Signals and Systems - Electrical Engineering

3.4 Inverse Laplace Transform 201

and

A 2 =X(s)(s+ 2 )|s=− 2 =

3 s+ 5

s+ 1

|s=− 2 = 1

Therefore,

X(s)=

2

s+ 1

+

1

s+ 2

and as such

x(t)=[2e−t+e−^2 t]u(t)

To check that the solution is correct one could use the initial or the final value theorems shown in

Table 3.2. According to the initial value theorem,x( 0 )=3 should coincide with

lim

s→∞

[

sX(s)=

3 s^2 + 5 s

s^2 + 3 s+ 2

]

=lim

s→∞

3 + 5 /s

1 + 3 /s+ 2 /s^2

= 3

as it does. The final value theorem indicates that limt→∞x(t)=0 should coincide with

lim

s→ 0

[

sX(s)=

3 s^2 + 5 s

s^2 + 3 s+ 2

]

= 0

as it does. Both of these validations seem to indicate that the result is correct. n

RemarksThe coefficients A 1 and A 2 can be found using other methods. For instance,

n We can compute

X(s)=

A 1

s+ 1

+

A 2

s+ 2

(3.23)

for two different values of s (as long as we do not divide by zero), such as s= 0 and s= 1 ,

s= 0 X( 0 )=

5

2

=A 1 +

1

2

A 2

s= 1 X( 1 )=

8

6

=

1

2

A 1 +

1

3

A 2

which gives a set of two linear equations with two unknows, and applying Cramer’s rule we find that
A 1 = 2 and A 2 = 1.
n We cross-multiply the partial expansion given by Equation (3.23) to get

X(s)=

3 s+ 5

s^2 + 3 s+ 2

=

s(A 1 +A 2 )+( 2 A 1 +A 2 )

s^2 + 3 s+ 2

Comparing the numerators, we have that A 1 +A 2 = 3 and 2 A 1 +A 2 = 5 , two equations with two

unknowns, which can be shown to have as unique solutions A 1 = 2 and A 2 = 1 , as before.