In this subsection we consider perhaps the most important properties of the z-transform. These properties
relate the z-transform
Y
(
z
)
of a sequence
{
y
n
}
to the z-transforms of
right shifted
or
delayed sequences
{
y
n
−
1
}
{
y
n
−
2
}
etc.
left shifted
or
advanced sequences
{
y
n
+
1
}
,
{
y
n
+
2
}
etc.
The results obtained, formally called shift theorems, are vital in enabling us to solve certain
types of difference equation and are also invaluable in the analysis of digital systems of
various types.
Let
{
v
n
}
=
{
y
n
−
1
}
i.e. the terms
of the sequence
{
v
n
}
are
the same as those of
{
y
n
}
but shifted one place to the right. The z-transforms are, by definition,
Y
(
z
)
=
y
0
+
y
1
z
−
1
+
y
2
z
−
2
+
y
j
z
−
3
+
…
V
(
z
)
=
v
0
+
v
1
z
−
1
+
v
2
z
−
2
+
v
3
z
−
3
+
…
=
y
−
1
+
y
0
z
−
1
+
y
1
z
−
2
+
y
2
z
−
3
+
…
=
y
−
1
+
z
−
1
(
y
0
+
y
1
z
−
1
+
y
2
z
−
2
+
…
)
i.e.
V
(
z
)
=
ℤ
{
y
n
−
1
}
=
y
−
1
+
z
−
1
Y
(
z
)
Obtain the z-transform of the sequence
{
w
n
}
=
{
y
n
−
2
}
using the method illustrated above.
Answer
The z-transform of
{
w
n
}
is
W
(
z
)
=
w
0
+
w
1
z
−
1
+
w
2
z
−
2
+
w
3
z
−
3
+
…
or,
since
w
n
=
y
n
−
2
,
W
(
z
)
=
y
−
2
+
y
−
1
z
−
1
+
y
0
z
−
2
+
y
1
z
−
3
+
…
=
y
−
2
+
y
−
1
z
−
1
+
z
−
2
(
y
0
+
y
1
z
−
1
+
…
)
i.e.
W
(
z
)
=
ℤ
{
y
n
−
2
}
=
y
−
2
+
y
−
1
z
−
1
+
z
−
2
Y
(
z
)
Clearly, we could proceed in a similar way to obtains a general result for
ℤ
{
y
n
−
m
}
where
m
is any
positive integer. The result is
ℤ
{
y
n
−
m
}
=
y
−
m
+
y
−
m
+
1
z
−
1
+
…
+
y
−
1
z
−
m
+
1
+
z
−
m
Y
(
z
)
For the particular case of causal sequences (where
y
−
1
=
y
−
2
=
…
=
0
) these
results are particularly simple:
ℤ
{
y
n
−
1
}
=
z
−
1
Y
(
z
)
ℤ
{
y
n
−
2
}
=
z
−
2
Y
(
z
)
ℤ
{
y
n
−
m
}
=
z
−
m
Y
(
z
)
(causal systems only)
You may recall from earlier in this Workbook that in a digital system we represented the right
shift operation symbolically in the following way:
Figure 6
The significance of the
z
−
1
factor inside the rectangles should now be clearer. If we replace the ‘input’ and ‘output’
sequences by their z-transforms:
ℤ
{
y
n
}
=
Y
(
z
)
ℤ
{
y
n
−
1
}
=
z
−
1
Y
(
z
)
it is evident that in the z-transform ‘domain’ the shift becomes a multiplication by the factor
z
−
1
.
N.B. This discussion applies strictly only to causal sequences.
Notational point:
A causal sequence is sometimes written as
y
n
u
n
where
u
n
is the
unit step sequence
u
n
=
0
n
=
−
1
,
−
2
,
…
1
n
=
0
,
1
,
2
,
…
The right shift theorem is then written, for a causal sequence,
ℤ
{
y
n
−
m
u
n
−
m
}
=
z
−
m
Y
(
z
)
Examples
Recall that the z-transform of the causal sequence
{
a
n
}
is
z
z
−
a
. It
follows, from the right shift theorems that
ℤ
{
a
n
−
1
}
=
ℤ
{
0
,
1
,
a
,
a
2
,
…
}
=
z
z
−
1
z
−
a
=
1
z
−
a
↑
ℤ
{
a
n
−
2
}
=
ℤ
{
0
,
0
,
1
,
a
,
a
2
,
…
}
=
z
−
1
z
−
a
=
1
z
(
z
−
a
)
↑
Write the following sequence
f
n
as a difference of two unit step sequences. Hence obtain its z-transform.
Answer
Since
{
u
n
}
=
1
n
=
0
,
1
,
2
,
…
0
n
=
−
1
,
−
2
,
…
and
{
u
n
−
5
}
=
1
n
=
5
,
6
,
7
,
…
0
otherwise
it follows that
f
n
=
u
n
−
u
n
−
5
Hence
F
(
z
)
=
z
z
−
1
−
z
−
5
z
z
−
1
=
z
−
z
−
4
z
−
1
Recall that the sequences
{
y
n
+
1
}
,
{
y
n
+
2
}
…
denote the sequences obtained by shifting the sequence
{
y
n
}
by
1
,
2
,
…
units to the left
respectively. Thus, since
Y
(
z
)
=
ℤ
{
y
n
}
=
y
0
+
y
1
z
−
1
+
y
2
z
−
2
+
…
then
ℤ
{
y
n
+
1
}
=
y
1
+
y
2
z
−
1
+
y
3
z
−
2
+
…
=
y
1
+
z
(
y
2
z
−
2
+
y
3
z
−
3
+
…
)
The term in brackets is the z-transform of the unshifted sequence
{
y
n
}
apart
from its first two terms:
thus
ℤ
{
y
n
+
1
}
=
y
1
+
z
(
Y
(
z
)
−
y
0
−
y
1
z
−
1
)
∴
Z
{
y
n
+
1
}
=
z
Y
(
z
)
−
z
y
0
Obtain the z-transform of the sequence
{
y
n
+
2
}
using the method illustrated above.
Answer
ℤ
{
y
n
+
2
}
=
y
2
+
y
3
z
−
1
+
y
4
z
−
2
+
…
=
y
2
+
z
2
(
y
3
z
−
3
+
y
4
z
−
4
+
…
)
=
y
2
+
z
2
(
Y
(
z
)
−
y
0
−
y
1
z
−
1
−
y
2
z
−
2
)
∴
ℤ
{
y
n
+
2
}
=
z
2
Y
(
z
)
−
z
2
y
0
−
z
y
1
These left shift theorems have simple forms in special cases:
if
y
0
=
0
ℤ
{
y
n
+
1
}
=
z
Y
(
z
)
if
y
0
=
y
1
=
0
ℤ
{
y
n
+
2
}
=
z
2
Y
(
z
)
if
y
0
=
y
1
=
…
y
m
−
1
=
0
ℤ
{
y
n
+
m
}
=
z
m
Y
(
z
)
The
right shift theorems
or delay theorems are:
ℤ
{
y
n
−
1
}
=
y
−
1
+
z
−
1
Y
(
z
)
ℤ
{
y
n
−
2
}
=
y
−
2
+
y
−
1
z
−
1
+
z
−
2
Y
(
z
)
⋮
⋮
⋮
⋮
ℤ
{
y
n
−
m
}
=
y
−
m
+
y
−
m
+
1
z
−
1
+
…
+
y
−
1
z
−
m
+
1
+
z
−
m
Y
(
z
)
The
left shift theorems
or advance theorems are:
ℤ
{
y
n
+
1
}
=
z
Y
(
z
)
−
z
y
0
ℤ
{
y
n
+
2
}
=
z
2
Y
(
z
)
−
z
2
y
0
−
z
y
1
⋮
⋮
ℤ
{
y
n
−
m
}
=
z
m
Y
(
z
)
−
z
m
y
0
−
z
m
−
1
y
1
−
…
−
z
y
m
−
1
Note carefully the occurrence of
positive
powers of
z
in the
left
shift theorems
and of
negative
powers of
z
in the
right
shift theorems.
Table 1: z-transforms
f
n
F
(
z
)
Name
δ
n
1
unit impulse
δ
n
−
m
z
−
m
u
n
z
z
−
1
unit step sequence
a
n
z
z
−
a
geometric sequence
e
α
n
z
z
−
e
α
sinh
α
n
z
sinh
α
z
2
−
2
z
cosh
α
+
1
cosh
α
n
z
2
−
z
cosh
α
z
2
−
2
z
cosh
α
+
1
sin
ω
n
z
sin
ω
z
2
−
2
z
cos
ω
+
1
cos
ω
n
z
2
−
z
cos
ω
z
2
−
2
z
cos
ω
+
1
e
−
α
n
sin
ω
n
z
e
−
α
sin
ω
z
2
−
2
z
e
−
α
cos
ω
+
e
−
2
α
e
−
α
n
cos
ω
n
z
2
−
z
e
−
α
cos
ω
z
2
−
2
z
e
−
α
cos
ω
+
e
−
2
α
n
z
(
z
−
1
)
2
ramp sequence
n
2
z
(
z
+
1
)
(
z
−
1
)
3
n
3
z
(
z
2
+
4
z
+
1
)
(
z
−
1
)
4
a
n
f
n
F
z
a
n
f
n
−
z
d
F
d
z
This table has been copied to the back of this Workbook (page 96) for convenience.