Page 17 - Servo Motors and Industrial Control Theory

1.6 First Order Transfer Function 9

Fig. 1.5 Ramp input 10
response of first order lag

input and response f (t) 5

g (t)

0
0 5 10
t
time for unity time constant

The Laplace Transform is
1
xs() = (1.31)
s 2

The output then becomes
1 A B C
() =
ys = + + (1.32)
s 2 ( s τ + 1) s s 2 s τ + 1
Calculating the coefficients A, B, C, by equating the common factors in s, the output
transfer equation becomes

τ 1 τ 2
() =− +
ys + (1.33)
s s 2 s τ + 1
Taking the inverse of Laplace Transform by referring to Table 1.1, the solution
becomes
− t (1.34)
y ()t =− (ττ− e τ )
t
The solution is shown graphically in Fig. 1.5.
It should be noted that both for step and ramp inputs, a unity gain was used. If
different gain is used, the solution will be multiplied by that factor. For a ramp in-
put, there is a shift between the output and input. This is known as following error.
The value of the following error depends on the transfer function, which for first
order system described in this section is τ. Later we will describe how to calculate
the steady state error without actually solving the differential (transfer function).
Throughout this book, the variables x, y are considered as input and output vari-
ables. From the equation, it will be clear whether they are in time or Laplace domain.