Digital Signal Processing Reference
In-Depth Information
Fig. 9.10. Time-domain
illustration of the zero-order
hold operation for a CT signal.
(a) Original signal
x
(
t
);
(b) zero-order hold output
x
s
(
t
).
x
(
t
)
x
s
(
t
)
t
t
0
−3
T
s
−2
T
s
−
T
s
0
T
s
2
T
s
3
T
s
(a)
(b)
zero-order hold operation is illustrated in Fig. 9.10. Unlike the pulse-train sam-
pling, the amplitude of the sampled signal is maintained constant for
T
s
seconds
until the next sample is taken.
For mathematical analysis, the zero-order hold operation can be modeled by
the following expression:
x
s
(
t
)
=
∞
x
(
kT
s
)
p
2
(
t
−
kT
s
)
(9.22a)
k
=−∞
or
∞
∞
x
s
(
t
)
=
p
2
(
t
)
∗
x
(
kT
s
)
δ
(
t
−
kT
s
)
=
p
2
(
t
)
∗
x
(
t
)
δ
(
t
−
kT
s
)
,
k
=−∞
k
=−∞
(9.22b)
where
p
2
(
t
) represents a rectangular pulse given by
t
−
0
.
5
T
s
T
s
p
2
(
t
)
=
rect
.
(9.23)
Equation (9.22b) models the zero-hold operation and is different from Eq. (9.18)
in two ways. First, the duration of the pulse
p
2
(
t
) in Eq. (9.22b) is the same as
the sampling interval
T
s
, whereas the duration of the pulse
p
1
(
t
) is much smaller
than
T
s
in pulse-train sampling. Secondly, the order of operation in the sampled
signal
x
s
(
t
) is different from that used in the corresponding sampled signal in
pulse-train sampling. In Eq. (9.22b), the sampled signal
x
s
(
t
) is obtained by
convolving
p
2
(
t
) with a periodic impulse train, which is scaled by the values
of the reference signal at the location of the impulse functions. In Eq. (9.18),
on the other hand,
x
s
(
t
) is obtained by multiplying the original signal directly
by the periodic pulse train
r
(
t
).
The CTFT of Eq. (9.22b) is given by
ω −
∞
X
s
(
ω
)
=
P
2
(
ω
)
1
2
π
2
π
T
s
2
k
π
T
s
X
(
ω
)
∗
δ
,
(9.24)
k
=−∞
where
P
2
(
ω
) denotes the CTFT of the rectangular pulse
p
2
(
t
). Based on entry
(16) of Table 5.2, the CTFT of
p
2
(
t
) is given by the following transform pair:
t
−
0
.
5
T
s
T
s
ω
T
s
2
π
CTFT
←→
T
s
sinc
e
−
j0
.
5
ω
T
s
.
rect
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