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2014 1 University of Technology Dept. Of Production Eng. & Metallurgy 2 nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures 1. Introduction Numerical Control (NC) refers to the method of controlling the manufacturing operation by means of directly inserted coded numerical instructions into the machine tool. It is important to realize that NC is not a machining method; rather, it is a concept of machine control. Although the most popular applications of NC are in machining, NC can be applied to many other operations, including welding, sheet metalworking, riveting, etc. NC machine tools can be classified as “cutting machines” and “non-cutting machines”. A cutting machine means a machine that performs a removal process to make a finished part like milling machines, turning machines and EDM machines. Whereas Non-cutting machine tools change the shape of the blank material by applying force and press machines are good examples of this. In addition, robot systems for welding, cutting and painting can be considered as NC system. The numerical controller was primarily applied to milling machines and boring machines. However, recently it has become popular to apply NC for increased productivity and the kinds of machine with NC have been varied to include machines such as turning machines, machining centers, and drill/tapping machines. Particularly, the application of NC has extended to non- conventional machine tools such as wire electro-discharge machines and laser cutting machines in addition to conventional metal-cutting machines. Figure 1 CNC Machines NC systems are used not only for machine tools but also all machines that need motion controlled by servo systems, such as cutting machines, drawing instruments, woodworking machines, Coordinate Measurement Machines (CMM) and embroidering machines and NC is the fundamental technology for factory automation. The major advantages of NC over conventional methods of machine control are as follows: 1. Higher precision: NC machine tool are capable of machining at very close tolerances, in some operations as small as 0.005 mm. 2. Machining of complex three-dimensional shapes. 3. Better quality: NC systems are capable of maintaining constant working conditions for all parts in a batch thus ensuring less spread of quality characteristics.

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Page 1: 1. Introduction · Numerical control Lectures 1. Introduction Numerical Control (NC) refers to the method of controlling the manufacturing operation by means of directly inserted

2014

1

University of Technology Dept. Of Production Eng. & Metallurgy

2nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures

1. Introduction

Numerical Control (NC) refers to the method of controlling the manufacturing operation by means of directly inserted coded numerical instructions into the machine tool. It is important to realize that NC is not a machining method; rather, it is a concept of machine control. Although the most popular applications of NC are in machining, NC can be applied to many other operations, including welding, sheet metalworking, riveting, etc.

NC machine tools can be classified as “cutting machines” and “non-cutting machines”. A cutting machine means a machine that performs a removal process to make a finished part like milling machines, turning machines and EDM machines. Whereas Non-cutting machine tools change the shape of the blank material by applying force and press machines are good examples of this.

In addition, robot systems for welding, cutting and painting can be considered as NC system. The numerical controller was primarily applied to milling machines and boring machines. However, recently it has become popular to apply NC for increased productivity and the kinds of machine with NC have been varied to include machines such as turning machines, machining centers, and drill/tapping machines. Particularly, the application of NC has extended to non- conventional machine tools such as wire electro-discharge machines and laser cutting machines in addition to conventional metal-cutting machines.

Figure 1 CNC Machines

NC systems are used not only for machine tools but also all machines that need

motion controlled by servo systems, such as cutting machines, drawing instruments,

woodworking machines, Coordinate Measurement Machines (CMM) and embroidering

machines and NC is the fundamental technology for factory automation.

The major advantages of NC over conventional methods of machine control are as follows: 1. Higher precision: NC machine tool are capable of machining at very close tolerances,

in some operations as small as 0.005 mm.

2. Machining of complex three-dimensional shapes.

3. Better quality: NC systems are capable of maintaining constant working conditions for

all parts in a batch thus ensuring less spread of quality characteristics.

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2nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures

4. Higher productivity: NC machine tools reduce drastically the non-machining time.

Adjusting the machine tool for a different product is as easy as changing the computer

program and tool turret with the new set of cutting tools required for the particular

part.

5. Multi-operational machining: some NC machine tools, for example machine centers,

are capable of accomplishing a very high number of machining operations thus

reducing significantly the number of machine tools in the workshops.

6. Low operator qualification: the role of the operation of a NC machine is simply to

upload the workpiece and to download the finished part. In some cases, industrial

robots are employed for material handling, thus eliminating the human operator.

2. Types of NC systems

Machine controls are divided into three groups 1- Traditional numerical control (NC).

2- Computer numerical control (CNC).

3- Distributed numerical control (DNC).

The original numerical control machines were referred to as NC machine tool. They have

“hardwired” control, whereby control is accomplished through the use of punched paper (or plastic) tapes or cards. Tapes tend to wear, and become dirty, thus causing misreadings. Many other problems arise from the use of NC tapes, for example the need to manual reload the NC tapes for each new part and the lack of program editing abilities, which increases the lead time. Figure 2 shows the punch tape and the punched tape reader. The end of NC tapes was the result of two competing developments, CNC and DNC.

|

Figure 2 Punch tape and punch tape reader in NC system

CNC refers to a system that has a local computer to store all required numerical data. While CNC was used to enhance tapes for a while, they eventually allowed the use of other storage media,

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2nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures

magnetic tapes and hard disks. The advantages of CNC systems include but are not limited to the possibility to store and execute a number of large programs (especially if a three or more dimensional machining of complex shapes is considered), to allow editing of programs, to execute cycles of machining commands, etc.

The development of CNC over many years, along with the development of local area networking, has evolved in the modern concept of DNC. Distributed numerical control is similar to CNC, except a remote computer is used to control a number of machines. An off-site mainframe host computer holds programs for all parts to be produced in the DNC facility. Programs are downloaded from the mainframe computer, and then the local controller feeds instructions to the hardwired NC machine. The recent developments use a central computer which communicates with local CNC computers (also called Direct Numerical Control)

3. NC System Components

Traditionally, NC systems have been composed of the following components:

3.1 Tape punch:

Converts written instructions into a corresponding hole pattern. The hole pattern is punched into tape which is passed through the tape punch. Much older units used a typewriter device called a Flexowriter, and later devices included a microcomputer coupled with a tape punch unit.

3.2 Tape reader:

Reads the hole pattern on the tape and converts the pattern to a corresponding electrical signal code.

3.3 Controller:

Receives the electrical signal code from the tape reader and subsequently causes the NC machine to respond.

3.4 NC machine:

Responds to programmed signals from the controller. Accordingly, the machine executes the required motions to manufacture a part (spindle rotation on/off, table and or spindle movement along programmed axis directions, etc.).

Figure 3 Components of traditional NC systems

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University of Technology Dept. Of Production Eng. & Metallurgy

2nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures

4. CNC System Components

As mentioned in advance the CNC machine is an NC machine with the added feature of an onboard computer. The onboard computer is often referred to as the machine control unit or MCU. Control units for NC machines are usually hardwired, which means that all machine functions are controlled by the physical electronic elements that are built into the controller. The onboard computer, on the other hand, is “soft” wired, which means the machine functions are encoded into the computer at the time of manufacture, and they will not be erased when the CNC machine is turned off. Computer memory that holds such information is known as ROM or read-only memory. The MCU usually has an alphanumeric keyboard for direct or manual data input (MDI) of part programs. Such programs are stored in RAM or the random-access memory portion of the computer. They can be played back, edited, and processed by the control. All programs residing in RAM, however, are lost when the CNC machine is turned off. These programs can be saved on auxiliary storage devices such as punched tape, magnetic tape, or magnetic disk. Newer MCU units have graphics screens that can display not only the CNC program but also the cutter paths generated and any errors in the program. The components found in many CNC systems are shown in Figure 4.

Figure 4 Components of modern CNC systems.

4.1 Machine control unit:

Generates, stores, and processes CNC programs. The MCU also contains the machine motion controller in the form of an executive software program. See Figure 1.3.

Figure 5 Machine control unit

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2nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures

4.2 NC machine:

Responds to programmed signals from the MCU and manufactures the part.

5. Control System fundamentals

5.1 Concept of a system

A system is different things to different people and can include purely physical system such as the machine table of Computer Numerically Controlled (CNC) machine tool.

However, all systems have certain things in common. They all, for example, require inputs and outputs to be specified. In the case of the CNC machine tool machine table, the input might be the power to the drive motor, and the outputs might be the position, velocity and acceleration of the table.

Figure 5-1 system concept

It is necessary to clearly define the boundary of a system, together with the inputs and outputs that cross that boundary. In general, a system may be defined as a collection of matter, parts, components or procedures which are included within some specified boundary as shown in Figure 5-1. A system may have any number of inputs and outputs.

In control engineering, the way in which the system outputs respond in changes to the system inputs (i.e. the system response) is very important. The control system design engineer will attempt to evaluate the system response by determining a mathematical model for the system. Knowledge of the system inputs, together with the mathematical model, will allow the system outputs to be calculated.

It is conventional to refer to the system being controlled as the plant, and a block diagram represents this, as with other elements. Some inputs, the engineer will have direct control over, and can be used to control the plant outputs. These are known as control inputs. There are other inputs over which the engineer has no control, and these will tend to deflect the plant outputs from their desired values. These are called disturbance inputs.

In the case of the ship shown in Figure 5-2, the rudder and engines are the control inputs, whose values can be adjusted to control certain outputs, for example heading and forward velocity. The wind, waves and current are disturbance inputs and will induce errors in the outputs (called controlled variables) of position, heading and forward velocity. In addition, the disturbances will introduce increased ship motion (roll, pitch and heave) which again is not desirable.

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2nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures

Figure 5-2 A ship control system

Generally, the relationship between control input, disturbance input, plant and controlled

variable is shown in Figure 5-3.

Figure 5-3 Plant inputs and outputs.

5.2 Open-loop control systems

Figure 5-3 represents an open-loop control system and is used for very simple applications. The main problem with open-loop control is that the controlled variable is sensitive to changes in disturbance inputs. So, for example, if a gas fire is switched on in a room, and the temperature climbs to 20 oC, it will remain at that value unless there is a disturbance. This could be caused by leaving a door to the room open, for example, or alternatively by a change in outside temperature. In either case, the internal room temperature will change. For the room temperature to remain constant, a mechanism is required to vary the energy output from the gas fire.

5.3 Closed-loop control systems

For a room temperature control system, the first requirement is to detect or sense changes in room temperature. The second requirement is to control or vary the energy output from the gas fire, if the sensed room temperature is different from the desired room temperature. In general, a system that is designed to control the output of a plant must contain at least one sensor and controller as shown in Figure 5-4.

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Figure 5-4 closed-loop control system.

Figure 5-4 shows the generalized schematic block-diagram for a closed-loop, or feedback

control system. The controller and plant lie along the forward path, and the sensor in the feedback path. The measured value of the plant output is compared at the summing point with the desired value. The difference or error is fed to the controller which generates a control signal to drive the plant until its output equals the desired value. Such an arrangement is sometimes called an error-actuated system.

5.4 Some examples of control systems

5.4.1 Room temperature control system

The physical realization of a system to control room temperature is shown in Figure 5-5. Here the output signal from a temperature sensing device such as a thermocouple or a resistance thermometer is compared with the desired temperature. Any difference or error causes the controller to send a control signal to the gas solenoid valve which produces a linear movement of the valve stem, thus adjusting the flow of gas to the burner of the gas fire. The desired temperature is usually obtained from manual adjustment of a potentiometer.

Figure 5-5 Room temperature control system.

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2nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures

Figure 5-6 Block diagram of room temperature control system.

A detailed block diagram is shown in Figure 5-6. The physical values of the signals around the control loop are shown in brackets.

Steady conditions will exist when the actual and desired temperatures are the same, and the heat input exactly balances the heat loss through the walls of the building.

The system can operate in two modes: 1. Proportional control: Here the linear movement of the valve stem is proportional to

the error. This provides a continuous modulation of the heat input to the room

producing very precise temperature control. This is used for applications where

temperature control, of say better than 1 oC, is required (i.e. hospital operating

theatres, industrial standards rooms, etc.) where accuracy is more important

than cost.

2. On-off control: Also called thermostatic or bang-bang control, the gas valve is either fully open or fully closed, i.e. the heater is either on or off. This form of control produces an oscillation of about 2 or 3 oC of the actual temperature about the desired temperature, but is cheap to implement and is used for low-cost applications (i.e. domestic heating systems).

5.4.2 Computer Numerically Controlled (CNC) machine tool

Many systems operate under computer control, and Figure 5-7 shows an example ofa CNC machine tool control system.

Information relating to the shape of the work-piece and hence the motion of the machine table is stored in a computer program. This is relayed in digital format, in a sequential form to the controller and is compared with a digital feedback signal from the shaft encoder to generate a digital error signal. This is converted to an analogue control signal which, when amplified, drives a d.c. servomotor. Connected to the output shaft of the servomotor (in some cases through a gearbox) is a lead-screw to which is attached the machine table, the shaft encoder and a tachogenerator. The purpose of this latter device, which produces an analogue signal proportional to velocity, is to form an inner, or minor control loop in order to dampen, or stabilize the response of the system.

The block diagram for the CNC machine tool control system is shown in Figure 5-8.

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Figure 5-7 Computer numerically controlled machine tool.

Figure 5-8 Block diagram of CNC machine-tool control system.

6. Control system design

In order to design and implement a control system the following essential generic elements are required:

1. Knowledge of the output or actual value: This must be measured by a feedback

sensor, again in a form suitable for the controller to understand. In addition, the

sensor must have the necessary resolution and dynamic response so that the

measured value has the accuracy required from the performance specification.

2. Knowledge of the controlling device: The controller must be able to accept

measurements of desired and actual values and compute a control signal in a

suitable form to drive an actuating element. Controllers can be a range of devices,

including mechanical levers, pneumatic elements, analogue or digital circuits or

microcomputers.

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3. Knowledge of the actuating device: This unit amplifies the control signal and

provides the 'effort' to move the output of the plant towards its desired value.

In the case of the room temperature control system the actuator is the gas solenoid

valve and burner, the 'effort' being heat input (W). For the ship autopilot

system the actuator is the steering gear and rudder, the 'effort' being turning

moment (Nm).

4. Knowledge of the plant: Most control strategies require some knowledge of the

static and dynamic characteristics of the plant. These can be obtained from

measurements or from the application of fundamental physical laws, or a

combination of both.

With all of these knowledge and information available to the control system designer, all that is left is to design the system. The first problem to be encountered is that the knowledge of the system will be uncertain and incomplete. In particular, the dynamic characteristics of the system may change with time (time-variant) and so a fixed control strategy will not work. Due to fuel consumption for example, the mass of an airliner can be almost half that of its take-off value at the end of a long haul flight.

Measurements of the controlled variables will be contaminated with electrical noise and disturbance effects. Some sensors will provide accurate and reliable data, others, because of difficulties in measuring the output variable may produce highly random and almost irrelevant information.

However, there is a standard methodology that can be applied to the design of most control systems. The steps in this methodology are shown in Figure 6-1.

The design of a control system is a mixture of technique and experience. This book explains some tried and tested, and some more recent approaches, techniques and methods available to the control system designer. Experience, however, only comes with time.

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Figure 6-1 Steps in the design of a control system.

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7. Control System modelling

7.1 Mathematical models

If the dynamic behavior of a physical system can be represented by an equation, or a set of equations, this is referred to as the mathematical model of the system. Such models can be constructed from knowledge of the physical characteristics of the system, i.e. mass for a mechanical system or resistance for an electrical system. Alternatively, a mathematical model may be determined by experimentation, by measuring how the system output responds to known inputs.

Example:

Assume that a mathematical model for a motor vehicle is required, relating the accelerator pedal angle 𝜃 to the forward speed 𝑢, a simple mathematical model might be

𝑢(𝑡) = 𝑎 𝜃(𝑡)

Eq. 1

Since 𝑢 and 𝜃 are functions of time, they are written 𝑢(𝑡)and 𝜃(𝑡). The constant 𝑎 could be calculated if the following vehicle data for engine torque 𝑇, wheel traction force 𝐹, and aerodynamic drag 𝐷 were available

𝑇 = 𝑏𝜃(𝑡) Eq. 2

𝐹 = 𝑐𝑇 Eq. 3

𝐷 = 𝑑𝑢(𝑡) Eq. 4

Now aerodynamic drag D must equal traction force F

𝐷 = 𝐹 Eq. 5

𝑑𝑢(𝑡) = 𝑐𝑇 Eq. 6

From Eq. 2 𝑑𝑢(𝑡) = 𝑐𝑏𝜃(𝑡)

Eq. 7

Giving

𝑢(𝑡) = (𝑐𝑏

𝑑)𝜃(𝑡)

Eq. 8

Hence the constant for the vehicle is

𝑎 = (𝑐𝑏

𝑑)

Eq. 9

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If the constants b, c and d were not available, then the vehicle model could be obtained by measuring the forward speed 𝑢(𝑡) for a number of different accelerator angles 𝜃(𝑡) and plotting the results, as shown in Figure 7-1.

Figure 7-1 Vehicle forward speed plotted against accelerator angle.

Since Figure 7-1 shows a linear relationship, the value of the vehicle constant a is the slope of the line.

More complex mathematical model

It can be seen that Eq. 1 gives a liner relationship, when there is a change in accelerator angle there is an instantaneous change in vehicle forward speed. In the real world, it takes time to build up to the new forward speed, so to model the dynamic characteristics of the vehicle accurately, this needs to be taken into account.

The mathematical model that represents the dynamic behavior of physical system can be constructed using differential equations. A more accurate representation of the motor vehicle would be

ⅇⅆ𝑢

ⅆ𝑡+ 𝑓𝑢 = 𝑔𝜃(𝑡)

Eq. 10

Here, 𝑑𝑢

𝑑𝑡 is the acceleration of the vehicle. When it travels at constant velocity, this term

becomes zero. So then 𝑓𝑢(𝑡) = 𝑔𝜃(𝑡)

Eq. 11

𝑢(𝑡) = (𝑔

𝒇) 𝜃(𝑡)

Eq. 12

Hence (𝑔

𝑓 )is again the vehicle constant, or parameter 𝑎 in equation Eq. 1.

7.2 Differential equations with constant coefficients

In general, consider a system whose output is 𝑥(𝑡), whose input is 𝑦(𝑡) and contains constant coefficients of values 𝑎, 𝑏, 𝑐, . . . , 𝑧. If the dynamics of the system produce a first-order differential equation, it would be represented as

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2nd CAD/CAM class Dr. Mohanned AL-Khafaji Numerical control Lectures

𝑎ⅆ𝑥

ⅆ𝑡+ 𝑏𝑥 = 𝑐𝑦(𝑡)

Eq. 13

If the system dynamics produced a second-order differential equation, it would be represented by

𝑎ⅆ2𝑥

ⅆ𝑡2+ 𝑏

ⅆ𝑥

ⅆ𝑓+ 𝑐𝑥 = ⅇ𝑦(𝑡)

Eq. 14

If the dynamics produce a third-order differential equation, i ts representation would be

𝑎ⅆ3𝑥

ⅆ𝑡3+ 𝑏

ⅆ2𝑥

ⅆ𝑡2+ 𝑐

ⅆ𝑥

ⅆ𝑡+ ⅇ𝑥 = 𝑓 𝑦(𝑡)

Eq. 15

Equations Eq. 13, Eq. 14 and Eq. 15 are linear differential equations with constant coefficients. Note that the order of the differential equation is the order of the highest derivative. Systems described by such equations called linear systems of the same order as the differential equation. For example, Eq. 13 describes a first-order linear system, Eq. 14 a second-order linear system and Eq. 15a third-order linear system.

Note for abbreviation the terms 𝒅𝒙

𝒅𝒕, 𝒅𝟐𝒙

𝒅𝒕𝟐 and

𝒅𝟑𝒙

𝒅𝒕𝟑 can be represented as �̇�, �̈� and �⃛�

respectively.

7.3 Mathematical models of mechanical systems

Mechanical systems usually considered to comprise of the linear lumped parameter elements of stiffness, damping and mass.

7.3.1 Stiffness in mechanical systems

An elastic element produces an extension proportional to the force applied to it. For the translational spring

𝐹𝑜𝑟𝑐𝑒 ∝ 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 If 𝑥𝑖(𝑡) > 𝑥0(𝑡), then

𝑃(𝑡) = 𝐾(𝑥𝑖(𝑡) − 𝑥𝑜(𝑡)) Eq. 16

Figure 7-2 Linear elastic elements.

And for the rotational spring, the angle proportional to the applied torque

𝑇𝑜𝑟𝑞𝑢𝑒 ∝ 𝑇𝑤𝑖𝑠𝑡

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If 𝜃𝑖(𝑡) > 𝜃𝑜(𝑡), then 𝑇 (𝑡) = 𝐾(𝜃𝑖(𝑡) − 𝜃𝑜(𝑡))

Eq. 17

Note that 𝐾 is the spring stiffness, has units of (𝑁/𝑚)in Eq. 16 and (𝑁𝑚/𝑟𝑎𝑑) in Eq. 17.

7.3.2 Damping in mechanical systems

A damping element (sometimes called a dashpot) is assumed to produces a velocity proportional to the force for linear motion (or torque) applied to it. For the translational damper

𝐹𝑜𝑟𝑐𝑒 ∝ 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑃(𝑡) = 𝐶𝑣(𝑡) = 𝐶𝑑x

𝑑𝑡= 𝑐�̇�

Eq. 18

And for the rotational damper 𝑇𝑜𝑟𝑞𝑢𝑒 ∝ 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑇 = 𝐶𝜔(𝑡) = 𝐶ⅆ𝜃

ⅆ𝑡= �̇�

Eq. 19

Figure 7-3 Linear damping elements.

7.3.3 Mass in mechanical systems

The force to accelerate a body is the product of its mass and acceleration (Newton's second law). For the translational system

𝐹𝑜𝑟𝑐𝑒 ∝ 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

𝑝(𝑡) = 𝑚𝑎(𝑡) = 𝑚ⅆ𝑣

ⅆ𝑡= 𝑚

ⅆ2𝑥0ⅆ𝑡2

= 𝑥0̈

Eq. 20

For the rotational system 𝑇𝑜𝑟𝑞𝑢𝑒 ∝ 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

𝑇(𝑡) = 𝐼𝛼(𝑡) = 𝐼ⅆ𝜔

ⅆ𝑡= 𝐼

ⅆ2𝜃0ⅆ𝑡2

= 𝜃0̈

Eq. 21

Where, 𝐼 is the moment of inertia about the rotational axis.

When analyzing mechanical systems, it is usual to identify all external forces by the use of a 'Free-body diagram', and then apply Newton's second law of motion in the form

∑𝐹 = 𝑚𝑎 Eq. 22

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for translational systems Or

∑𝑀 = 𝐼𝛼 Eq. 23

for rotational systems

Example:

Find the differential equation relating the displacements 𝑥𝑖(𝑡) and 𝑥𝑜(𝑡) for the spring-mass-damper system shown in Figure 7-4. What would be the effect of neglecting the mass?

Figure 7-4 Spring - mass - damper system.

Solution:

Figure 7-5 shows the free-body diagram of the system by using Eq. 16 and Eq. 18. Applying Eq. 22, the equation of motion is

Figure 7-5 Free-body diagram for spring - mass - damper system.

∑𝐹𝑥 = 𝑚𝑎𝑥

𝐾(𝑥𝑖 − 𝑥𝑜) − 𝐶ⅆ𝑥𝑜𝑑𝑡

= 𝑚ⅆ2𝑥𝑜ⅆ𝑡2

𝐾𝑥𝑖 − 𝐾𝑥0 = 𝑚ⅆ2𝑥𝑜ⅆ𝑡2

+ 𝐶ⅆ𝑥𝑜ⅆ𝑡

From the equations, it can be seen that the highest degree of the differential equation is the second degree. So rearrange the equation to take the form of Eq. 14.

𝑚ⅆ2𝑥0ⅆ𝑡2

+ 𝐶ⅆ𝑥0ⅆ𝑡

+ 𝐾𝑥0 = 𝐾𝑥𝑖(𝑡)

Note: by using the abbreviation terms the equation can be rewritten as 𝑚𝑥�̈� + 𝐶𝑥�̇� + 𝐾𝑥𝑜 = 𝐾𝑥𝑖

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If the mass is zero then

∑𝐹𝑥 = 0

𝐾(𝑥𝑖 − 𝑥𝑜) − 𝐶ⅆ𝑥0𝑑𝑡

= 0

𝐾𝑥𝑖 − 𝐾𝑥0 = 𝐶ⅆ𝑥0ⅆ𝑡

Hence

𝐶ⅆ𝑥0ⅆ𝑡

+ 𝐾𝑥0 = 𝐾𝑥𝑖

Thus if the mass is neglected, the system becomes a first-order system.

7.3.4 Gear trains

𝑟1𝑟2=𝑁1𝑁2

Eq. 24

𝑟1𝑟2=𝜃2𝜃1

Eq. 25

The work done by one gear is same as the other 𝑇1𝜃1 = 𝑇2𝜃2

Eq. 26

Hence: 𝑇1𝑇2=𝜃2𝜃1=𝑟1𝑟2=𝑁1𝑁2

Eq. 27

Figure 6 Gear system

TT2

TT1

𝜽𝟐, 𝝎𝟐

𝜽𝟏, 𝝎𝟏

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Where𝑇1,𝑟1, 𝑁1 and 𝜃1are torque, radius, teeth number and angular displacement for gear one, respectively. 𝑇2,𝑟2, 𝑁2 and 𝜃2are torque, radius, teeth number and angular displacement for gear two, respectively.

7.3.5 Gear Trains with inertia and friction

In practice, gears have inertia and frictions, which cannot neglected. Figure 7 shows the practical gears arrangement connected to the load.

Figure 7 Gear train practical configuration

Where: 𝑇 is the applied torque 𝜃1, 𝜃2 angular displacement 𝑇1, 𝑇2 are the torques transmitted to gears 𝐽1, 𝐽2 inertia of the gears 𝑁1, 𝑁2 𝑛umber of teeth 𝐵1, 𝐵2 friction coefficient The torque equation of side 1 is:

𝑇 = 𝐽1𝑑2𝜃1(𝑡)

𝑑𝑡2+ 𝐵1

𝑑𝜃1(𝑡)

𝑑𝑡+ 𝑇1(𝑡)

Eq. 28

The torque equation of side 2 is:

𝑇2 = 𝐽2𝑑2𝜃2(𝑡)

𝑑𝑡2+ 𝐵2

𝑑𝜃2(𝑡)

𝑑𝑡+ 𝑇𝐿(𝑡)

Eq. 29

From Eq. 27

𝑇2 =𝑁2𝑁1 𝑇1

Substituting in Eq. 29

𝑁2𝑁1 𝑇1 = 𝐽2

𝑑2𝜃2(𝑡)

𝑑𝑡2+ 𝐵2

𝑑𝜃2(𝑡)

𝑑𝑡+ 𝑇𝐿(𝑡)

𝑇1 =𝑁1𝑁2𝐽2𝑑2𝜃2(𝑡)

𝑑𝑡2+𝑁1𝑁2𝐵2𝑑𝜃2(𝑡)

𝑑𝑡+𝑁1𝑁2𝑇𝐿(𝑡)

Eq. 30

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Substituting Eq. 30 into Eq. 28 the applied torque will be

𝑇 = 𝐽1𝑑2𝜃1𝑑𝑡2

+ 𝐵1𝑑𝜃1𝑑𝑡

+𝑁1𝑁2𝐽2𝑑2𝜃2𝑑𝑡2

+𝑁1𝑁2𝐵2𝑑𝜃2𝑑𝑡

+𝑁1𝑁2𝑇𝐿

Eq. 31

Substituting 𝜃2 =𝑁1

𝑁2𝜃1

𝑇 = 𝐽1𝑑2𝜃1𝑑𝑡2

+ 𝐵1𝑑𝜃1𝑑𝑡

+𝑁1𝑁2𝐽2𝑁1𝑁2

𝑑2𝜃1𝑑𝑡2

+𝑁1𝑁2𝐵2𝑁1𝑁2

𝑑𝜃1𝑑𝑡

+𝑁1𝑁2𝑇𝐿

𝑇 = [𝐽1 + 𝐽2 (𝑁1𝑁2)2

]𝑑2𝜃1𝑑𝑡2

+ [𝐵1 + 𝐵2 (𝑁1𝑁2)2

]𝑑𝜃1𝑑𝑡

+𝑁1𝑁2𝑇𝐿

𝐽1𝑒 = 𝐽1 + 𝐽2 (𝑁1𝑁2)2

𝐽1𝑒 The equivalent inertia of the primary side

𝐵1𝑒 = 𝐵1 + 𝐵2 (𝑁1𝑁2)2

𝐵1𝑒 The equivalent friction of primary side

𝑇 = 𝐽1𝑒𝑑2𝜃1𝑑𝑡2

+ 𝐵1𝑒𝑑𝜃1𝑑𝑡

+𝑁1𝑁2𝑇𝐿

Eq. 32

7.3.6 Belt or chain drives

The belt or chain drives perform same function as that of gear train. Assuming there is no slipping between belt and pulleys we can write.

Eq. 33 Belt or chain drives

𝑇1𝑇2=𝜃2𝜃1=𝑟1𝑟2

Eq. 34

7.3.7 Levers

Lever transmit translational motion and forces similar to gear trains. Figure 8 shows the lever system. Using the moment law

𝑓1𝑙1 = 𝑓2𝑙2 Eq. 35

The work done is the same in both sides

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𝑓1𝑥1 = 𝑓2𝑥2 Eq. 36

Hence 𝑓1𝑓2=𝑙2𝑙1=𝑥2𝑥1

Eq. 37

Figure 8 lever system

7.4 Mathematical Model of Electrical system

Similar to the mechanical system, very commonly used systems are of electrical type. Ohm’s law governs the behavior of the electrical systems. The dominant elements of an electrical system are

1- Resistor

2- Inductor

3- Capacitor

Figure 9 Passive elements of an electrical system

For a resistive element, Ohm’s Law can be written 𝑉1(𝑡) − 𝑉2(𝑡) = 𝑅 𝑖(𝑡)

Eq. 38

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For an inductive element, the relationship between voltage and current is

𝑉1(𝑡) − 𝑉2(𝑡) = 𝐿 𝑑𝑖(𝑡)

𝑑𝑡

Eq. 39

For a capacitive element, the electrostatic equation is

𝑄(𝑡) = 𝑐[𝑉1(𝑡) − 𝑉2(𝑡)] Eq. 40

Differentiating both sides with respect to t

𝑑𝑄(𝑡)

𝑑𝑡= 𝑖(𝑡) = 𝑐

𝑑

𝑑𝑡[𝑉1(𝑡) − 𝑉2(𝑡)]

Eq. 41

Note that if both sides of equation Eq. 41 are integrated then

𝑉1(𝑡) − 𝑉2(𝑡) =1

𝑐∫ 𝑖 𝑑𝑡

Eq. 42

Example:

Find the differential equation relating 𝑣1(𝑡) and 𝑣2(𝑡) for the RC network shown in the following figure.

Solution:

From Eq. 38 and Eq. 42 𝑣1(𝑡) − 𝑣2(𝑡) = 𝑅𝑖(𝑡) … (1)

𝑣2(𝑡) =1

𝑐∫ 𝑖(𝑡)𝑑𝑡

Or

𝑖(𝑡) = 𝑐𝑑𝑣2(𝑡)

𝑑𝑡 … (2)

Substituting equation 2 in 1 then

𝑣1(𝑡) − 𝑣2(𝑡) = 𝑅𝑐𝑑𝑣2(𝑡)

𝑑𝑡

𝑅𝑐𝑑𝑣2(𝑡)

𝑑𝑡+ 𝑣2(𝑡) = 𝑣1(𝑡)

So the system is represented by first order differential equation.

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Example:

Find the differential equation relating 𝑣1(𝑡) and 𝑣2(𝑡) for the RC network shown in the following figure.

𝑣1(𝑡) − 𝑣2(𝑡) = 𝑅 𝑖(𝑡) + 𝐿𝑑𝑖(𝑡)

𝑑𝑡 … (1)

𝑖(𝑡) = 𝐶𝑑𝑣2(𝑡)

𝑑𝑡 … (2)

Taking the derivative for equation 2 𝑑𝑖(𝑡)

𝑑𝑡= 𝐶

𝑑2𝑣(𝑡)

𝑑𝑡2 … (3)

Substituting equation 2 and 3 in equation 1

𝑣1(𝑡) − 𝑣2(𝑡) = 𝑅 𝐶𝑑𝑣2(𝑡)

𝑑𝑡+ 𝐿𝐶

𝑑2𝑣(𝑡)

𝑑𝑡2 … (4)

Rearranging equation 4

𝑅 𝐶𝑑𝑣2(𝑡)

𝑑𝑡+ 𝐿𝐶

𝑑2𝑣(𝑡)

𝑑𝑡2+ 𝑣2(𝑡) = 𝑣1(𝑡) … (5)

Equation 5 is a second order differential equation

7.5 Mathematical model of thermal system

It is convenient to consider thermal systems as being analogous to electrical systems so that they contain both resistive and capacitive elements.

7.5.1 Thermal resistance

Fourier’s Law gives heat flow by conduction

𝑄𝑇 =𝐾𝐴(𝑇1 − 𝑇2)

𝑙

Eq. 43

Where:

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QT is the heat flow through a flat plate in (𝐽

𝑠= 𝑊), A is the normal cross sectional area in (𝑚2),

𝑙 is the plate thickness, K is the thermal conductivity of plat’s material in (𝑊

𝑚 𝐾) and (𝑇1 − 𝑇2) is the

temperature different (𝐾). The Eq. 43 can be rewritten to be similar to Ohm’s law

𝑇1(𝑡) − 𝑇2(𝑡) = 𝑅𝑡 𝑄𝑇(𝑡) Eq. 44

Where 𝑅𝑇 =𝑙

𝐾𝐴 which is the thermal resistance

Figure 10 Heat flow through a flat plate.

7.5.2 Thermal capacitance

The heat stored by a body is 𝐻(𝑡) = 𝑚 𝐶𝑝 𝑇(𝑡)

Eq. 45

Where: 𝐻(𝑡) is the heat in 𝐽, 𝑚 is the body mass (𝑘𝑔), 𝐶𝑝 is the specific heat at constant pressure

(𝐽

𝑘𝑔 𝐾) and 𝑇(𝑡) is the temperature rise.

If Eq. 45 is compared with the electrostatic equation 𝑄(𝑡) = 𝐶 𝑣(𝑡)

Eq. 46

Then the thermal capacitance 𝐶𝑝 is

𝐶𝑇 = 𝑚 𝐶𝑝

Eq. 47

To obtain the heat flow 𝑄𝑇, Eq. 45 is differentiated with respect to time

𝑑𝐻(𝑡)

𝑑𝑡= 𝑚 𝐶𝑝

𝑑𝑇(𝑡)

𝑑𝑡

Eq. 48

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Or can be written as

𝑄𝑇(𝑡) = 𝐶𝑇 𝑑𝑇(𝑡)

𝑑𝑡

Eq. 49

8. Laplace Transforms

The Laplace transform is defined by the linear transformation

𝐹(𝑠) = 𝐿{𝑓(𝑡)} = ∫ 𝑒−𝑠𝑡 𝑓(𝑡) 𝑑𝑡 ∞

0

Eq. 50

Where 𝐹(𝑠) represents the Laplace transform of the function 𝑓(𝑡), and 𝑓(𝑡) = 0 𝑓𝑜𝑟 𝑡 < 0

𝑠 = 𝜎 + 𝑖𝜔

The Laplace Transform is one of the mathematical tools used to solve linear ordinary

differential equations.

In practice, one usually does not need to perform a contour integration in the complex plane. Instead, a "dictionary" of Laplace transform pairs [𝑓(𝑡) ⟺ 𝐹(𝑠) ] is generated and some simple rules allow one convert between the time domain solution,𝑓(𝑡) and the frequency or s-plane solution, 𝐹(𝑠).

8.1 Some Laplace Transform Pairs

Unit Step Function

𝑓(𝑡) = 𝑢(𝑡) = {0 𝑡 < 01 𝑡 ≥ 0

Eq. 51

𝐹(𝑠) = ∫ (1) 𝑒−𝑠𝑡 𝑑𝑡 ∞

0

= −1

𝑠𝑒−𝑠𝑡|

0

= −1

𝑠(0 − 1)

=1

𝑠

Eq. 52

Exponential Function 𝑓(𝑡) = 𝑒−𝑎𝑡

Eq. 53

𝐹(𝑠) = ∫ 𝑒−𝑎𝑡 𝑒−𝑠𝑡 𝑑𝑡 ∞

0

= ∫ 𝑒−(𝑠+𝑎)𝑡 𝑑𝑡 ∞

0

= −1

𝑠 + 𝑎𝑒−(𝑠+𝑎)𝑡|

0

= −1

𝑠 + 𝑎(0 − 1)

𝐹(𝑠) =1

𝑠 + 𝑎

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8.2 Properties of the Laplace transform

Superposition (Sum and Difference):

𝑓(𝑡) = 𝑎𝑓1(𝑡) ∓ 𝑏𝑓2(𝑓) 𝑎𝑛𝑑 𝐹(𝑠) = 𝑎𝐹1(𝑠) ∓ 𝑏𝐹2(𝑠) Eq. 54

Differentiation:

𝑓(𝑡) =𝑑𝑓1(𝑡)

𝑑𝑡 𝑎𝑛𝑑. 𝐹(𝑠) = 𝑠𝐹1(𝑆) − 𝑓1(0)

For the 2nd derivative:

𝑓(𝑡) =𝑑2𝑓1𝑑2𝑡

𝒂𝒏𝒅 𝑓(𝑠) = 𝑠2𝐹1(𝑠) − 𝑠𝑓1(𝑡)|0 −𝑑𝑓1(𝑡)

𝑑𝑡|0

In general I the nth derivative is:

𝑓(𝑡) =𝑑𝑛𝑓1(𝑡)

𝑑𝑡 𝑎𝑛𝑑 𝐹(𝑠) = 𝑠𝑛𝐹1(𝑠) − 𝑠

𝑛−1𝑓1(𝑡)|0 − 𝑠𝑛−2

𝑑𝑓1(𝑡)

𝑑𝑡|0

−⋯−𝑑𝑛−1𝑓1(𝑡)

𝑑𝑡𝑛−1|0

Eq. 55

Table 1 Some Laplace Transforms

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Time Function f(t) Laplace

Transform F(s) Function Plot

Step function 𝑓(𝑡) = ℎ

where: h=constant

𝐹(𝑠) =ℎ

𝑠

Pulse function

𝐹(𝑡) = {ℎ 𝑤ℎ𝑒𝑛 0 < 𝑡 < 𝑡𝑜0 𝑤ℎ𝑒𝑛 𝑡 > 𝑡𝑜

𝐹(𝑠)

=ℎ

𝑠(1 − 𝑒−𝑡0𝑠)

Exponential function 𝑓(𝑡) = 𝑒−𝑎𝑡

𝐹(𝑠) =1

𝑠 + 𝑎

f(t)

t, time

𝑓(𝑡) = 𝑒𝑎𝑡 𝐹(𝑠) =1

𝑠 − 𝑎

Sine wave function 𝑓(𝑡) = sin𝜔𝑡

𝐹(𝑠) =𝜔

𝑠2 +𝜔2

Polynomial function

𝑓(𝑡) = 𝑡𝑛 (n=positive integer)

𝐹(𝑠) =𝑛!

𝑠𝑛+1

𝑓(𝑡) = 𝑡𝑛𝑒−𝑎𝑡 𝐹(𝑠) =𝑛!

(𝑠 + 𝑎)𝑛+1

𝑓(𝑡) = 𝑡𝑛𝑒𝑎𝑡 𝐹(𝑠) =𝑛!

(𝑠 − 𝑎)𝑛−1

Multiplication by a Constant Kf(t) (K= constant)

𝐾 𝐹(𝑠)

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Figure 11 complex plane ( Z plane)

9. Inverse Transforms:

Laplace transform solutions to nth-order parameter linear time-invariant systems are typically of the form:

𝐹(𝑠) =(𝑠 − 𝑎1)(𝑠 − 𝑎2)⋯ (𝑠 − 𝑎𝑚)

(𝑠 − 𝑏1)(𝑠 − 𝑏2) ⋅ ⋯ (𝑠 − 𝑏𝑛)=𝑍(𝑠)

𝑃(𝑠)

Eq. 56

Where 𝑎1, 𝑎2, … , 𝑎𝑚 are referred to as the zerosof F(s) and the 𝑏1, 𝑏2, … , 𝑏𝑛 are the poles of F(s).

The main principal method for finding 𝐿−1{𝐹(𝑠)} when F(s) is a ratio of polynomials is Partial Fraction Expansion

Non-Repeated Linear Factors

10. Transfer function

Physically, a control system is a collection of components and circuits connected together to perform a useful function. Each component in the system converts energy from one form to another; for example, it can be thinking of a temperature sensor as converting degrees to volts or a motor as converting volts to revolutions per minute. To describe the performance of the entire control system, the designers must have some common language so that they can calculate the combined effects of the different components in the system. This need is behind the transfer function concept.

A transfer function (TF) is a mathematical relationship between the input and output of a control system component. Specifically, the transfer function is defined as the output divided by the input, expressed as

𝑇𝐹 =𝑂𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡

Eq. 57

EXAMPLE

A potentiometer is used as a position sensor. The pot is configured in such a way that 0° of rotation yields 0 V and 300° yields 10 V. Find the transfer function of the pot.

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SOLUTION

The transfer function is output divided by input. In this case, the input to the pot is “position in degrees,” and output is volts:

𝑇 ⋅ 𝐹 =𝑂𝑢𝑡𝑝𝑢𝑡

i𝑛𝑝𝑢𝑡=10𝑉

3000= 0 ⋅ 3333

𝑉

𝑑ⅇ𝑔

The transfer function of a component is an extremely useful number. It allows you to calculate the output of a component if you know the input. The procedure is simply to multiply the transfer function by the input.

EXAMPLE

For a temperature-measuring sensor, the input is temperature, and the output is voltage. The sensor transfer function is given as 0.01 V/deg. Find the sensor output voltage if the temperature is 600°F.

Solution:

From Eq. 57 the output can be calculated 𝑂𝑢𝑡𝑝𝑢𝑡 = 𝑖𝑛𝑝𝑢𝑡 × 𝑇𝐹

=600 × 0.01𝑉

𝑑𝑒𝑔= 6𝑉

The transfer functions can be used to analyze an entire system of components. One common

situation involves a series of components where the output of one component becomes the input to the next and each component has its own transfer function. Figure 12 shows the block diagram for this situation. This diagram can be reduced into a single block that has a 𝐹𝑡𝑜𝑡 , which is the product of all the individual transfer functions.

Figure 12 A series of transfer functions reduced to a single transfer function

𝑇𝐹𝑡𝑜𝑡 = systⅇm gain = 𝑇𝐹1 × 𝑇𝐹2 × 𝑇𝐹3 Eq. 58

𝑇𝐹𝑡𝑜𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑠𝑡𝑒𝑎𝑑𝑦 − 𝑠𝑡𝑎𝑡𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑒𝑛𝑡𝑖𝑟𝑒 (𝑜𝑝𝑒𝑛 − 𝑙𝑜𝑜𝑝) 𝑠𝑦𝑠𝑡𝑒𝑚 𝑇𝐹1, 𝑇𝐹2, . . . = 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠

EXAMPLE

Consider the system shown in Figure 13. It consists of an electric motor driving a gear train, which is driving a winch. Each component has its own characteristics: The motor (under these conditions) turns at 100 rpmm for each volt (Vm) supplied; the output shaft of the gear train rotates at one-half of the motor speed; the winch (with a 3-inch shaft circumference) converts the rotary motion (rpmw) to linear speed. The individual transfer functions are given as follows:

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Motor:

𝑇𝐹𝑚 = 𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡 =100 𝑟𝑝𝑚𝑚

1𝑉𝑚= 100

𝑟𝑝𝑚𝑚

𝑉

Gear train:

𝑇𝐹𝑔 = 𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡 =1 𝑟𝑝𝑚𝑤

2 𝑟𝑝𝑚𝑚= 0.5 𝑟𝑝𝑚𝑤/rpmm

Winch:

𝑇𝐹𝑤 = 𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡 =

3𝑖𝑛min

1 𝑟𝑝𝑚𝑤= 3 in./× min/rpm𝑤

Using Eq. 58, we can calculate the system transfer function. If everything is correct, all units will cancel except for the desired set:

𝑇𝐹𝑡𝑜𝑡 = 𝑇𝐹𝑚 × 𝑇𝐹𝑔 × 𝑇𝐹𝑤

=100 𝑟𝑝𝑚𝑚

1𝑉𝑚×1 𝑟𝑝𝑚𝑤

2 𝑟𝑝𝑚𝑚×

3𝑖𝑛min

1 𝑟𝑝𝑚𝑤

= 150 𝑖𝑛./𝑚𝑖𝑛/𝑉𝑚

Figure 13A system with three transfer

As shown that the transfer function of the complete system is 150 in./min/Vm. Knowing this value, we can calculate the system output for any system input. For example, if the input to the this system is 12 V (to the motor), the output speed of the winch is calculated as follows:

𝑂𝑢𝑡𝑝𝑢𝑡 = 𝑖𝑛𝑝𝑢𝑡 × 𝑇𝐹 =12𝑉 × 150 𝑖𝑛./min

1𝑉𝑚= 1800 𝑖𝑛./min

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11. State-space representation

State variables The smallest possible set of independent variables, which completely describes the state of the system is referred to as the set of state variables. These variables at some fixed time

(t = to) and system inputs for all (t ≥ t0) will provide a complete description of system behavior at any time (t ≥ t0) because independence is essential, state variable cannot expressible as algebraic function of each other and the inputs. Moreover, the set of state variables for a certain system is not unique.

Note: the objective of the state space method is to rewrite system's equations of motion as a large system of first order differential equations. Each of these differential equations consists of the time derivative of one of the state variables on the left-hand side and algebraic function of the state variable, as well as the system input, on the right-hand side. These differential equations are referred to as state-variable equation.

Methodology

Given a set of differential equations describing a certain dynamic system, the two key questions are

Q1. How many state variables are there?

The number of state variables is equal to the number of initial conditions required in order

to completely solve the differential equations of motion. For instance, if a single second-

order differential equation describes dynamic system two initial conditions are required.

Hence, there are two stale variables.

Q2. What are the state variables?

Those variables for which initial conditions are required in Q1 are chosen as state variables.

Example1:

For the mechanical system of (spring, damper and mass) given by the following equation. 𝑚�̈� + 𝑏�̇� + 𝑘𝑥 = 𝑓(𝑡) to solve this second order differential equation, knowledge of the two initial conditions.

𝑥(0) 𝑎𝑛𝑑 �̇�(0) Hence, there are two state variables (according Q1). Furthermore, because initial conditions

corresponding to 𝑥 and �̇�, state variables should be chosen, according to Q2, as 𝑥 and �̇� ; i.e.; 𝑥1 = 𝑥 𝑥2 = �̇�

Example2:

Suppose that the governing equation for a certain dynamic system are found to be

{�̈�1 + �̇�1 + 𝜃1 − 𝜃2 = 0

�̇�2 − 2𝜃1 + 𝜃2 = 0

where 𝜃1and 𝜃2represent angular displacements, and �̇�1and �̇�2 denote angular velocities. Determine the state variables.

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Solution:

Note, because the highest order of the differentiation is two for 𝜃1and one for 𝜃2, the required

initial conditions are 𝜃1(0), 𝜃2(0) and �̇�1(𝑜). Therefore, there are a total of three state variables, and they are chosen to be 𝑥1, 𝑥2 and �̇�1.

It is important to distinguish between state variables and physical variables such as displacement and velocity. State variables are mathematical quantities that are employed to represent a system's governing equations in a convenient form.

So the physical quantities have chosen as stale variables.

𝑥1 = 𝜃1 𝑥2 = 𝜃2 𝑥3 = �̇�1

General formulation

The next task is to construct the state space equations, which are first order differential equations, each of which contains the first derivative of one of the state variables on the left hand side, and an algebraic function of the state variables, inputs, and (possibly) time on the right hand side.

Assume, there is a system of multi input/ multi output (MIMO) with n state variables 𝑥1, 𝑥2, … , 𝑥𝑛, m inputs,𝑢1, 𝑢2, … , 𝑢𝑚,and p output,𝑦1, 𝑦2, … , 𝑦𝑝.

𝑇ℎ𝑒 𝑠𝑡𝑎𝑡𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 {

�̇�1 = 𝑓1(𝑥1, … , 𝑥𝑛; 𝑢1, … , 𝑢m; 𝑡)

�̇�2 = 𝑓2(𝑥1, … , 𝑥𝑛; 𝑢1, … , 𝑢m; 𝑡)⋮

�̇�𝑛 = 𝑓𝑛(𝑥1, … , 𝑥𝑛; 𝑢1, … , 𝑢m; 𝑡)

Eq. 59

Where 𝑓1, 𝑓2, … , 𝑓𝑛 are Nonlinear, in general. Similarly, system outputs may also be expressed

as follows:

𝑇ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑢𝑡𝑝𝑢𝑡 {

y = ℎ1(𝑥1, … , 𝑥𝑛; 𝑢1, … , 𝑢m𝑡)

y2 = ℎ2(𝑥1, … , 𝑥𝑛; 𝑢1, … , 𝑢m𝑡)⋮

y𝑝 = ℎ𝑛(𝑥1, … , 𝑥𝑛; 𝑢1, … , 𝑢m𝑡)

Eq. 60

These functions are nonlinear. If all elements of the system are linear the equations will be

rewriter with the following form:

𝑇ℎ𝑒 𝑙𝑖𝑛𝑒𝑎𝑟 𝑠𝑡𝑎𝑡𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 {

�̇�1 = 𝑎11𝑥1 +⋯+ 𝑎1𝑛𝑥𝑛 + 𝑏11𝑢1 +⋯+ 𝑏1𝑚𝑢m�̇�2 = 𝑎21𝑥1 +⋯+ 𝑎2𝑛𝑥𝑛 + 𝑏21𝑢1 +⋯+ 𝑏2𝑚𝑢m

⋮�̇�𝑛 = 𝑎𝑛1𝑥1 +⋯+ 𝑎𝑛𝑛𝑥𝑛 + 𝑏𝑛1𝑢1 +⋯+ 𝑏𝑛𝑚𝑢m

Eq. 61

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𝑇ℎ𝑒 𝑙𝑖𝑛𝑒𝑎𝑟 𝑠𝑦𝑠𝑡𝑒𝑚 𝑜𝑢𝑡𝑝𝑢𝑡𝑠 {

�̇�1 = 𝑐11𝑥1 +⋯+ 𝑐1𝑛𝑥𝑛 + 𝑑11𝑢1 +⋯+ 𝑑1𝑚𝑢m�̇�2 = 𝑐21𝑥1 +⋯+ 𝑐2𝑛𝑥𝑛 + 𝑑21𝑢1 +⋯+ 𝑑2𝑚𝑢m

⋮�̇�𝑝 = 𝑐𝑝1𝑥1 +⋯+ 𝑐𝑝𝑛𝑥𝑛 + 𝑑𝑝1𝑢1 +⋯+ 𝑑𝑝𝑚𝑢m

Eq. 62

Representing equations Eq. 61and Eq. 62 in matrix form we have:

𝒙 = {

𝑥1𝑥2⋮𝑥𝑛

}

𝑛×1

= 𝑠𝑡𝑎𝑡𝑒 𝑣𝑒𝑐𝑡𝑜𝑟

𝒖 = {

𝑢1𝑢2⋮𝑢𝑚

}

𝑚×1

= 𝑖𝑛𝑝𝑢𝑡 𝑣𝑒𝑐𝑡𝑜𝑟

𝒚 = {

𝑦1𝑦2⋮𝑦𝑝

}

𝑝×1

= 𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑒𝑐𝑡𝑜𝑟

𝑨 = [

𝑎11 𝑎12 … 𝑎1𝑛𝑎21 𝑎22 … 𝑎2𝑛⋮ ⋮ ⋮𝑎𝑛1 𝑎𝑛2 … 𝑎𝑛𝑛

]

𝑛×𝑛

𝑠𝑡𝑎𝑡𝑒 𝑚𝑎𝑡𝑟𝑖𝑥

𝑩 = [

𝑏11 𝑏12 … 𝑏1𝑚𝑏21 𝑏22 … 𝑏2𝑚⋮ ⋮ ⋮𝑏𝑚1 𝑏𝑚2 … 𝑏𝑚𝑚

]

𝑚×𝑚

𝑖𝑛𝑝𝑢𝑡 𝑚𝑎𝑡𝑟𝑖𝑥

𝑪 = [

𝑐11 𝑐12 … 𝑐1𝑛𝑐21 𝑐22 … 𝑐2𝑛⋮ ⋮ ⋮𝑐𝑝1 𝑐𝑝2 … 𝑐𝑝𝑛

]

𝑝×𝑛

𝑜𝑢𝑡𝑝𝑢𝑡 𝑚𝑎𝑡𝑟𝑖𝑥

𝑫 = [

𝑑11 𝑑12 … 𝑑1𝑚𝑑21 𝑑22 … 𝑑2𝑚⋮ ⋮ ⋮𝑑𝑝1 𝑑𝑝2 … 𝑑𝑝𝑚

]

𝑝×𝑚

𝑑𝑖𝑟𝑒𝑐𝑡 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑚𝑎𝑡𝑟𝑖𝑥

So the state space representation in matrix form is

{�̇� = 𝐴𝑥 + 𝐵𝑢𝑦 = 𝐶𝑥 + 𝐷𝑢

Eq. 63

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Example 3:

Formulate the state space equation for the system of mass, spring, and damper, which is governed with the following equation.

𝑚�̈� + 𝑏�̇� + 𝑘𝑥 = 𝑓(𝑡)

Solution:

The equation of the system has one variable with second derivative so the two initial conditions should be known

𝑥(0) 𝑎𝑛𝑑 �̇�(0) Hence, there are two state variables these are

𝑥1 = 𝑥 𝑎𝑛𝑑 𝑥2 = �̇� And accordingly, because there are two state variables, two first order differential equations

are expected. The first of these equations simply relates the two state variables.

�̇�1 = 𝑥2 𝑓𝑖𝑟𝑠𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 Which does not depends on the dynamic of the system. The second equation is directly

obtained from the equation of motion by substituting the state variables, as follows:

𝑚�̇�2 + 𝑏𝑥2 + 𝑘𝑥1 = 𝑓(𝑡) Solving for �̇�2

�̇�2 =−𝑏𝑥2 − 𝑘𝑥1 + 𝑓(𝑡)

𝑚 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛

𝑠𝑡𝑎𝑡𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 {

�̇�1 = 𝑥2

�̇�2 =−𝑏𝑥2 − 𝑘𝑥1 + 𝑓(𝑡)

𝑚

As it is known for this type of system the displacement is the system output,

𝑦 = 𝑥 ⇐⇒ 𝑦 = 𝑥1 Rewriting the equations in matrix form

{�̇�1�̇�2} = [

0 1

−𝑘

𝑚−𝑏

𝑚

] {𝑥1𝑥2} + {

01

𝑚

} 𝑓

From this

�̇� = 𝐴𝑥 + 𝐵𝑢

𝒙 = {𝑥1𝑥2} , 𝐴 = [

0 1

−𝑘

𝑚−𝑏

𝑚

] , 𝐵 = {01

𝑚

} , 𝑎𝑛𝑑 𝑢 = 𝑓(𝑡)

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And for the output

𝑦 = [1 0] {𝑥1𝑥2} + 0. 𝑢

𝑦 = 𝐶𝑥 + 𝐷𝑢

𝐶 = [1 0] 𝑎𝑛𝑑 𝐷 = 0

Note that the two state variable are independent, because it is impossible to express displacement as an algebraic function of velocity.

Example 4:

An electrical circuit has an input which is the voltage 𝑒(𝑡) and two output which are 𝑞1 and �̇�1. The governing equation of the system is given

𝐿1�̈�1 + 𝑅(�̇�1 − �̇�2) +1

𝐶1𝑞1 = 𝑒

𝐿2�̈�2 + 𝑅(�̇�2 − �̇�1) +1

𝐶2𝑞2 = 0

Find state space representation of the system

Solution:

Because each of the differential equations is second order so four initial conditions are required, form that there are four state variables

𝑥1 = 𝑞1 , 𝑥2 = 𝑞2 , 𝑥3 = �̇�1 , 𝑎𝑛𝑑 𝑥4 = �̇�2

Substituting the State variables in the governing equations

𝐿1𝑥3̇ + 𝑅(𝑥3 − 𝑥4) +1

𝐶1𝑥1 = 𝑒

𝐿2𝑥4̇ + 𝑅(𝑥4 − 𝑥3) +1

𝐶2𝑥2 = 0

The state variables equations are: �̇�1 = 𝑥3 �̇�2 = 𝑥4

�̇�3 = �̈�1 =1

𝐿1[−𝑅(𝑥3 − 𝑥4) −

1

𝐶1𝑥1 + 𝑒(𝑡)]

�̇�4 = �̈�2 =1

𝐿2[−𝑅(𝑥4 − 𝑥3) −

1

𝐶2𝑥2]

The state equation

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{

�̇�1�̇�2�̇�3�̇�4

} =

[

0 0 1 00 0 0 1

−1

𝐿1𝐶10 −

𝑅

𝐿1

𝑅

𝐿1

0 −1

𝐿2𝐶2

𝑅

𝐿2 −𝑅

𝐿2 ]

{

𝑥1𝑥2𝑥3𝑥4

} +

{

001

𝐿10 }

ⅇ(t)

𝑦 = {𝑞1�̇�1} ⟹ y = {

𝑥1𝑥3}

y = Cx + Du

C = [1 0 0 00 0 1 0

] anⅆ D = 0

12. Transient and Steady-State Response Analyses

In practice, the input signal to a control system is not known ahead of time but is random in nature, and the instantaneous input cannot be expressed analytically. Only in some special cases is the input signal known in advance and expressible analytically or by curves, such as in the case of the automatic control of cutting tools.

In analyzing and designing control systems, we must have a basis of comparison of performance of various control systems. This basis may be set up by specifying particular test input signals and by comparing the responses of various systems to these input signals.

Many design criteria are based on the response to such test signals or on the response of systems to changes in initial conditions (without any test signals). The use of test signals can be justified, because of a correlation existing between the response characteristics of a system to a typical test input signal and the capability of the system to cope with actual input signals.

Typical Test Signals : The commonly used test input signals are step functions, ramp functions,

acceleration functions, impulse functions, sinusoidal function. With these test signals,

mathematical and experimental analyses of control systems can be carried out easily, since

the signals are very simple functions of time.

Which of these typical input signals to use for analyzing system characteristics may be determined by the form of the input that the system will be subjected to most frequently under normal operation. If the inputs to a control system are gradually changing functions of time, then a ramp function of time may be a good test signal. Similarly, if a system is subjected to sudden disturbances, a step function of time may be a good test signal; and for a system subjected to shock inputs, an impulse function may be best. Once a control system is designed on the basis of test signals, the performance of the system in response to actual inputs is generally satisfactory. The use of such test signals enables one to compare the performance of many systems on the same basis.

Transient Response and Steady-State Response:

The time response of a control system consists of two parts: the transient response and the steady-state response. By transient response, we mean that which goes from the initial state to the final state. By steady-state response, we mean the manner in which the system output behaves as 𝑡 approaches infinity. Thus the system response 𝑐(𝑡) may be written as

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𝑐(𝑡) = 𝑐𝑡𝑟(𝑡) + 𝑐𝑠𝑠(𝑡)

Eq. 64

Where the first term on the right-hand side 𝑐𝑡𝑟(𝑡) of the equation is the transient response and the second term 𝑐𝑠𝑠(𝑡) is the steady-state response.

Absolute Stability, Relative Stability, and Steady-State Error:

In designing a control system, we must be able to predict the dynamic behavior of the system from a knowledge of the components. The most important characteristic of the dynamic behavior of a control system is absolute stability—that is, whether the system is stable or unstable. A control system is in equilibrium if, in the absence of any disturbance or input, the output stays in the same state. A linear time-invariant control system is stable if the output eventually comes back to its equilibrium state when the system is subjected to an initial condition. A linear time-invariant control system is critically stable if oscillations of the output continue forever. It is unstable if the output diverges without bound from its equilibrium state when the system is subjected to an initial condition. Actually, the output of a physical system may increase to a certain extent but may be limited by mechanical “stops,” or the system may break down or become nonlinear after the output exceeds a certain magnitude so that the linear differential equations no longer apply.

Important system behavior (other than absolute stability) to which we must give careful consideration includes relative stability and steady-state error. Since a physical control system involves energy storage, the output of the system, when subjected to an input, cannot follow the input immediately but exhibits a transient response before a steady state can be reached. The transient response of a practical control system often exhibits damped oscillations before reaching a steady state. If the output of a system at steady state does not exactly agree with the input, the system is said to have steady state error. This error is indicative of the accuracy of the system. In analyzing a control system, we must examine transient-response behavior and steady-state behavior.

12.1 First-Order Systems Response

Consider the first-order system shown in Figure 14a. Physically, this system may represent an RC circuit, thermal system, or the like. A simplified block diagram is shown in Figure 14b.The input-output relationship (the transfer function) is given by

𝐶(𝑠)

𝑅(𝑠)=

1

𝑇𝑠 + 1

Eq. 65

Figure 14 First-Order Systems

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In the following, the system responses will be analyzed to such inputs as the unit-step, unit-ramp, and unit-impulse functions. The initial conditions are assumed zero. Note that all systems having the same transfer function will exhibit the same output in response to the same input. For any given physical system, the mathematical response can be given a physical interpretation.

12.1.1 Unit-Step Response of First-Order Systems

Since the Laplace transform of the unit-step function is 1

𝑠, substituting 𝑅(𝑠) =

1

𝑠 into Eq. 65

𝐶(𝑠) =1

𝑇𝑠 + 1

1

𝑠

Eq. 66

Expanding 𝐶(𝑠) into partial fractions gives

𝐶(𝑠) =1

𝑠−

𝑇

𝑇𝑠 + 1=1

𝑠−

1

𝑠 +1𝑇

Eq. 67

Taking the invers Laplace Transform

𝑐(𝑡) = 1 − 𝑒−𝑡𝑇 , 𝑓𝑜𝑟 𝑡 ≥ 0

Eq. 68

Eq. 68 states that initially the output 𝑐(𝑡) is zero and finally it becomes unity. One important characteristic of such an exponential response curve 𝑐(𝑡) is that at 𝑡 = 𝑇 the value of 𝑐(𝑡) is 0.632, or the response 𝑐(𝑡) has reached 63.2% of its total change. This may be easily seen by substituting 𝑡 = 𝑇 in 𝑐(𝑡).That is,

𝑐(𝑡) = 1 − 𝑒−1 = 0.632

Figure 15 exponential response for first-order system

Note that the smaller the time constant T, the faster the system response. Another important

characteristic of the exponential response curve is that the slope of the tangent line at 𝑡 = 0 is 1

𝑇,

since 𝑑𝑐

𝑑𝑡|𝑡=0

=1

𝑇𝑒−

𝑡𝑇|𝑡=0

=1

𝑇

Eq. 69

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The output would reach the final value at 𝑡 = 𝑇 if it maintained its initial speed of response. From Eq. 69 it can be seen that the slope of the response curve 𝑐(𝑡) decreases monotonically from 1/𝑇 at 𝑡 = 0 to zero at 𝑡 = ∞.

The exponential response curve 𝑐(𝑡) given by Eq. 68 is shown in Figure 15. In one time constant, the exponential response curve has gone from 0 to 63.2% of the final value. In two time constants, the response reaches 86.5% of the final value. At 𝑡 = 3𝑇, 4𝑇, and 5𝑇, the response reaches 95%, 98.2%, and 99.3%, respectively, of the final value.Thus, for 𝑡 ≥ 4𝑇, the response remains within 2% of the final value. As seen from Eq. 68, the steady state is reached mathematically only after an infinite time. In practice, however, a reasonable estimate of the response time is the length of time the response curve needs to reach and stay within the 2% line of the final value, or four time constants.

12.1.2 Unit-Ramp Response of First-Order Systems

Since the Laplace transform of the unit-ramp function is 1

𝑠2 , the same system analyzed in

section 12.1.1 will be analyzed

𝐶(𝑠) =1

𝑇𝑠 + 1

1

𝑠2

Eq. 70

Expanding 𝐶(𝑠) into partial fractions gives

𝐶(𝑠) =1

𝑠2−𝑇

𝑠+

𝑇2

𝑇𝑠 + 1

Eq. 71

Taking the inverse Laplace transform of

𝑐(𝑡) = 𝑡 − 𝑇 + 𝑇 𝑒−𝑡𝑇 𝑓𝑜𝑟 𝑡 ≥ 0

Eq. 72

The error signal 𝑒(𝑡) is then

𝑒(𝑡) = 𝑟(𝑡) − 𝑐(𝑡) = 𝑇 (1 − 𝑒−𝑡𝑇)

Eq. 73

Figure 16 The unit ramp respose of the system

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As 𝑡 approaches infinity, 𝑒–𝑡

𝑇 approaches zero, and thus the error signal 𝑒(𝑡) approaches 𝑇 or 𝑒(∞) = 𝑇

The unit-ramp input and the system output are shown in Figure 16. The error in following the unit-ramp input is equal to 𝑇 for sufficiently large 𝑡. The smaller the time constant 𝑇, the smaller the steady-state error in following the ramp input.

12.1.3 Unit-Impulse Response of First-Order Systems

For the unit-impulse input 𝑅(𝑠) = 1 and the output of the system in section 12.1.1 can be obtained as

𝐶(𝑠) =1

𝑇𝑠 + 1

Eq. 74

Taking the inverse Laplace transform

𝑐(𝑡) =1

𝑇𝑒−

𝑡𝑇 𝑓𝑜𝑟 𝑡 ≥ 0

Eq. 75

The response curve given by Eq. 75 is shown in Figure 17.

Figure 17 Unit-impulse response of the system

12.1.4 An Important Property of Linear Time-Invariant Systems

I n the analysis of the first order system above, it has been shown that for the unit-ramp input the output 𝑐(𝑡) is

𝑐(𝑡) = 𝑡 − 𝑇 + 𝑇 𝑒−𝑡𝑇 𝑓𝑜𝑟 𝑡 ≥ 0

For the unit-step input, which is the derivative of unit-ramp input, the output 𝑐(𝑡) is

𝑐(𝑡) = 1 − 𝑒−𝑡𝑇 𝑓𝑜𝑟 𝑡 ≥ 0

Finally, for the unit-impulse input, which is the derivative of unit-step input, the output 𝑐(𝑡) is

𝑐(𝑡) =1

𝑇𝑒−

𝑡𝑇 𝑓𝑜𝑟 𝑡 ≥ 0

Comparing the system responses to these three inputs clearly indicates that the response to the derivative of an input signal can be obtained by differentiating the response of the system to the original signal. It can also be seen that the response to the integral of the original signal can be obtained by integrating the response of the system to the original signal and by determining the

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integration constant from the zero-output initial condition. This is a property of linear time-invariant systems. Linear time-varying systems and nonlinear systems do not possess this property.

12.2 Second-Order Systems

An example for second order system is the servo system. Servo System. The servo system shown in Figure 18a consists of a proportional controller and

load elements (inertia and viscous-friction elements). Suppose that the objective is to control the output position c in accordance with the input position r.

Figure 18 (a) Servo system; (b) block diagram; (c) simplified block diagram.

𝐽�̈� + 𝐵�̇� = 𝑇 Eq. 76

Where T is the torque produced by the proportional controller whose gain is K. By taking Laplace transforms of both sides of this last equation, assuming the zero initial conditions,

𝐽𝑠2𝐶(𝑠) + 𝐵𝑠𝐶(𝑠) = 𝑇(𝑠) Eq. 77

So the transfer function between C(s) and T(s) is 𝐶(𝑠)

𝑇(𝑠)=

1

𝑠(𝐽𝑠 + 𝐵)

Eq. 78

By using this transfer function, Figure 18(a) can be redrawn as in Figure 18(b), which can be modified to that shown in Figure 18(c).The closed-loop transfer function is then obtained as

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𝐶(𝑠)

𝑅(𝑠)=

𝐾

𝐽𝑠2 + 𝐵𝑠 + 𝐾=

𝐾 𝐽⁄

𝑠2 + (𝐵 𝐽⁄ )𝑠 + (𝐾 𝐽⁄ )

Eq. 79

Such a system where the closed-loop transfer function possesses two poles is called a second-order system. (Some second-order systems may involve one or two zeros).

12.2.1 Step Response of Second-Order System

The closed-loop transfer function of the system shown in Figure 18(c) is 𝐶(𝑠)

𝑅(𝑠)=

𝐾

𝐽𝑠2 + 𝐵𝑠 + 𝐾

Eq. 80

Which can be rewritten as

𝐶(𝑠)

𝑅(𝑠)=

𝐾𝐽

[𝑠 +𝐵2𝐽 +

√(𝐵2𝐽)

2

−𝐾𝐽 ] [𝑠 +

𝐵2𝐽 −

√(𝐵2𝐽)

2

−𝑘𝐽]

The closed-loop poles are complex conjugates if 𝐵2 − 4𝐽𝐾 < 0 and they are real if 𝐵2 − 4𝐽𝐾 ≥ 0. In the transient-response analysis, it is convenient to write

𝐾

𝐽= 𝜔𝑛

2 , 𝐵

𝐽= 2𝜁𝜔𝑛 = 2𝜎

Where 𝜎 is called the attenuation; 𝜔𝑛, the undamped natural frequency; and 𝜁, the damping ratio of the system. The damping ratio 𝜁 is the ratio of the actual damping B to the critical damping

𝐵𝑐 = 2√𝐽𝐾 or

𝜁 =𝐵

𝐵𝑐=

𝐵

2√𝐽 𝐾

In terms of 𝜁 and 𝜔𝑛, the system shown in Figure 18(c) can be modified to that shown in Figure 19,

and the closed-loop transfer function 𝐶(𝑠)

𝑅(𝑠) given by Eq. 80 can be written

Figure 19 Second-order system.

𝐶(𝑠)

𝑅(𝑠)=

𝜔𝑛2

𝑠2 + 2 𝜁 𝜔𝑛 𝑠 + 𝜔𝑛2

Eq. 81

The Eq. 81 is called the standard form of the second-order system.

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The dynamic behavior of the second-order system can then be described in terms of two

parameters 𝜻 and 𝝎𝒏. If 𝟎 < 𝜻 < 𝟏, the closed-loop poles are complex conjugates and lie in

the left-half s plane. The system is then called underdamped, and the transient response is

oscillatory. If 𝜻 = 𝟎, the transient response does not die out. If 𝜻 = 𝟏, the system is called

critically damped. Over damped systems correspond to 𝜻 > 𝟏.

𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 {

𝑛𝑜𝑡 𝑑𝑎𝑚𝑝𝑒𝑑 ( 𝑡ℎ𝑒 𝑟𝑒𝑠𝑝𝑜𝑛𝑠𝑒 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑑𝑖𝑒 𝑜𝑢𝑡) 𝜁 = 0𝑢𝑛𝑑𝑒𝑟 𝑑𝑎𝑚𝑝𝑒𝑑 1 > 𝜁 > 0critically ⅆampⅇⅆ ζ = 1Ovⅇr ⅆampⅇⅆ ζ > 1

In order to understand the system response for unit step, three different cases will be

considered: the underdamped (0 < 𝜁 < 1), critically damped (𝜁 = 1), and overdamped (𝜁 > 1) cases.

1) Underdamped case (0 < 𝜁 < 1): In this case, 𝐶(𝑠)

𝑅(𝑠) can be written

𝐶(𝑠)

𝑅(𝑠)=

𝜔𝑛2

(s + 𝜁 𝜔𝑛 + j ωd) (s + 𝜁 𝜔𝑛 − j ωd)

Eq. 82

Where 𝜔𝑑 = 𝜔𝑛√1 − 𝜁2. The frequency 𝜔𝑑 is called the damped natural frequency. For a unit-

step input 𝑅(𝑠) =1

𝑠, 𝐶(𝑠) can be written

𝐶(𝑠) =𝜔𝑛2

(𝑠2 + 2 𝜁 𝜔𝑛 𝑠 + 𝜔𝑛2)𝑠

Eq. 83

The inverse Laplace transform of Eq. 83 can be obtained easily if 𝐶(𝑠) is written in the following

form:

𝐶(𝑠) =1

𝑠−

𝑠 + 2𝜁 𝜔𝑛𝑠2 + 2𝜁 𝜔𝑛𝑠 + 𝜔𝑛2

𝐶(𝑠) =1

𝑠−

𝑠 + 𝜁 𝜔𝑛(𝑠 + 𝜁 𝜔𝑛)2 + 𝜔𝑑

2 −𝜁 𝜔𝑛

(𝑠 + 𝜁 𝜔𝑛)2 + 𝜔𝑑2

Referring to the Laplace transform table in Table 1 (see page 25), it can be shown that

ℒ−1 [𝑠 + 𝜁 𝜔𝑛

(𝑠 + 𝜁 𝜔𝑛)2 + 𝜔𝑑2] = 𝑒

−𝜁𝜔𝑛𝑡 cos(𝜔𝑑𝑡)

ℒ−1 [𝜔𝑑

(𝑠 + 𝜁 𝜔𝑛)2 + 𝜔𝑑2] = 𝑒

−𝜁𝜔𝑛𝑡 sin(𝜔𝑑𝑡)

Hence the inverse Laplace transform of Eq. 83 is obtained as

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ℒ−1[𝐶(𝑠)] = 𝑐(𝑡)

𝑐(𝑡) = 1 − 𝑒−𝜁𝜔𝑛𝑡 (cos(𝜔𝑑𝑡) +𝜁

√1 − 𝜁2sin(𝜔𝑑𝑡)) 𝑓𝑜𝑟 𝑡 ≥ 0 Eq. 84

From Eq. 84 it can be seen that the frequency of transient oscillation is the damped natural

frequency 𝜔𝑑 and thus varies with the damping ratio 𝜁. The error signal for this system is the difference between the input and output and is

𝑒(𝑡) = 𝑟(𝑡) − 𝑐(𝑡)

𝑒(𝑡) = 𝑒−𝜁𝜔𝑛𝑡 (cos(𝜔𝑑𝑡) +𝜁

√1 − 𝜁2sin(𝜔𝑑𝑡)) 𝑓𝑜𝑟 𝑡 ≥ 0

This error signal exhibits a damped sinusoidal oscillation. At steady state, or at 𝑡 = ∞, no error exists between the input and output.

If the damping ratio 𝜁 is equal to zero, the response becomes undamped and oscillations continue indefinitely. The response 𝑐(𝑡) for the zero damping case may be obtained by substituting 𝜁 = 0 in Eq. 84, yielding

𝑐(𝑡) = 1 − cos(𝜔𝑛𝑡) 𝑓𝑜𝑟 𝑡 ≥ 0 Eq. 85

Thus, from Eq. 85, it can be seen that 𝜔𝑛 represents the undamped natural frequency of the system. That is, 𝜔𝑛 is that frequency at which the system output would oscillate if the damping were decreased to zero. If the linear system has any amount of damping, the undamped natural frequency cannot be observed experimentally. The frequency that may be observed is the damped natural frequency 𝜔𝑑, which is equal to this frequency is always lower than the undamped natural frequency. An increase in 𝜁 would reduce the damped natural frequency 𝜔𝑑. If 𝜁 is increased beyond unity, the response becomes over damped and will not oscillate.

2) Critically damped case (𝜁 = 1): If the two poles of C(s)/R(s) are equal, the system is said to be

a critically damped one.

For a unit-step input, 𝑅(𝑠) =1

𝑠 and 𝐶(𝑠) can be written

𝐶(𝑠) =𝜔𝑛2

(𝑠 + 𝜔𝑛)2𝑠

Eq. 86

The inverse Laplace transform of Eq. 86 may be found as 𝑐(𝑡) = 1 − 𝑒−𝜔𝑛𝑡(1 + 𝜔𝑛𝑡) 𝑓𝑜𝑟 𝑡 ≥ 0

Eq. 87

3) Overdamped case (𝜁 > 1): In this case, the two poles of 𝐶(𝑠)

𝑅(𝑠) are negative real and unequal.

For a unit-step input, 𝑅(𝑠) =1

𝑠 and 𝐶(𝑠) can be written

𝐶(𝑠) =𝜔𝑛2

(𝑠 + 𝜁𝜔𝑛 + 𝜔𝑛√𝜁2 − 1)(𝑠 + 𝜁𝜔𝑛 − 𝜔𝑛√𝜁

2 − 1)𝑠

Eq. 88

The inverse Laplace transform of Eq. 88 is

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𝑐(𝑡) = 1 +1

2√𝜁2 − 1 (𝜁 + √𝜁2 − 1)𝑒−(𝜁+

√𝜁2−1)𝜔𝑛𝑡

−1

2√𝜁2 − 1 (𝜁 − √𝜁2 − 1)𝑒−(𝜁−

√𝜁2−1)𝜔𝑛𝑡

𝑐(𝑡) = 1 +𝜔𝑛

2√𝜁2 − 1(𝑒−𝑠1𝑡

𝑠1−𝑒−𝑠2𝑡

𝑠2)

Eq. 89

Where 𝑠1 = (𝜁 + √𝜁2 − 1)𝜔𝑛 and 𝑠2 = (𝜁 − √𝜁2 − 1)𝜔𝑛. Thus, the response 𝑐(𝑡) includes

two decaying exponential terms.

When 𝜁 is appreciably greater than unity, one of the two decaying exponentials decreases

much faster than the other, so the faster-decaying exponential term (which corresponds to a

smaller time constant) may be neglected. That is, if – 𝑠2 is located very much closer to the

𝑗𝜔 axis than – 𝑠1 (which means |𝑠2| ≪ |𝑠1|), then for an approximate solution we may neglect

– 𝑠1.This is permissible because the effect of – 𝑠1 on the response is much smaller than that of

– 𝑠2 , since the term involving 𝑠1 in Eq. 89 decays much faster than the term involving 𝑠2 . Once

the faster-decaying exponential term has disappeared, the response is similar to that of a first-

order system, and 𝐶(𝑠)

𝑅(𝑠) may be approximated by

𝐶(𝑠)

𝑅(𝑠)=

𝜁𝜔𝑛 − 𝜔𝑛√𝜁2 − 1

(𝑠 + 𝜁𝜔𝑛 − 𝜔𝑛√𝜁2 − 1)=

𝑠2𝑠 + 𝑠2

Eq. 90

This approximate form is a direct consequence of the fact that the initial values and final values

of both the original 𝐶(𝑠)

𝑅(𝑠) and the approximate one agree with each other.

With the approximate transfer function 𝐶(𝑠)

𝑅(𝑠) , the unit-step response can be obtained as

𝐶(𝑠) = 𝜁𝜔𝑛 − 𝜔𝑛√𝜁2 − 1

(𝑠 + 𝜁𝜔𝑛 − 𝜔𝑛√𝜁2 − 1)𝑠

The time response 𝑐(𝑡) is then

𝑐(𝑡) = 1 − 𝑒−(𝜁−√𝜁2−1)𝜔𝑛𝑡 𝑓𝑜𝑟 𝑡 ≥ 0

This gives an approximate unit-step response when one of the poles of 𝐶(𝑠)

𝑅(𝑠) can be neglected.

A family of unit-step response curves 𝑐(𝑡) with various values of 𝜁 is shown in Figure 20, where the abscissa is the dimensionless variable 𝜔𝑛𝑡.The curves are functions only of 𝜁. These curves are obtained from Eq. 84, Eq. 87, and Eq. 89. The system described by these equations was initially at rest.

Note that two second-order systems having the same 𝜻 but different 𝝎𝒏 will exhibit the same

overshoot and the same oscillatory pattern. Such systems are said to have the same relative

stability.

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From Figure 20, it can be seen that an underdamped system with 𝜁 between 0.5 and 0.8 gets close to the final value more rapidly than a critically damped or overdamped system. Among the systems responding without oscillation, a critically damped system exhibits the fastest response. An overdamped system is always sluggish in responding to any inputs.

Figure 20 Unit-step response curves of the system Figure 19.

It is important to note that, for second-order systems whose closed-loop transfer functions are different from that given by Eq. 81, the step-response curves may look quite different from those shown in Figure 20

12.2.2 Definitions of Transient-Response Specifications

Frequently, the performance characteristics of a control system are specified in terms of the transient response to a unit-step input, since it is easy to generate and is sufficiently drastic. (If the response to a step input is known, it is mathematically possible to compute the response to any input).

The transient response of a system to a unit-step input depends on the initial conditions. For convenience, in comparing transient responses of various systems. A common practice to use the standard initial condition that the system is at rest initially with the output and all time derivatives thereof zero. Then the response characteristics of many systems can be easily compared.

The transient response of a practical control system often exhibits damped oscillations before reaching steady state. In specifying the transient-response characteristics of a control system to a unit-step input, it is common to specify the following:

1- Delay time, 𝒕𝒅: The delay time is the time required for the response to reach half the final

value the very first time.

2- Rise time, 𝒕𝒓 : The rise time is the time required for the response to rise from 10% to 90%,

5% to 95%, or 0% to 100% of its final value. For underdamped second order systems, the

0% to 100% rise time is normally used. For overdamped systems, the 10% to 90% rise time

is commonly used.

3- Peak time, 𝒕𝒑: The peak time is the time required for the response to reach the first peak

of the overshoot.

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4- Maximum (percent) overshoot, 𝑴𝒑: The maximum overshoot is the maximum peak value

of the response curve measured from unity. If the final steady-state value of the response

differs from unity, then it is common to use the maximum percent overshoot. It is defined

by

𝑀𝑝 =

𝑐(𝑡𝑝) − 𝑐(∞)

𝑐(∞)× 100%

Eq. 91

The amount of the maximum (percent) overshoot directly indicates the relative stability of the system. 5. Settling time, 𝒕𝒔: The settling time is the time required for the response curve to reach and stay within a range about the final value of size specified by absolute percentage of the final value (usually 2% or 5%). The settling time is related to the largest time constant of the control system. Which percentage error criterion to use may be determined from the objectives of the system design in question.

Figure 21 Unit-step response curve showing td, tr , tp,Mp, and ts .

These specifications shown graphically in Figure 21. The time-domain specifications just given are quite important, since most control systems are time-domain systems; that is, they must exhibit acceptable time responses. (This means that, the control system must be modified until the transient response is satisfactory).

Note that not all these specifications necessarily apply to any given case. For example, for an overdamped system, the terms peak time and maximum overshoot do not apply. (For systems that yield steady-state errors for step inputs, this error must be kept within a specified percentage level.

A Few Comments on Transient-Response Specifications. Except for certain applications where oscillations cannot be tolerated, it is desirable that the transient response be sufficiently fast and be sufficiently damped. Thus, for a desirable transient response of a second-order system, the

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damping ratio must be between 0.4 and 0.8. Small values of 𝜁(that is, 𝜁 < 0.4) yield excessive overshoot in the transient response, and a system with a large value of 𝜁(that is, 𝜁 > 0.8) responds sluggishly. It should be notes that the maximum overshoot and the rise time conflict with each other. In other words, both the maximum overshoot and the rise time can not be made smaller simultaneously. If one of them is made smaller, the other necessarily becomes larger.

12.2.3 Second-Order Systems and Transient-Response Specifications

computation

In this section the rise time, peak time, maximum overshoot, and settling time of the second-order system given by Eq. 81 will be obtained. These values will be obtained in terms of 𝜁and 𝜔𝑛. The system is assumed to be underdamped.

Rise time 𝒕𝒓 : Referring to Eq. 84, we obtain the rise time 𝑡𝑟 by letting 𝐶(𝑡𝑟) = 1.

𝑐(𝑡) = 1 = 1 − 𝑒−𝜁𝜔𝑛𝑡𝑟 (cos(𝜔𝑑𝑡𝑟) +𝜁

√1 − 𝜁2sin(𝜔𝑑𝑡𝑟))

Eq. 92

Since 𝑒−𝜁𝜔𝑛𝑡𝑟 ≠ 0 we obtain from Eq. 92 the following equation:

cos(𝜔𝑑𝑡𝑟) +𝜁

√1 − 𝜁2sin(𝜔𝑑𝑡𝑟) = 0

Since 𝜔𝑛 √1 − 𝜁2 = 𝜔𝑑 and 𝜁𝜔𝑛 = 𝜎, hence

tan𝜔𝑑𝑡𝑟 = −√1 − 𝜁2

𝜁= −

𝜔𝑑𝜎

Thus, the rise time 𝑡𝑟 is

𝑡𝑟 =1

𝜔𝑑tan−1 (

𝜔𝑑−𝜎) =

𝜋 − 𝛽

𝜔𝑑 Eq. 93

Where angle 𝛽 is defined in Figure 22. Clearly, for a small value of 𝑡𝑟 , 𝜔𝑑 must be large.

Figure 22 Definition of the angle 𝜷.

Peak time 𝒕𝒑: Referring to Eq. 84, the peak time can be obtained by differentiating 𝑐(𝑡) with respect

to time and letting this derivative equal zero. Since 𝑑𝑐(𝑡)

𝑑𝑡= 𝜁𝜔𝑛𝑒

−𝜁𝜔𝑛𝑡 (cos(𝜔𝑑𝑡) +𝜁

√1 − 𝜁2sin(𝜔𝑑𝑡))

+ 𝑒−𝜁𝜔𝑛𝑡 (𝜔𝑑 sin(𝜔𝑑𝑡) −𝜁𝜔𝑑

√1 − 𝜁2cos(𝜔𝑑𝑡))

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And the cosine terms in this last equation cancel each other, 𝑑𝑐(𝑡)

𝑑𝑡, evaluated at 𝑡 = 𝑡𝑝, can be

simplified to 𝑑𝑐(𝑡)

𝑑𝑡|𝑡=𝑡𝑝

= (sin𝜔𝑑𝑡𝑃)𝜔𝑛

√1 − 𝜁2𝑒−𝜁𝜔𝑛𝑡𝑝 = 0

This last equation yields the following equation: sin𝜔𝑑𝑡𝑃 = 0

Or 𝜔𝑑𝑡𝑃 = 0, 𝜋, 2𝜋, 3𝜋,…

Since the peak time corresponds to the first peak overshoot, 𝜔𝑑𝑡𝑃 = 𝜋 Hence

𝑡𝑃 =𝜋

𝜔𝑑 Eq. 94

The peak time 𝑡𝑝 corresponds to one-half cycle of the frequency of damped oscillation.

Maximum overshoot 𝑴𝒑: The maximum overshoot occurs at the peak time or at 𝑡 = 𝑡𝑝 =𝜋

𝜔𝑑.

Assuming that the final value of the output is unity,𝑀𝑝 is obtained from Eq. 84 as

𝑀𝑝 = 𝑐(𝑡𝑝) − 1

𝑀𝑝 = 𝑒−𝜁𝜔𝑛(

𝜋𝜔𝑑

)(cos(𝜋) +

𝜁

√1 − 𝜁2sin(𝜋))

= 𝑒−(

𝜎𝜔𝑑

)𝜋= 𝑒

−(𝜁

√1−𝜁2)𝜋

Eq. 95

The maximum percent overshoot is 𝑒−(

𝜎

𝜔𝑑)𝜋× 100% .

If the final value 𝑐(∞) of the output is not unity, then we need to use the following equation:

𝑀𝑝 =𝑐(𝑡𝑝) − 𝑐(∞)

𝑐(∞)

Settling time 𝒕𝒔 ∶ For an underdamped second-order system, the transient response is obtained from Eq. 84 as

𝑐(𝑡) = 1 −ⅇ−𝜁𝜔𝑛𝑡

√1 − 𝜁2𝑠𝑖𝑛 (𝜔𝑑𝑡 + tan

−1√1 − 𝜁2

𝜁) 𝑓𝑜𝑟 𝑡 ≥ 0

The curves 1 ∓ (𝑒−𝜁𝜔𝑛𝑡

√1−𝜁2) are the envelope curves of the transient response to a unit-step

input.The response curve 𝑐(𝑡) always remains within a pair of the envelope curves, as shown in

Figure 23.The time constant of these envelope curves 1

𝜁𝜔𝑛.

The speed of decay of the transient response depends on the value of the time constant 1/𝜁𝜔𝑛. For a given 𝜔𝑛, the settling time 𝑡𝑠 is a function of the damping ratio 𝜁. From Figure 20, it can be seen that for the same 𝜔𝑛 and for a range of 𝜁 between 0 and 1 the settling time 𝑡𝑠 for a very lightly damped system is larger than that for a properly damped system.For an overdamped system, the settling time 𝑡𝑠 becomes large because of the sluggish response.

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The settling time corresponding to a ±2% or ±5% tolerance band may be measured in terms

of the time constant 𝑇 =1

𝜁𝜔𝑛 from the curves of Figure 20 for different values of 𝜁. The results are

shown in Figure 24. For 0 < 𝜁 < 0.9, if the 2% criterion is used, 𝑡𝑠 is approximately four times the time constant of the system. If the 5% criterion is used, then 𝑡𝑠 is approximately three times the time constant. Note that the settling time reaches a minimum value around 𝜁 = 0.76 (for the 2% criterion) or𝜁 = 0.68 (for the 5% criterion) and then increases almost linearly for large values of 𝜁. The discontinuities in the curves of Figure 24 arise because an infinitesimal change in the value of 𝜁 can cause a finite change in the settling time.

Figure 23 Pair of envelope curves for the unit step response curve of the 2nd order system

Figure 24 Settling time 𝒕𝒔 versus ζ curves.

For convenience in comparing the responses of systems, the settling time 𝑡𝑠 is commonly defined to be

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𝑡𝑆 = 4𝑇 =

4

𝜎=

4

𝜁𝜔𝑛 𝑓𝑜𝑟 2% 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑜𝑛 Eq. 96

𝑡𝑆 = 3𝑇 =

3

𝜎=

3

𝜁𝜔𝑛 𝑓𝑜𝑟 5% 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑜𝑛 Eq. 97

Note that the settling time is inversely proportional to the product of the damping ratio and the undamped natural frequency of the system. Since the value of 𝜁 is usually determined from the requirement of permissible maximum overshoot, the settling time is determined primarily by the undamped natural frequency 𝜔𝑛. This means that the duration of the transient period may be varied, without changing the maximum overshoot, by adjusting the undamped natural frequency 𝜔𝑛.

From the preceding analysis, it is evident that for rapid response 𝜔𝑛 must be large. To limit the maximum overshoot 𝑀𝑝 and to make the settling time small, the damping ratio 𝜁 should not be

too small. The relationship between the maximum percent overshoot 𝑀𝑝 and the damping ratio 𝜁

is presented in Figure 25. Note that if the damping ratio is between 0.4 and 0.7, then the maximum percent overshoot for step response is between 25% and 4%.

Figure 25 𝑴𝒑 versus 𝜻 curve

It is important to note that the equations for obtaining the rise time, peak time, maximum

overshoot, and settling time are valid only for the standard second-order system defined by

Eq. 81. If the second-order system involves a zero or two zeros, the shape of the unit-step

response curve will be quite different from those shown in Figure 20.

Example:

Consider the system shown in the following figure, where 𝜁 = 0.6 and 𝜔𝑛 = 5𝑟𝑎𝑑

𝑠𝑒𝑐. Let us obtain the

rise time 𝑡𝑟 , peak time 𝑡𝑝, maximum overshoot 𝑀𝑝, and settling time 𝑡𝑠 when the system is

subjected to a unit-step input.

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Solution :

𝜔𝑑 = 𝜔𝑛√1 − 𝜁2 = 5√1 − (0.6)2 = 4 𝜎 = 𝜁𝜔𝑛 = 0 ⋅ 6 × 5 = 3

The rising time:

𝑡𝑟 =𝜋 − 𝛽

𝜔𝑑=3.14 − 𝛽

4

Where

𝛽 = tan−1𝜔𝑑𝜎= tan−1

4

3= 0.93 𝑟𝑎𝑑

𝑡𝑟 =3.14 − 0.93

4= 0.55 𝑠𝑒𝑐

Peak time:

𝑡𝑝 =𝜋

𝜔𝑑=3.14

4= 0.785 𝑠𝑒𝑐

Maximum overshoot:

𝑀𝑝 = 𝑒−(

𝜎𝜔𝑑

)𝜋 = 𝑒−(34)×3.14 = 0.095

The maximum percent overshoot is thus 9.5%. Settling time: For the 2% criterion, the settling time is

𝑡𝑠 =4

𝜎=4

3= 1.33 𝑠𝑒𝑐

For the 5% criterion,

𝑡𝑠 =3

𝜎=3

3= 1 𝑠𝑒𝑐

12.2.4 Impulse Response of Second-Order Systems

For a unit-impulse input 𝑟(𝑡), the corresponding Laplace transform is unity, or 𝑅(𝑠) = 1.The unit-impulse response 𝐶(𝑠) of the second-order system shown in Figure 18 is

𝐶(𝑠) =𝜔𝑛2

𝑠2 + 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2

The inverse Laplace transform of this equation yields the time solution for the response 𝑐(𝑡)

as follows: For 1 > 𝜁 ≥ 0

𝐶(𝑡) =𝜔𝑛

√1 − 𝜁2𝑒−𝜁𝜔𝑛𝑡 sin𝜔𝑛√1 − 𝜁2𝑡 𝑓𝑜𝑟 𝑡 ≥ 0 Eq. 98

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For 𝜁 = 1 𝐶(𝑡) = ω𝑛

2 𝑡𝑒−𝜁𝜔𝑛𝑡 𝑓𝑜𝑟 𝑡 ≥ 0

Eq. 99

For 𝜁 > 1 𝐶(𝑡) =

𝜔𝑛

2√1 − 𝜁2𝑒−(𝜁−√1−𝜁

2)𝜔𝑛𝑡

−𝜔𝑛

2√1 − 𝜁2𝑒−(𝜁+√1−𝜁

2)𝜔𝑛𝑡 𝑓𝑜𝑟 𝑡 ≥ 0

Eq. 100

Note that without taking the inverse Laplace transform of 𝐶(𝑠) we can also obtain the time response 𝑐(𝑡) by differentiating the corresponding unit-step response, since the unit-impulse function is the time derivative of the unit-step function. A family of unit-impulse response curves

given by Eq. 98 to Eq. 100 with various values of 𝜁is shown in Figure 26. The curves 𝑐(𝑡)

𝜔𝑛 are plotted

against the dimensionless variable 𝜔𝑛 𝑡, and thus they are functions only of 𝜁. For the critically damped and overdamped cases, the unit-impulse response is always positive or zero; that is, 𝑐(𝑡) ≥0. This can be seen from Eq. 99 and Eq. 100. For the underdamped case, the unit-impulse response 𝑐(𝑡) oscillates about zero and takes both positive and negative values.

Figure 26 Unit-impulse response curves of the system