Signals and Systems - Electrical Engineering

84 C H A P T E R 1: Continuous-Time Signals

which is again a sum of harmonically related frequency sinusoids, so thatx^2 (t)is periodic of period

T 0 =1. As in the previous examples, we have

Px=

1

T 0

∫T^0

0

x^2 (t)dt= 1

which is the integral of the constant since the other integrals are zero. In this case, we used the

periodicity ofx(t)andx^2 (t)to calculate the power directly. That is not possible when computing

the power ofy(t)because it is not periodic, so we have to consider each of its components. We

have that

y^2 (t)=cos^2 ( 2 πt)+cos^2 ( 2 t)+2 cos( 2 πt)cos( 2 t)

= 1 +

1

2

cos( 4 πt)+

1

2

cos( 4 t)+cos( 2 (π+ 1 )t)+cos( 2 (π− 1 )t)

and the power ofy(t)is

Py= lim

T→∞

1

2 T

∫T

−T

y^2 (t)dt

= 1 +

1

2 T 4

∫T^4

0

cos( 4 πt)dt+

1

2 T 5

∫T^5

0

cos( 4 t)dt

+

1

T 6

∫T^6

0

cos( 2 (π+ 1 )t)dt+

1

T 7

∫T^7

0

cos( 2 (π− 1 )t)dt= 1

whereT 4 ,T 5 ,T 6 , andT 7 are the periods of the sinusoidal components ofy^2 (t). Fortunately, only

the first integral is not zero and the others are zero (the average over a period of the sinusoidal

components ofy^2 (t)). Fortunately, too, we have that the power ofx(t)and the power ofy(t)are the

sum of the powers of its components. That is if

x(t)=cos( 2 πt)+cos( 4 πt)=x 1 (t)+x 2 (t)

y(t)=cos( 2 πt)+cos( 2 t)=y 1 (t)+y 2 (t)

then as in previous examplesPx 1 =Px 2 =Py 1 =Py 2 =0.5, so that

Px=Px 1 +Px 2 = 1

Py=Py 1 +Py 2 = 1

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