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Samplingï EEE 507 - Lecture 3
Most signals are either explicitly or implicitly sampled.
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Sampling has both similarities and differences in the 1-D and 2-Dcases.
Sampling
Continuous-domain
x
a
(t
1
,t
2
Discrete-domain
x(n
1
,n
2
Sampling
Continuous-domain
x
a
(t
1
,t
2
1-D Sampling Theorem (Review)ï EEE 507 - Lecture 3
A bandlimited signal can be periodically sampled with no loss ofinformation provided it is sampled often enough.
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The minimum sampling rate (Nyquist) rate is proportional to thesignalís bandwidth.
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The DTFT of the samples can be related to the Fourier transform ofthe continuous signal that is sampled.
Sampling EEE 507 - Lecture 3
2D
signals
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There are many ways to sample
2D
signals
Sampling example
Sampling example
Analog Image
Sampling EEE 507 - Lecture 3
2D
signals
ï
The simplest and most common of this is rectangular sampling
x
a
(t
1
,t
2
x
a
(n
1
T
1
,n
2
T
2
) = x(n
1
,n
2
T
1
: Horizontal sampling interval
F.T. replicated with period
/T
1
in horizontal direction
T
2
: Vertical sampling interval
F.T. replicated with period 2
/T
2
in vertical direction
indices ofsamples
discrete time atregular intervals
Continuous-domain
2D signal
Sample
t
1
and t
2
Derivation (Sampling Theorem)ï EEE 507 - Lecture 3
Define
p
a
(t
,t
to be a field of impulses- one at each sampling
location (bed of nails)
t
1
t
2
T
1
T
2
Sampling EEE 507 - Lecture 3
2D
signals
ï
The Bed of Nails
t
1
t
2
(
)
(
)
∞
−∞
∞
−∞
−
−
=
1
2
2 2 2 1 1 1 2 1
,
,
n
n
a
T n t T n t t t p
δ
ï EEE 507 - Lecture 3
The bed-of-nails (field of impulses):
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The Fourier transform of a bed-of-nails is another bed-of-nails with reciprocal spacing
(
)
(
)
∑
∑
∞
−∞
∞
−∞
−
−
∆
1
2
2 2 2 1 1 1 2 1
,
,
n
n
a
T n t T n t t t p
δ
(
)
∑ ∑
∞
−∞
∞
−∞
− Ω − Ω = Ω Ω
1
2
2
2
2
1
1
1
2
1
2
2
1
2 , 2 ) 2 ( ,
k
k
a
T
k
T
k
T
T
p
π
π
δ
π
Sampling 2-D signals
Sampling EEE 507 - Lecture 3
2D
signals
ï
x
a
(t
,t
) can be
rectangularly
sampled as follows:
∑ ∑∑ ∑
∑ ∑
∞
−∞
∞
−∞
∞
−∞
∞
−∞
∞
−∞
∞
−∞
−
−
=
−
−
=
−
−
=
1
2
2
1
1
2
1
2
2 2 2 1 1 1 ) , (
2
2
1
1
2 2 2 1 1 1 2 1 2 2 2 1 1 1 2 1 2 1
,
,
,
,
,
,
,
n
n
n
n
x
a
n
n
a
n
n
a
s
T n t T n t T n T n x
T n t T n t t t x T n t T n t t t x t t x
δ
δ δ
4
43
4 42
1
)
,
(
).
,
(
)
,
(
t t p t t x t t x
a
a
s
=
Sampledbut stillcontinuous
EEE 507 - Lecture 3
)
,
(
Ω
Ω
a
X
1
2
W
2
W
1
Sampling
2D
signals
ï
In the frequency domain:
Period
Vertical
T
Period
Horizontal
T
π^1 π^2
)
2 , 0 (
2 T
π
) 0 ,
2 (
1 T
π
1
)
,
(
Ω
Ω
s
X
2
Sampling EEE 507 - Lecture 3
2D
signals
ï
Let
W
and
W
be the horizontal and vertical bandwidth of
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Then exact recovery requires
or, equivalently,
or, equivalently,
)
,
(
t
t
x
a
condition
aliasing
no
2
2
1
1
W
T
W
T
condition
aliasing
no
2
2
2
1
1
1
W
T
W
T
s s
π^ π
condition
aliasing
no
1 1
2
2
2
1
1
1
=>
>
=
>
=
π π^ W
T
f
W
T
f
s s
ï EEE 507 - Lecture 3
Comparing, one can easily see that:
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As in the 1-D case, we have a normalization relation between (
and
) and (
and
) given by:
T
or
/ T
T
/ T
ï
If
x(n
1
n
2
)=x
a
(n
1
T
1
, n
2
T
2
then
X(
1
2
DTFT{
x(n
1
n
2
} is related to
X
a
1
2
) = FT{x
a
(t
1
, t
2
by:
( 1. Periodicity (aliasing, repetition)2. Amplitude scaling3. Frequency normalization
2
2
(^22)
1
1
(^11)
2
1
(^22)
(^11)
2
1
1
2
T
k
T
T
k T X T T T T X X
k
k
a
s
π ω π ω ω ω ω ω
Relating the Discrete and Continuous FrequencyDomains
radians
radians/sec
EEE 507 - Lecture 3
Period
Vertical
T
Period
Horizontal
T
π^1 π^2
)
,
(
Ω
Ω
a
X
1
2
W
2
W
1
) 0 ,
2 (
1 T
π
)
2 , 0 (
2 T
π
Sampling
2D
signals
ï
In the frequency domain:
)
,
(
Ω
Ω
s
X
1
2
Sampling EEE 507 - Lecture 3
2D
signals
ï
Aliasing occurs if we sample below the Nyquist rate
critical sampling
rate = Nyquist rate
no aliasing
oversampling
rate > Nyquist rate
no aliasing
"
usually oversample since nonideal antialiasing filter (does not have to havesharp cutoff)
⇒
replicas more spaced out
undersampling
rate < Nyquist rate
Aliasing
cannot recover
signal
Sampling EEE 507 - Lecture 3
2D
signals
ï
Oversampling Example: no aliasing
2
T
1
1
T
2
ω
ω =
1
1
1
T
1
2
2
T
2
