0195136047.pdf

16.2 FEEDBACK CONTROL SYSTEMS 803

or

C(s)

R(s)

=

68

s^2 + 8 s+ 80

(13)

Noting that the denominator of Equation (13) is the same as the left side of Equation (7),

it follows that the damping ratio and the natural frequency will have the same values as

found in part (b). The oscillatory dynamic behavior can be described with the two figures

of meritξandωn, along with percent maximum overshoot and settling time.

EXAMPLE 16.2.5

A feedback control system with the configuration of Figure 16.2.7 has the following parameters:
Kp = 0 .5 V/rad,Ka =100 V/V,Km = 2. 7 × 10 −^4 N·m/V,J = 1. 5 × 10 −^5 kg·m^2 , and
F= 2 × 10 −^4 kg·m^2 /s.

(a) With an applied step-input command, describe the dynamic response of the system,

assuming the system to be initially at rest.

(b) Find the position lag error in radians, if the load disturbance of 1. 5 × 10 −^3 N·m is present

on the system when the step command is applied.

(c) In order that the position lag error of part (b) be no greater than 0.025 rad, determine the

new amplifier gain.

(d) Evaluate the damping ratio for the gain of part (c) and the corresponding percent

maximum overshoot.

(e) With the gain of part (c), in order to have the percent maximum overshoot not to exceed

25%, find the value of the output-rate gain factor.

Solution

(a)K=KpKaKm= 0. 5 × 100 × 2. 7 × 10 −^4 = 0 .0135 N·m/rad

ξ=

F

2

√

KJ

=

2 × 10 −^4

2

√

0. 0135 × 1. 5 × 10 −^5

=

10 −^4

0. 45 × 10 −^3

= 0. 222

From Figure 16.2.10, the percent maximum overshoot is 48%,

ωn=

√

K

J

=

√

0. 0135

1. 5 × 10 −^5

=30 rad/s

The commanded value of the controlled variable reaches within 1% of its final value in

ts=

5

ξωn

=

5

0. 222 × 30

= 0 .75 s

The roots of the characteristic equation

s^2 +

F

J

s+

K

J

= 0