Inverse Laplace Transform 201

3.4 Inverse Laplace Transform 201

and

3s + 5

A 2 = X(s)(s + 2)| s=−2 =

X(s) =

s+1

s+2

and as such

x(t) = [2e −t +e −2t ]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

3s 2 + 5s

3 + 5/s

lim sX(s) =

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

sX(s) =

s 2 + 3s + 2 =0

s→0

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

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

We can compute

X(s) =

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) =

=A 1 + A 2

s = 1 X(1) =

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.

We cross-multiply the partial expansion given by Equation (3.23) to get

3s + 5

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

X(s) =

+ 3s + 2

s 2 + 3s + 2

Comparing the numerators, we have that A 1 +A 2 = 3 and 2A 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.

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

Simple Complex Conjugate Poles

The partial fraction expansion of a proper rational function

N(s)

N(s)

X(s) =

( s + α) 2 + 2 0 ( s + α − j 0 )( s + α + j 0 )

with complex conjugate poles {s 1,2 = −α ± j 0 } is given by A A ∗

X(s) =

0 s + α − j + s + α + j 0

where

A = X(s)(s + α − j 0 )

| jθ

s=−α+j 0 = |A|e

so that the inverse is the function

x(t) = 2|A|e −αt cos( 0 t + θ)u(t)

Because the numerator and the denominator polynomials of X(s) have real coefficients, the zeros and poles whenever complex appear as complex conjugate pairs. One could thus think of the case of a pair of complex conjugate poles as similar to the case of two simple real poles presented above. Notice that the numerator N(s) must be a first-order polynomial for X(s) to be proper rational. The

poles of X(s), s 1,2 = −α ± j 0 , indicate that the signal x(t) will have an exponential e −αt , given that the real part of the poles is −α, multiplied by a sinusoid of frequency  0 , given that the imaginary

parts of the poles are ± 0 . We have the expansion

X(s) =

0 s + α − j + s + α + j 0

where the expansion coefficients are complex conjugate of each other. From the pole information, the general form of the inverse is

x(t) = Ke −αt cos( 0 t + 8)u(t)

for some constants K and 8. As before, we can find A as

A = X(s)(s + α − j jθ 0 ) | s=−α+j

0 = |A|e

and that X(s)(s + α + j 0 ) | s=−α−j 0 =A ∗ can be easily verified. Then the inverse transform is given by

x(t) = Ae −(α−j 0 +A ∗ e −(α+j 0 u(t)

−αt ( e j( = |A|e 0 t+θ)

−j( +e 0 t+θ) ) u(t) = 2|A|e −αt cos( 0 t + θ)u(t).