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86 CHAPTER 3 The Laplace Transform

EXAMPLE 3.7
Compute

3s − 1
−1
L .
2
s − 6s + 2
Follow the strategy of Example 3.6. Manipulate F(s) to a function of s − a for some a:
3s − 1 3s − 1
=
2
2
s − 6s + 2 (s − 3) − 7
3(s − 3) + 8
=
(s − 3) − 7
2
3(s − 3) 8
= +
2
2
(s − 3) − 7 (s − 3) − 7
= G(s − 3) + K(s − 3)
where
3s 8
G(s) = and K(s) = .
s − 7 s − 7
2
2
By equation (3.5),

3s − 1
−1 −1 −1
L = L [G(s − 3)]+ L [K(s − 3)]
2
s − 6s + 2
3t −1 3t −1
= e L [G(s)]+ e L [K(s)]
3s 8

3t −1 3t −1
= e L + e L
2
s − 7 s − 7
2
√ 8 √
= 3e cosh( 7t) + √ e sinh( 7t).
3t
3t
7
3.3.2 The Heaviside Function and Pulses
Functions having jump discontinuities are efficiently treated by using the unit step function,
or Heaviside function H, defined by

0for t < 0
H(t) =
1for t ≥ 0.

H is graphed in Figure 3.8. We will also use the shifted Heaviside function H(t − a) of
Figure 3.9. This is the Heaviside function shifted a units to the right:

0for t < a
H(t − a) =
1for t ≥ a.

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October 14, 2010 14:14 THM/NEIL Page-86 27410_03_ch03_p77-120