Signals and Systems - Electrical Engineering

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

The Laplace transform forcc(t)=e−atu(t), as seen before, is

Cc(s)=

1

s+a

with a region of convergence{(σ,):σ >−a,−∞<  <∞}. The Laplace transform of the anti-

causal part is

L[cac(−t)u(t)](−s)=

1

−s+a

and since it is anti-causal and has a pole ats=a, its region of convergence is{(σ,):σ <a,−∞<

 <∞}.

We thus have that

C(s)=

1

s+a

+

1

−s+a

=

2 a

a^2 −s^2

with a region of convergence the intersection ofσ >−awithσ <aor{(σ,):−a< σ <a,−∞<

 <∞}. This region contains thejaxis, which permits us to compute the distribution of the

power over frequencies or the power spectrum of the random signals|C()|^2 (see in Figure 3.8).

n

nExample 3.6

Consider a noncausal LTI system with impulse response

h(t)=e−tu(t)+e^2 tu(−t)

=hc(t)+hac(t)

Let us compute the system functionH(s)for this system, and find out whether we could compute

H(j)from its Laplace transform.

Solution

As from before, the Laplace transform of the causal component,hc(t), is

Hc(s)=

1

s+ 1

provided thatσ >−1. For the anti-causal component

L[hac(t)]=L[hac(−t)u(t)](−s)=

1

−s+ 2

which converges whenσ− 2 <0 orσ <2, or its region of convergence is{(σ,):σ <2,−∞<

 <∞}.