Databases Reference
In-Depth Information
In order to find the Z-transform of this sequence, we go through a two-step process. Define
the sequence
1
2
(
y
1
,
n
=
e
jn
π
)
1
+
y
1
,
n
(26)
y
1
,
n
n
even
=
(27)
0
otherwise
We could also have written Equation (
26
)as
1
2
(
y
1
,
n
=
n
1
+
(
−
1
)
)
y
1
,
n
however, writing the relationship as in Equation (
26
) makes it easier to extend this development
to the case where we divide the source output into more than two bands.
The Z-transform of
y
1
,
n
is given as
∞
1
2
(
Y
1
(
e
jn
π
)
y
1
,
n
z
−
n
)
=
+
(28)
z
1
n
=−∞
Assuming all summations converge,
∞
∞
1
2
1
2
Y
1
(
y
1
,
n
z
−
n
ze
−
j
π
)
−
n
z
)
=
+
y
1
,
n
(
(29)
n
=−∞
n
=−∞
1
2
Y
1
(
1
2
Y
1
(
−
=
z
)
+
z
)
(30)
where we have used the fact that
e
−
j
π
=
(π)
−
π
=−
cos
j
sin
1
Note that
y
1
,
2
n
w
1
,
n
=
(31)
n
=−∞
w
1
,
n
z
−
n
∞
−∞
y
1
,
2
n
z
−
n
W
1
(
z
)
=
=
(32)
Substituting
m
=
2
n
,
∞
−∞
y
1
,
m
z
−
2
W
1
(
z
)
=
(33)
Y
1
z
2
=
(34)
2
Y
1
z
2
2
Y
1
z
2
1
1
=
+
−
(35)
Why didn't we simply write the Z-transform of
w
1
,
n
directly in terms of
y
1
,
n
and use
the substitution
m
2
n
? If we had, the equivalent equation to (
33
) would contain the odd
indexed terms of
y
1
,
n
, which we know do not appear at the output of the downsampler. In
=
