residue method] ] 43. Z 0 ( )( 2) ( 44. Z ) ] ( + )( ) 2 x(n) = 0 2 n ] x(n) = ( ) n + n

8 282 Unit V : Z-Transform and difference equations(z.t.&d.e.) 5.4 Convolution theorem Z X () Y ()] = Z X()] Z Y ()] Note : Convolution of x(n) & y(n): n x(n) y(n) = x(k)y(n k) k=0 Here one n = k, another n = (n k)] x n+ Formula : + x + x x n = x, x > xn+ x, x < 45. Z 2 ] bn+ a n+ ( a)( b) b a 46. Z 2 ] (n + )( a) n ( + a) Z 8 2 ] (2) n 4 (2 )(4 ) n Difference equations : Difference equations classified as (i) Formation of difference equation (ii) Solution of difference equation 5.5 Formation of difference equation Note : Recall the following : x = (x + ) x = (x + ) = (x + 2) (x + ) = 2 (x) = x] = (x + ) x] = () = 0 (x 2 ) = (x + ) 2 x 2 = 2x + x(n)] = x(n + ) n x(n + )] = x(n + 2) x(n + ) 2 x(n)] = x(n)] = x(n + ) x(n)] = x(n + )] x(n)] = x(n + 2) x(n + ) x(n + ) x(n)] = x(n + 2) 2x(n + ) + x(n)

9 MA635 Transforms and Partial Differential Equations by K A Niranjan Kumar Problems under Formation of difference equation by eliminating arbitrary constants 48. y n = a2 n + b( 2) n y n+2 4y n = y = ax + bx 2 ] y = y 2 y (2x + ) x + 2 y 2 2 x2 5.6 Solution of difference equation using Z - transform Procedure for solution of difference equation (i) : Apply Z-transform on both sides of given difference equation. (ii) : Expand formulae. (iii) : Take Zx(n)] as one side of the equation and remaining other side. (iv) : Take Z on both sides. (v) : Apply partial fraction (or) residue method to find Z. To solve difference equation, recall the following:(property 5) If Z x (n)] = X (), Z y (n)] = Y (), then (i)z x (n + )] = Z x (n)] x (0) = X () x (0)] (ii)z x (n + 2)] = 2 Z x (n)] 2 x (0) x () = 2 X () x (0) x () ] (iii)z x (n + 3)] = 3 Z x (n)] 3 x (0) 2 x () x (2) = 3 X () x (0) x () x (2) ] 2 (iv)z x (n k)] = k Z x (n)] = k X ()... In general 5.6. Problems under solution of difference equation using Z-transform 50. y(n + 2) 4y(n + ) + 4y(n) = 0 with y(0) =, y() = 0 y(n) = 2 n ( n) 5. y(n + 2) + y(n) = 2 with y(0) = y() = 0 y(n) = in + i + ( i)n i 52. y(n + 3) 3y(n + 2) + 2y(n) = 0 with y(0) = 4, y() = 0, y(2) = 8 y(n) = ( 2)n ]

10 284 Unit V : Z-Transform and difference equations(z.t.&d.e.) 53. y(n + 2) + 6y(n + ) + 9y(n) = 2 n with y(0) = y() = 0 y(n) = 2 n 5n( 3) n ( 3) n] u n+2 + 4u n+ + 9u n = 3 n with u 0 = u = 55. x n+2 2x n+ + x n = 3n + 5 with x 0 = x = 56. y(n) + 6y(n ) + 9y(n 2) = 2 n with y(0) = y() = 0 y(n) = 2 n 5n( 3) n ( 3) n] Assignment VZ-Transform and difference equations]. Find the Z-transform of cos nθ and sin nθ. Hence deduce the Z-transform of a n cos nθ and a n sin nθ. 2. Find Z (na n sin nθ). ( ( 3. Find ) ) ( + ) ( ) 2 by using method of partial fraction. ( 4. Find Z ) ] ] and Z ( + )( ) 2 ( )( 2) 5. Using convolution theorem, find the Z 2 ] ( 4) ( 3) 6. Using convolution theorem find the inverse Z-transform of 2 2 (3 ) (4 + ) 7. Find the inverse Z-transform of ( + ) by residue method. ( ) 3 8. Derive the difference equation from y n = (A + B n )( 3) n. 9. Solve the difference equation using Z-transform y (n+3) 3y (n+2) + 2y (n) = 0 given that y 0 = 4, y = 0, y 2 = Solve y (n+2) + 3y (n+) + 9y (n) = 2 n given that y 0 = y = 0.