Modern Control Engineering

366 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method

A–6–12. Obtain the transfer function of the mechanical system shown in Figure 6–75. Assume that the dis-

placementxiis the input and displacement xois the output.

Solution.The equations of motion for this system are

By taking the Laplace transforms of these two equations, assuming zero initial conditions, we

obtain

If we eliminate Y(s)from the last two equations, the transfer function can be

obtained as

Define

Ifk 1 ,k 2 , b 1 ,andb 2 are chosen such that there exists a ıthat satisfies the following equation:

(6–30)

then can be obtained as

[Note that depending on the choice of k 1 , k 2 , b 1 , and b 2 ,there does not exist a ı that satisfies

Equation (6–30).]

If such a ı exists and if for a given s 1 (wheres=s 1 is one of the dominant closed-loop poles

of the control system to which we wish to use this mechanical device) the following conditions are

satisfied:

then the mechanical system shown in Figure 6–75 acts as a lag–lead compensator.

- 5 ° 6

n

s 1 +

1

T 2

s 1 +

1

bT 2

4 60 °

s 1 +

1

T 2

s 1 +

1

bT 2

4 1,

Xo(s)

Xi(s)

=

AT 1 s+ 1 BAT 2 s+ 1 B

a

T 1

b

s+ 1 bAbT 2 s+ 1 B

=

as+

1

T 1

bas+

1

T 2

b

as+

b

T 1

bas+

1

bT 2

b

Xo(s)Xi(s)

b 1

k 1

+

b 2

k 2

+

b 1

k 2

=

T 1

b

+bT 2 (b 71 )

T 1 =

b 1

k 1

, T 2 =

b 2

k 2

,

Xo(s)

Xi(s)

=

a

b 1

k 1

s+ 1 ba

b 2

k 2

s+ 1 b

a

b 1

k 1

s+ 1 ba

b 2

k 2

s+ 1 b+

b 1

k 2

s

Xo(s)Xi(s)

b 1 CsXo(s)-sY(s)D=k 1 Y(s)

b 2 CsXi(s)-sXo(s)D+k 2 CXi(s)-Xo(s)D=b 1 CsXo(s)-sY(s)D

b 1 Ax#o-y#B=k 1 y

b 2 Ax#i-x#oB+k 2 Axi-xoB=b 1 Ax#o-y#B

b 1

b 2

y

xi

xo

k 2

k 1

Figure 6–75

Mechanical system.

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