Signals and Systems - Electrical Engineering

3.3 The One-Sided Laplace Transform 177

Solution

Even thoughδ(t)is not a regular signal, its Laplace transform can be easily obtained:

L[δ(t)]=

∫∞

−∞

δ(t)e−stdt=

∫∞

−∞

δ(t)e−s^0 dt=

∫∞

−∞

δ(t)dt= 1

Since there are no conditions for the integral to exist, we say thatL[δ(t)]=1 exists for all values
ofs, or that its ROC is the wholes-plane. This is also indicated by the fact thatL[δ(t)]=1 has no
poles.

The Laplace transform ofu(t)can be found as

U(s)=L[u(t)]=

∫∞

−∞

u(t)e−stdt=

∫∞

0

e−stdt=

∫∞

0

e−σte−jtdt

where we replaced the variables=σ+j. Using Euler’s equation, the above equation becomes

U(s)=

∫∞

0

e−σt[cos(t)−jsin(t)]dt

and since the sine and the cosine are bound, then we need to find a value forσso that the expo-
nentiale−σtdoes not grow astincreases. Ifσ <0, the exponentiale−σtfort≥0 will grow and
the integral will not converge. On the other hand, ifσ >0, the integral will converge ase−σtfor
t≥0 decays, and it is not clear what happens whenσ=0. Thus, the integral exists in the region
defined byσ >0 and all frequencies−∞<  <∞(the frequency values do not interfere in the
convergence). Such a region is the open right-hands-plane, and is called the ROC ofU(s).

In the region of convergence, the integral is found to be

U(s)=

e−st

−s

|∞t= 0 =

1

s

where the limit fort=∞is zero sinceσ >0. So the Laplace transformU(s)= 1 /sconverges in
the region defined by{(σ,):σ >0,−∞<  <∞}, or the open (i.e., thejaxis is not included)
right-hands-plane. This ROC can also be obtained by considering that the pole ofU(s)is ats= 0
and thatu(t)is casual.

We can find the Laplace transform of signals using symbolic computations in MATLAB. For the
unit-step and the delta functions, once the symbolic parameters are defined, the MATLAB function
laplacecomputes their Laplace transforms as indicated by the following script.

%%%%%%%%%%%%%%%%%

% Example 3.2

%%%%%%%%%%%%%%%%%

syms t s

% Unit-step function