Signals and Systems - Electrical Engineering

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

For the Laplace transform off(t)to exist we need that

∣∣

∣∣

∣∣

∫∞

−∞

f(t)e−stdt

∣∣

∣∣

∣∣=

∣∣

∣∣

∣∣

∫∞

−∞

f(t)e−σte−jtdt

∣∣

∣∣

∣∣

≤

∫∞

−∞

|f(t)e−σt|dt<∞

or thatf(t)e−σtbe absolutely integrable. This may be possible by choosing an appropriateσeven in the case

whenf(t)is not absolutely integrable. The value chosen forσdetermines the ROC ofF(s); the frequency

does not affect the ROC.

3.2.2 Poles and Zeros and Region of Convergence

Theregion of convergence(ROC) can be obtained from the conditions for the integral in the Laplace

transform to exist. The ROC is related to thepolesof the transform, which is in general a complex

rational function.

For a rational functionF(s)=L[f(t)]=N(s)/D(s), its zeros are the values ofsthat make the functionF(s)= 0 ,

and its poles are the values ofsthat make the functionF(s)→∞. Although only finite zeros and poles are

considered, infinite zeros and poles are also possible.

Typically,F(s)is rational, a ratio of two polynomialsN(s)andD(s), orF(s)=N(s)/D(s), and as such

its zeros are the values ofsthat make the numerator polynomialN(s)=0, while the poles are the

values ofsthat make the denominator polynomialD(s)=0. For instance, for

F(s)=

2 (s^2 + 1 )

s^2 + 2 s+ 5

=

2 (s+j)(s−j)

(s+ 1 )^2 + 4

=

2 (s+j)(s−j)

(s+ 1 + 2 j)(s+ 1 − 2 j)

we have the zeros are ats=±j, roots ofN(s)=0, sinceF(±j)=0, and a pair of complex conjugate

poles− 1 ± 2 j, the roots of the equationD(s)=0 and such thatF(− 1 ± 2 j)→∞. Geometrically,

zeros can be visualized as those values that make the function go to zero, and poles as those val-

ues that make the function approach infinity (looking like the main “pole” of a circus tent). See

Figure 3.4.

Not all rational functions have poles or a finite number of zeros. Consider the Laplace transform

P(s)=

1

s

(

es−e−s

)

P(s)seems to have a pole ats=0. Its zeros are obtained by lettinges−e−s=0, which when

multiplied byesgives

e^2 s= 1 =ej^2 πk