The z-transform - Continuous and Discrete Time Signals and Systems - page 586

Fig. 13.4. (a) Original DT
sequence x [ k ] = α

x [ k ] = a | k |

1

1

k

. Parts
(b)-(d) show sequences
obtained by time shifting the
sequence in part (a):
(b) x [ k − 2] u [ k − 2];
(c) x [ k − 2] u [ k ];
(d) x [ k + 2] u [ k ].

x [ k − 2] u [ k − 2]

a

a a

a 2

a 2 a 2

a 3

a 3 a 3

a 5

a 4 a 5

a 4

k

k

−5 −4 −3 −2 −1 0

−5 −4 −3 −2 −1 0

(a) (b)

x [ k − 2] u [ k ] 1

x [ k + 2] u [ k ]

a

a

a 2 a 2 a 2

a 3

a 3

a 4 a 5 a 6 a 7

k k

−5 −4 −3 −2 −1 0 −5 −4 −3 −2 −1 0

(c) (d)

Substituting p = k − m , the above summation reduces to

∞ ∞

− ( p + m ) − m − p , = z − m X ( z ) .

Z x [ k − m ] u [ k − m ] = x [ p ] z = z x [ p ] z

p = 0 p = 0

Equation (13.20)

∞ ∞

− k − k .

Z x [ k + m ] u [ k ] = x [ k + m ] u [ k ] z = x [ k + m ] z

k = 0 k = 0

Substituting p = k + m , the above summation reduces to

∞ ∞ m − 1

− ( p − m ) = z m − p − z m − p ,

Z x [ k + m ] u [ k ] = x [ p ] z x [ p ] z x [ p ] z

p = m p = 0 p = 0

m − 1

= z m X ( z ) − z m

− k .

x [ k ] z

k = 0

Equation (13.21)

∞ ∞

− k − k .

Z x [ k − m ] u [ k ] = x [ k − m ] u [ k ] z = x [ k − m ] z

k = 0 k = 0

Substituting p = k − m , the above summation reduces to

∞

− ( p + m )

Z x [ k − m ] u [ k ] =

x [ p ] z

p =− m

∞ − 1

− m − p + z − m − p .

= z x [ p ] z x [ p ] z

p = 0

p =− m

m

− m x ( z ) + z − m

x [ − k ] z k .

= z

k = 1