The z-Transform - Digital Signal Processing: Fundamentals and Applications - page 141

EXAMPLE 5.4
Find the z-transform of the sequence defined by
x n ¼ u n ð0:5Þ

n u n

Solution:
Applying the linearity of the z-transform discussed above, we have

n

X ðzÞ¼ZðxðnÞÞ ¼ ZðuðnÞÞ Zð0:5 ðnÞÞ

Using Table 5.1 yields

z
z 1

ZðuðnÞÞ ¼

and

z
z 0:5

n uðnÞÞ ¼

Zð0:5

Substituting these results in X ðzÞ leads to the final solution,

z
z 1

z
z 0:5

X ðzÞ¼

Shift theorem: Given X ðzÞ , the z-transform of a sequence xðnÞ , the z-transform of xðn mÞ , the
time-shifted sequence, is given by

Zðxðn mÞÞ ¼ z m X z (5.3)

Note that if m 0, then xðn mÞ is obtained by right shifting xðnÞ by m samples. Since the shift
theorem plays a very important role in developing the transfer function from a difference equation, we
verify the shift theorem for the causal sequence. Note that the shift theorem also works for the
noncausal sequence.
Verification: Applying the z-transform to the shifted causal signal xðn mÞ leads to

n¼ 0 x n m z n
¼ xðmÞz 0

Zðxðn mÞÞ ¼ N

þ/þ xð 1 Þz ðm 1 Þ þ xð 0 Þz m þ x 1 z m 1

þ.

Since xðnÞ is assumed to be a causal sequence, this means that

xð mÞ¼xð m þ 1 Þ¼/ ¼ xð 1 Þ¼ 0

Then we achieve

Zðxðn mÞÞ ¼ xð 0 Þz m þ xð 1 Þz m 1 þ xð 2 Þz m 2 þ. (5.4)