0195136047.pdf

634 SIGNAL PROCESSING

Hence, the givenx(t) is a power-type signal with its power given byA^2 /2.

(b) For any periodic signal with periodT 0 , the energy is given by

Ex= lim

T→∞

∫+T/ 2

−T/ 2

|x(t)|^2 dt=lim

n→∞

∫+nT 0 / 2

−nT 0 / 2

|x(t)|^2 dt

=lim

n→∞

n

∫+T 0 / 2

−T 0 / 2

|x(t)|^2 dt=∞

Therefore, periodic signals are not typically energy type. The power content of any

periodic signal is

Px=lim

T→∞

1

T

∫T/ 2

−T/ 2

|x(t)|^2 dt=lim

n→∞

1

nT 0

∫nT 0 / 2

−nT 0 / 2

|x(t)|^2 dt

=lim

n→∞

n

nT 0

∫T 0 / 2

−T 0 / 2

|x(t)|^2 dt=

1

T 0

∫T 0 / 2

−T 0 / 2

|x(t)|^2 dt

which shows that the power content of a periodic signal is equal to the average power in

one period.

TheFourier-series representationstates that almost any periodic signal can be decomposed

into an infinite series of the form

x(t)=a 0 +

∑∞

n= 1

(ancosnωt+bnsinnωt) (14.1.11)

wherea 0 ,a 1 ,b 1 ,... are the Fourier coefficients, andωis the fundamental angular frequency

related to the periodTbyω= 2 π/T= 2 πf. The integer multiples ofωare known as harmonics:

2 ωbeing the second harmonic that is even, 3ωbeing the third harmonic that is odd, and so forth.

The dc component is given by

a 0 =

1

T

∫T

0

x(t)dt (14.1.12)

which is seen to be the average value ofx(t). The remaining coefficients can be computed from

the following integrals:

an=

2

T

∫T

0

x(t)cosnωt dt, forn= 1 , 2 ,... (14.1.13)

bn=

2

T

∫T

0

x(t)sinnωt dt, forn= 1 , 2 ,... (14.1.14)

It can be seen thatbn=0 forn=1, 2, 3,... for even symmetry. Similarly, for odd symmetry,

an=0 forn= 0, 1, 2, 3,.... For half-wave symmetry,an=bn=0 forn=2, 4, 6,... so that

the series contains only the odd-harmonic components. For relatively smooth signals, the higher

harmonic components tend to be smaller than the lower ones. Discontinuous signals have more

significant high-frequency content than continuous signals.