Physics, asked by pjd3123, 9 months ago

Determine if the discrete signal is periodic x(t)= cos(πn/6+π/3)​

Answers

Answered by ItzParth14
2

Explanation:

Determine if the discrete signal is periodic x(t)= cos(πn/6+π/3)

mark at brainliest please ❣️❣️

Answered by nitashachadha84
0

Answer:

mark me as brainliest

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

.

Similar questions