Determine if the discrete signal is periodic x(t)= cos(πn/6+π/3)
Answers
Explanation:
Determine if the discrete signal is periodic x(t)= cos(πn/6+π/3)
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Answer:
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Explanation:
Determine the values of P∞ and E∞ for each of the following signals:
(a) x1(t) = e
−2tu(t)
(b) x2(t) = e
j(2t+π/4)
(c) x3(t) = cos(t)
(d) x1[n] = ( 1
2
)
nu[n]
(e) x2[n] = e
j(π/2n+π/8)
(f) x3[n] = cos(
π
4
n)
Solution
(a) E∞ =
R ∞
0
e
−2tdt =
1
4
. P∞ = 0, because E∞ < ∞.
(b) x2(t) = e
j(2t+π/4)
, |x2(t)| = 1. Therefore,
E∞ =
Z ∞
−∞
|x2(t)|
2
dt =
Z ∞
−∞
dt = ∞.
P∞ = lim
T→∞
1
2T
Z T
−T
|x2(t)|
2
dt = lim
T→∞
1
2T
Z T
−T
dt = lim
T→∞
1 = 1.
(c) x3(t) = cos(t). Therefore,
E∞ =
Z ∞
−∞
|x3(t)|
2
dt =
Z ∞
−∞
cos2
(t)dt = ∞.
P∞ = lim
T→∞
1
2T
Z T
−T
|x3(t)|
2
dt = lim
T→∞
1
2T
Z T
−T
cos2
(t)dt = lim
T→∞
1
2T
Z T
−T
(
1 + cos(2t)
2
)dt =
1
2
.
(d) x1[n] = ( 1
2
)
nu[n], |x1[n]|
2 = ( 1
4
)
nu[n]. Therefore,
E∞ =
X∞
n=−∞
|x1[n]|
2 =
X∞
n=0
(
1
4
)
n =
4
3
.
P∞ = 0, because E∞ < ∞.
(e) x2[n] = e
j(π/2n+π/8)
, |x2[n]|
2 = 1. Therefore,
E∞ =
X∞
n=−∞
|x2[n]|
2 =
X∞
n=−∞
1 = ∞.
P∞ = lim
N→∞
1
2N + 1
X
N
n=−N
|x2[n]|
2 = lim
N→∞
1
2N + 1
X
N
n=−N
1 = 1.
(f) x3[n] = cos(
π
4
n). Therefore,
E∞ =
X∞
n=−∞
|x3[n]|
2 =
X∞
n=−∞
cos2
(
π
4
n) = ∞.
P∞ = lim
N→∞
1
2N + 1
X
N
n=−N
|x3[n]|
2 = lim
N→∞
1
2N + 1
X
N
n=−N
cos2
(
π
4
n) = lim
N→∞
1
2N + 1
X
N
n=−N
(
1 + cos(
π
2
n)
2
) = 1
2
.