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EnergyPower DT

The document determines whether a given signal x(n)=(1/4)nU(n) is an energy or power signal. It calculates the energy of the signal, which is finite at 1.066, and the power of the signal, which approaches 0 as N approaches infinity. Therefore, the signal is an energy signal, not a power signal, since its energy is finite but its power is zero.

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Keerthi Varman
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83 views5 pages

EnergyPower DT

The document determines whether a given signal x(n)=(1/4)nU(n) is an energy or power signal. It calculates the energy of the signal, which is finite at 1.066, and the power of the signal, which approaches 0 as N approaches infinity. Therefore, the signal is an energy signal, not a power signal, since its energy is finite but its power is zero.

Uploaded by

Keerthi Varman
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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1.

Dt whether the given signal is energy or power signal

X(n)= (1/4)n U(n)

-∞ ∞

0 n 

E∞=  ( x(n)u (n)) +  ( x(n)u(n))


n  
2

n 0
2

n 

E =0+ 
2
( x(n)1)
∞
n 0
n  n 

E =
( x(n))  (1 / 4)2 2n
∞ =
n 0 n 0

By using the formula


1
E∞= 1 =4/3= 1.066
1−(4)2

E∞=1.066
Now for Power:

N
1
2𝑁+1 
P∞=Nlim ∞ ( x ( n) 2

⥱ n N

P∞=Nlim ∞
⥱
1
2𝑁+1
{ 0

n N
( x ( n) 2 u ( n) +
N

 ( x ( n)
n 0
2
u(n) }
0

 ( x(n) u (n) =0
n N
2

Then

P∞=Nlim ∞
⥱
1
2𝑁+1
{ 0+
N

 ( x ( n)
n 0
2
1 }
P∞=Nlim ∞
⥱
1
2𝑁+1
{ 0+
N

 ((1 / 4)
n 0
2
}
1

By using this formula

N N
1 1−(0.25)
P∞=Nlim ∞
⥱ 2𝑁+1  ((1 / 4) 2 = ∑𝑁𝑛=0
n 0
1−(0.25)
=0

So Energy is Finite Power is Zero


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