HOMEWORK 6 Solution Due Date: 27.04.2015

Prepared by

M.Uğur DİLBEROĞLU (B-183)

You should submit your homework on its due time (17:00). No extension is given afterward. Problem 1:

An electrical system is illustrated below in Figure 1.

Figure 1: Electrical system

The current source supplies the input current 𝐼𝐼(𝑡𝑡) to the electrical system in the direction of

the arrow indicated in the figure. The transformer shown in the figure can be considered as

an ideal transmission element having a turn ratio of 𝑁𝑁1:𝑁𝑁2. The inductances and the

resistances of the windings are separated out as pure elements with coefficients 𝑅𝑅1,𝑅𝑅2, 𝐿𝐿1, 𝐿𝐿2.

Voltage differences on either side of the transformer are represented as 𝑉𝑉1 and 𝑉𝑉2 in the

figure. Moreover, there are two ideal capacitors connected to the electrical system that are

MIDDLE EAST TECHNICAL UNIVERSITY

MECHANICAL ENGINEERING DEPARTMENT ME 304 CONTROL SYSTEMS

SPRING 2015

denoted as 𝐶𝐶1 and 𝐶𝐶2. The charge on the second capacitance 𝑄𝑄2 is to be taken as the output

of the system.

a) Write down all the elemental equations together with the connectivity equations.

b) Identify the unknown variables in the system. Then, check whether the number of the

unknown variables is equal to the number of the equations.

c) In order to express the input-output relationship for the selected output 𝑄𝑄2(𝑡𝑡) ,

determine the transfer function 𝐺𝐺(𝑠𝑠) = 𝐺𝐺𝑄𝑄𝑄𝑄(𝑠𝑠).

SOLUTION:

a) The electrical system is illustrated below:

Figure 1: Electrical system

The elemental equations for the passive elements can be written as follows by defining the

voltage drops across the elements as indicated below:

Elemental equations of the resistances are written as follows:

𝑉𝑉𝑅𝑅1(𝑡𝑡) = 𝑉𝑉𝐴𝐴(𝑡𝑡) − 𝑉𝑉𝐵𝐵(𝑡𝑡) = 𝑅𝑅1𝑖𝑖1(𝑡𝑡) (1)

𝑉𝑉𝑅𝑅2(𝑡𝑡) = 𝑉𝑉𝐸𝐸(𝑡𝑡) − 𝑉𝑉𝐹𝐹(𝑡𝑡) = 𝑅𝑅2𝑖𝑖3(𝑡𝑡) (2)

Elemental equations of the capacitors are expressed as in the following equations:

�̇�𝑉𝐶𝐶1(𝑡𝑡) = �̇�𝑉𝐴𝐴(𝑡𝑡) − �̇�𝑉𝐷𝐷(𝑡𝑡) =1𝐶𝐶1𝑖𝑖2(𝑡𝑡) (3)

�̇�𝑉𝐶𝐶2(𝑡𝑡) = �̇�𝑉𝐹𝐹(𝑡𝑡) − �̇�𝑉𝐺𝐺(𝑡𝑡) =1𝐶𝐶2𝑖𝑖3(𝑡𝑡) (4)

Elemental equations of the inductances are written as follows:

𝑉𝑉𝐿𝐿1(𝑡𝑡) = 𝑉𝑉𝐶𝐶(𝑡𝑡) − 𝑉𝑉𝐷𝐷(𝑡𝑡) = 𝐿𝐿1𝑑𝑑𝑖𝑖1(𝑡𝑡)𝑑𝑑𝑡𝑡

(5)

𝑉𝑉𝐿𝐿2(𝑡𝑡) = 𝑉𝑉𝐺𝐺(𝑡𝑡) − 𝑉𝑉𝐻𝐻(𝑡𝑡) = 𝐿𝐿2𝑑𝑑𝑖𝑖3(𝑡𝑡)𝑑𝑑𝑡𝑡

(6)

𝐴𝐴 𝐵𝐵

𝐶𝐶 𝐷𝐷

𝐸𝐸 𝐹𝐹

𝐺𝐺 𝐻𝐻

𝑖𝑖1

𝑖𝑖2 𝑖𝑖3

Two companion elemental equations for the ideal transformer (i.e. the 2-port transmission

element) is written as follows:

𝑉𝑉1(𝑡𝑡) =𝑁𝑁1𝑁𝑁2

𝑉𝑉2(𝑡𝑡) = 𝑛𝑛𝑉𝑉2(𝑡𝑡) (7)

𝑖𝑖3(𝑡𝑡) =𝑁𝑁1𝑁𝑁2

𝑖𝑖1(𝑡𝑡) = 𝑛𝑛𝑖𝑖1(𝑡𝑡) (8)

Note that in equation (7), voltages is expressed as follows:

𝑉𝑉1(𝑡𝑡) = 𝑉𝑉𝐵𝐵(𝑡𝑡) − 𝑉𝑉𝐶𝐶(𝑡𝑡) & 𝑉𝑉2(𝑡𝑡) = 𝑉𝑉𝐸𝐸(𝑡𝑡) − 𝑉𝑉𝐻𝐻(𝑡𝑡)

The Kirchhoff's current law equation for the node A is written as below:

𝐼𝐼(𝑡𝑡) = 𝑖𝑖1(𝑡𝑡) + 𝑖𝑖2(𝑡𝑡) (9)

The Kirchhoff's voltage law equations for the two loops (indicated on the circuit diagram) are

as written below:

𝑉𝑉𝑅𝑅1(𝑡𝑡) + 𝑉𝑉1(𝑡𝑡) + 𝑉𝑉𝐿𝐿1(𝑡𝑡) − 𝑉𝑉𝐶𝐶1(𝑡𝑡) = 0 (10)

𝑉𝑉2(𝑡𝑡) − 𝑉𝑉𝑅𝑅2(𝑡𝑡) − 𝑉𝑉𝐶𝐶2(𝑡𝑡) − 𝑉𝑉𝐿𝐿2(𝑡𝑡) = 0 (11)

Note also that the circuit is normally grounded. Therefore, the voltage at the terminal D

becomes as 𝑉𝑉𝐷𝐷(𝑡𝑡) = 0.

In the question, the output is taken as the charges accumulated in the capacitors. Therefore,

an additional equation for the current on the second capacitor is expressed as follows:

�̇�𝑄2(𝑡𝑡) = 𝑖𝑖3(𝑡𝑡) (12)

b) The system has 12 unknown variables, which are listed below:

1 2 3 4 5 6 7 8 9 10 11 12

𝑉𝑉𝑅𝑅1 𝑉𝑉𝑅𝑅2 𝑉𝑉𝐶𝐶1 𝑉𝑉𝐶𝐶2 𝑉𝑉𝐿𝐿1 𝑉𝑉𝐿𝐿2 𝑉𝑉1 𝑉𝑉2 𝑖𝑖1 𝑖𝑖2 𝑖𝑖3 𝑄𝑄2

TABLE 1: Unknown variables of the electrical system

Note that the unknown variables are matched with the 12 equations found in part (a)

previously.

c) At the beginning, take the Laplace transform of all the equations as follows:

𝑉𝑉𝑅𝑅1(𝑠𝑠) = 𝑅𝑅1𝐼𝐼1(𝑠𝑠) (1)′

𝑉𝑉𝑅𝑅2(𝑠𝑠) = 𝑅𝑅2𝐼𝐼3(𝑠𝑠) (2)′

𝑉𝑉𝐶𝐶1(𝑠𝑠) =1𝐶𝐶1𝑠𝑠

𝐼𝐼2(𝑠𝑠) (3)′

𝑉𝑉𝐶𝐶2(𝑠𝑠) =1𝐶𝐶2𝑠𝑠

𝐼𝐼3(𝑠𝑠) (4)′

𝑉𝑉𝐿𝐿1(𝑠𝑠) = 𝐿𝐿1𝑠𝑠𝐼𝐼1(𝑠𝑠) (5)′

𝑉𝑉𝐿𝐿2(𝑠𝑠) = 𝐿𝐿2𝑠𝑠𝐼𝐼3(𝑠𝑠) (6)′

𝑉𝑉1(𝑠𝑠) = 𝑛𝑛𝑉𝑉2(𝑠𝑠) (7)′

𝐼𝐼3(𝑠𝑠) = 𝑛𝑛𝐼𝐼1(𝑠𝑠) (8)′

𝐼𝐼(𝑠𝑠) = 𝐼𝐼1(𝑠𝑠) + 𝐼𝐼2(𝑠𝑠) (9)′

𝑉𝑉𝑅𝑅1(𝑠𝑠) + 𝑉𝑉1(𝑠𝑠) + 𝑉𝑉𝐿𝐿1(𝑠𝑠) − 𝑉𝑉𝐶𝐶1(𝑠𝑠) = 0 (10)′

𝑉𝑉2(𝑠𝑠) − 𝑉𝑉𝑅𝑅2(𝑠𝑠) − 𝑉𝑉𝐶𝐶2(𝑠𝑠) − 𝑉𝑉𝐿𝐿2(𝑠𝑠) = 0 (11)′

𝑄𝑄2(𝑡𝑡) =𝐼𝐼3(𝑠𝑠)𝑠𝑠

(12)′

Substitute eqn.’s (1)’, (3)’ and (5)’ into eqn. (10)’:

𝑉𝑉1(𝑠𝑠) + {𝐿𝐿1𝑠𝑠 + 𝑅𝑅1}𝐼𝐼1(𝑠𝑠)−1𝐶𝐶1𝑠𝑠

𝐼𝐼2(𝑠𝑠) = 0 (13)

Substitute eqn.(9)’ into eqn. (13) to eliminate 𝐼𝐼2(𝑠𝑠) :

𝑉𝑉1(𝑠𝑠) + �𝐿𝐿1𝑠𝑠 + 𝑅𝑅1 +1𝐶𝐶1𝑠𝑠

� 𝐼𝐼1(𝑠𝑠) = �1𝐶𝐶1𝑠𝑠

� 𝐼𝐼(𝑠𝑠) (14)

Substitute eqn.’s (2)’, (4)’ and (6)’ into eqn. (11)’ as follows:

𝑉𝑉2(𝑠𝑠) = �𝐿𝐿2𝑠𝑠 + 𝑅𝑅2 +1𝐶𝐶2𝑠𝑠

� 𝐼𝐼3(𝑠𝑠) (15)

To make the derivation easier, let us define some polynomials as below:

𝐴𝐴(𝑠𝑠) = 𝐿𝐿1𝑠𝑠 + 𝑅𝑅1 +1𝐶𝐶1𝑠𝑠

𝐵𝐵(𝑠𝑠) = 𝐿𝐿2𝑠𝑠 + 𝑅𝑅2 +1𝐶𝐶2𝑠𝑠

Substitute eqn. (7)’ and (8)’ into eqn. (14) to find the following expression:

𝑛𝑛𝑉𝑉2(𝑠𝑠) + �𝐴𝐴(𝑠𝑠)𝑛𝑛 � 𝐼𝐼3(𝑠𝑠) = �

1𝐶𝐶1𝑠𝑠

� 𝐼𝐼(𝑠𝑠) (16)

Substitute eqn. (15) into eqn. (16) as follows:

{𝐴𝐴(𝑠𝑠) + 𝑛𝑛2𝐵𝐵(𝑠𝑠)}𝐼𝐼3(𝑠𝑠) = �𝑛𝑛𝐶𝐶1𝑠𝑠

� 𝐼𝐼(𝑠𝑠) (17)

Substitute eqn. (12)’ into eqn. (17):

{𝐴𝐴(𝑠𝑠) + 𝑛𝑛2𝐵𝐵(𝑠𝑠)}𝑠𝑠𝑄𝑄2(𝑠𝑠) = �𝑛𝑛𝐶𝐶1𝑠𝑠

� 𝐼𝐼(𝑠𝑠) (18)

Rearrange eqn. (18) to find the transfer function between the input current 𝐼𝐼(𝑠𝑠) and the

output charge 𝑄𝑄2(𝑠𝑠):

𝑄𝑄2(𝑠𝑠) =𝑛𝑛

𝐶𝐶1𝑠𝑠2{𝐴𝐴(𝑠𝑠) + 𝑛𝑛2𝐵𝐵(𝑠𝑠)} 𝐼𝐼(𝑠𝑠)

Back substitute the defined polynomial as follows:

𝐺𝐺(𝑠𝑠) =𝑄𝑄2(𝑠𝑠)𝐼𝐼(𝑠𝑠)

=𝑛𝑛𝐶𝐶2

{𝐶𝐶1𝐶𝐶2𝐿𝐿1 + 𝑛𝑛2𝐶𝐶1𝐶𝐶2𝐿𝐿2}𝑠𝑠3 + {𝑅𝑅1𝐶𝐶1𝐶𝐶2 + 𝑛𝑛2𝑅𝑅2𝐶𝐶1𝐶𝐶2}𝑠𝑠2 + {𝐶𝐶2 + 𝑛𝑛2𝐶𝐶1}𝑠𝑠

Problem 2:

Operational amplifiers are commonly used in electrical control system. An operational

amplifier circuit is shown below in Figure 2:

Figure 2: An operational amplifier

This op-amp circuit has four electrical resistances represented as 𝑅𝑅1 , 𝑅𝑅2 , 𝑅𝑅3 , 𝑅𝑅4 and two

capacitances denoted as 𝐶𝐶1 and 𝐶𝐶2. The input and output voltages are 𝑉𝑉𝑖𝑖(𝑡𝑡) and 𝑉𝑉𝑜𝑜(𝑡𝑡).

Obtain the transfer function between 𝑉𝑉𝑖𝑖(𝑡𝑡) and 𝑉𝑉𝑜𝑜(𝑡𝑡).

SOLUTION:

For the ease of derivation, divide the given operational amplifier into two parts as in the

following figure:

Figure 1: An operational amplifier

First part of the operational amplifier is illustrated below:

Figure 2: First part of the operational amplifier

Using impedances 𝑍𝑍1(𝑠𝑠) and 𝑍𝑍2(𝑠𝑠) , relation of the voltages is expressed as in the eqn. (1):

𝑉𝑉𝑖𝑖(𝑡𝑡)𝑉𝑉1(𝑡𝑡)

= −𝑍𝑍2(𝑠𝑠)𝑍𝑍1(𝑠𝑠) (1)

𝑉𝑉1

𝑍𝑍1(𝑠𝑠)

𝑍𝑍2(𝑠𝑠)

In eqn. (1), corresponding impedances are represented as follows:

𝑍𝑍1(𝑠𝑠) = 𝑅𝑅1

𝑍𝑍2(𝑠𝑠) =1

𝐶𝐶1𝑠𝑠 + 1𝑅𝑅2

=𝑅𝑅2

𝑅𝑅2𝐶𝐶1𝑠𝑠 + 1

After substituting the related impedances indicated above, one can find the following

expression:

𝑉𝑉𝑖𝑖(𝑡𝑡)𝑉𝑉1(𝑡𝑡)

= −𝑍𝑍2𝑍𝑍1

= −𝑅𝑅2

𝑅𝑅1𝑅𝑅2𝐶𝐶1𝑠𝑠 + 𝑅𝑅1 (2)

For the second part of the operational amplifier, following figure illustrates the corresponding

impedances:

Figure 3: Second part of the operational amplifier

In that case, voltages relation is written as follows:

𝑉𝑉1(𝑡𝑡)𝑉𝑉𝑜𝑜(𝑡𝑡)

= −𝑍𝑍4(𝑠𝑠)𝑍𝑍3(𝑠𝑠) (3)

where the corresponding impedances are represented as

𝑍𝑍3 = 𝑅𝑅3

𝑍𝑍4 =𝑅𝑅4𝐶𝐶2𝑠𝑠 + 1

𝐶𝐶2𝑠𝑠

Therefore, one should substitute the related impedances into eqn. (3) to find the following

expression:

𝑉𝑉1(𝑡𝑡)𝑉𝑉𝑜𝑜(𝑡𝑡)

= −𝑅𝑅4𝐶𝐶2𝑠𝑠 + 1𝑅𝑅3𝐶𝐶2𝑠𝑠

(4)

𝑍𝑍3(𝑠𝑠) 𝑍𝑍4(𝑠𝑠)

Finally, combine the equations (2) and (4) to find the transfer function as follows:

𝑉𝑉𝑖𝑖(𝑡𝑡)𝑉𝑉0(𝑡𝑡)

=𝑉𝑉𝑖𝑖(𝑡𝑡)𝑉𝑉1(𝑡𝑡)

𝑉𝑉1(𝑡𝑡)𝑉𝑉0(𝑡𝑡)

= �−𝑍𝑍2𝑍𝑍1� �−

𝑍𝑍4𝑍𝑍3� (5)

In order to express the transfer function properly, rearrange the eqn. (5) as follows:

𝐺𝐺(𝑠𝑠) =𝑉𝑉𝑖𝑖(𝑡𝑡)𝑉𝑉0(𝑡𝑡)

=𝑍𝑍2𝑍𝑍1𝑍𝑍4𝑍𝑍3

={𝑅𝑅2𝑅𝑅4𝐶𝐶2}𝑠𝑠 + 𝑅𝑅2

{𝑅𝑅1𝑅𝑅2𝑅𝑅3𝐶𝐶1𝐶𝐶2}𝑠𝑠2 + {𝑅𝑅1𝑅𝑅3𝐶𝐶2}𝑠𝑠

Problem 3:

A hydraulic servo actuator is illustrated below in Figure 3:

Figure 3: Hydraulic linear actuator

A hydraulic actuator works with a hydraulic fluid pressurized by a supplier pump. The pilot

valve controls the flow rate of the hydraulic fluid from the supply to the actuator cylinder with

the help of the spool. When the spool is moved to the right of its neutral position, the fluid

enters the right-hand piston chamber and pushes the piston to the left. The fluid displaced by

this motion exits through the left-hand drain port. The action is reversed for a spool

displacement of the pilot valve to the left. Both return lines are connected to a tank from

which the pump pressurizes fluid to deliver to the supply line.

The flow rate (𝑄𝑄𝐿𝐿) into the actuator cylinder is related to the spool displacement (𝑥𝑥) and the

pressure difference (𝑃𝑃𝐿𝐿) on the actuator piston by the following linearized equation.

𝑄𝑄𝐿𝐿 = 𝐾𝐾1𝑥𝑥 − 𝐾𝐾2𝑃𝑃𝐿𝐿

In this equation,

𝑄𝑄𝐿𝐿 = 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑟𝑟𝑟𝑟𝑡𝑡𝑟𝑟 𝑡𝑡ℎ𝑟𝑟𝑓𝑓𝑟𝑟𝑟𝑟ℎ 𝑡𝑡ℎ𝑟𝑟 ℎ𝑦𝑦𝑑𝑑𝑟𝑟𝑟𝑟𝑟𝑟𝑓𝑓𝑖𝑖𝑦𝑦 𝑓𝑓𝑖𝑖𝑛𝑛𝑟𝑟𝑠𝑠

𝑥𝑥 = 𝑑𝑑𝑖𝑖𝑠𝑠𝑑𝑑𝑓𝑓𝑟𝑟𝑦𝑦𝑟𝑟𝑑𝑑𝑟𝑟𝑛𝑛𝑡𝑡 𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑟𝑟 𝑠𝑠𝑑𝑑𝑓𝑓𝑓𝑓𝑓𝑓

𝑃𝑃𝐿𝐿 = 𝑑𝑑𝑟𝑟𝑟𝑟𝑠𝑠𝑠𝑠𝑟𝑟𝑟𝑟𝑟𝑟 𝑑𝑑𝑖𝑖𝑓𝑓𝑓𝑓𝑟𝑟𝑟𝑟𝑟𝑟𝑛𝑛𝑦𝑦𝑟𝑟 𝑏𝑏𝑟𝑟𝑡𝑡𝑓𝑓𝑟𝑟𝑟𝑟𝑛𝑛 𝑡𝑡ℎ𝑟𝑟 𝑡𝑡𝑓𝑓𝑓𝑓 𝑠𝑠𝑖𝑖𝑑𝑑𝑟𝑟𝑠𝑠 𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑟𝑟 𝑑𝑑𝑖𝑖𝑠𝑠𝑡𝑡𝑓𝑓𝑛𝑛

𝐾𝐾1 ,𝐾𝐾2 = 𝑦𝑦𝑓𝑓𝑟𝑟𝑟𝑟𝑟𝑟𝑠𝑠𝑑𝑑𝑓𝑓𝑛𝑛𝑑𝑑𝑖𝑖𝑛𝑛𝑟𝑟 𝑦𝑦𝑓𝑓𝑟𝑟𝑓𝑓𝑓𝑓𝑖𝑖𝑦𝑦𝑖𝑖𝑟𝑟𝑛𝑛𝑡𝑡𝑠𝑠

For further information about the linearization of the preceding equation, you may read the

related part of the textbook.

A Typical Usage of a Hydraulic Servo Actuator:

A simplified sketch of a force-feedback linear hydraulic servo actuator driving a load of mass

M is illustrated below in figure 4:

Figure 4: Hydraulic servo actuator with force feedback

The position of the spool of the pilot valve having a mass 𝑀𝑀𝑣𝑣 is commanded by a force 𝐹𝐹 as

seen above in Figure 4. The supply pressure and the return-to-tank pressure are denoted as

𝑃𝑃𝑠𝑠 and 𝑃𝑃𝑇𝑇 , respectively. The fluid flow into the actuator cylinder is 𝑄𝑄𝐿𝐿. It obeys the equation

given in the previous part of the question. The pressure difference between the two sides of

the piston is denoted as 𝑃𝑃𝐿𝐿. The force feedback is obtained through a spring with stiffness 𝑘𝑘𝑓𝑓.

The hydraulic servo actuator drives a load of mass 𝑀𝑀 connected to the ground with a spring

and a damper, whose parameters are 𝑘𝑘 and 𝑏𝑏.Moreover, the net area of the piston is 𝐴𝐴, which

is equal on both sides.

In the question, the input is the command force 𝐹𝐹(𝑡𝑡) and the output is selected as the

displacement 𝑦𝑦(𝑡𝑡) of the load.

Assume that the fluid is incompressible and neglect any possible leakage in the piston-cylinder

system.

a) Draw the necessary free body diagrams corresponding to the elements of the system.

b) Write down all the elemental equations together with the connectivity equations.

c) Identify the unknown variables in the system and check if the number of the unknown

variables is equal to the number of the equations.

d) In order to express the input-output relationship for the selected output 𝑦𝑦(𝑡𝑡),

determine the transfer function 𝐺𝐺(𝑠𝑠) = 𝐺𝐺𝑌𝑌𝐹𝐹(𝑠𝑠).

SOLUTION:

a) Free body diagrams of the system components are drawn as follows:

Figure 1: FBD of the spool

Figure 2: FBD of the feedback spring

Figure 3: FBD of the load mass

𝑀𝑀𝑣𝑣 𝐹𝐹 𝐹𝐹𝑓𝑓

𝑥𝑥

𝐹𝐹𝑓𝑓 𝐹𝐹𝑓𝑓

𝑥𝑥

𝑘𝑘𝑓𝑓

𝑦𝑦

M

𝑦𝑦

𝐹𝐹𝑠𝑠

𝐹𝐹𝑏𝑏

𝐹𝐹𝑝𝑝

Figure 4: FBD of the piston of the linear actuator

b) The necessary equations are as written below.

The elemental equation for the spool:

𝑀𝑀𝑣𝑣�̈�𝑥(𝑡𝑡) = 𝐹𝐹(𝑡𝑡) − 𝐹𝐹𝑓𝑓(𝑡𝑡) (1)

The elemental equation of the feedback spring:

𝐹𝐹𝑓𝑓(𝑡𝑡) = 𝑘𝑘𝑓𝑓[𝑥𝑥(𝑡𝑡) + 𝑦𝑦(𝑡𝑡)] (2)

The characteristic equation of the pilot valve:

𝑄𝑄𝐿𝐿(𝑡𝑡) = 𝐾𝐾1𝑥𝑥(𝑡𝑡) − 𝐾𝐾2𝑃𝑃𝐿𝐿(𝑡𝑡) (3)

The elemental equation for the mass M:

𝑀𝑀�̈�𝑦(𝑡𝑡) = 𝐹𝐹𝑝𝑝(𝑡𝑡) − 𝐹𝐹𝑠𝑠(𝑡𝑡) − 𝐹𝐹𝑏𝑏(𝑡𝑡) (4)

The elemental equations of the spring and damper connected to M:

𝐹𝐹𝑠𝑠(𝑡𝑡) = 𝑘𝑘𝑦𝑦(𝑡𝑡) (5)

𝐹𝐹𝑏𝑏(𝑡𝑡) = 𝑏𝑏�̇�𝑦(𝑡𝑡) (6)

The elemental equations of the piston-cylinder pair as a 2-port element:

𝑄𝑄𝐿𝐿(𝑡𝑡) = 𝐴𝐴�̇�𝑦(𝑡𝑡) (7)

𝑃𝑃𝐿𝐿(𝑡𝑡) =1𝐴𝐴

[𝐹𝐹𝑝𝑝(𝑡𝑡) + 𝐹𝐹𝑓𝑓(𝑡𝑡)] (8)

A

𝑦𝑦

𝐹𝐹𝑝𝑝 𝐹𝐹𝑓𝑓

𝑃𝑃𝐿𝐿𝐴𝐴

c) The distinct unknown variables are listed as in the following table:

1 2 3 4 5 6 7 8

𝑥𝑥(𝑡𝑡) 𝑦𝑦(𝑡𝑡) 𝐹𝐹𝑓𝑓(𝑡𝑡) 𝐹𝐹𝑠𝑠(𝑡𝑡) 𝐹𝐹𝑏𝑏(𝑡𝑡) 𝐹𝐹𝑝𝑝(𝑡𝑡) 𝑄𝑄𝐿𝐿(𝑡𝑡) 𝑃𝑃𝐿𝐿(𝑡𝑡)

TABLE 1: Unknown Variables

There are 8 equations corresponding to the 8 unknown variables. Therefore, one can obtain

the input-output relation properly.

d) First of all, take the Laplace transform of all the equations as follows:

𝑀𝑀𝑣𝑣𝑠𝑠2𝑋𝑋(𝑠𝑠) = 𝐹𝐹(𝑠𝑠) − 𝐹𝐹𝑓𝑓(𝑠𝑠) (1)′

𝐹𝐹𝑓𝑓(𝑠𝑠) = 𝑘𝑘𝑓𝑓[𝑋𝑋(𝑠𝑠) + 𝑌𝑌(𝑠𝑠)] (2)′

𝑄𝑄𝐿𝐿(𝑠𝑠) = 𝐾𝐾1𝑋𝑋(𝑠𝑠) − 𝐾𝐾2𝑃𝑃𝐿𝐿(𝑠𝑠) (3)′

𝑀𝑀𝑠𝑠2𝑌𝑌(𝑠𝑠) = 𝐹𝐹𝑝𝑝(𝑠𝑠) − 𝐹𝐹𝑠𝑠(𝑠𝑠) − 𝐹𝐹𝑏𝑏(𝑠𝑠) (4)′

𝐹𝐹𝑠𝑠(𝑠𝑠) = 𝑘𝑘𝑌𝑌(𝑠𝑠) (5)′

𝐹𝐹𝑏𝑏(𝑠𝑠) = 𝑏𝑏𝑠𝑠𝑌𝑌(𝑠𝑠) (6)′

𝑄𝑄𝐿𝐿(𝑠𝑠) = 𝐴𝐴𝑠𝑠𝑌𝑌(𝑠𝑠) (7)′

𝐴𝐴𝑃𝑃𝐿𝐿(𝑠𝑠) = 𝐹𝐹𝑝𝑝(𝑠𝑠) + 𝐹𝐹𝑓𝑓(𝑠𝑠) (8)′

Substitute eqn. (2)’ into the eqn. (1)’:

�𝑀𝑀𝑣𝑣𝑠𝑠2 + 𝑘𝑘𝑓𝑓�𝑋𝑋(𝑠𝑠) + 𝑘𝑘𝑓𝑓𝑌𝑌(𝑠𝑠) = 𝐹𝐹(𝑠𝑠) (9)

Substitute eqn. (5)’ and (6)’ into eqn. (4)’:

{𝑀𝑀𝑠𝑠2 + 𝑏𝑏𝑠𝑠 + 𝑘𝑘}𝑌𝑌(𝑠𝑠) = 𝐹𝐹𝑝𝑝(𝑠𝑠) (10)

Substitute eqn. (2)’ and (10) into the eqn. (8)’:

𝐴𝐴𝑃𝑃𝐿𝐿(𝑠𝑠) = �𝑀𝑀𝑠𝑠2 + 𝑏𝑏𝑠𝑠 + 𝑘𝑘 + 𝑘𝑘𝑓𝑓�𝑌𝑌(𝑠𝑠) + 𝑘𝑘𝑓𝑓𝑋𝑋(𝑠𝑠) (11)

Let us define the polynomials 𝐴𝐴(𝑠𝑠) and 𝐵𝐵(𝑠𝑠) as follows:

𝐴𝐴(𝑠𝑠) = 𝑀𝑀𝑣𝑣𝑠𝑠2 + 𝑘𝑘𝑓𝑓

𝐵𝐵(𝑠𝑠) = 𝑀𝑀𝑠𝑠2 + 𝑏𝑏𝑠𝑠 + 𝑘𝑘 + 𝑘𝑘𝑓𝑓

Substitute eqn. (7)’ and (11) into eqn. (3)’:

{𝐴𝐴2𝑠𝑠 + 𝐾𝐾2𝐵𝐵(𝑠𝑠)}𝑌𝑌(𝑠𝑠) = �𝐴𝐴𝐾𝐾1 − 𝐾𝐾2𝑘𝑘𝑓𝑓�𝑋𝑋(𝑠𝑠) (12)

Substitute eqn. (12) into eqn. (9):

𝑌𝑌(𝑠𝑠) =�𝐴𝐴𝐾𝐾1 − 𝐾𝐾2𝑘𝑘𝑓𝑓�

𝐴𝐴(𝑠𝑠){𝐴𝐴2𝑠𝑠 + 𝐾𝐾2𝐵𝐵(𝑠𝑠)} + 𝑘𝑘𝑓𝑓�𝐴𝐴𝐾𝐾1 − 𝐾𝐾2𝑘𝑘𝑓𝑓�𝐹𝐹(𝑠𝑠) (13)

Finally, back substitution of the polynomials 𝐴𝐴(𝑠𝑠) and 𝐵𝐵(𝑠𝑠) and rearranging will result in the

following transfer function:

𝐺𝐺(𝑠𝑠) =𝑌𝑌(𝑠𝑠)𝐹𝐹(𝑠𝑠)

=𝐴𝐴𝐾𝐾1 − 𝐾𝐾2𝑘𝑘𝑓𝑓

𝐴𝐴4𝑠𝑠4 + 𝐴𝐴3𝑠𝑠3 + 𝐴𝐴2𝑠𝑠2 + 𝐴𝐴1𝑠𝑠 + 𝐴𝐴0

where

𝐴𝐴4 = 𝐾𝐾2𝑀𝑀𝑣𝑣𝑀𝑀

𝐴𝐴3 = 𝑀𝑀𝑣𝑣𝐴𝐴2 + 𝐾𝐾2𝑀𝑀𝑣𝑣𝑏𝑏

𝐴𝐴2 = 𝑀𝑀𝑣𝑣𝐾𝐾2𝑘𝑘 + 𝑀𝑀𝑣𝑣𝐾𝐾2𝑘𝑘𝑓𝑓 + 𝑀𝑀𝐾𝐾2𝑘𝑘𝑓𝑓

𝐴𝐴1 = 𝐴𝐴2𝑘𝑘𝑓𝑓 + 𝐾𝐾2𝑏𝑏𝑘𝑘𝑓𝑓

𝐴𝐴0 = 𝐾𝐾2𝑘𝑘𝑘𝑘𝑓𝑓 + 𝑘𝑘𝑓𝑓𝐴𝐴𝐾𝐾1