Assignment 1
ELEC442/ELEC6601
Concordia University
1
1
‐
Consider following two discrete time systems having input and output signals of
]
[
n
x
and
]
[
n
y
,
respectively.
System 1:
]
2
[
]
[
n
x
n
y
=
System 2:
⎪
⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
odd
n
even
n
n
x
n
y
0
2
]
[
a
‐
If
]
[
n
x
is periodic, determine whether
]
[
n
y
is periodic or not. Answer this question for both
systems.
b
‐
If
]
[
n
y
is periodic, determine whether
]
[
n
x
is periodic or not. Answer this question for both
systems.
Solution 1:
a
‐
System 1: The answer is true. See the proof below:
If
]
[
n
x
is periodic with period of “
N
” then we have
]
[
]
[
N
n
x
n
x
+
=
for all “
n
” and therefore
[
]
]
[
]
2
[
]
2
[
]
2
2
[
n
y
n
x
N
n
x
N
n
x
N
n
y
=
=
+
=
+
=
+
Having
[
]
]
[
n
y
N
n
y
=
+
proves that
]
[
n
y
is periodic with period of
N
. However if
N
is even
the smallest period is
2
N
since
]
[
]
2
[
]
2
[
2
n
y
n
x
N
n
x
N
n
y
=
=
+
=
⎥
⎦
⎤
⎢
⎣
⎡
+
System 2: The answer is true. See the proof below:
If
]
[
n
x
is periodic with period of “
N
” then we have
]
[
]
[
N
n
x
n
x
+
=
for all “
n
” and therefore
[
]
]
[
0
2
0
2
2
n
y
odd
n
even
n
n
x
odd
n
even
n
N
n
x
N
n
y
⎪
⎩
⎪
⎨
⎧
=
⎥
⎦
⎤
⎢
⎣
⎡
⎪
⎩
⎪
⎨
⎧
=
⎥
⎦
⎤
⎢
⎣
⎡
+
=
+
and therefore
]
[
n
y
is periodic with period of
N
2
.
b
‐
System 1: The answer is false . We just have to give an example.
Consider
(
)
⎩
⎨
⎧
=
odd
n
even
n
n
x
n
2
/
1
1
]
[
which is not periodic. Having this input will result in
output of
1
]
2
[
]
[
=
=
n
x
n
y
which has period of 1 and therefore periodic.
System 2: The answer is true. See the proof below:
If
]
[
n
y
is periodic with period of “
N
” then we have
]
[
]
[
N
n
y
n
y
+
=
for all “
n
” and therefore
⎪
⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
+
⎪
⎩
⎪
⎨
⎧
=
⎥
⎦
⎤
⎢
⎣
⎡
odd
n
even
n
N
n
x
odd
n
even
n
n
x
0
2
0
2
. This shows that
]
[
n
y
can be only periodic with an
even period and so
0
2
N
N
=
. Therefore:
⎪
⎩
⎪
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
+
⎪
⎩
⎪
⎨
⎧
=
⎥
⎦
⎤
⎢
⎣
⎡
odd
n
even
n
N
n
x
odd
n
even
n
n
x
0
2
0
2
0
and
]
[
]
[
0
N
n
x
n
x
+
=
and
]
[
n
x
is periodic with period of
0
N

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