Signals and Systems - Electrical Engineering

186 C H A P T E R 3: The Laplace Transform

f(t)=Ae−atu(t)whereain general can be a complex number is

F(s)=

A

s+a

ROC:σ >−a

The location of the poles=−aclosely relates to the signal. For instance, ifa=5,f(t)=Ae−^5 tu(t)is

a decaying exponential and the pole ofF(s)is ats=−5 (in left-hands-plane); ifa=−5, we have an

increasing exponential and the pole is ats=5 (in right-hands-plane). The larger the value of|a|the

faster the exponential decays (fora>0) or increases (fora<0); thus,Ae−^10 tu(t)decays a lot faster

thatAe−^5 tu(t), andAe^10 tu(t)grows a lot faster thanAe^5 tu(t).

The Laplace transformF(s)= 1 /(s+a)off(t)=e−atu(t), for any real value ofa, has a pole on the real

axisσof thes-plane, and we have the following three cases:

n Fora=0, the pole at the origins=0 corresponds to the signalf(t)=u(t), which is constant for

t≥0 (i.e., it does not decay).

n Fora>0, the signal isf(t)=e−atu(t), a decaying exponential, and the poles=−aofF(s)is in

the real axisσof the left-hands-plane. As the pole is moved away from the origin toward the left,

the faster the exponential decays, and as it moves toward the origin, the slower the exponential

decays.

n Fora<0, the poles=−ais on the real axisσof the right-hands-plane, and corresponds to

a growing exponential. As the pole moves to the right the exponential grows faster, and as it is

moved toward the origin it grows at a slower rate—clearly this signal is not useful, as it grows

continuously.

The conclusion is that theσaxis of the Laplace plane corresponds to damping. A single pole on this axis

and in the left-hands-plane corresponds to a decaying exponential, and a single pole on this axis and in the

right-hands-plane corresponds to a growing exponential.

Suppose then we consider

g(t)=Acos( 0 t)u(t)=A

ej^0 t

2

u(t)+A

e−j^0 t

2

u(t)

and leta=j 0 to expressg(t)as

g(t)=0.5[Aeatu(t)+Ae−atu(t)]

Then, by the linearity of the Laplace transform and the previous result we obtain

G(s)=

A

2

1

s−j 0

+

A

2

1

s+j 0

=

As

s^2 +^20

(3.9)

with a zero ats=0, and the poles are values for which

s^2 +^20 = 0 ⇒s^2 =−^20 or s1,2=±j 0