196 Transient analysis
Transforming to the time domain from the s-domain
From the transform circuit of Fig. 8.32 we have that
V/s + Li(O)- Vc(O)/s- i(s)[R + Ls + 1/Csl
Assuming that the initial conditions are zero (i = 0 and Vc = 0) then Li(O) = 0
and Vc(O)/s - O. Thus
i(s) = V/s/[R + Ls + 1/Cs]In order to proceed from here to obtain i as a function of time we have to use
the inverse Laplace transform table. Very often the expression for i(s) is of the
form i(s) = M(s)/N(s) where M(s) and N(s) are polynomials. These must be
reduced to a series of partial fractions in order to identify a suitable transform
pair from the table.
Partial fractions
By using a common denominator the expression [A/(s + 1)] + [B/(s + 2)1 can
be written {A(s + 2) + B(s + 1)}/{(s + 1)(s + 2)}. It follows that the inverse
process is possible so that the expression with a common denominator may be
converted to a series of terms with separate denominators. These are called
partial fractions. There are a number of techniques for finding the partial
fractions of an expression, one of which is known as 'equating coefficients'. The
following example illustrates the method.Example 8.11
Find the time response of a circuit for which the current in the transform circuit
is given by i(s) = (V/R)(s + 3)/[(s + 1)(s + 2)].Solution
To identify suitable transform pairs we must find the partial fractions of
(s + 3)/[(s + 1)(s + 2)]
Let
(s + 3)//[(s + 1)(s + 2)1 = [A//(s + 1)1 + [B/(s + 2)1
Since
[A/(s + 1)1 + [B/(s + 2)1 = [A(s + 2) + B(s + 1)]/[(s + 1)(s + 2)1
then
s + 3 - A(s + 2) + B(s + 1)
i.e.
ls 1 + 3s ~ (A + B)s 1 + (2A + B)s ~ remembering that s o= 1.