Modern Control Engineering

aa

238 Chapter 5 / Transient and Steady-State Response Analyses

or

Hence

A–5–7. In the system shown in Figure 5–55, the numerical values of m, b, and kare given as m=1kg,

b=2N-secm, and k=100Nm. The mass is displaced 0.05 m and released without initial ve-

locity. Find the frequency observed in the vibration. In addition, find the amplitude four cycles later.

The displacement xis measured from the equilibrium position.

Solution.The equation of motion for the system is

Substituting the numerical values for m, b, and kinto this equation gives

where the initial conditions are x(0)=0.05 and From this last equation the undamped

natural frequency vnand the damping ratio zare found to be

The frequency actually observed in the vibration is the damped natural frequency vd.

In the present analysis, is given as zero. Thus, solution x(t)can be written as

It follows that at t=nT, where T=2pvd,

Consequently, the amplitude four cycles later becomes

A–5–8. Obtain both analytically and computationally the unit-step response of tbe following higher-order

system:

[Obtain the partial-fraction expansion of C(s)with MATLAB when R(s)is a unit-step function.]

C(s)

R(s)

=

3s^3 +25s^2 +72s+ 80

s^4 +8s^3 +40s^2 +96s+ 80

=0.05e-2.526=0.05*0.07998=0.004 m

x(4T)=x(0)e-zvn^ 4T=x(0)e-(0.1)(10)(4)(0.6315)

x(nT)=x(0)e-zvn^ nT

x(t)=x(0)e-zvn^ tacosvd t+

z

21 - z^2

sinvd tb

x#(0)

vd=vn 21 - z^2 = 1011 - 0.01=9.95 radsec

vn=10, z=0.1

x#(0)=0.

x$+2x#+100x= 0

mx$+bx#+kx= 0

z=

1

n- 1

aln

x 1

xn

b

B

4 p^2 + c

1

n- 1

aln

x 1

xn

bd

2

ln

x 1

xn

=(n-1)

2 zp

21 - z^2

k

m

b

x

Figure 5–55

Spring-mass-damper

system.

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