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92 CHAPTER 3 The Laplace Transform

4
2

3 1

–1 0 1 2 3 4 5
2
t
–1
1 –2

–3

0 4 8 12 –4
t
FIGURE 3.17 f (t) in Example 3.12.
FIGURE 3.16 Graph of the solution in
Example 3.11.

EXAMPLE 3.12
Sometimes we need to deal with a function having several jump discontinuities. Here is an
example of writing such a function in terms of step functions. Let
⎧
⎪0 for t < 2
⎨
f (t) = t − 1 for 2 ≤ t < 3
⎪
−4 for t ≥ 3.
⎩
Figure 3.17 shows a graph of f . There are jump discontinuities of magnitude 1 at t = 2 and
magnitude 6 at t = 3.
Think of f (t) as consisting of two nonzero parts: the part that is t − 1for 2 ≤ t < 3 and the
part that is −4for t ≥ 3. We want to turn on t − 1 at time 2 and turn it off at time 3, then turn −4
on at time 3 and leave it on.
The first effect is achieved by multiplying t − 1 by the pulse H(t − 2) − H(t − 3).The
second is achieved by multiplying −4by H(t − 3). Thus, write
f (t) =[H(t − 2) − H(t − 3)](t − 1) − 4H(t − 3).

EXAMPLE 3.13
Suppose the capacitor in the circuit of Figure 3.18 initially has a charge of zero and there is no
initial current. At time t = 2 seconds, the switch is thrown from position B to A, held there for 1
second, and then switched back to B. We want the output voltage E out on the capacitor.
From the circuit, write
E(t) = 10[H(t − 2) − H(t − 3)].
By Kirchhoff’s voltage law,
1
Ri(t) + q(t) = E(t)
C
or
6

250,000q (t) + 10 q(t) = E(t).

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October 14, 2010 14:14 THM/NEIL Page-92 27410_03_ch03_p77-120