Z-TRANSFORMATION Inverse Z-Transform:

If Z {u n } = U(z) , then inverse Z-Transform of

U(z) is defined by Z −1{U(z)} = u n .

To obtain inverse Z-Transform there are three methods i.e. (i)

Power series method,

(ii)

Partial fraction method and

(iii)

Inversion integral method or Residues method.

Inversion method or Residues Method: The inverse Z-Transform of

un =

1

U(z) is given by the formula

U (z).z 2π i ∫

n −1

C

= Sum of residues of Poles: If

dz

U (z).z n −1 at the poles of U(z) .

U(z) is rational function i. e. U(z) =

P (z) , then poles of U(z) is the values of z for which Q(z) is Q(z)

zero. Now let a and b are two poles of U(z) , then residues of U (z).z n −1 is defined as

Re s (z= a)= lim z →a (z − a)[U (z).z n −1 ] Similarly at b

Re s (z= b= ) lim z →b (z − b)[U (z).z n −1 ]

2 z2 + 3 z Que. Find the inverse Z-transform of . (z + 2) (z − 4)

Dr. Jogendra Kumar

Page 1

Z-TRANSFORMATION Soln. Let

2 z2 + 3 z = U (z) . Clearly, z = −2 and z = 4 are the poles of U(z) . (z + 2) (z − 4)

∴ Z−1{U(z)} = Sum of residues of U (z).z n −1 at the poles of U(z) . Now,

 2 z2 + 3 z  n −1 Re = −2) lim z →−2 (z + 2)[U (z).z = ] lim z →−2 (z + 2)  . z n −1  s (z =  (z + 2) (z − 4)  2 n +1 n 2 z + 3 z n −1 2. z + 3 z .z lim z →−2 = lim = z →−2 (z − 4) z−4 1 1 n = −  2.(−2) n +1 + 3 ( −2 )  = − (−2) n [2.(−2) + 3]  6 6 1 (−2) n = 6 and

 2 z2 + 3 z  Re s (z = 4) = lim z →4 (z − 4)[U (z).z n −1 ] = lim z →4 (z − 4)  . z n −1   (z + 2) (z − 4)  2 n +1 n 2 z + 3 z n −1 2z +3z = lim = .z lim z →4 z →4 (z + 2) (z + 2) 2.4n +1 + 3.4n 1 n = .4 (2.4 + 3) (4 + 2) 6 11 = .4n 6

=

1 11 ∴ Z−1{U(z)} = (−2) n + (4) n 6 6

Dr. Jogendra Kumar

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Z-TRANSFORMATION

Note: (i) This method is very useful but applies this method when Q(z) has only linear facto otherwise use partial fraction method.

(ii) In partial fraction method we write U(z) in terms of known result of inverse Z-Transform. We find partial fraction of

U(z) in place of partial fraction of U(z) . z

Que. Find the inverse Z-Transform of Soln. Let U ( z ) =

2 z ( z − 1)( z 2 + 1) .

2z . It has three poles at z = 1, − i, i . ( z − 1)( z 2 + 1)

Now, we will find the residues of U ( z ). z n −1 at these poles.

2z 2 z.z n −1 zn n −1 Re s ( z = 1) = lim z →1 ( z − 1). . z lim lim 2. = = z →1 z →1 ( z − 1)( z 2 + 1) ( z 2 + 1) ( z 2 + 1) (1) n = 2.= 1, (1 + 1) 2z 2 z.z n −1 i) = Re s ( z = lim z →i ( z − i ). .z n −1 = lim z →i ( z − 1)( z − i ) ( z + i ) ( z − 1) ( z + i ) = lim z →i 2. = −

zn (i) n (i) n (i) n = 2. = = ( z − 1)( z + i ) (i − 1)(2i) (i 2 − i) (−1 − i)

(i) n (1 + i)

and

Dr. Jogendra Kumar

Page 3

Z-TRANSFORMATION Re s ( z =−i ) =lim z →− i ( z + i ). = lim z →− i 2.

2z 2 z.z n −1 .z n −1 =lim z →− i ( z − 1)( z − i ) ( z + i ) ( z − 1) ( z − i )

zn ( z − 1)( z − i )

(− i) n (− i) n = 2.= (−i − 1)(−2i) (− i 2 + i) =

(− i) n (− i) n = (−1 + i) (i − 1)

(i) n (− i) n ∴ Z −1{U(z)} = + 1− un = Sume of residues = (1 + i ) (i − 1)

Application of Z-Transform (Difference Equations): Z-Transform plays an important role in the analysis and representation of discrete time system. For such systems, the solution of difference equations is required for which Z-Transform is very useful.

Working Rule: To solve a linear difference equation with constant coefficient by Z-Transform (i) Take the Z-Transform of both sides of the difference equation. (ii) Transpose all terms without

U (z) to the right.

(iii) Divide by the coefficient of U (z) , getting

U (z) in terms of z .

(iv) Now take inverse Z-Transform of both sides in result of which we get

un as a function n which is desired

solution. Que. Using the Z-Transform, solve

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

Soln. Let Z {u n } = U (z) . Now taking Z-Transform of

un + 2 + 4 un +1 + 3 un = 3n , i. e.

Z {un + 2 } + 4 Z { un +1} + 3Z { un } = Z {3n } , we get

Dr. Jogendra Kumar

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Z-TRANSFORMATION

z z 2 [U(z) − u 0 − u1 z −1 ] + 4z[U(z) − u 0 ] + 3 U(z) = ,(By using Left Shifting property of Z-Transform) z −3

Which on simplifying and using the given conditions, gives?

z z ⇒ (z + 3) (z + 1) U (z) = z + z −3 z −3 2 z −2z U (z) = (z + 1)(z + 3)(z − 3)

( z 2 + 4 z + 3) U(z) − z = ⇒ Thus,

U (z) have three poles at z = −1, z = −3, z = 3 . Now, residues of U (z).z n −1 at these poles are

Re s (z = −1) = lim z →−1 (z + 1).

z2 − 2 z z n +1 − 2 z n .z n −1 = lim z →−1 (z + 1)(z + 3)(z − 3) (z + 3)(z − 3)

 −1 − 2  3 (−1) n +1 − 2 (−1) n = = = (−1) n .  .( −1) n ,  (−1 + 3)(−1 − 3)  2 (.−4 ) 8 z2 − 2 z z n +1 − 2 z n n −1 Re s (z = −3) = lim z →−3 (z + 3). .z = lim z →−3 (z + 1)(z + 3)(z − 3) (z + 1)(z − 3)  −3 − 2 (−3) n +1 − 2 (−3) n = = (−3) n .  (−3 + 1)(−3 − 3)  (−2 .() −6

 5 − .( −3) n , ) = 1 2 

And

z2 − 2 z z n +1 − 2 z n Re s (z = 3) = lim z →3 (z − 3). .z n −1 = lim z →3 (z + 1)(z + 3)(z − 3) (z + 3)(z + 1) (3) n +1 − 2 (3) n 1 n  3− 2  = = (3) = . .(3) n  (3 + 3)(3 + 1)  (6).(4)  24 3 1 5 ∴ Z −1{U(z)} = un = Sume of residues = .( −1) n + .(3) n − .( −3) n , which is desired solution. 8 24 12 Que. Using the Z-Transform, solve

yn + 2 − 5 yn +1 + 6 yn = un , with y0 = 0, y1 = 1 and un = 1for n = 0,1, 2,3,...... . Dr. Jogendra Kumar

Page 5

Z-TRANSFORMATION Soln. Let Z {y n } = Y (z) . Now, taking Z-Transform of

yn + 2 − 5 yn +1 + 6 yn = un i. e.

Z {y n + 2 } − 5 Z { yn +1} + 6 Z { yn } = Z {u n } , we get

z    Z {u n } =  z −1  

z z 2 [Y(z) − y0 − y1 z −1 ] − 5 z[Y(z) − y0 ] + 6Y (z) = z −1 Which on simplifying and using given conditions, we get

( z 2 − 5 z + 6) Y(z) = −z

z ⇒ ( z 2 − 5 z + 6) Y(z) = −z z −1

z z −1

z z2 ⇒ ( z − 5 z + 6) Y(z) = z + = z −1 z −1 z2 ⇒ Y(z) = ( z − 1) (z − 2) (z − 3) 2

Thus,

Y(z) have three poles at= z 1,= z 2,= z 3 . Now, residues of Y(z).z n−1 at these poles are

z2 z n +1 Re s (z = 1) = lim z →1 (z − 1). .z n −1 = lim z →1 (z − 1)(z − 2)(z − 3) (z − 2)(z − 3) (1) n +1 1 , = = (1 − 2)(1 − 3) 2 z2 z n +1 Re s (z = 2) = lim z →2 (z − 2). .z n −1 = lim z →2 (z − 1)(z − 2)(z − 3) (z − 1)(z − 3) =

(2) n +1 n = − 2. ( 2 ) , (2 − 1)(2 − 3)

And

Dr. Jogendra Kumar

Page 6

Z-TRANSFORMATION

z2 z n +1 Re s (z = 3) = lim z →3 (z − 3). .z n −1 = lim z →3 (z − 1)(z − 2)(z − 3) (z − 2)(z − 1) =

(3) n +1 3 .(3) n = (3 − 2)(3 − 1) 2

1 3 ∴ Z −1{Y(z)} = yn = Sume of residues = − 2.(2) n + .(3) n , which is desired solution. 2 2 Que. Using the Z-Transform, solve

un + 2 − 2 un +1 + un = 3 n + 5.

Soln. Let Z {u n } = U (z) . Now taking Z-Transform of

un + 2 − 2 un +1 + un = 3 n + 5 , i. e.

Z {un + 2 } − 2 Z { un +1} + Z { un } =Z {3 n + 5} ⇒ Z {un + 2 } − 2 Z { un +1} + Z { un } = 3 Z { n} + 5 Z {1} Which on simplifying, gives

z z + 5. 2 z −1 (z − 1) z z u1 z 3. ⇒ ( z 2 − 2 z + 1) U(z) − ( z 2 − 2 z ) u 0 −= + 5. 2 z −1 (z − 1) z 2 [U(z) − u 0 − u1 z −1 ] − 2 z [U(z) − u 0 ] += U(z) 3.

5 z2 − 2 z ⇒ (z − 1) 2 U (z) − ( z 2 − 2 z ) u 0 − u1 z = 2 (z − 1) ⇒ U (z)=

z 5 z2 − 2 z ( z 2 − 2 z) u . + + u1. 0 4 2 (z − 1) (z − 1) (z − 1) 2

On inversion, we get 2 5 z2 − 2 z   z  −1  z − 2 z  , + + u1.Z−1  Z −1{U (z)} = un = Z −1  u .Z  0 4  2  2 − − − (z 1) (z 1) (z 1)      

Note. We cannot use residues method since there are no only linear factors in denominator, so we will

use partial fraction method.

Dr. Jogendra Kumar

Page 7

Z-TRANSFORMATION Since we know that

z z z2 + z z3 + 4 z 2 + z 2 3 = Z {1} = , Z {n} = , Z {n } = , Z {n } z −1 (z − 1) 2 (z − 1)3 (z − 1) 4 We can write ,

5 z2 − 2 z z3 + 4 z 2 + z z2 + z z z = A. + B. + C. + D. 4 4 3 2 (z − 1) (z − 1) (z − 1) (z − 1) z −1 On equating the coefficient of like powers, we get

1 3 A= , B= 1, C = − and D = 0 2 2



5 z 2 − 2 z 1 z3 + 4 z 2 + z z 2 + z 3 z and therefore . = + − . 4 4 3 (z − 1) 2 (z − 1) (z − 1) 2 (z − 1) 2

2  5 z 2 − 2 z  1 −1  z 3 + 4 z 2 + z   z  3 −1  z + z  Z −1  = .Z + Z − . Z −1     4  4 3 2  (z − 1)   (z − 1)  2  (z − 1)   (z − 1)  2 1 3 1 = . n3 + n 2 − . n= . n (n 2 + 2 n − 3) 2 2 2 1 . n (n − 1)(n + 3) = 2

Now, we can write

z2 − 2 z = (z − 1) 2 =

z (z − 2) z{(z − 1) − 1} = = 2 (z − 1) (z − 1) 2 z z − (z − 1) (z − 1) 2

z (z − 1) z − 2 (z − 1) (z − 1) 2

 z2 − 2 z  z   z  −1  Z −1  ∴ Z−1  = −Z  2  2  z− 1  (z − 1)   (z − 1)  = 1− n Dr. Jogendra Kumar

Page 8

Z-TRANSFORMATION And obviously,

 z  = n . Therefore, Z−1  2  (z − 1) 

2 5 z2 − 2 z   z  −1  z − 2 z  Z −1{U (z)} = un = Z −1  u .Z + + u1.Z−1    0 4 2  2  (z − 1)   (z − 1)   (z − 1)  1 = . n (n − 1) (n + 3) + u 0 .(1 − n) + u1 . n 2 1 = . n (n − 1) (n + 3) + c0 + c1. n, where c0 = u0 and c1 = u1 − u0 2

n

Que. Using Z-Transformation solve

1 1 0. yn +1 + y= n 0   (n ≥ 0), y= 4 4 n

1 1 Soln. Let Z {y n } = Y (z) . Now taking Z-Transform of yn +1 + yn =   , i.e. 4 4 n 1  1   Z {y n +1} + Z { yn } = Z    4  4  

⇒ ⇒

{ }

1 Z {y n +1} + Z { yn } = Z 4− n 4 1 4z [Using Damping rule i.e. Z (a − n .u n ) U(a z)] = z [Y(z) − y0 ] + Y (z) 4 4 z −1

Now using given condition, we have

Dr. Jogendra Kumar

Page 9

Z-TRANSFORMATION 1 4z z [Y(z) − y0 ] + Y (z) = 4 4 z −1

⇒ ⇒

(

)

1 4z  = ⇒ z + 1 Y (z) =  z +  Y (z) 4 4 4 z −1  z Y (z) = z− 1 z+ 1 4 4

(

)(

z z− 1

4

)

1 . Now, residues of Y(z).z n−1 at these poles are Clearly, Y(z) have two poles at z = −1 , z= 4

(

)

4

1 ) lim s (z −= = Re z+ 1 . z →− 1 4 4 z+ 1 4 lim z →− 1 =

( − 14 )

4

(

z

4

(

= .z n −1 lim z →− 1 z + 1

)( z− 14 )

4

4

)

.

zn

( z+ 14 )( z− 14 )

n

( − 14 − 14 )

(

= − 2. − 1

4)

n

And

(

)

1 )= Re s (z = lim z → 1 z − 1 . 4 4 z+ 1 4

(

1 ) ( 4 lim = ( 14 + 14 )

z

z− 1 ) 4 )( 4

(

)

.z n −1 = lim z → 1 z − 1 . 4 z+ 1 4

(

zn 4

)( z− 14 )

n

=

z→ 1

4

( 4)

2. 1

n

(

∴ Z −1{Y(z)} =yn =Sume of residues =−2. − 1 Que. Solve using Z-Transformation yn + 3

4)

n

( 4)

+ 2. 1

n

( ) − ( − 1 4 ) 

=2.  1  4

n

n

− 3 yn +1 + 2 yn = 0 , given that= y0 4,= y1 0 and= y 2 8.

Soln. Let Z {y n } = Y (z) . Now taking Z-Transform of

yn +3 − 3 yn +1 + 2 yn = 0 , i.e.

Dr. Jogendra Kumar

Page 10

Z-TRANSFORMATION Z {y n +3 } − 3 Z { yn +1} + 2 Z{y n } = Z {0}



z 3 [Y(z) − y0 − y1 z −1 − y 2 z −2 ] − 3z [Y (z) − y0 ] + 2 Y (z) = 0

Now using given condition, we have

0 z 3 [Y(z) − 4 − 8 z −2 ] − 3z [Y (z) − 4] + 2 Y (z) = ⇒

( z 3 − 3 z + 2) Y(z) − 4 z 3 − 8 z + 12 z = 0



( z 3 − 3 z + 2) Y(z)= 4 ( z 3 − z)



= 4 z ( z 2 − 1) (z − 1) ( z 2 + z − 2) Y(z)

⇒ ( z 2 + z − 2) Y(z) = 4 z ( z + 1) ⇒ Y(z) =

4 z ( z + 1) 4 ( z 2 + z) = ( z 2 + z − 2) (z − 1) (z + 2)

Clearly, Y (z) has two poles at z = 1, z = −2 .Thus the residues of Y (z).z n −1 are

4( z 2 + z ) n −1 4( z n +1 + z n ) Re s (z = 1) = lim z →1 (z − 1). .z = lim z →1 (z − 1)(z + 2) (z + 2) 4 z n ( z + 1) 4 (1) n (1 + 1) 8 = lim z →1 = = , (z + 2) (1 + 2) 3 And

Re s (z = lim z →−2 (z + 2). −2) = = lim z →1

4( z 2 + z ) n −1 4( z n +1 + z n ) .z = lim z →−2 (z − 1)(z + 2) (z − 1)

4 z n ( z + 1) 4 (−2) n (−2 + 1) 4 .(−2) n = = (z − 1) (−2 − 1) 3

∴ Z −1{Y(z)} = yn = Sume of residues =

8 4 4  2 + (−2) n  + .(−2) n = 3 3 3

Suggestion: Learn all formulae and solve Tutorial of Z-Transform which is uploaded in Tutorials/Assignment section.

Dr. Jogendra Kumar

Page 11

z-transformation

Partial fraction method and. (iii). Inversion integral method or Residues method. Inversion method or Residues Method: The inverse Z-Transform of U(z) is given ...

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