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480 Chapter 9 SAMPLING RATE CONVERSION
¯x(mD)z −m
as shown in (9.6). Figure 9.4 shows an example of sequences x(n), ¯x(n),
and y(m) defined in (9.7)–(9.10).
Now the z-transform of the output sequence y(m) is
Y (z) =
∞∑
m=−∞
y(m)z −m =
∞∑
m=−∞
(9.11)
∞∑
Y (z) = ¯x(m)z −m/D
m=−∞
where the last step follows from the fact that ¯x(m) =0, except at multiples
of D. Bymaking use of the relations in (9.7) and (9.8) in (9.11), we
obtain
Y (z) =
= 1 D
= 1 D
∞∑
m=−∞
D−1
∑
x(m)
∞∑
[
k=0 m=−∞
D−1
∑
k=0
1
D
D−1
∑
k=0
e j2πmk/D ]
z −m/D
x(m) ( e −j2πk/D z 1/D) −m
X ( e −j2πk/D z 1/D) (9.12)
The key steps in obtaining the z-transform representation (9.12), for the
(D ↓ 1) downsampler, are as follows:
• the introduction of the high-rate sequence ¯x(n), which has (D−1) zeros
in between the retained values x(nD), and
• the impulse-train representation (9.8) for the periodic sampling series
that relates x(n) to¯x(n).
By evaluating Y (z) onthe unit circle, we obtain the spectrum of the
output signal y(m). Since the rate of y(m) isF y =1/T y, the frequency
variable, which we denote as ω y ,isinradians and is relative to the sampling
rate F y ,
ω y = 2πF =2πFT y (9.13)
F y
Since the sampling rates are related by the expression
F y = F x
D
it follows that the frequency variables ω y and
(9.14)
ω x = 2πF
F x
2πFT x (9.15)
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