FeedConunit-2

Feedback and Control Systems

System Modeling Page 1

SYSTEM MODELING

At the end of this chapter, the students shall be able to:

2.1 Discuss the two methods of modeling dynamic systems.

2.2 (a) Find the Laplace transform of a given function in time domain, and obtain the time domain

function given its Laplace transform; (b) Solve differential equations using Laplace transform

methods.

2.3 (a) Write a transfer function associated with a given differential equation and vice versa; (b) Solve

the output of a system with a given differential equation or transfer function for a given input.

2.4 Obtain transfer functions for single and multiple loops, passive and active (op-amp) electrical

networks.

2.5 Determine the transfer function of translational, rotational and translational-rotational mechanical

systems, including systems with gears.

2.6 Obtain transfer function relating the output displacement to the input voltage of electromechanical

systems.

2.7 Define terms associated with state-space modeling of systems and enumerate steps in obtaining

the state-space representation of a system.

2.8 Obtain the state-space model of electrical and mechanical systems.

2.9 Convert a transfer function into state-space representation and vice versa.

2.1 Introduction

Intended Learning Outcome: Discuss the two methods of modeling dynamic systems.

In the previous discussion, the analysis and design sequence that included obtaining the system’s

schematic and demonstrated this step for a position control system. The next step is to develop

mathematical models from schematics of physical systems. Two methods will be discussed: (1) transfer

functions in the frequency domain and (2) state equations in the time domain.

It should be remembered that in any case, the first step in developing a mathematical model is to apply the

fundamental physical laws of science and engineering. When electric networks are being modeled, Ohm’s

law and Kirchhoff’s laws are applied initially. For mechanical systems, Newton’s laws will be used.



From previous courses, it is seen that a differential equation can describe the relationship between the

input and the output of the system. The form of the differential equation and its coefficients are a

formulation or description of the system. Although the differential equation relates the system to its input

and output, it is not a satisfying representation from a system perspective. It is much preferred that a

mathematical representation such as shown in figure 2.1(a) where the input, output and the system are

distinct and separate parts. Also, a representation where several interconnected subsystems like the

cascade connection of figure 2.1(b) can also be conveniently written as a single system like that of figure

2.1(a). It is then preferred that a single mathematical function, called the transfer function, will represent the

system.

Figure 2.1. Block diagram representation of a system, showing the relationship between the input r(t) and the output c(t)

Note in figure 2.1, r�t� represents the reference input while c�t� represents the controlled variable. A major advantage of frequency-domain modeling is that they rapidly provide stability and transient

response information. Thus, we can immediately see the effects of varying system parameters until an

acceptable design is met.

The primary disadvantage of this approach, also called the classical approach is its limited applicability. It

can be applied only to linear, time-invariant systems or systems that can be approximated as such.

With an increasing design scope, the modern or time-domain approach of modeling systems is devised.

The state-space approach is a unified method for modeling, analyzing, and designing a wide range of

system. It can model nonlinear systems and systems with nonzero initial conditions. Time-varying systems

and systems with multiple inputs and multiple outputs can be compactly represented in state-space model

which is similar to the form of single input, single output systems.



The time-domain approach can also be used for the same class of systems modeled by classical approach.

This alternate model gives the control systems designer another perspective from which to create the

design. The main drawback of the time-domain approach is its lack of intuition compared to the classical

approach. A few calculations must be made before the physical interpretation of the model becomes

apparent.

2.2 The Laplace Transform: A Review

Intended Learning Outcomes: (a) Find the Laplace transform of a given function in time domain, and obtain

the time domain function given its Laplace transform; (b) Solve differential equations using Laplace

transform methods.

The Laplace transform satisfies the requirements for the convenient mathematical representation of

systems using the transfer function as discussed previously. Furthermore, the Laplace transform makes the

relationships between systems algebraic.

The Laplace transform is defined as

ℒ�f�t� = F�s� = f�t�e��dt�� (2.1)

where s = σ + jω , a complex variable. The function f�t� is in t -domain (or time domain) and has a Laplace transform F�s� in s-domain (or the complex frequency domain) if the integral of Equation 2.1 exists.

The inverse Laplace transform, is defined as

ℒ��F�s� = f�t�u�t� = 1j2π F�s�e��ds� !��!� (2.2)

where

u�t� = "0, t < 01, t > 0' (2.3)

called the unit step function. Multiplication of a function with a unit step function yields a function that is zero

for negative values of t.



Tables 2.1 and 2.2 summarize the Laplace transform of commonly used functions and the properties of the

Laplace transform, respectively.



Example 2.1

Find the Laplace transform of the function f�t� = e�(� cos�ωt + θ�

Answer: �s + a� cos θ − ω sin θ�s + a�/ + ω/

Example 2.2

Find the inverse Laplace transform of F�s� = �� /�� 0�1.

Answer:

f�t� = 59 − 5e�/� + 409 e�0� + 103 te�0�

Example 2.3

Solve the differential equation d/ydt/ + 12 dydt + 32y = 32u�t�

using Laplace transform when all the initial conditions are zero.

Answer: y�t� = �1 + e�7� − 2e�8�u�t�

Drill Problem 2.1

1. Find the Laplace transform of the function f�t� = 3te�/�sin�4t + 60°�.

2. Find the Laplace transform of the following functions

a. f�t� = sinh/at



b. f�t� = t cos 5t c. f�t� = sin8t

3. Determine the inverse Laplace transform of the following functions in s-domain. a. F�s� = 2s−56s2−4s−12. b. F�s� = <�1 0� ��=�� >�� 0�� 8��1 /� �� c. F�s� = �? 8�1 /� @�� 7��1 7� 0��1 >� A�

4. What is the time domain function that has F�s� = 1s2<s2+ ω2= as its Laplace transform?

5. Determine the solution of the following differential equations using Laplace transform.

a. yCC − yC − 6y = 0, with initial conditions y�0� = 6 and yC�0� = 13. b. yCC − 4yC + 4y = 0, with initial conditions y�0� = 2.1 and yC�0� = 3.9. c. yCC + kyC − 2k/y = 0, with initial conditions y�0� = 2 and yC�0� = 2k.

2.3. The Transfer Function

Intended Learning Outcomes: (a) Write a transfer function associated with a given differential equation and

vice versa; (b) Solve the output of a system with a given differential equation or transfer function for a given

input.

The Laplace transform, as stated before, can be used to establish algebraic input-system-output

relationship as depicted in figure 2.1(a). Furthermore, it will allow algebraic combination of mathematical

representations of subsystems to yield a total system representation.

For an nth-order, linear, time-invariant differential equation, aF dFc�t�dtF + aF�� dF��c�t�dtF�� + ⋯ + a�c�t� = bI dIr�t�dtI + bI�� dI��r�t�dtI�� + ⋯ + b�r�t� (2.4)



where c�t� is the output, r�t� is the input, and aJ ’s, bJ ’s and the form of the differential equation represents the system. Taking the Laplace transform of equation 2.4, and solving for the ratio of

K��L�� with zero initial conditions yield

C�s�R�s� = G�s� = bIsI + bI��sI�� + ⋯ + b�aFsF + aF��sF�� + ⋯ + a� (2.5)

with G�s�, called the system’s transfer function, evaluated at zero initial conditions. Thus, a system can now be represented by a block diagram as seen in figure 2.2.

Figure 2.2. System representation using transfer function

From here, it can be seen that equation 2.5 can be used to determine the output when the system’s

transfer function and the input are known, as

C�s� = R�s�G�s� (2.6)

Example 2.4

Find the transfer function of the system represented by the differential equation dc�t�dt + 2c�t� = r�t� assuming zero initial conditions. Then find the system response to a step input r�t� = u�t�.

Answer:

The transfer function is

G�s� = 1s + 2 and the system response for the specified input is

c�t� = 12 − 12 e�/�



Drill Problems 2.2

1. Find the transfer function G�s� = C�s�/R�s� corresponding to the differential equation d0cdt0 + 3 d/cdt/ + 7 dcdt + 5c = d/rdt/ + 4 drdt + 3r

2. A system is described by the following differential equation d0ydt0 + 3 d/ydt/ + 5 dydt + y = d0xdt0 + 4 d/xdt/ + 6 dxdt + 8x Find the expression for the transfer function

T��U��.

3. Find the differential equation corresponding to the transfer functions

a. G�s� = K��L�� = /� ��1 @� / b. G�s� = K��L�� = A�1 >� �� c. G�s� = K��L�� = �>�� d. G�s� = K��L�� = � 0�? ��1 �/� �7

4. For each of the systems described by the transfer functions in number 3, find the response of the

system to a step input.

5. Find the ramp response for a system whose transfer function is G�s� = s�s + 4��s + 8�

In general, a physical system than can be represented by a linear, time-invariant differential equation can

be modeled as a transfer function. The subsequent sections will demonstrate the use of the transfer

functions to model electrical networks, mechanical systems, and electro-mechanical systems.



2.4 Electrical Network Transfer Functions

Intended Learning Outcome: Obtain transfer functions for single and multiple loops, passive and active (op-

amp) electrical networks.

Passive Networks. In this section, mathematical models of electric circuits including passive networks and

operational amplifiers will be obtained. Equivalent circuits for the electric networks will first consist of

resistors, inductors and capacitors. Table 2.3 summarizes the components and the relationships between

voltage and current and between voltage and charge under zero initial conditions.

As stated before, the first step in formulating the mathematical model of a system is to use fundamental

laws governing the system. Since the systems being modeled in this section are electrical networks, Ohm’s

Law and Kirchhoff’s Laws will be used.

Example 2.5

Find the transfer function relating the capacitor voltage VK�s� to the input voltage V�s� for the circuit shown below.

Answer:



G�s� = VK�s�V�s� = �WKs/ + LW s + �WK

Example 2.6

Determine the transfer function H��s� = Y1��Z�� using mesh analysis. Obtain the transfer function H/�s� =Z[��Z�� for the network shown below using nodal analysis.

Answers:

H��s� = I/�s�V�s� = LCs/�R� + R/�LCs/ + �R�R/C + L�s + R�

H/�s� = VK�s�V�s� = ^_^1K s�G� + G/�s/ + ^_^1W KWK s + ^1WK



From the foregoing examples, a technique in which mesh equations can be written can be developed. For

two loop electrical network as shown in example 2.6, the mesh equations can be written as

`Sum of impedancesaround mesh 1 d I��s� − eSum of impedancescommon tothe two meshes g I/�s� = e Sum of applied voltagesaround mesh 1 g (2.7a)

− eSum of impedancescommon tothe two meshes g I��s� + `Sum of impedancesaround mesh 2 d I/�s� = e Sum of applied voltagesaround mesh 2 g (2.7b)

This technique can be expanded to three-loop electrical network. The use of the technique is illustrated in

the following example.

Example 2.7

Write, but do not solve, the mesh equations for the network shown below

Answer:



Example 2.8

Find the transfer function G�s� = VW�s�/V�s� for the circuit shown below. Solve the problem in two ways – mesh analysis and nodal analysis.

Answer:

G�s� = VW�s�V�s� = s/ + 2s + 1s/ + 5s + 2

Operational Amplifiers. An operational amplifier is an electronic amplifier used as a basic building block to

implement transfer functions. It has the following characteristics:

• Differential input v/�t� − v��t� • High input impedance, ZJ = ∞ (ideal)

• Low output impedance, Zm = 0 (ideal) • High constant gain amplification, A = ∞ (ideal)

The output of the operational amplifier vm�t� is vm�t� = A<v/�t� − v��t�= (2.8)



Figure 2.3. Operational amplifier

Example 2.9

Determine the transfer function Zo��Zp�� for each of the op-amp configurations shown below. Assume that the

op-amp has ideal characteristics.

Answers:

a. Zo��Zp�� = − q1��q_��, an inverting amplifier

b. Zo��Zp�� = 1 + q1��q_��, a non-inverting amplifier

Example 2.10

Find the transfer function Zo��Zp�� for the circuit given below.

(a) (b)



Answer: Vm�s�VJ�s� = −1.232 s/ + 45.95s + 22.55s

Example 2.11

Find the transfer function Zo��Zp�� for the op-amp configuration shown below:

Answer: Vm�s�VJ�s� = R�R/C�C/s/ + �R/C/ + R/C� + R�C��s + 1R�R/C�C/s/ + �R/C/ + R�C��s + 1

Drill Problems 2.3

1. Find the transfer function G�s� = Zo��Zp�� for each of the networks shown below:



2. Find the transfer function G�s� = Zr��Z�� for each of the networks shown below:

3. Using mesh and nodal analysis, find the transfer function G�s� = Zo��Z�� for each of the networks shown below:

4. For each of the operational amplifiers shown, find the transfer function G�s� = Zo��Zp�� .

5. Determine the transfer function G�s� = Zo��Zp�� for each of the networks shown below:



2.5 Mechanical System Transfer Function

Intended Learning Outcome: Determine the transfer function of translational, rotational and translational-

rotational mechanical systems, including systems with gears.

In the previous section, the transfer function for electrical networks was obtained. In this section, transfer

function for translational and rotational mechanical systems will be obtained. It will be evident later on that

the transfer functions of both systems are mathematically indistinguishable. Hence, an electrical network

can be interfaced to a mechanical system by cascading their transfer functions provided that one system is

not loaded by the other.

Translational Mechanical Systems. Table 2.4 below shows the components of mechanical systems and

relationships between force, velocity and displacement.

In mechanical systems there are three components:

• Spring, with the spring constant K as the parameter • Viscous damper, with the coefficient of friction ft as the parameter • Mass, M as its parameter

These components can be compared to the passive electrical components:

• Capacitor

• Resistor

• Inductor



Comparing Tables 2.3 and 2.4, the following analogy between the mechanical and electrical systems can

be formulated.

• The capacitor and inductor of the electrical system are energy-storage devices, whereas for

mechanical system, these are analogous to the spring and the mass.

• Both the resistor of the electrical system and viscous damper of the mechanical system dissipate

energy.

• The mechanical force is analogous to electrical voltage and the mechanical velocity is analogous to

electrical current.

• Summing forces in terms of velocity is analogous to summing voltages written in terms of current

and the resulting mechanical differential equations are analogous to mesh equations.

Example 2.12

Find the transfer function U��v�� for the system shown below.



Answer:

G�s� = X�s�F�s� = 1Ms/ + fts + K

Many mechanical systems are similar to multiple-loop and multiple-node electrical networks, where more

than one simultaneous differential equation is required to describe the system. In mechanical systems, the

number of equations of motion required is equal to the number of linearly independent motions. Linear

independence implies that a point of motion in a system can still move if all other points of motion are held

still; the number of these linearly independent motions is called the degrees of freedom of the system. For

mechanical systems with multiple degrees of freedom, use superposition: for each free-body diagram, hold

all other points of motion still and find the forces acting on the body only due to its own motion. Do this for

all of the bodies and the result will be a system of simultaneous equation of motion.

Example 2.13

Find the transfer function U1��v�� for the system shown below.

Answer:

G�s� = X/�s�F�s� = ft?�s�D

where

D = yzM�s/ + <ft_ + ft? =s + �K� + K/�{ −<ft?s + K/=−<ft? s + K/= zM/s/ + <ft1 + ft?=s + �K/ + K0�{y



The method by which the equation in each degree of freedom is obtained similar to the one done with

electrical networks, as

| Sum of impedancesconnected to the motion at x�} X��s� − |Sum of impedancesbetween x� and x/ } X/�s� = |Sum of appliedforces at x� } (2.9a)

− |Sum of impedancesbetween x� and x/ } X��s� + | Sum of impedancesconnected to the motion at x/} X/�s� = |Sum of appliedforces at x� } (2.9b)

This can also be expanded to systems with multiple degrees of freedom, as illustrated in the example

below.

Example 2.14

Write, but do not solve, the equations of motions for the mechanical network below.

Answer:



Example 2.15

Find the transfer function G�s� = X/�s�/F�s� for the translational mechanical system shown below.

Answer:

G�s� = 3s + 1s�s0 + 7s/ + 5s + 1�

Drill Problems 2.4

1. Find the transfer function G�s� = U1��v�� for the translational mechanical system shown below.

2. Find the transfer function G�s� = X/�s�/F�s� for the translational mechanical network shown below.



3. Find the transfer function G�s� = X0�s�/F�s� for the translational mechanical system shown below.

4. Find the transfer function G��s� = U?��v�� for the translational mechanical system shown below.

5. Write, but do not solve, the equations of motion for the translational mechanical system shown below

Rotational Mechanical Systems. The rotational mechanical systems are handled the same way as

translational mechanical systems, except that the torque replaces force and angular displacement replaces

translational displacement. The mechanical components for rotational systems are the same as those for



translational systems, except that the components undergo rotation instead of translation. Table 2.5 shows

the components along with the relationships between torque, angular velocity and angular displacement.

For rotational systems, the mass is replaced by inertia, and its parameter, called the moment of inertia, J. The concept of degrees of freedom carries over to rotational systems, except that we test a point of motion

by rotating it while holding still all other points of motion. Writing the equations of motion for rotational

systems is similar to writing them for translational systems; the only difference is that the free-body diagram

consists of torques rather than forces.

Example 2.16

Find the transfer function G�s� = θ/�s�/T�s� for rotational systems shown.

Answer:

G�s� = θ/�s�T�s� = KD

where

D = ��J�s/ + D�s + K� −K−K �J�s/ + D�s + K��



Notice that the above system has that now well-known form

| Sum of impedancesconnected to the motion at θ�} θ��s� − |Sum of impedancesbetween θ�and θ/ } θ/�s� = |Sum of appliedtorques at θ� } (2.10a)

− |Sum of impedancesbetween θ�and θ/ } θ��s� + | Sum of impedancesconnected to the motion at θ/} θ/�s� = |Sum of appliedtorques at θ/ } (2.10b)

Example 2.17

Write, but do not solve, the Laplace transform of the equations of motion for the system shown below.

Answer:

Example 2.18

Find the transfer function G�s� = θ/�s�/T�s� for the rotational mechanical system shown below.

Answer:

G�s� = 12s/ + s + 1



Systems with Gears. Gears provide mechanical advantage to rotational systems. Gears allow a system to

match the driving torque and the load – a trade-off between speed and torque. The linearized interaction

between gears is seen in figure 2.4.

Figure 2.4. A gear system

An input gear with radius r� and N� teeth is rotated through angle θ��t� due to a torque T��t�. An output gear with radius r/ and N/ teeth responds by rotating through angle θ/�t� and delivering a torque T/�t�. The distance traveled along each gear’s circumference is the same, thus

r�θ� = r/θ/ (2.11)

or

θ/θ� = r�r/ = N�N/ (2.12)

since the ratio of the number of teeth along the circumference is in the same proportion as the ratio of the

radii. Thus, the ratio of the angular displacement of the gears is inversely proportional to the ratio of the

number of teeth. For the input and output torques, assuming that the gears are lossless (they neither

absorb nor store energy),

T/T� = θ�θ/ = N/N� (2.13)

Figure 2.5. Transfer functions for angular displacement and torques in lossless gears.



Rotational mechanical impedances can be reflected through gear trains by multiplying the mechanical

impedance by the ratio

� Number of teeth of gear on the �� shaftNumber of teeth of gear on the �� shaft �/

Example 2.19

Represent the rotational mechanical system shown to an equivalent system without gears.

Answer:

Example 2.20

Find the transfer function G�s� = θ/�s�/T��s� for the system shown below

Answer:

G�s� = θ/�s�T��s� = N//N�J�s/ + D�s + K�



where

J� = J� �N/N��/ + J/ D� = D� �N/N��/ + D/

K� = K/

In order to eliminate gears with large radii, a gear train is used to implement large gear ratios by cascading

smaller gear ratios. Also, gears may exhibit inertia and damping. Handling of cases when the gears are

non-ideal is illustrated in the next example.

Example 2.21

Find the transfer function G�s� = θ��s�/T��s� for the system shown below.

Answer:

G�s� = θ��s�T��s� = 1J�s/ + D�s where

J� = J� + �J/ + J0� �N�N/�/ + �J8 + J>� �N�N0N/N8�/

and

D� = D� + D/ �N�N/�/



Example 2.22

Find the transfer function G�s� = θ/�s�/T�s� for the rotational mechanical system with gears shown in the figure below.

Answer:

G�s� = 1/2s/ + s + 1

Example 2.23

For the combined translational and rotational mechanical system shown below, find the transfer function G�s� = X�s�/T�s�

Answer: X�s�T�s� = 859s/ + 13s + 6



Drill Problems 2.5

1. For each of the rotational mechanical systems shown below, determine the transfer function G�s� =θ��s�/T�s�.

2. For the rotational mechanical system shown below, find the transfer function G�s� = θ/�s�/T�s�.

3. Find the transfer function G�s� = θ/�s�/T�s� for the rotational mechanical system shown in the figure below.

4. For the rotational mechanical system shown, find the transfer function G�s� = θ/�s�/T�s�



5. Given the combined translational and rotational system shown in the figure below, find the transfer

function G�s� = X�s�/T�s�.

2.6 Electromechanical System Transfer Function

Intended Learning Outcome: Obtain transfer function relating the output displacement to the input voltage

of electromechanical systems.

A motor is an electromechanical component that yields a displacement output for a voltage input, that is, a

mechanical output generated by an electrical input. The transfer function for the armature-controlled dc

servomotor will be derived. Its schematic and block diagram is shown in the figure below.

Figure 2.6. The schematic and block diagram representation of a DC motor.

In the above figure, a magnetic field is developed by stationary permanent magnets or a stationary

electromagnet called the fixed field. A rotating circuit, called the armature, through which the current i(�t� flows, passes through this magnetic field at right angles and feels a force F = Bℓi(�t�, where B is the



magnetic field strength and ℓ is the length of the conductor. The resulting torque turns the rotor, the rotating member of the motor.

There is another phenomenon that occurs in the motor: A conductor moving at right angles to a magnetic

field generates a voltage at the terminals of the conductor, equal to e = Bℓv, where e is the voltage and v is the velocity of the conductor normal to the magnetic field. Since the current-carrying armature is rotating

in a magnetic field, its voltage is proportional to speed. Thus,

v��t� = K� dθI�t�dt (2.13)

wherev��t� is the back electromotive force, K� is the back emf constant (in volt-second/radian) and �� = ωI�t� is the angular velocity of the motor. Taking Laplace transform, V��s� = K�sθI�s� (2.14)

The relationship between the armature current and, i(�t�, the applied armature voltage, e(�t� and the back emf v��t� is found by writing a loop equation around the Laplace transformed armature circuit.

R(I(�s� + L(sI(�s� + V��s� = E(�s� (2.15)

The torque developed by the motor is proportional to the armature current;

TI�s� = K�I(�s� (2.16)

whereK� is the motor torque constant (in Newton-meters/ampere), which depends on the motor and magnetic field characteristics. Solving equation 2.16 in terms of the armature current and substituting this

and the back emf of Equation 2.14 to equation 2.15,

�R( + L(s�TI�s�K� + K�sθI�s� = E(�s� (2.17)



To solve for the transfer function of the motor G�s� = θI�s�/E(�s�, an expression for TI in terms of θI�s� must be obtained. The equivalent mechanical loading on a motor is shown in a figure below.

Figure 2.7. Equivalent mechanical loading on a DC motor.

JI is the equivalent inertia at the armature and includes both the armature inertia and the load inertia

reflected to the armature. DI is the equivalent viscous damping at the armature and includes both the

armature viscous damping and the load viscous damping reflected to the armature. From figure 2.7,

TI�s� = �JIs/ + DIs�θI�s� (2.18)

Substituting this to equation 2.17 and assuming the armature inductance L( is small compared to the armature resistance R( which is very usual for a DC motor, the transfer function of the motor is solved as

G�s� = θI�s�E(�s� = K�/�R(JI�s `s + �� DI + ��L� d (2.19)

Note that equation 2.19 is of the form

G�s� = Ks�s + α� (2.20)

To find the constants JI and DI, refer to figure 2.8, with a motor with inertia J( and damping D( at the armature driving a load consisting of inertia JW and damping DW.



Figure 2.8. DC motor driving a rotational load

If all of the inertias and damping values are known, the load inertia and load damping can be reflected back

to the armature as some equivalent inertia and damping to be added to armature inertia and armature

damping. Thus,

JI = J( + JW �N�N/�/ (2.21a)

DI = D( + DW �N�N/�/ (2.21b)

The electrical constants of the motor can be obtained through a dynamometer test of the motor, where a

dynamometer measures the torque and speed of a motor under the condition of a constant applied voltage.

It can be shown that the relationship between motor torque TI when a voltage e( is applied and the motor speed ωI is

TI = − K�K�R( ωI + K�R( e( (2.22)

which is linear. The y-intercept is called the stall torque and is the torque of the motor when the angular

velocity is zero. Thus,

T��(¢¢ = K�R( e( (2.23)

The x-intercept is called the no-load speed and is the angular velocity of the motor at zero torque. Thus,



ωFm�¢m(� = e(K� (2.24)

Thus if T��(¢¢ and the ωFm�¢m(� is known, through the dynamometer test, the electrical constants can be found as

K�R( = T��(¢¢e( (2.25)

and

K� = e(ωFm�¢m(� (2.26)

The figure bellow shows the plot as a result of dynamometer test.

Figure 2.9. Torque-speed curves with an armature voltage as a parameter

Example 2.24

Given the system and torque-speed curve below respectively, find the transfer function G�s� =θW�s�/E(�s�.



Answer:

G�s� = θW�s�E(�s� = 0.0417s�s + 1.667�

Example 2.25

Find the transfer functionG�s� = θW�s�/E(�s� for the motor and the load shown below. The torque-speed curve is given by TI = −8ωI + 200 when the input voltage is 100 volts.

Answer:

G�s� = 1/20s�s + �15/2�



Drill Problems 2.6

1. For the motor, load, and torque-speed curve shown, find the transfer function G�s� = θW�s�/E(�s�.

2. The motor whose torque-speed characteristics and the load drives are shown. Some of the gears have

inertia. Find the transfer function G�s� = θ/�s�/E(�s�.

3. A dc motor 55 N-m of torque at a speed of 600 rad/s when 12 volts are applied. It stalls out at this

voltage with 100 N-m of torque. If the inertia and damping of the armature are 7 kg-m2 and 3 N-m-

s/rad, respectively, find the transfer function G�s� = θW�s�/E(�s� of this motor if it drives an inertia

load of 105 kg-m2 through a gear train as shown below.



4. In this chapter, the transfer function of a dc motor relating the angular displacement output to armature

input voltage. Often, we want to control the output torque rather than the displacement. Derive the

transfer function of the motor that relates output torque to input armature voltage.

5. Find the transfer function G�s� = X�s�/E(�s� for the system shown below.

2.7. The General State-Space Representation and Application

Intended Learning Outcome: Define terms associated with state-space modeling of systems and

enumerate steps in obtaining the state-space representation of a system.

In previous articles, the frequency-domain, or the classical technique of modeling physical systems was

discussed. This approach is based on converting a system’s differential equation to a transfer function, thus

generating a mathematical model of the system that algebraically relates the representation of the output to



a representation of the input. Replacing a differential equation with an algebraic equation not only simplifies

the representation of individual subsystems but also simplifies modeling interconnected subsystems.

This section presents a method by which an alternative to transfer functions in modeling dynamic systems

is presented; the method is called state-space modeling. It can be used for the same class of systems that

the transfer functions can model. The discussions presented here are on the introductory level; advanced

techniques for state-space modeling is beyond the scope of this course. Also, since state-space

representations of systems are expressed as matrices, the student may want to review matrices and linear

algebra.

The following are the steps in writing a state-space model for physical systems:

1. Select a particular subset of all possible system variables and call the variables in this subset state

variables.

2. For an nth-order system, write n simultaneous, first-order differential equations in terms of the state variables. These systems of simultaneous differential equations are called state equations.

3. If the initial condition of all the state variables at t� as well as the system input for t ≥ t� is known, the simultaneous differential equations for the state variables can be solved for t ≥ t�.

4. The state variables with the system’s input will be algebraically combined and all of the other

system variables for t ≥ t� will be found. The algebraic equations that will arise are called the output equations.

5. The state equations and the output equations of the system are its state-space representation or

state-space model.

The application of such process is demonstrated using an RL and an RLC circuit.

Example 2.26

Find a possible state-space representation for the RL circuit shown below. Follow the steps enumerated

above.



Example 2.27

Find a possible state-space representation for the RLC circuit shown below.

If the system is linear, the state and output equations can be written in vector-matrix form. But before the

general state-space representation is presented, some definitions are in order.

• A linear combination of n variables, xJ, for i = 1 to n is given by the following sum, S, S = kFxF + kF��xF�� + ⋯ + K�x�

where each KJ is a constant. • A set of variables is said to be linearly independent if none of the variables can be written as a

linear combination of the others.

• A system variable is any variable that responds to an input or initial condition in a system.

• The state variables are the smallest set of linearly independent system variables such that the

values of the members of the set at time t� along with known forcing functions completely determine the value of all system variables for all t ≥ t�.

• A state vector is a vector whose elements are the state variables.

• The state space is the n-dimensional space whose axes are the state variables. • The state equations are a set of n simultaneous, first order differential equations with n variables

where the n variables to be solved are the state variables. • The output equation is the algebraic equation that expresses the output variables of a system as

linear combinations of the state variables and the inputs.

The general state-space representation of a linear system is given as

¤¥ = ¦¤ + §� (2.27a)



¨ = ©¤ + ª� (2.27b)

for t ≥ t� and initial conditions, ¤�t��, where: ¤ = state vector

¤¥ = �¤�� = derivatives of the state vectors with respect to time ¨= output vector � = input or control vector ¦ = system matrix

§ = input matrix

©= output matrix ª = feedforward matrix

The state equation is given by equation 2.27a and the vector ¤, the state vector, contains the state variables. As discussed previously, if the initial conditions of the state variables and the input vector � are known, this equation can be solved for the state variables. Equation 2.27b, the output equation, can be

used to solve for other system variables.

The choice of state variables for a given system is not unique. However, they must satisfy the following

requirements:

1. They must be linearly independent.

2. The minimum number of state variables to completely describe the system is equal to the order of

the differential equation representing the system.

Although the rule specifies the minimum number of state variables to be identified, the number can exceed

this minimum as long as all the chosen state variables are linearly independent. In some cases, this can

simplify the writing of state equations and output equations.



Example 2.28

Write the state-space representation of the systems in examples 2.26 and 2.27 in vector matrix form.

Answers:

For the RL network, the state equation is

didt = − RL i + v�t� and an output equation is vL�t� = Ri�t�

Thus, letting x¥ = `�J��d , x = �i , A = `− LWd , B = �1 , y = �vL , C = �R , D = �0 , and u = v�t� . Therefore

x¥ = |− RL} x + v�t� y = �Rx

For the RLC network, the state equations are dqdt = i didt = − 1LC q − RL i + 1L v

and an output equation is

vW�t� = − 1C q − Ri + v

Thus, letting ¤¥ = |dq/dtdi/dt }, ¤ = `qtd, ¦ = | 0 1−1/LC −R/L}, § = | 01/L}, y = vW, © = �−1/C −R, D = �1, u = v. Therefore

|dq/dtdi/dt } = | 0 1−1/LC −R/L} `qtd + | 01/L} v



vW = �−1/C −R `qtd + v

2.8 State-Space Representation of Electrical and Mechanical Systems

Intended Learning Outcome: Obtain the state-space model of electrical and mechanical systems.

This section discusses the methods by which electrical and mechanical systems will be modeled through

state-space. In the previous section, it is stated that the choice of state variables are arbitrary. This section

describes a systematic technique by which these state variables can be selected. The approach is done as

follows:

1. Write a simple derivative equation for each energy-storage element and solve for each derivative

term as a linear combination of any of the system variables and input that are present in the

equation.

2. Select the differentiated variable as a state variable.

3. Express all other system variables in the equations in terms of the state variables and the input.

4. The output variables are written as linear combinations of the state variables and the input.

These steps will be demonstrated for electrical and mechanical systems through the following examples.

Example 2.29

Find a state-space representation of the network shown below if the output is the current through the

resistor.

Answer:



For mechanical systems, it is more convenient to obtain the state-equations directly from the equations of

motion rather than from the energy-storage elements. Thus, in mechanical systems, the position and

velocity of each point of linearly independent motions are selected as state variables. The procedure is

illustrated in the following example.

Example 2.30

Find the state and output equations of the translational mechanical system shown below

Answers:

¨ = ¤ = `x�x/d = `1 0 0 00 0 1 0d «x�v�x¬v¬

Example 2.31

Find the state-space representation of the electrical network shown. The output is vm�t�.

Answers:



Example 2.32

Represent the translational mechanical system shown below in state-space, where x0�t� is the output.

Answers:

Drill Problems 2.7

1. Represent the electrical network shown in the figure below in state-space, where vm�t� is the output.

2. Find the state-space representation of the network shown below if the output is vm�t�.



3. Represent the system shown below in state-space where the output is x0�t�.

4. Represent the rotational mechanical system shown below in state-space where θ��t� is the output.

5. Represent the system shown below in state-space where the output is θW�t�.

2.9 The Transfer Function and the State-Space Relationship

Intended Learning Outcome: Convert a transfer function into state-space representation and vice versa.

To convert a transfer function into state-space representation, state variables must be chosen. One choice

would be that of phase variables, that is, each subsequent state variable is defined to be the derivative of

the previous state variable.



Consider the differential equation

dFydtF + aF�� dF��ydtF�� + ⋯ + a� dydt + a�y = b�u (2.28)

A convenient way to choose state variables is to choose the output, y�t� and its �n − 1� derivatives. Choosing the state variables, xJ,

x� = y x/ = dydt

x0 = d/ydt/ ⋮

xF = dF��ydtF�� Differentiating both sides yields

x¥ � = dydx = x/

x¥ / = d/ydx/ = x0

x¥ 0 = d0ydx0 = x8 ⋮

x¥ F�� = dF��ydxF�� = xF x¥ F = −a�x� − a�x/ − ⋯ − aF��xF + b�u



In vector-matrix form,

(2.28)

and the output equation is

(2.29)

Thus, to convert a transfer function into state equations, convert the transfer function to a differential

equation by cross-multiplication and taking the inverse Laplace transform, assuming zero initial conditions.

Then the differential equation is represented in state-space in phase-variable form.

Example 2.33

Find the state-space representation in phase-variable form for the transfer function

G�s� = C�s�R�s� = 24s0 + 9s/ + 26s + 24

Answer:



Generating an equivalent block diagram based on the phase variable form of the state-space

representation of the previous example will yield

which means that the output is generated by integrators and constant-gain blocks. The previous example

has a constant numerator. The next example illustrates the procedure when the numerator is a polynomial

of lower degree than the denominator.

Example 2.34

Find the state-space representation of the transfer function

G�s� = s/ + 7s + 2s0 + 9s/ + 26s + 24 Answer:



In the previous example, it can be seen that the denominator yields the state equations, while the

numerator yields the output equation. A similar equivalent block diagram can also be obtained for this

example, illustrating how the state-space representation is implemented using block diagrams, as in

Example 2.35

Find the state equations and the output equation for the phase-variable representation of the transfer

function

G�s� = 2s + 1s/ + 7s + 9

Answer:

To convert a state-space representation into its equivalent transfer function, given the state and output

equations,

¤¥ = ¦¤ + §� (2.30a) ¨ = ©¤ + ª� (2.30b)



Take the Laplace transform, assuming zero initial conditions,

s¯�s� = ¦¯�s� + §°�s�

±�s� = ©¯�s� + ª°�s�

For the first equation, solving for ¯�s�, ¯�s� = �s² − ¦��³§°�s�

Substituting this to the second equation

±�s� = �©�s² − ¦��³§ + ª°�s� (2.31)

in which the quantity ©�s² − ¦��³§ + ª is called the transfer function matrix since it relates the output ±�s� to the input °�s�. However, if the input and the output are scalars, then G�s� = Y�s�U�s� = ©�s² − ¦��³§ + ª (2.32)

Note that this equation evaluates an inverse of a matrix. A review on how to evaluate this must be done by

the student. One method is by

�s² − ¦��³ = adj�s² − ¦�det �s² − ¦� The method is illustrated in the following examples.

Example 2.36

Given the system defined as

Find the transfer function G�s� = Y�s�/U�s�.



Answer:

G�s� = 10�s/ + 3s + 2�s0 + 3s/ + 2s + 1

Example 2.37

Convert the state and output equations to transfer function.

Answer:

G�s� = 3s + 5s/ + 4s + 6

Drill Problems 2.8

1. Find the state-space representation in phase-variable form for the system shown below.

2. Write the state equations and the output equation for the phase-variable representation.

3. Represent the transfer function in state-space. Give the answer in vector-matrix form.

G�s� = s/ + 3s + 8�s + 1��s/ + 5s + 5�

4. Find the transfer function G�s� = Y�s�/R�s� for the following system represented in state-space.



5. Determine the transfer function G�s� = Y�s�/R�s� equivalent to the state and output equations shown below.

References:

N. Nise. (2011). Control Systems Engineering 6th Edition. United States of America: John Wiley & Sons.

R. Dorf& R. Bishop. (2011). Modern Control Systems 12th Edition. New Jersey: Prentice Hall.

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