Inverse Laplace Transform 199
3.4 Inverse Laplace Transform 199
Table 3.1 One-Sided Laplace Transforms
Function of Time
Function of s, ROC
1. δ( t)
1, whole s-plane
2. u(t)
s , Re[s] > 0
3. r(t)
s 2 , Re[s] > 0
4. e −at u(t), a > 0
1 s+a , Re[s] > −a
5. cos( 0 t)u(t)
s 2 + 2 0 , Re[s] > 0
6. sin( 0 t)u(t)
s 2 + 2 0 , Re[s] > 0
7. e −at cos( 0 t)u(t), a > 0
s+a
( s+a) 2 + 2 0 , Re[s] > −a
8. e −at sin( 0 t)u(t), a > 0
( s+a) 2 + 2 0 , Re[s] > −a
9. 2A e −at cos( 0 t + θ)u(t), a > 0
A∠θ
s+a−j 0 + A∠−θ s+a+j 0 , Re[s] > −a
10. ( 1 N−1)! t N−1 u(t)
1 s N N an integer, Re[s] > 0
11. 1 t N−1 e −at u(t)
( N−1)!
( s+a) N N an integer, Re[s] > −a
12. ( 2A N−1)! t N−1 e −at cos( 0 t + θ)u(t)
A∠θ
( s+a−j 0 ) N + A∠−θ ( s+a+j 0 ) N , Re[s] > −a
Table 3.2 Basic Properties of One-Sided Laplace Transforms Causal functions and constants α f (t), βg(t)
α F(s), βG(s)
Linearity
α f (t) + βg(t)
α F(s) + βG(s)
Time shifting
f (t − α)
e −αs F(s)
Frequency shifting
e α t f (t)
F(s − α)
Multiplication by t
t f (t)
− dF(s)
sF(s) − f (0−)
Second derivative 2 d f (t) 2 s 2 ( dt 1) F(s) − sf (0−) − f ( 0) Integral R t
Expansion/contraction
f (αt) α 6= 0
|α| F α s
Initial value
f (0+) = lim s→∞ sF(s)
Final value
lim t→∞ f (t) = lim s→0 sF(s)
Simple Real Poles
If X(s) is a proper rational function
N(s)
N(s)
X(s) = D(s) = Q
k ( s−p k )
C H A P T E R 3: The Laplace Transform
where the {p k } are simple real poles of X(s), its partial fraction expansion and its inverse are given by
X X(s) = p ⇔ x(t) = A
k e k t u(t)
k s−p k
where the expansion coefficients are computed as
A k = X(s)(s − p k ) s=p k
According to Laplace transform tables the time function corresponding to A k /( s−p k ) is A k e p k t u(t), thus the form of the inverse x(t). To find the coefficients of the expansion, say A j , we multiply both sides of the Equation (3.22) by the corresponding denominator (s − p j ) so that
X A k ( s−p j ) X(s)(s − p j ) =A j +
s−p k
k6=j
If we let s = p j , or s − p j = 0, in the above expression, all the terms in the sum will be zero and we find that
A j = X(s)(s − p j ) s=p j
■ Example 3.14
Consider the proper rational function
3s + 5
3s + 5
X(s) =
s 2 + 3s + 2
( s + 1)(s + 2)
Find its causal inverse.
Solution
The partial fraction expansion is
X(s) =
s+1
s+2
Given that the two poles are real, the expected signal x(t) will be a superposition of two decaying exponentials, with damping factors −1 and −2, or
x(t) = [A 1 e −t +A 2 e −t ]u(t)
where as indicated above,
3s + 5
A 1 = X(s)(s + 1)| s=−1 =
| s+2 s=−1 =2