Page 86 - Mechatronics with Experiments

72 MECHATRONICS
The maximum overshoot occurs at the first time instant, t , where the derivative of x is
p
zero. Once t is found, then the maximum value of response at that time can be evaluated
p
and the percent overshoot can be determined.
dx(t) 2.5
= 0 ⇒ find t : t =
p
p
dt
n
Then
x(t ) − 1 √ −
p
PO%= × 100 = e 1− 2
1
The settling time is the time it takes for the response to settle within ±1or ±2% of the final
value, and it can be shown that

⎧ 4.6
; ±1%
⎪ n
t = ⎨ 4.0
s
⎪ ; ±2%
n
⎩
Therefore, given a (PO%, t ) specification, the corresponding second-order system pole
s
√
2
locations (− w ± (1 − )w can be directly obtained.
n
n
Effect of an additional zero Let us consider a second-order system with a real
zero (Figure 2.22). The system is the same as a standard second-order system with two
complex conjugate poles and d.c. gain of 1, with an additional zero on the real axis.
( ) 2
s n
G(s) = + 1
a s + 2 s + 2 n
2
n
Let = 1; a = , the transfer function can be expressed as
n n
( )
s
+ 1 1 1 s
G(s) = = +
2
2
2
s + 2 s + 1 s + 2 s + 1 s + 2 s + 1
Notice that the effect of zero is to add the derivative of the step response to the second-order
1
system response by an amount proportional to . Clearly if is large, a is to the left of

, and the influence of the addition of zero is not much. As gets smaller, a gets closer
n
to area, and (1∕ ) grows. Hence, the influence of zero on the response increases. The
n
main effect of zero as it gets close to the value is to increase the percent overshoot. If
n
the zero is on the right half s-plane (non-mininum phase transfer function) the initial value
of step response goes in the opposite direction. This is illustrated in Figures 2.23 and 2.24.
j 1−ξ 2 ω n

–a −ξ ω n
− j 1−ξ 2 ω n
FIGURE 2.22: Second-order system with
two complex conjugate poles and a real
zero.