Signals and Systems - Electrical Engineering

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

Att=0 we assume thatu( 0 )=0.5 to getf( 0 )from the sumfc( 0 )+fac( 0 ). The Laplace transform of

the two-sided signalf(t)can then be computed as

F(s)=

∫∞

0

f(−t)u(t)estdt+

∫∞

0

f(t)u(t)e−stdt

=L[fac(−t)u(t)](−s)+L[fc(t)u(t)] (3.7)

with an ROC the intersection of the ROCs of the causal and the anti-causal Laplace transforms.

3.3 The One-Sided Laplace Transform

Theone-sided Laplace transformis defined as

F(s)=L[f(t)u(t)]=

∫∞

0 −

f(t)u(t)e−stdt (3.8)

wheref(t)is either a causal function or made into a causal function by the multiplication byu(t). The one-

sided Laplace transform is of significance given that most of the applications deal with causal systems and

signals, and that any signal or system can be decomposed into causal and anti-causal components requiring

only the computation of one-sided Laplace transforms.

Remarks

n If f(t)is causal the multiplication by u(t)is redundant but harmless, but if f(t)is not causal the multi-

plication by u(t)makes f(t)u(t)causal. Notice that when f(t)is causal, the two-sided and the one-sided

Laplace transforms of f(t)coincide.

n The lower limit of the integral in the one-sided Laplace transform is set to 0 −= 0 −ε, which corresponds

to a value on the left side of 0 for an infinitesimal valueε. The reason for this is to make sure that an

impulse functionδ(t), only defined at t= 0 , is included when we are computing its Laplace transform.

For any other signal this limit can be taken as 0 with no effect on the transform.

n As we will see, the advantage of the one-sided Laplace transform is that it can be used in the solution

of differential equations with initial conditions. In fact, the two-sided Laplace transform by starting at

t=−∞(lower bound of the integral) ignores initial conditions at t= 0 , and thus it is not useful in

solving differential equations unless the initial conditions are zero.

nExample 3.2

Find the Laplace transforms ofδ(t),u(t), and a pulsep(t)=u(t)−u(t− 1 ). Use MATLAB to verify

the transforms.