Modern Control Engineering

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Example Problems and Solutions 237

Hence

Note that Mp=9.5%corresponds to z=0.6. The peak time tpis given by

The experimental curve shows that tp=2sec. Therefore,

Sincev^2 n=km=20m, we obtain

(Note that 1 slug=1lbf-sec^2 ft.) Then bis determined from

or

A–5–6. Consider the unit-step response of the second-order system

The amplitude of the exponentially damped sinusoid changes as a geometric series. At time

t=tp=pvd, the amplitude is equal to After one oscillation, or at

t=tp+2p (^) d=3pvd, the amplitude is equal to after another cycle of oscillation, the
amplitude is The logarithm of the ratio of successive amplitudes is called the logarithmic
decrement. Determine the logarithmic decrement for this second-order system. Describe a method
for experimental determination of the damping ratio from the rate of decay of the oscillation.
Solution.Let us define the amplitude of the output oscillation at t=tito be xi, where
ti=tp+(i-1)T(T=period of oscillation). The amplitude ratio per one period of damped
oscillation is
Thus, the logarithmic decrement dis
It is a function only of the damping ratio z. Thus, the damping ratio zcan be determined by use
of the logarithmic. decrement.
In the experimental determination of the damping ratio zfrom the rate of decay of the oscil-
lation, we measure the amplitude x 1 att=tpand amplitude xnatt=tp+(n-1)T. Note that
it is necessary to choose nlarge enough so that the ratio x 1 /xnis not near unity. Then
x 1
xn
=e(n-^1 )^2 zp^21 - z
2
d=ln
x 1
x 2

=

2 zp

21 - z^2

x 1

x 2

=

e-AsvdBp

e-AsvdB^3 p

=e^2 AsvdBp=e^2 zp^21 - z

2

e-AsvdB^5 p.

e-AsvdB^3 p;

e-AsvdBp.

C(s)

R(s)

=

v^2 n

s^2 + 2 zvn s+v^2 n

b= 2 zvn m= 2 *0.6*1.96*5.2=12.2 lbfftsec

2 zvn=

b

m

m=

20

v^2 n

=

20

1.96^2

=5.2 slugs=167 lb

vn=

3.14

2 *0.8

=1.96 radsec

tp=

p

vd

=

p

vn 21 - z^2

=

p

0.8vn

k= 20 lbfft