Modern Control Engineering

aa

236 Chapter 5 / Transient and Steady-State Response Analyses

or

The peak time tpis specified as 2 sec. And so

or

Then the undamped natural frequency vnis

Therefore, we obtain

A–5–5. Figure 5–54(a) shows a mechanical vibratory system. When 2 lb of force (step input) is applied to

the system, the mass oscillates, as shown in Figure 5–54(b). Determine m, b, and kof the system

from this response curve. The displacement xis measured from the equilibrium position.

Solution.The transfer function of this system is

Since

we obtain

It follows that the steady-state value of xis

x(q)=slimS 0 sX(s)=

2

k

=0.1 ft

X(s)=

2

sAms^2 +bs+kB

P(s)=

2

s

X(s)

P(s)

=

1

ms^2 +bs+k

k =

2 zvn

K

=

2 0.4041.72

2.95

=0.471 sec

K =v^2 n=1.72^2 =2.95 N-m

vn=

vd

21 - z^2

=

1.57

21 - 0.404^2

=1.72

vd=1.57

tp=

p

vd

= 2

z=0.404

k

b

x

(a) (b)

P(2-lb force)

x(t)

ft

0.1

012345 t

m 0.0095 ft

Figure 5–54

(a) Mechanical

vibratory system;

(b) step-response

curve.

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