Sampling and Reconstruction:
•
Until now, we have considered continuoustime signals &
systems
separately
from discretetime signals & systems.
•
We will now connect them through
uniform sampling
:
x
[
n
] =
x
(
nT
)
for
n
∈
Z
,
where
T
(in seconds/sample) is the
sampling interval
and
so
1
/T
is the
sampling rate
.
x
(
t
)
x
[
n
]
t
=
nT
•
Key question:
–
Do we always lose information in going from
{
x
(
t
)
}
∀
t
to
{
x
[
n
]
}
∞
n
=
∞
?
–
Under what conditions can we reconstruct
{
x
(
t
)
}
∀
t
from
{
x
[
n
]
}
∀
n
?
1
2
Aliasing — Nyquist zones:
•
So far, we have seen that a sampled sinusoid
x
[
n
]
can be
associated with
many
continuoustime sinusoids
x
(
t
)
.
•
Can visualize this for
x
(
t
) =
e
j
Ω
k
t
via “Nyquist zones”:
Defn: The
k
th
Nyquist zone is
Ω
∈
(
2
πk

π
T
,
2
πk
+
π
T
]
.
Ω
Ω
Ω
ω
2
π
2
π
T
2
π
T
2
π
T
0
0
0
0
−
2
π
−
2
π
T
−
2
π
T
−
2
π
T
···
···

X
c
(
j
Ω)


X
c
(
j
Ω)


X
c
(
j
Ω)


X
(
e
jω
)

ω

1
ω
0
ω
1
Ω

1
Ω
0
Ω
1
ω
k
=
Ω
k
T
zone 1
zone 1
zone 1
zone 0
zone 0
zone 0
zone 1
zone 1
zone 1
1
/T
rate
sampling
Recall the DTFT is
2
π
periodic!
•
Due to the linearity of sampling, the same phenomenon
applies to generic signals:
.
Ω
Ω
Ω
ω
2
π
2
π
T
2
π
T
2
π
T
0
0
0
0
−
2
π
−
2
π
T
−
2
π
T
−
2
π
T
···
···

X
c
(
j
Ω)


X
c
(
j
Ω)


X
c
(
j
Ω)


X
(
e
jω
)

zone 1
zone 1
zone 1
zone 0
zone 0
zone 0
zone 1
zone 1
zone 1
1
/T
rate
sampling
Spectral overlap due to “aliasing”
will prevent perfect reconstruction!
•
Nyquist zone locations determined by sampling rate
1
T
!
3
Nyquist sampling:
•
As we have seen, sampling can introduce ambiguity:
Can’t distinguish one Nyquist zone from another!
•
How can we avoid this ambiguity?
Sample in such a way that the CT signal is
contained within a particular Nyquist zone.
There are two basic ways to do this:
1.
Nyquist sampling
:
–
Ensure that all frequency components of
x
(
t
)
are in
the
0
th
Nyquist zone. Equivalently. . .
–
Sample more than twice as fast as the largest
absolute signal frequency:
1
T
>

Ω

max
π
= 2

f

max
.
Ω
ω
2
π
2
π
T
0
0
−
2
π
−
2
π
T
···
···

X
c
(
j
Ω)


X
(
e
jω
)

zone 1
zone 0
zone 1

Ω

max
−
Ω

max
1
/T
rate
sampling
reconstructtion
2.
Bandpass sampling
:
–
Ensure than all frequency components of
x
(
t
)
are
contained within some Nyquist zone
k
, for
k
negationslash
= 0
.
–
Note: works only for bandpass
x
(
t
)
.
–
But how does one choose
1
T
and
k
in this case?
•
Reconstruction then boils down to optimal interpolation.
Stay tuned for details...
4