1. Obtain mathematical models ( ) of the mechanical systems shown

below.  

Solution  

a) 

Here, the input is u(t) and the output is the displacement x as shown in the figure.  

 

the rollers under the mass means there is no 

friction.  

Converting the above equation to Laplace domain.  

 

Taking X(s) out of the brackets: 

 

Then the transfer function (T.F) equal to: 

 

b) 

The solution will be the same procedure, but the spring factor will be: 

 

Then the T.F would be: 

 

 

Note: 

The second derivative of x could be presented as:  

” 

 

86 Chapter 3 / Mathematical Modeling of Mechanical Systems and Electrical Systems

EXAMPLE PROBLEMS AND SOLUTIONS

A–3–1. Figure 3–20(a) shows a schematic diagram of an automobile suspension system.As the car movesalong the road, the vertical displacements at the tires act as the motion excitation to the auto-mobile suspension system.The motion of this system consists of a translational motion of the cen-ter of mass and a rotational motion about the center of mass. Mathematical modeling of thecomplete system is quite complicated.

A very simplified version of the suspension system is shown in Figure 3–20(b).Assuming thatthe motion xi at point P is the input to the system and the vertical motion xo of the body is theoutput, obtain the transfer function (Consider the motion of the body only in the ver-tical direction.) Displacement xo is measured from the equilibrium position in the absence ofinput xi.

Solution. The equation of motion for the system shown in Figure 3–20(b) is

or

Taking the Laplace transform of this last equation, assuming zero initial conditions, we obtain

Hence the transfer function Xo(s)/Xi(s) is given by

Xo(s)

Xi(s)=

bs + k

ms2 + bs + k

Ams2 + bs + kBXo(s) = (bs + k)Xi(s)

mx$

o + bx#o + kxo = bxi

# + kxi

mx$

o + bAx# o - x#iB + kAxo - xiB = 0

Xo(s)�Xi(s).

(a)

k

(b)

xi

Center of mass

Auto body

b

P

xo

m

Figure 3–20(a) Automobilesuspension system;(b) simplifiedsuspension system.

Example Problems and Solutions 87

A–3–2. Obtain the transfer function Y(s)/U(s) of the system shown in Figure 3–21. The input u is adisplacement input. (Like the system of Problem A–3–1, this is also a simplified version of anautomobile or motorcycle suspension system.)

Solution. Assume that displacements x and y are measured from respective steady-statepositions in the absence of the input u. Applying the Newton’s second law to this system, weobtain

Hence, we have

Taking Laplace transforms of these two equations, assuming zero initial conditions, we obtain

Eliminating X(s) from the last two equations, we have

which yields

Y(s)

U(s)=

k1Abs + k2B

m1 m2 s4 + Am1 + m2Bbs3 + Ck1 m2 + Am1 + m2Bk2 Ds2 + k1 bs + k1 k2

Am1 s2 + bs + k1 + k2Bm2 s2 + bs + k2

bs + k2Y(s) = Abs + k2BY(s) + k1 U(s)

Cm2 s2 + bs + k2 DY(s) = Abs + k2BX(s)

Cm1 s2 + bs + Ak1 + k2B DX(s) = Abs + k2BY(s) + k1 U(s)

m2 y$ + by# + k2 y = bx

# + k2 x

m1 x$ + bx# + Ak1 + k2Bx = by

# + k2 y + k1 u

m2 y$ = -k2(y - x) - b(y# - x

#)

m1 x$ = k2(y - x) + b(y# - x

#) + k1(u - x)

y

b

x

u

m2

m1

k2

k1

Figure 3–21Suspension system.

 

 

2. Obtain the transfer function / of the electrical circuit shown in.

The equations for the given circuit are as follow:

The Laplace transforms of these three equations, with zero initial conditions, are

(1)

(2)

(3)

From Equation (2) we obtain.

or

(4)

Or

Substituting equation (4) into equation (1), we get

Or

(5)

From Equation (3) and (4), we get

(6)

From equation (5) and (6), we obtain

3. Consider the system shown in Figure below. An armature-controlled dc servomotor drives a load consisting of the moment of inertia .The torque developed by the motor is T. The moment of inertia of the motor rotor is Jm. The angular displacements of the motor rotor and the load element are , respectively. The gear ratio is / Obtain the transfer function / .

Solution

Define the current in the armature circuit to be then, we have

الزخم الزاوي

العزم الدوران الصافي  الذي یسلط على الحمل

العزم الكھربائي الناتج من مرور التیار في

المحرك

عزم الحمل

Where 

Or Taking Laplace transforms

(1)

Where Kb is the back emf constant of the motor. We also have

(2)

Where K is the motor torque constant and ia is the armature current. Equation (2) can be rewritten as.

(

Or

Taking Laplace transforms

(3)

By substituting equation (3) into equation (1), we obtain.

Or

یة القوة الدافعة الكھربائتیجة تیار المتولدة ن

الحمل وتكون دائما ةمعاكس  

Hence