ECE 2200 Signals and Information – Section 3 (Week 4 Feb 1518)
Page 1
1. The continuoustime sinusoid
x
(
t
) = 4
.
5 sin(16
.
6
πt
)
is to be sampled at
f
s
to produce the discretetime sinusoid
x
[
k
] =
M
cos(
γk
+
θ
)
(i) For
f
s
= 15 Hz, determine
M
,
γ
, and
θ
. Has
x
(
t
) been undersampled or oversampled?
(ii) For
f
s
= 7
.
5 Hz, determine
M
,
γ
, and
θ
. Has
x
(
t
) been undersampled or oversampled?
(iii) For
f
s
= 22
.
5 Hz, determine
M
,
γ
, and
θ
. Has
x
(
t
) been undersampled or oversampled?
Solution
(read “Section 41 Sampling”):
x
(
t
)
=
4
.
5 sin(16
.
6
πt
)
=
4
.
5 cos
2
π
8
.
3
t

π
2
This means that the discretetime signal
x
[
k
] =
x
(
kT
s
) =
x
(
k/f
s
) is parameterized as
M
=
4
.
5,
γ
= 2
π
8
.
3
/f
s
and
θ
=

π/
2.
(i)
M
= 4
.
5,
γ
= 2
π
8
.
3
/
15 (undersampled, 15 Hz
<
2
×
8
.
3 Hz ) and
θ
=

π/
2
(ii)
M
= 4
.
5,
γ
= 2
π
8
.
3
/
7
.
5 (undersampled, 7
.
5 Hz
<
2
×
8
.
3 Hz ) and
θ
=

π/
2
(iii)
M
= 4
.
5,
γ
= 2
π
8
.
3
/
22
.
5 (oversampled, 22
.
5 Hz
>
2
×
8
.
3 Hz ) and
θ
=

π/
2
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
−
4
−
2
0
2
4
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
−
4
−
2
0
2
4
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
−
4
−
2
0
2
4
(i)
f
s
= 15 Hz
(ii)
f
s
= 7
.
5 Hz
(iii)
f
s
= 22
.
5 Hz
t
(sec)
Continuous Waveform:
x
(
t
) = 4
.
5 sin(16
.
6
π
t
)
Figure 1: A continuoustime 8
.
3 Hz sinusoid sampled at (top)
f
s
= 15 Hz, (middle)
f
s
= 7
.
5 Hz
and
f
s
= 22
.
5 Hz.
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ECE 2200 Signals and Information – Section 3 (Week 4 Feb 1518)
Page 2
2. Consider the discretetime sinusoid
y
[
n
] = 11
.
8 cos(0
.
21
πn

π/
5)
generated by sampling
y
(
t
) =
M
cos(2
παt
+
θ
)
at a sampling rate of 4200 Hz.
Determine three different continuoustime sinusoids
y
(
t
)
with frequencies all less than 5
.
8 kHz that could have produced
y
[
n
].
Specify your three
possibilities with unique triples of
M
,
α
and
θ
.
Solution
(read “Section 41 Sampling”): the sampled signal
y
[
n
] is defined as
y
[
n
]
=
y
(
n/f
s
)
=
M
cos(2
πα
(
n/f
s
) +
θ
)
⇒
=
M
cos(ˆ
ω
‘
n
+
θ
‘
)
M
=
11
.
8
ˆ
ω
‘
=
2
πα
‘
/f
s
+ 2
π‘
θ
‘
=

π/
5 + 2
π‘
for
‘
= 0
,
±
1
,
±
2
, . . .
, which can be repeated for various choices of the parameter
‘
:
‘
= 0
:
ˆ
ω
0
= 2
πα
0
/
4200 + 2
π
(0) = 0
.
21
π
⇒
α
0
= 441
,
θ
0
=

π
5
‘
=

1
:
ˆ
ω
1
= 2
πα
1
/
4200 + 2
π
(

1) = 0
.
21
π
⇒
α
1
= 4641
,
θ
1
=

π
5

2
π
‘
=

2
:
ˆ
ω
2
= 2
πα
2
/
4200 + 2
π
(

2) = 0
.
21
π
⇒
α
2
= 8841
,
θ
2
=

π
5

4
π
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
x 10
−
3
−
10
−
5
0
5
10
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
x 10
−
3
−
10
−
5
0
5
10
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
x 10
−
3
−
10
−
5
0
5
10
α
= 441 ,
θ
=

π
5
α
= 4641 ,
θ
=

π
5

2
π
α
= 8841 ,
θ
=

π
5

4
π
Continuous Waveform:
y
(
t
) = 11
.
8 cos(2
πα
t
+
θ
)
t
(sec)
Figure 2: Plots of
y
[
n
] = 11
.
8 cos
(
0
.
21
πn

π
5
)
for various parameterizations of
y
(
t
).