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is given by
(5.49)
(5.50)
It is convenient to use the method of partial fractions to factor the right-hand side.
Let
s+b
s(s + a)
  A
- -+
   s
Multiply throughout by s(s + a) so that
s+a
s +b = A(s +a)+ Bs
(5.51)
(5.52)
This ident.ity is. supposed to hold for all values of s. Therefore, with s - -a, we
get B - (a - b)la and, with s = O, we get A = b/a. Then,
 ...      b      (a - b)
y(s) = b +(.(,
          a!      z(s + a)
(5.5 3)
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW      447
Taking the inverse Laplace transform,
'     b
y(t) = b + (  . b)e_ur
      a
(5.54)
We note that y(0) = 1 and y(oo) = b/a.
   The first term on the right-hand side represents forced response and the sec-
ond term represents the natural response. The forced response is also known as
 steady-state response, and the natural response is also known as transient response.
Observe that the pole at s = 0 corresponding to the input unit-step function gen-
erates the forced response. The transient response is generated by the system pole
at s - -a and is of the form e-"r. Thus, the farther to the left the pole is located
on the negative real axis, the faster the transient response will decay to zero. On
the other hand, if this pole is located on the positive real axis, then the response
 will be of diverging nature because the out,put will increase steadily with time.
    The zeros of the system and the inpue function influence the amplitude of both
 the steady-state and the transient response. In this case, we have only one system
zero at s - -b, and there is no zero because of the input function. The effect of
this system zero on the amplitude of the response can be seen in Eq. (5.54).
   The quantity l/a is called the time constant of the given first-order system.
The time constant is a measure of the speed with which a system responds to an
 external input. The lower the value of the time constant (or l:igher the value of a),
 the faster will be the system response. Sometimes, a is also called the exponential
decay frequency. For t = lla, y(t) = 0.63 times its final rise above the initial
 value. In other words, at time equal to one tim.e constant. the output rises to 63%
 ofits steady-state value above the initial value.
   The rise time Tr is the time for the output to increase from 0.1 to 0.9 times its
 final or steady-state value. However, it may be noted that some authors define it as
 the time for the output to rise from 0.1 to l00o/o of the final value. However, this
alternative definition is not used in this text.
For a first-order system,
  2.2
Tr - N
     a
(5.55)
  The settling time Tx is defined as the time required for the output to reach, for
the first occurrence, within 2% ofits final or steady-state value. For the first-order
system, this value is approximately given by
     4
Ts - -
    a
5-4-2  Response of Second-Order Systems
    Now let us consider a second-order system given by
                                       b
                                G(s) = -
                                                  = s~+ as +b
The response to a unit-step input is
                                                           ),(s) = G(s)r (s)
                                   b
　
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