596 Higher Engineering Mathematics
9. (a)
s + 1
s 2 + 2s + 10
(b)
3
s 2 + 6s + 13
(a) e −t cos 3t (b)
3
2
e −3t sin 2t
10. (a)
2(s − 3)
s 2 − 6s + 13
(b)
7
s 2 − 8s + 12
(a) 2e 3t cos 2t (b)
7
2
e 4t sinh 2t
11. (a)
2s + 5
s 2 + 4s − 5
(b)
3s + 2
s 2 − 8s + 25
⎡
⎢
⎣
(a) 2e −2t cosh 3t +
1
3
e −2t sinh 3t
(b) 3e 4t cos 3t +
14
3
e 4t sin 3t
⎤
⎥
⎦
63.3 Inverse Laplace transforms using
partial fractions
Sometimes the function whose inverse is required is not
recognisable as a standard type, such as those listed in
Table 63.1. In such cases it may be possible, by using
partial fractions, to resolve the function into simpler
fractions which may be inverted on sight. For example,
the function,
F(s) =
2s − 3
s(s − 3)
cannot be inverted on sight from Table 63.1. However,
by using partial fractions,
2s − 3
s(s − 3)
≡
1
s
+
1
s − 3
which
may be inverted as 1 + e 3t from (i) and (iii) of Table 61.1.
Partial fractions are discussed in Chapter 2, and a summary of the forms of partial fractions is given in Table 2.1
on page 13.
Problem 7. Determine L −1
4s − 5
s 2 − s − 2
4s − 5
s 2 − s − 2
≡
4s − 5
(s − 2)(s + 1)
≡
A
(s − 2)
+
B
(s + 1)
≡
A(s +1) + B(s −2)
(s − 2)(s + 1)
Hence 4s − 5 ≡ A(s + 1) + B(s − 2).
When s = 2, 3 =3 A, from which, A = 1.
When s =−1, −9 =−3B, from which, B = 3.
Hence L −1
4s − 5
s 2 − s − 2
≡ L
−1
1
s − 2
+
3
s + 1
= L
−1
1
s − 2
+ L
−1
3
s + 1
= e
2t
+ 3e
−t
, from (iii) of Table 63.1
Problem 8. Find L −1
3s 3 + s 2 + 12s + 2
(s − 3)(s + 1) 3
3s 3 + s 2 + 12s + 2
(s − 3)(s + 1) 3
≡
A
s − 3
+
B
s + 1
+
C
(s + 1) 2 +
D
(s + 1) 3
≡
A(s + 1) 3 + B(s − 3)(s + 1) 2
+ C(s − 3)(s + 1) + D(s − 3)
(s − 3)(s + 1) 3
Hence
3s
3
+ s
2
+ 12s + 2 ≡ A(s + 1)
3
+ B(s − 3)(s + 1)
2
+ C(s − 3)(s + 1) + D(s − 3)
When s = 3, 128 =64 A, from which, A = 2.
When s =−1, −12 =−4D, from which, D = 3.
Equating s 3 terms gives: 3 = A + B, from which, B = 1.
Equating constant terms gives:
2 = A − 3B − 3C − 3D,
i.e.
2 = 2 − 3 − 3C − 9,
from which, 3C =−12 and C =−4
Hence
L −1
3s
3
+ s
2
+ 12s + 2
(s − 3)(s + 1) 3
≡ L
−1
2
s − 3
+
1
s + 1
−
4
(s + 1) 2 +
3
(s + 1) 3
= 2e
3t
+ e
−t
− 4e
−t t +
3
2
e
−t t
2
,
from (iii) and (xi) of Table 63.1