Second-order Systems and a Zero - Time Response

H. R. Pota


G(s) =  p1 p2
z
  s+z
(s+p1)(s+p2)
(1)
Step-response:
Y(s)
=
 1
s
  p1 p2
z
  s+z
(s+p1)(s+p2)
=
 1
s
 +  p2
p2 - p1
  p1 - z
z
  1
s+p1
 -  p1
p2 - p1
  p2 - z
z
  1
s+p2
(2)
This gives,
y(t) = 1 +  p2
p2 - p1
  p1 - z
z
 e-p1 t -  p1
p2 - p1
  p2 - z
z
 e-p2 t
(3)
To see the effect of the zero location on time-response, let's see if there is any qualitative change in time-response y(t) as zero location changes from far left of the imaginary axis to near it. The behaviour we would look at first is if there is any zero location for which there is an overshoot. The time-response at t=0 is zero and as t® ¥ it's one. We say that there is an overshoot if there is a time interval for which y(t) > 1. Since y(t) = 0 at t=0 and y(t) = 1 as t ® ¥, so if it overshoots there would be a maximum in time-response and at that point the rate of change of y(t) will be zero. If such a point exists then we can say that there is an overshoot otherwise not. The rate of change of y(t) is,
 dy
dt
 =  p1 p2
z(p2 - p1)
 ((p2-z) e-p2 t - (p1-z) e-p1 t )
(4)
The derivative [ dy/dt] = 0 when
(p2-z) e-p2 t = (p1-z) e-p1 t
Þ e(p2-p1) t =  p2-z
p1-z
Þ tp =  1
p2-p1
 ln æ
è 		 p2-z
p1-z
 ö
ø
(5)

(a) z > p2 > p1

This means that
0 <  p2-z
p1-z
 < 1
and the from equation (5) it can be seen that tp < 0 meaning that there is no tp > 0 for which a maximum exists. From this we can conclud that whenever z > p2 > p1, the step-response will be less than unity for all finite time.

(b) p2 > z > p1

For these conditions (p2 > z > p1) we see that
p2 - z > 0 and p1 - z < 0.
Under these conditions we can see that the rate of change in equation (4) will always be positive, i.e., the step-response y(t) is monotonic and it steadily increases from zero to one as time goes from zero to infinity. Hence no overshoot.

(c) p2 > p1 > z

This (p2 > p1 > z) is the most interesting case. For this case
 p2-z
p1-z
 > 1
and equation (5) gives a positive value of time at which the maximum is attained. This implies an overshoot.

Example

Figure 1 shows step-response for the system in (1) for different zero locations with p2 = 10 and p1 = 1, i.e.,
G(s) =  10
z
  s+z
(s+1)(s+10)
 .
(6)
secondZero.jpg
Figure 1: Second-order System with a zero
From equation (5), tp = 0.32 s when z=0.5. It can be seen from Figure 1 that the step-response for this case does peak at 0.32 s.

File translated from TEX by TTH, version 3.40.
On 22 Mar 2005, 11:59.