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ECEN 314: Signals and Systems
Solutions to HW 2
Problem 1.21
0.5
0
1
1
2
3
1
1
2
0
1
2
3
4
x(t1)
x(2t)
1
2
1
1.5
1
0.5
0
0.5
1
2
1
x(2t+1)
4
6
8
10
12
2
1
0
1
x(4t/2)
[x(t)+x(t)]u(t)
3
1.5
1.5
2
1
1
0.5
Problem 1.23
3t/2
2
1
2
1/2
1/2
1/2
1/2
2
1
0
1
2
e
x (t)
o
x (t)
(a)
1
2
1
0
1
0
2
2
1
0
1
2
x (t)
e
(b)
1/2
1/2
1/2
1
1
1/2
x (t)
o
0
x (t)
e
3t/2
(c)
x (t)
o
t/2
1
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View Full DocumentProblem 1.25
(a)
x
(
t
) = 3 cos(4
t
+
π
3
) is periodic with period:
π
2
. To see why this is true, note that we
need to ﬁnd a
T
so that
x
(
t
+
T
) =
x
(
t
)
Note that
x
(
t
) = 3 cos(4
t
+
π
3
) = 3cos
(
4(
t
+
π
2
) +
π
3
)
, which yields the desired answer.
(b)
x
(
t
) =
e
j
(
πt

1)
is periodic with period =
2
π
π
= 2.
(c)
x
(
t
) = [cos(2
t

π
3
)]
2
[cos(2
t

π
3
)]
2
=
1 + cos(4
t

2
π
3
)
2
which is periodic with period
T
=
π
2
.
(d)
x
(
t
) =
E
(cos(4
πt
)
u
(
t
))
=
cos(4
πt
)
u
(
t
) + cos(4
πt
)
u
(

t
)
2
=
1
2
cos(4
πt
)
 ∞
< t
≤ ∞
which is periodic with period
1
2
. (Here,
E
is the even part).
(e)
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 Spring '08
 HALVERSON

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