This simple property states that if
{
v
n
}
and
{
w
n
}
have z-transforms
V
(
z
)
and
W
(
z
)
respectively then
ℤ
{
a
v
n
+
b
w
n
}
=
a
V
(
z
)
+
b
W
(
z
)
for any constants
a
and
b
.
(In particular if
a
=
b
=
1
this property tells us that adding sequences corresponds to adding their z-transforms).
The proof of the linearity property is straightforward using obvious properties of the summation operation. By the z-transform definition:
ℤ
{
a
v
n
+
b
w
n
}
=
∑
n
=
0
∞
(
a
v
n
+
b
w
n
)
z
−
n
=
∑
n
=
0
∞
(
a
v
n
z
−
n
+
b
w
n
z
−
n
)
=
a
∑
n
=
0
∞
v
n
z
−
n
+
b
∑
n
=
0
∞
w
n
z
−
n
=
a
V
(
z
)
+
b
V
(
z
)
We can now use the linearity property and the exponential sequence
{
e
−
α
n
}
to obtain the z-transforms of hyperbolic and of trigonometric sequences relatively easily. For example,
sinh
n
=
e
n
−
e
−
n
2
Hence, by the linearity property,
ℤ
{
sinh
n
}
=
1
2
ℤ
{
e
n
}
−
1
2
ℤ
{
e
−
n
}
=
1
2
z
z
−
e
−
z
z
−
e
−
1
=
z
2
z
−
e
−
1
−
(
z
−
e
)
z
2
−
(
e
+
e
−
1
)
z
+
1
=
z
2
e
−
e
−
1
z
2
−
(
2
cosh
1
)
z
+
1
=
z
sinh
1
z
2
−
2
z
cosh
1
+
1
Using
α
n
instead of
n
in this calculation, where
α
is a constant, we obtain
ℤ
{
sinh
α
n
}
=
z
sinh
α
z
2
−
2
z
cosh
α
+
1
Using
cosh
α
n
≡
e
α
n
+
e
−
α
n
2
obtain the z-transform of the sequence
{
cosh
α
n
}
=
{
1
,
cosh
α
,
cosh
2
α
,
…
}
Answer
We have, by linearity,
ℤ
{
cosh
α
n
}
=
1
2
ℤ
{
e
α
n
}
+
1
2
ℤ
{
e
−
α
n
}
=
z
2
1
z
−
e
α
+
1
z
−
e
−
α
=
z
2
2
z
−
(
e
α
+
e
−
α
)
z
2
−
2
z
cosh
α
+
1
=
z
2
−
z
cosh
α
z
2
−
2
z
cosh
α
+
1
If we use the result
ℤ
{
a
n
}
=
z
z
−
a
|
z
|
>
|
a
|
with, respectively,
a
=
e
i
ω
and
a
=
e
−
i
ω
where
ω
is a constant and
i
denotes
−
1
we obtain
ℤ
{
e
i
ω
n
}
=
z
z
−
e
+
i
ω
ℤ
{
e
−
i
ω
n
}
=
z
z
−
e
−
i
ω
Hence, recalling from complex number theory that
cos
x
=
e
i
x
+
e
−
i
x
2
we can state, using the linearity property, that
ℤ
{
cos
ω
n
}
=
1
2
ℤ
{
e
i
ω
n
}
+
1
2
ℤ
{
e
−
i
ω
n
}
=
z
2
1
z
−
e
i
ω
+
1
z
−
e
−
i
ω
=
z
2
2
z
−
(
e
i
ω
+
e
−
i
ω
)
z
2
−
(
e
i
ω
+
e
−
i
ω
)
z
+
1
=
z
2
−
z
cos
ω
z
2
−
2
z
cos
ω
+
1
(Note the similarity of the algebra here to that arising in the corresponding hyperbolic case. Note also the similarity of the results for
ℤ
{
cosh
α
n
}
and
ℤ
{
cos
ω
n
}
.)
By a similar procedure to that used above for
ℤ
{
cos
ω
n
}
obtain
ℤ
{
sin
ω
n
}
.
Answer
We have
ℤ
{
sin
ω
n
}
=
1
2
i
ℤ
{
e
i
ω
n
}
−
1
2
i
ℤ
{
e
−
i
ω
n
}
(Don’t miss the
i
factor here!)
∴
ℤ
{
sin
ω
n
}
=
z
2
i
1
z
−
e
i
ω
−
1
z
−
e
−
i
ω
=
z
2
i
−
e
−
i
ω
+
e
i
ω
z
2
−
2
z
cos
ω
+
1
=
z
sin
ω
z
2
−
2
z
cos
ω
+
1
ℤ
{
cos
ω
n
}
=
z
2
−
z
cos
ω
z
2
−
2
z
cos
ω
+
1
ℤ
{
sin
ω
n
}
=
z
sin
ω
z
2
−
2
z
cos
ω
+
1
Notice the same denominator in the two results in Key Point 6.
ℤ
{
cosh
α
n
}
=
z
2
−
z
cosh
α
z
2
−
2
z
cosh
α
+
1
ℤ
{
sinh
α
n
}
=
z
sinh
α
z
2
−
2
z
cosh
α
+
1
Again notice the denominators in Key Point 7. Compare these results with those for the two trigonometric sequences in Key Point 6.
Use Key Points 6 and 7 to write down the z-transforms of
sin
n
2
cos
3
n
{
sinh
2
n
}
{
cosh
n
}
Answer
ℤ
sin
n
2
=
z
sin
1
2
z
2
−
2
z
cos
1
2
+
1
ℤ
{
cos
3
n
}
=
z
2
−
z
cos
3
z
2
−
2
z
cos
3
+
1
ℤ
{
sinh
2
n
}
=
z
sinh
2
z
2
−
2
z
cosh
2
+
1
ℤ
{
cosh
n
}
=
z
2
−
z
cosh
1
z
2
−
2
z
cosh
1
+
1
Use the results for
ℤ
{
cos
ω
n
}
and
ℤ
{
sin
ω
n
}
in Key Point 6 to obtain the z-transforms of
{
cos
(
n
π
)
}
sin
n
π
2
cos
n
π
2
Write out the first few terms of each sequence.
Answer
With
ω
=
π
ℤ
{
cos
n
π
}
=
z
2
−
z
cos
π
z
2
−
2
z
cos
π
+
1
=
z
2
+
z
z
2
+
2
z
+
1
=
z
z
+
1
{
cos
n
π
}
=
{
1
,
−
1
,
1
,
−
1
,
…
}
=
{
(
−
1
)
n
}
We have re-derived the z-transform of the unit alternating sequence. (See Task on page 17).
With
ω
=
π
2
ℤ
sin
n
π
2
=
z
sin
π
2
z
2
−
2
z
cos
π
2
+
1
=
z
z
2
+
1
where
sin
n
π
2
=
{
0
,
1
,
0
,
−
1
,
0
,
…
}
With
ω
=
π
2
ℤ
cos
n
π
2
=
z
2
−
cos
π
2
z
2
+
1
=
z
2
z
2
+
1
where
cos
n
π
2
=
{
1
,
0
,
−
1
,
0
,
1
,
…
}
(These three results can also be readily obtained from the definition of the z-transform. Try!)