Page 113 - Advanced engineering mathematics

3.3 Shifting and the Heaviside Function 93

250,000 Ω

1 micro F
A
B E out
10 V

FIGURE 3.18 The circuit of Example
3.13.

We want to solve for q(t) subject to the condition q(0) = 0. Take the Laplace transform of the
differential equation to get
6
250,000[sQ(s) − q(0)]+ 10 Q(s) = L[E(t)].
Now
L[E(t)](s) = 10L[H(t − 2)](s) − 10L[(t − 3)](s)
10 −2s 10 −3s
= e − e .
s s
Now we have an equation for Q:
10 10
5 6 −2s −3s
2.5(10 )sQ(s) + 10 Q(s) = e − e .
s s
Then
1 −2s −5 1 −3s
−5
Q(s) = 4(10 ) e − 4(10 ) e .
s(s + 4) s(s + 4)
Use a partial fractions decomposition to write
1 1 1 1

Q(s) = 10 −5 e −2s − e −2s − 10 −5 e −3s − e −3s .
s s + 4 s s + 4
Applying the second shifting theorem, we get
−5
−5
q(t) = 10 H(t − 2)[1 − e −4(t−2) ]− 10 H(t − 3)[1 − e −4(t−3) ].
Finally, the output voltage is E out (t) = 10 q(t). Figure 3.19 shows a graph of E out (t).
6
3.3.3 Heaviside’s Formula
There is a formula due to Heaviside that can be used to take the inverse transform of a quotient
of polynomials.
Suppose F(s) = p(s)/q(s) with p and q polynomials and q of higher degree than p.We
assume that q can be factored into linear factors and has the form

q(s) = c(s − a 1 )(s − a 2 )···(s − a n ),
with c a nonzero constant and the a j ’s n distinct numbers (which may be real or complex). None
of the a j ’s are roots of p(s).

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October 14, 2010 14:14 THM/NEIL Page-93 27410_03_ch03_p77-120