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Syllabus

0777837368

Introduction to Control Systems . Laplace transforms. Modeling of mechanical, liquid-level and electrical control systems. Block diagram Reduction and obtaining transfer function. Calculation of time response of first-order and second-order .Applying different types of controllers (P, I, D, PI.PD and PID).Analyzing system stability using Routh’s and Nyquist criteria. Drawing Bode Plot and Nyquist plot for a given transfer function. Applying root-locus technique for control system analysis

Introduction to Control Systems��Text book: K. Ogata Modern control Engineering. 5^th edition

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applications

• Robotic systems
• Missile guidance systems
• Industrial processes
• Transportations

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Liquid-level system

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plant

Feedback Element

Feedback Signal

Definitions.

• controlled variable is the quantity or condition that is measured and controlled. the controlled variable is the output of the system.
• The manipulated variable is the quantity or condition that is varied by the controller so as to affect the value of the controlled variable.

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• Control means measuring the value of the controlled variable of the system and applying the manipulated variable to the system to correct or limit deviation of the measured value from a desired value.
• Plants. A plant may be a piece of equipment, perhaps just a set of machine parts functioning together, the purpose of which is to perform a particular operation.

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• Processes. defines a process to be a natural, progressively continuing operation or development marked by a series of gradual changes that succeed one another in a relatively fixed way and lead toward a particular result or end;
• Systems. A system is a combination of components that act together and perform a certain objective.

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• Disturbances. A disturbance is a signal that tends to adversely affect the value of the output of a system. If a disturbance is generated within the system, it is called internal, while an external disturbance is generated outside the system and is an input.

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• Feedback Control Feedback control refers to an operation that, in the presence of disturbances, tends to reduce the difference between the output of a system and some reference input and that does so on the basis of this difference.

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Closed-Loop control Versus Open-Loop control

• Feedback control systems. A system that maintains a prescribed relationship between the output and the reference input by comparing them and using the difference as a means of control is called a feedback control system

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Closed-loop control systems

• Closed-loop control systems. Feedback control systems are often referred to as closed-loop control systems.
• In practice, the terms feedback control and closed-loop control are used interchangeably.
• In a closed-loop control system the actuating error signal, which is the difference between the input signal and the feedback signal , is fed to the controller so as to reduce the error and bring the output of the system to a desired value.�

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open-loop control systems

• open-loop control systems. Those systems in which the output has no effect on the control action are called open-loop control systems.
• In other words, in an open-loop control system the output is neither measured nor fed back for comparison with the input.

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• In the presence of disturbances, an open-loop control system will not perform the desired task.
• Open-loop control can be used, in practice, only if the relationship between the input and output is known and if there are neither internal nor external disturbances. �

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• Closed-loop versus open-loop control systems.

possible to use relatively inaccurate and inexpensive components to obtain the accurate control of a given plant, whereas doing so is impossible in the open-loop case.

�From the point of view of stability, the open-loop control system is easier to build because system stability is not a major problem. On the other hand, stability is a major problem in the closed-loop control system,

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• The amount or fuel admitted to the engine is adjusted according to the difference between the desired and the actual engine speeds. �The sequence of actions may be stated as follows:
• The speed governor is adjusted such that, at the desired speed, no pressured oil will flow into either side of the power cylinder.
• If the actual speed drops below the desired value due to disturbance, then the decrease in the centrifugal force of the speed governor causes the control valve to move downward, supplying more fuel, and the speed of the engine increases until the desired value is reached,
• On the other hand, if the speed of the engine increases above the desired value, then the increase in the centrifugal force of the governor causes the control valve to move upward. This decreases the supply of fuel, and the speed of the engine decreases until the desired value is reached. �

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Robot control system

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• the computerized robot recognizes the presence and orientation of each mechanical part by a pattern-recognition process that consists of reading the code numbers attached to it. Then the robot picks up the part and moves it to an appropriate place for assembling, and there it assembles several parts into a component. A well-programmed digital computer acts as a controller.

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Temperature control system

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• The temperature in the electric furnace is measured by a thermometer, which is an analog device.
• The analog temperature is converted to a digital temperature by an AID converter.
• The digital temperature is fed to a controller through an interface.
• This digital temperature is compared with the programmed input temperature, and if there is any error, the controller sends out a signal to �the heater

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UNIT 2

Laplace Transform

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• Why Laplace Transform
• Definition of Laplace Transform
• Existence of Laplace transform
• Laplace transform Table
• Laplace Transform properties
• Inverse Laplace Transform

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Why Laplace Transform�

• Solving linear differential equations

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• Obtaining transfer functions

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• Converts many functions into algebraic functions of s

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• Converts LDEs into algebraic equations of s

• Use graphical techniques for predicting a system performance with out solving DE

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Definition of Laplace Transform

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EXAMPLES

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Damped exponential function

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Laplace Transform Pairs

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INVERSE LAPLACE TRANSFORMATION

• the inverse Laplace transform can be obtained using the inversion integral

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• However, the inversion integral is complicated

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The inverse Laplace Transform can be calculated in a few ways. If the function whose inverse Laplace Transform you are trying to calculate is in the table, you are done. Otherwise we will use

partial fraction expansion (PFE); it is also called partial fraction decomposition. If you have never used partial fraction

expansions you may wish to read a background article, but you can probably continue without it.

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Inverse Laplace Transform by Partial Fraction Expansion

This technique uses Partial Fraction Expansion to split up a complicated fraction into forms that are in the Laplace Transform table.  As you read through this section, you may find it helpful to refer to the review section on partial fraction expansion techniques.  The text below assumes you are familiar with that material.

• For problems in control systems analysis, F(s), the Laplace transform of f(t), frequently occurs in the form � � �
• where A(s) and B(s) are polynomials in s. In the expansion of F(s) into a partial-fraction form, it is important that the highest power of s in A(s) be greater than the highest power of s in B(s),
• If such is not the case, the numerator B(s) must be divided by the denominator A(s) in order to produce a polynomial ins plus a remainder

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If F(s) is broken up into components,

and if the inverse Laplace transforms of F1(s),F2(s), . . . , Fn(s) are read available, then

-z₁,-z₂,…-z_m are the zeros of F(s)

-p₁,-p₂,…-p_n are the poles of F(s)

Case #1 Partial-fraction expansion when F(s) involves distinct poles only

• Consider

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Partial Fraction Expansion

There are three cases to consider in doing the partial fraction expansion of F(s).

Case 1: F(s) has all non repeated simple roots.(distinct poles)

Case 2: F(s) has complex poles:

Case 3: F(s) has repeated poles. (multiple poles)

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Cancelling common terms and set s= -P₁

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Multiplying both sides by s

Canceling the common terms and replacing s by zero

S=0

We have

S=0

And A1=1

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S=-2

A2=-2

A3=+1

S=-1

If all poles of the function F(s) are distinct except zero pole. The time function f(t) will be an exponential

Partial-Fraction Expansion

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Check.

Case #2 Partial-fraction expansion when F(s) involves complex poles

• Since the initial equations are real, complex poles will always appear as complex conjugate pole pairs
• Complex poles pairs in the partial fraction expansion of the Laplace transform

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Method of completing the square

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: Complex Conjugate Poles (Method 2)

Taking common denominators

Equating both numerators

Given:

Partial Fraction

Equating the coefficients of same power

Solving we obtain A=2 ; B=-2; C=-8

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And

Finally the inverse Laplace transform is

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If all poles of the function F(s) are complex conjugate, t except zero pole. The time function f(t) will be an damped sinusoidal

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Special Case :Imaginary Conjugate Poles (Method 2)

If all poles of the function F(s) are imaginary

conjugate except zero pole. The time function

f(t) will be sinusoidal

Case #3 Partial-fraction expansion when F(s) involves multiple poles

• Instead of discussing the general case, we shall use an example to show how to obtain the partial- fraction expansion of F(s).

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• Consider the following

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Taking common denominators and equating numerators

A=1 ; B=-1; C=-1

Equating like powers of "s" gives us:

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The inverse Laplace

If all poles of the function F(s) are multiple except zero pole. The time function f(t) will be damped ramp

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• In solving linear, time-invariant, differential equations by the Laplace transform method, two steps are involved.

�1. By taking the Laplace transform of each term in the given differential equation, convert the differential equation into an algebraic equation in s and obtain the expression for the Laplace transform of the dependent variable by rearranging the algebraic equation.

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SOLVING LINEAR DIFFERENTIAL EQUATIONS

• 2. The time solution of the differential equation is obtained by finding the inverse Laplace transform of the dependent variable. �
• Example

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UNIT#3A

Mathematical Modeling �of Dynamic Systems

.Mathematical Models

• linear and nonlinear systems.
• transfer function
• block diagrams
• Examples

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The dynamics of many systems, whether they are mechanical, electrical, thermal, economic, biological, and so on, may be described in terms of differential equations.

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Such differential equations may be obtained by using physical laws governing a particular system, for example, Newton's laws for mechanical systems and Kirchhoff's laws for electrical systems.

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deriving reasonable mathematical models is the most important part of the entire analysis of control systems.

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A mathematical model of a dynamic system is defined as a set of equations that represents the dynamics of the system accurately or, at least, fairly well

Steps in Modeling

• Understand the physical systems and its components
• Make appropriate simplifying assumptions
• Use basic principles to formulate the mathematical model
• Write differential and algebraic equations describing the model
• Check the model for validity 

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What will the model be used for

• Solution of the differential and algebraic equations allows system response and performance to analyzed and designed
• The Laplace transform will be applied to the model to allow convenient manipulations and dynamic analysis
• Input-output relationships for system and components will be obtained
• Controller models will be designed that can be implemented in hardware

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Simplicity versus accuracy

• . It is possible to improve the accuracy of a mathematical model by increasing its complexity.
• we must make a compromise between the simplicity of the model and the accuracy of the results of the analysis.
• In deriving a reasonably simplified mathematical model, we frequently find it necessary to ignore certain inherent physical properties of the system.
• In particular, it is always necessary to ignore certain nonlinearities and distributed parameters that may be present in the physical system.

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Linear systems

• A system is linear if the principle of superposition applies.
• The principle of superposition states that the response produced by th concurrent application of two different input functions is the sum of the two individual responses
• The system is linear if it is represented by linear differential equation

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Linear time-invariant systems

• A differential equation is linear if the coefficients are constants or not functions only of the independent variable.
• Dynamic systems that are composed of linear time-invariant lumped-parameter components may be described by linear time-invariant (constant- coefficient) differential equations, Such systems are called linear time-invariant (or linear constant-coefficient) systems.

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linear time-varying systems

• Systems that are represented by differential equations whose coefficients are functions of time are called linear time-varying systems. An example of a time-varying control system is a spacecraft control system. (The mass of a spacecraft changes due to fuel consumption.) ��

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Nonlinear systems

• A system is nonlinear if the principle of superposition does not apply.
• Thus, for a nonlinear system the response to two inputs cannot be calculated by treating one input at a time and adding the results. �

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Examples of nonlinear differential equations

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Examples of nonlinearities

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TRANSFER FUNCTION

• transfer functions are commonly used to characterize the input—output relationships of components or systems that can be described by linear, time-invariant, differential equations.

• The transfer function of a linear, time-invariant, differential equation system is defined as the ratio of the Laplace transform of the output (response function) to the Laplace transform of the input (driving function) under the assumption that all initial conditions are zero. �

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�Consider a system defined : ��

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where y is the output of the system and x is the input

The transfer function of this system is obtained by taking the Laplace transforms of both sides of Equation under the assumption that all initial conditions are zero, or ��

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Comments on transfer function

• The transfer function of a system is a mathematical model in that it is an operational method of expressing the differential equation that relates the output variable to the input variable. �
• The transfer function is a property of a system itself, independent of the magnitude and nature of the input or driving function.

• The transfer function includes the units necessary to relate the input to the output; however, it does not provide any information concerning the physical structure of the system. (The transfer functions of many physically different systems can be identical.) �

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• If the transfer function of a system is known, the output or response can be studied for various forms of inputs with a view toward understanding the nature of the system.

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• If the transfer function of a system is unknown, it may be established experimentally by introducing known inputs and studying the output of the system.

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• a transfer function gives a full description of the dynamic characteristics of the system, as distinct from its physical description. ��

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IMPULSE- �RESPONSE FUNCTION

• Impulse-response function. Consider the output (response) of a system to a unit impulse input when the initial conditions are zero, Since the Laplace transform of the unit-impulse function is unity, the Laplace transform of the output of the system is
• �
• The inverse Laplace transform of the output gives the impulse response of the system. The inverse Laplace transform of G(s), or �
• is called the impulse-response function. This function g(t) is also called the weighting function of the system. �

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To derive the transfer function

�1. Write the differential equation for the system.

�2. Take the Laplace transform of the differential equation, assuming all initial conditions are zero.

�3. Take the ratio of the output C(s) to the input R(s). This ratio is the transfer function. ��

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� BLOCK DIAGRAMS �

• A block diagram of a system is a pictorial representation of the functions performed by each component and of the flow of signals.
• Such diagram depicts the interrelationships that exist among the various components. Differing from a purely abstract mathematical representation,
• a block diagram has the advantage of indicating more realistically the signal flows of the actual system. �

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• In a block diagram all system variables are linked to each other through functional blocks.

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• The functional block or simply block is a symbol for the mathematical operation on the input signal to the block that produces the output.
• The transfer functions of the components are usually entered in the corresponding blocks, which are connected by arrows to indicate the direction of the flow of signals.

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�Element of a block diagram. �

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The arrowhead pointing toward the block indicates the input.

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the arrowhead leading away from the block represents the output.

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Such arrows are referred to as signals. ��

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Summing Point

The summing point is represented with a circle having cross (X) inside it. It has two or more inputs and single output. It produces the algebraic sum of the inputs. It also performs the summation or subtraction or combination of summation and subtraction of the inputs based on the polarity of the inputs.

Branch Point

• branch point (take off point) is a point from which the signal from a block goes concurrently to other blocks or summing points. ��

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Block diagram of a closed-loop system.

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B(s) the feedback signal

E(s) actuating error signal

G(S) Feed forward transfer function

G(s)H(s) Open-loop transfer function

C(S)/R(s) Closed-loop transfer function

Transfer function only used for linear time invariant systems mainly single input single output(SISO)

• Modeling in State Space

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Modern control theory versus conventional control theory.

• Modern control theory is applicable to multiple-input—multiple-output systems, which may be linear or nonlinear, time invariant or time varying,

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• conventional control theory is applicable only to linear time- invariant single-input—single-output systems.

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• modern control theory is essentially a time-domain approach, while conventional control theory is a complex frequency- domain approach.

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definitions

• State. The state of a dynamic system is the smallest set of variables (called state variables) such that the knowledge of these variables at t = to, together with the knowledge of the input for t≥ t0, completely determines the behavior of the system for any time t ≥to. �

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• State variables. The state variables of a dynamic system are the variables making up the smallest set of variables that determine the state of the dynamic system. If at least n variables x1, x2,. . . , xn, are needed to completely describe the behavior of a dynamic system (so that once the input is given for t ≥ to and the initial state at t = t0 is specified, the future state of the system is completely determined), then such n variables are a set of state variables. ��

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• State vector. If n state variables are needed to completely describe the behavior of a given system, then these n state variables can be considered the n components of a vector x. Such a vector is called a state vector.
• State space. The n-dimensional space whose coordinate axes consist of the x1 axis, x2 axis,.. . , xn, axis is called a state space.

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• State-space equations. In state-space analysis we are concerned with three types of variables that are involved in the modeling of dynamic systems:
• input variables,
• output variables,
• and state variables

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STATE-SPACE REPRESENTATION �OF DYNAMIC SYSTEMS ��

• By use of vector-matrix notation, an nth-order differential equation may be expressed by a first- order vector-matrix differential equation. If n elements of the vector are a set of state variables, then the vector-matrix differential equation is a state equation. In this section we shall present methods for obtaining state-space representations of continuous- time systems. ��

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Case #1 The input does not involve derivative terms

• Assume that a multiple-input—multiple-output system involves
• n state variables x1(t), x2(t),..., xn(t).
• Assume also that there are r inputs u1(t), u2(t), . . . , ur(t)
• and m outputs y1(t), y2(t),…, ym(t).
• Then the system may be described by �

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• then the state equations and the output equations become �

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And the linearized equations

• Where:
• A(t) the state matrix, (n*n)
• B(t) the input matrix, (n*r)
• C(t) the output matrix,(m*n)
• D(t) the direct transmission matrix.(m*r)
• A block diagram representation of These Equations

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EXAMPLE Mechanical system

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u(t) is the input to the system,

the displacement y(t) of the mass is the output.

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. This system is a single-input—single-output system.

�From the diagram, the system equation is

Let us define state variables x1 (t) and x2(t) as

x1(t)=y(t)

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Then we obtain

• Or

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block diagram for the system

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6 0 0

C=

0 0 9

0 4

D=

0 2

From the given state space model , obtain the matrices A,B,C and D

2 0

B=0 5

0 0

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Solving Differential equation by Simulation

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Correlation between transfer functions and state-space equations

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• The Laplace transform

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assume that x(0) is zero. Then we have

Or

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By substituting this Equation into the output equation, we get �

EXAMPLES�ELECTRICAL SYSTEMS

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RC Circuit

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Parameters: R,C

Independent variables (inputs): ei

dependent variables (outputs): eo, i

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Laplace Transformed equations

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RL Series Circuit

LRC circuit

• Consider the electrical circuit shown . The circuit consists of an inductance L (henry), a resistance R (ohm), and a capacitance C (farad). Applying Kirchhoff’s voltage law to the system, we obtain the following equations: �

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Transfer functions of cascaded elements

Taking forward Laplace transform

Rearrange results in

First loop equation

Transfer function

In block diagram form

G1(s)

I1(s)

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Taking forward Laplace transform and rearrange we obtain the

transfer function of the second loop

And the block diagram

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Taking forward Laplace transform and rearrange we obtain the transfer function

The block diagram

Finally ,connection all blocks together results in block diagram of the system

The transfer function of the system

Liquid-level system

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• For linear or linearized system

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Modeling Mechanical Systems

Translational System

• take the Laplace transform, assuming zero initial conditions,

• solving for the transfer
• function yields

Block Diagram

Rotational Mechanical System

• Consider the satellite attitude control system . The diagram shows the control of only the yaw angle θ.

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• Small jets apply reaction forces to rotate the satellite body into the desired position. The two skew symmetrically placed jets denoted by A or B operate in pairs.

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• Assume that each jet thrust is F/2 and a torque T = Fl is applied to the system. The jets are applied for a certain time duration and thus the torque can be written as T(t).

• The moment of inertia about the axis of rotation at the center of mass is J. �Let us obtain the transfer function of this system by assuming that torque T(t) is the input, and the angular displacement θ(t) of the satellite is the output.

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• MECHANICAL SYSTEM

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• Applying Newton’s second law to the present system and noting that there is no friction in the environment of the satellite, we have ��

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Taking the Laplace transform of both sides of this last equation and assuming all initial conditions to be zero yields ��

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• An electric circuit that is analogous to a system from another discipline is called an electric circuit analog.
• The mechanical systems with which we worked can be represented by equivalent electric circuits.
• Analogs can be obtained by comparing the equations of motion of a mechanical system, with either electrical mesh or nodal equations.
• When compared with mesh equations, the resulting electrical circuit is called a series analog.
• When compared with nodal equations, the resulting electrical circuit is called a parallel analog.

Electrical and Mechanical Systems Analogs

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RLC series circuit equation

Translational mechanical system equation

Replacing displace met x(t) by velocity v(t) = dx(t)/dt the equation of mechanical system is

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Series Analog(Force -Voltage Analog)

Equation of motion of the above translational mechanical system is;

Kirchhoff’s mesh equation for the above simple series RLC network is;

For a direct analogy b/w Eq (1) & (2), convert displacement to velocity by divide and multiply the left-hand side of Eq (1) by s, yielding;

Comparing Eqs. (2) & (3), we recognize the sum of impedances & draw the circuit shown in Figure (c). The conversions are summarized in Figure (d).

Replacing the position by the speed

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Parallel Analog(Force-Current Analog)

• Equation of motion of the above translational mechanical system is;

• Kirchhoff’s nodal equation for the simple parallel RLC network shown above is;

• Comparing Eqs. (1) & (2), we identify the sum of admittances & draw the circuit shown in Figure (c).

• The conversions are summarized in Figure 2.43(d).

(1)

(2)

Block Diagram reduction (Manipulation)�

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• We often represent control systems using block diagrams. A block diagram consists of blocks that represent transfer functions of the different variables of interest.
• If a block diagram has many blocks, not all of which are in cascade, then it is useful to have rules for rearranging the diagram such that you end up with only one block.
• Reduction of Block Diagram means replacing all blocks and summing points by a single block relating the output and the input of the system

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For example, we would want to transform the following diagram

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To the next form

Block Diagram Transformations Rules

1. Combining blocks in cascade(series connections)

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2.Parallel Connection:

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3.Feedback Connection :

Solving We obtain ,the equivalent or closed-loop

transfer function is

Negative feedback

Positive feedback

C(s)=G(s)E(s)

B(s)=H(s)C(s)

E(s)=R(s)+B(s)

Solving We obtain ,the equivalent or closed-loop

transfer function is

4. Moving summing�A.point behind a block

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4. Moving a summing point �B.ahead of a block

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5. Moving a pickoff point �A.ahead of a block

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5. Moving a pickoff point �B.behind a block

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8. replacing summing points

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C=R-A-B

C=R-A-B

10. Combining summing points

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Attention Don't use

thisMoving branch point through

summing point

Example�reduce the following block diagram

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Example

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Signal-flow graph�

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• Alternative method to block diagram representation, developed by S.J.Mason.
• Advantage: the availability of a flow graph gain formula, also called Mason’s gain formula.
• A signal-flow graph consists of a network in which nodes are connected by directed branches.
• It depicts the flow of signals from one point of a system to another and gives the relationships among the signals. Note that the signal flows in only one direction.

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Definitions

• Node - a point representing a signal or variable.
• Branch – unidirectional line segment joining two nodes.
• Path – a branch or a continuous sequence of branches that can be traversed from one node to another node.
• Loop – a closed path that originates and terminates on the same node and along the path no node is met twice.
• Nontouching loops – two loops are said to be nontouching if they do not have a common node.

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Signal-flow graph of two algebraic equations

Signal-flow graph of interconnected system

Corresponding block diagram

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Mason’s gain formula

The linear dependence T_ij between the independent variable x_i (also called the input variable) and a dependent variable x_j is

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1.Calculate forward path transfer function Pk for each forward path k.

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2.Calculate all loop TF’s.

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3.Consider nontouching loops 2 at a time.

Loops L1 and L2 do not touch Loops L3 and L4

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4. Consider nontouching loops 3 at a time.

None.

5. Calculate Δ from steps 2,3,4.

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• Calculate Δ_k as portion of Δ not touching forward path k.

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The TF of the system is

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الى هنا مادة امتحان منتصف الفصل

UNIT FOUR �Time response of Control Systems

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Transient response and steady-state response

• The time response of a control system consists of two parts:
• transient response means that which goes from the initial state to the final state.(free or natural response)
• steady-state response, we mean the manner in which the system output behaves as t approaches infinity.(forced response) �

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typical input signals

• typical input signals to use for analyzing system characteristics may he 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.
• 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.

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• 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 all systems on the same basis. �

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FIRST-ORDER SYSTEMS

• Consider the first-order system shown . Physically, this system may represent an RC circuit, thermal system, or the like. �

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• A simplified block diagram is shown in Figure (b). The input—output relationship is given by

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In the following, we shall analyze the system responses to such inputs as the unit-step, unit-ramp, and unit-impulse functions. The initial conditions are assumed to be zero. �

Unit-step response of first-order systems

• Since the Laplace transform of the unit-step function is 1/s, substituting R(s) , we obtain ��
• Expanding C(s) into partial fractions gives �

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COMMENTS

• smaller the time constant T, 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 ��

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• The output would reach the final value at t = T if it maintained its initial speed of response

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• 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 the 2% line of the final value, or four time constants. �

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• The first order system given below.

• D.C Gain of the system is 3/5 and time constant is 1/5 seconds.

Unit-ramp time response of first-order systems

• Since the Laplace transform of the unit-ramp function is 1/s^2, we obtain the output of the system as

�

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Unit-impulse response of first-order systems

• Since the Laplace transform of the unit-impulse function is R(s)=1, we obtain the output of the system

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• Comparison of 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.
• the response to the integral of the original signal can be obtained by integrating the response of the system to the original signal
• This is a property of linear time-invariant systems.

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SECOND-ORDER SYSTEMS �

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Example: servo system

168

The closed-loop transfer function of a second order system

and the block diagram are

• The dynamic behavior of the second-order system can then be described in terms of two parameters
• undamped natural frequency
• damping ratio
• We shall now solve for the response of the system a unit-step input. We shall consider three different cases:
• the underdamped (0 < ζ < 1),
• critically damped (ζ = 1),
• overdamped (ζ> 1)
• ζ = 0, the transient response does not die out. ��

169

Underdamped case (0 < ζ < I):

170

un-damped natural frequency of the second order system, which is the frequency of oscillation of the system without damping.

damping ratio of the second order system, which is a measure of the degree of resistance to change in the system output.

• Two poles of the system are

For unit step input

-

171

Φ

Φ

If (0 < ζ < I) the poles of the system are

complex conjugate and the time response

is underdamped

3.Critically damped case (ζ = 1):

172

2.undamped case ζ=0

the response becomes undamped and oscillations continue indefinitely. The response c(t) for the zero damping case can be obtained by substituting zeta = 0 in forgoing Equation

If (ζ =0) the poles of the system are imagenary conjugate and the time

response is undamped

If (ζ = I) the poles of the system are multiple and the time

response is criticaly damped

4.overdamped case (ζ> 1):

• In this case, the two poles are negative real and udequal. For a unit-step input

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unit-step time response

3

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Given the transfer function below, find natural frequency ω_n and damping ratio ζ.

​

Solution

​

The general form of the second-order transfer function is

​

​

​

​

​

​

Comparing the equations we will get :ω_n² = 36, or natural frequency ω_n = 6

​

2ζω_n= 4.2, substituting the value of ω_n we will get, damping ratio ζ = 0.35.

Time domain specifications of control systems

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Time domain specifications

180

• Peak time Referring to the same output, we may obtain the peak time by differentiating c(t) with respect to time and letting this derivative equal zero. Since

​

​

181

and the cosine terms in this last equation cancel each other, dc/dt, evaluated

at t = tp, can be simplified to

This last equation yields the following equation:

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Since the peak time corresponds to the first peak overshoot,

.Maximum overshoot

The maximum overshoot occurs at the peak time or at t = tp . Thus, Mp is obtained as

​

���

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Example: Servo system with velocity feedback

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For the system, determine the values of gain K and velocity feedback constant Kh so that the maximum overshoot in the unit-step response is 0.2 and the peak time is 1 sec. With these values of K and Kh, obtain the rise time and settling time. Assume that J 1 kg-m2 and B = 1 N-m/rad/sec. �

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Higher order systems & Stability

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189

• Higher order systems

190

Where

Unit step response of higher

order system

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191

-σ

If

2.

3. There are no zeros nearby the complex conjugate poles

1, There are two complex conjugate poles

The transfer function is

Reduction The Order Of Higher Order System

The order of higher system can be reduced to a second order system if it has dominant closed-loop poles

• Definition of stability. A linear system is said to be stable if the output response is bounded for all bounded inputs. Otherwise, it is said to be unstable.
• For negative pole F(s)= 1/(s+a) f(t)= e^-at stable term
• For positive pole F(s)= 1/(s-a) f(t)= e^+at unstable term
• Instead of poles we use roots of characteristic equation
• stable

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1 critically stable

2.Stable

3.unstable

Routh’s Stability Criterion

• A necessary condition for stability of the system is that all of the roots of its characteristic equation have negative real parts, which in turn requires that all the coefficients be positive.

​

193

A necessary (but not sufficient) condition for stability

is that all the coefficients of the polynomial

characteristic equation be positive.

​

A necessary and sufficient condition �for Stability

• Routh’s formulation requires the computation of a triangular array that is a function of the coefficients of the polynomial characteristic equation.

​

​

194

​

A system is stable if and only if all the elements of

the first column of the Routh array are positive

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Characteristic Equation

• Consider an nth-order system whose the characteristic equation (which is also the denominator of the transfer function) is:

​

195

Method for determining the Routh array

• Consider the characteristic equation:

​

​

• First arrange the coefficients of the characteristic equation in two rows, beginning with the first and second coefficients and followed by the even-numbered and odd-numbered coefficients:

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Routh array: method (cont’d)

• Then add subsequent rows to complete the Routh array:

​

​

​

​

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?!

Routh array: method (cont’d)

• Compute elements for the 3^rd row:

​

​

​

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Routh array: method (cont’d)

• Compute elements for the 4^th row:

​

​

​

​

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Example 1:

Given the characteristic equation,

​

is the system described by this characteristic equation stable?

​

Answer:

• One coefficient (-2) is negative.

​

• Therefore, the system does not satisfy the necessary condition for stability.

200

Example 2:

Given the characteristic equation,

​

is the system described by this characteristic equation stable?

​

Answer:

• All the coefficients are positive and nonzero.
• Therefore, the system satisfies the necessary condition for stability.
• We should determine whether any of the coefficients of the first column of the Routh array are negative.

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Example 2 (cont’d): Routh array

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Example 2 (cont’d): Routh array

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Example 2 (cont’d): Routh array

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Example 2 (cont’d): Routh array

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The elements of

the 1^st column

are not all positive:

​

the system is unstable

Example 3: Stability versus Parameter Range

Consider a feedback system such as:

​

​

​

​

​

The stability properties of this system are a function of the proportional feedback gain K. Determine the range of K over which the system is stable.

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Example 3 (cont’d)

​

​

​

​

​

​

• The characteristic equation for the system is given by:

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Example 3 (cont’d)

​

​

​

​

• Expressing the characteristic equation in polynomial form, we obtain:

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Example 3 (cont’d)

• The corresponding Routh array is:

• Therefore, the system is stable if and only if

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Example 4: Stability versus Two Parameter Range

Consider a Proportional-Integral (PI) control such as:

​

​

​

​

​

​

Find the range of the controller gains so that the PI feedback system is stable.

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Example 4 (cont’d)

​

​

​

​

​

​

• The characteristic equation for the system is given by:

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Example 4 (cont’d)

​

​

​

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• Expressing the characteristic equation in polynomial form, we obtain:

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Example 4 (cont’d)

• The corresponding Routh array is:

• For stability, we must have:

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• Zero 1^st column element
• Replace element by ε > 0
• Proceed as usual
• Example

⇒ 2 roots in RHP

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Basic control actions

217

• A controller compares the actual value of

output with the reference input, determines the deviation, and produces a control signal that will reduce the deviation to zero or to a small value.

• The manner in which the controller produces the control signal is called the control action.

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Open loop system

Closed loop uncontrolled system

​

Closed loop controlled system

block diagram of an industrial control system

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Classifications of Industrial Controllers. �

​

• 1. Two-position or on-off controllers
• 2. Proportional controllers
• 3. Integral controllers
• 4. Proportional-plus-integral controllers
• 5. Proportional-plus-derivative controllers
• 6. Proportional-plus-integral-plus-derivative controllers

221

First order systems: Proportional control�

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For open loop system

For uncontrolled closed loop system

For closed loop system

Proportional control of first order system reduces the steady state error and the time constant both advantages

The proportional controller is defined as : m(t)=K e(t)

Or M(s)/E(s)=K

first Order System Integral Control

223

Integral control of the system eliminates the steady-state error in the response to the step input and rise the order of the system by one . This

Oscillates the output

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Time response of first order system:1. open loop system, 2.uncontrolled

Closed loop system, 3. proportional control, 4.integral control

*Proportional controller reduce the steady state error and the time constant

*Integral controller eliminates the steady state error , rise the order of

The system by one and increase the oscillations

Second Order System�Proportional Control

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For open loop

For uncontrolled

For controlled

Second Order System�Integral Control

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For controlled system

The controlled system is stable if

227

1. Uncontrolled system 2. proportional control 3. open loop system
2. 4. Integral control

Time response of second order system

*Proportional controller reduces the steady state error and the damping ratio

*Integral controller eliminates the steady state error , rise the order of

The system by one and the system may be unstable

Response to Torque Disturbances (Proportional Control)

228

Assume that R=0, then

And

The proportional controller reduces the steady state error caused By disturbance

T(s)/E(s)= Kp+Ki/s=Kp(1+1/Tis), Ti =Ki/Kp

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Proportional plus Integral control

Differential Control Action

​

​

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Proportional-Plus-Derivative Control

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. Thus derivative control introduces a damping effect. A typical response curve c ( t ) to a unit step

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Change in gain in P controller

• Increase in gain:

​

→ Upgrade both steady-

state and transient

responses

→ Reduce steady-state

error

​

→ Reduce stability! Of second orser system and the higer order system may be unstable

​

Integral Controller

• Integral of error with a constant gain
• increase the system type by 1
• eliminate steady-state error for

a unit step input

• amplify overshoot and oscillations
• Second and higher order systems may be unstable

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Change in gain for PI controller

234

• Increase in gain:

​

→ Do not upgrade steady-

state responses

→ Increase oscillations

and overshoot! But less than I controller

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Effect of change for gain PD controller

235

The PD controller is mainly used to improve the transient response of a control system. Reduce the max. peak and settling time

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Changes in gains for PID Controller

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Tuning PID controller Using trial error method

1.Set Ti to infinity and Td to zero

2.Increase Kp to reduce initial steady state error to half it's value

3. Add small Ti and tune for min. overshoot

4.Add a Td so that max. peak than 20%

• These rules are used to determine Kp, Ti and Td for PID controllers
• Firs Method: The response is obtained experimentally to a unit step input. The plant involves neither integrators nor differentiators

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Second Method

• Set Ti= inf and Td=0, increase Kp from 0 t a critical value Kcr where the output exhibits sustained oscillations.
• Use Kcr , Pcr and Table 10-2 to determine the parameters of the controller

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�Root Locus Design Method

244

Goal

Learn a specific technique which shows hochanges in one ofa system’s parameter (usually the controller gain, K)

will modify the location of the closed-loop poles in the s-domain.

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Definition

• The closed-loop poles of the negative feedback control system: are the roots of the characteristic equation:

249

The root locus is the locus of the closed-loop poles

when a specific parameter (usually gain, K)

is varied from 0 to infinity.

Root Locus Method Foundations

• The value of s in the s-plane that make the loop gain KG(s)H(s) equal to -1 are the closed-loop poles

(i.e. )

​

• KG(s)H(s) = -1 can be split into two equations by equating the magnitudes and angles of both sides of the equation.

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Angle and Magnitude Conditions

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Independent of K

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Magnitude and phase of distinct zero assume

G(s) =s+3 and s=i4 then

G(i4)=3+i4

​

Magnitude and phase of distinct pole assume

G)s)=1/(s+1) and s=i1

Then G(i1)=1/(1+i1) =0.5 –i0.5

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Magnitude and Phase

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angle and phase conditions

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Rules for Construction of Root Locus�

Follow these rules for constructing a root locus.

Rule 1 −

Starting points G(s)H(s)=1/K

0,-10

Terminating points -4, or infinity

Rule 2 − Locate the open loop poles and zeros in the ‘s’ plane.

Find the number of root locus branches.

We know that the root locus branches start at the open loop poles and end at open loop zeros. So, the number of root locus branches N is equal to the number of finite open loop poles P or the number of finite open loop zeros Z, whichever is greater.

Mathematically, we can write the number of root locus branches N as

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Example: Plot a root locus of a system given as

Rule 4 − Find the centroid and the angle of asymptotes.

If P=Z  , then all the root locus branches start at finite open loop poles and end at finite open loop zeros.

If P>Z  , then Z  number of root locus branches start at finite open loop poles and end at finite open loop zeros and P−Z number of root locus branches start at finite open loop poles and end at infinite open loop zeros.

If P<Z, then P number of root locus branches start at finite open loop poles and end at finite open loop zeros and Z−P  number of root locus branches start at infinite open loop poles and end at finite open loop zeros.

So, some of the root locus branches approach infinity, when P≠Z  .

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Asymptotes give the direction of these root locus branches. The intersection point of asymptotes on the real axis is known as centroid.

We can calculate the centroid α by using this formula,

​

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The formula for the angle of asymptotes θ is

Where

Determine the n - m asymptotes:

• Locate s = α on the real axis:

​

​

​

• Compute and draw angles:

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​

​

​

​

​

• Draw the asymptotes using dash lines.

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Rule 5 − Find the intersection points of root locus branches with an imaginary axis.

We can calculate the point at which the root locus branch intersects the imaginary axis and the value of K at that point by using the Routh array method and special case (ii).

If all elements of any row of the Routh array are zero, then the root locus branch intersects the imaginary axis and vice-versa.

Identify the row in such a way that if we make the first element as zero, then the elements of the entire row are zero. Find the value of K for this combination.

Substitute this K value in the auxiliary equation. You will get the intersection point of the root locus branch with an imaginary axis.

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Routh table

1. 2
3. K

(6-K)/3 0

K 0

Critical gain K cr=6

Auxiliary equation

• Rule 6 − Find Break-away and Break-in points.
• If there exists a real axis root locus branch between two open loop poles, then there will be a break-away point in between these two open loop poles.
• If there exists a real axis root locus branch between two open loop zeros, then there will be a break-in point in between these two open loop zeros.

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265

Find the breakpoints.

​

• Express K such as:

​

​

​

​

• Set dK/ds = 0 and solve for the roots.

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Rule 7 :Angles of departure and arrival

*If there are starting points at complex conjugate

poles there will be departure angels

• If there are terminating points at complex

conjugate zeros there will be arrival angels

​

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Frequency Response Analysis

270

What is frequency response

So far we have described the response and performance of a system in terms of complex frequency variable s=σ+jω and the location of poles and zeros in the s-plane. An important alternative approach to system analysis and design is the frequency response method.

​

The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input signal with constant amplitude and varying frequency . We will investigate the steady-state response of the system to the sinusoidal input as the frequency varies.

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When the input signal is a sinusoid, the resulting output signal for LTI systems is sinusoidal in the steady state, it differs from the input only in amplitude and phase.

where p₁, p₂,…,p_n are distinctive poles,

then in partial fraction expansion form, we have

Taking the inverse Laplace transform yields

Suppose the system is stable, then all the poles are located in the left half plane and thus the exponential terms decay to zero as t→∞. Hence, the steady-state response of the system is

273

That is, the steady-state response depends only on the magnitude and phase of T(jω).

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Plot the frequency response of the following system

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• Bode Plot

​

​

• The Bode plot is a most useful technique for hand plotting was developed by H.W. Bode at Bell Laboratories between 1932 and 1942.
• This technique allows plotting that is quick and yet sufficiently accurate for control systems design.
• The idea in Bode’s method is to plot magnitude curves using a logarithmic scale and phase curves using a linear scale.

​

The Bode plot consists of two graphs:

​

i. A logarithmic plot of the magnitude of a transfer function.

ii. A plot of the phase angle.

​

• Both are plotted against the frequency on a logarithmic scale.
• The standard representation of the logarithmic magnitude of G(jw) is 20log|G(jw)| where the base of the logarithm is 10, and the unit is in decibel (dB).

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277

• The log-magnitude and phase frequency response curves as functions of log ω are called Bode plots or Bode diagrams.

​

• Sketching Bode plots can be simplified because they can be approximated as a sequence of straight lines.

​

• Straight-line approximations simplify the evaluation of the magnitude and phase frequency response.

​

• We call the straight-line approximations as asymptotes.

​

• The low-frequency approximation is called the low-frequency asymptote, and the high-frequency approximation is called the high-frequency asymptote.

278

Classes of Factors of Transfer Functions

• Basic factors of G(jw)H(jw) that frequently occur in an arbitrarily transfer function are

279

Class-I: The Constant Gain Factor (K)

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The slope intersects with 0 dB line at frequency ω =1

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The frequency response of the function G(s) = s, is shown in the Figure.

G(s) = s has only a high-frequency asymptote, where s = jω.

​

The Bode magnitude plot is a straight line with a +20 dB/decade slope passing through 0 dB at ω = 1.

​

​

​

​

​

The Bode phase plot is equal to a constant +90^o.

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Low-Frequency Asymptote (letting frequency S 0

High-Frequency Asymptote (letting frequency s INF

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Low-Frequency

High-Frequency Asy

Corner-Frequency

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Magnitude of a real pole: The piecewise linear asymptotic Bode plot for magnitude is at 0 dB until the break frequency and then drops at 20 dB per decade as frequency increases (i.e., the slope is -20 dB/decade).

Phase of a real pole: The piecewise linear asymptotic Bode plot for phase follows the low frequency asymptote at 0° until one tenth the break frequency (0.1·ω_c) then decrease linearly to meet the high frequency asymptote at ten times the break frequency (10·ω_c). This line is shown above.  Note that there is no error at the break frequency and about 5.7° of error at 0.1·ω_c and 10·ω_c the break frequency.

290

Again, as with the case of the real pole, there are three cases:

1. At  low frequencies, ω<<ω₀, the gain is approximately 1 (or 0 dB).
2. At high frequencies, ω>>ω₀, the gain increases at 20 dB/decade and goes through the break frequency at 0 dB.
3. At the break frequency, ω=ω_c, the gain is about 3 dB.

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The phase of a single real zero also has three cases (which can be derived similarly to those for the real pole, given above):

1. At  low frequencies, ω<<ω_c, the phase is approximately zero.
2. At high frequencies, ω>>ω_c, the phase is +90°.
3. At the break frequency, ω=ω_c, the phase is +45°.

292

Key Concept: Bode Plot of Real Zero:

• The plots for a real zero are like those for the real pole but mirrored about 0dB or 0°.
• For a simple real zero the piecewise linear asymptotic Bode plot for magnitude is at 0 dB until the break frequency and then rises at +20 dB per decade (i.e., the slope is +20 dB/decade).   An n ^th order zero has a slope of +20·n dB/decade.
• The phase plot is at 0° until one tenth the break frequency and then rises linearly to +90° at ten times the break frequency.  An n^th order zero rises to +90°·n.

�

293

Example :The first example is a simple pole at 5 radians per second. .G(s)=1/(1+s/5)

�

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Key Concept: Bode Plot for Real Pole

• For a simple real pole the piecewise linear asymptotic Bode plot for magnitude is at 0 dB until the break frequency and then drops at 20 dB per decade (i.e., the slope is -20 dB/decade).   An n^th order pole has a slope of -20·n dB/decade.
• The phase plot is at 0° until one tenth the break frequency and then drops linearly to -90° at ten times the break frequency.  An n^th order pole drops to -90°·n.

�

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Example-5: Obtain the Bode plot of the system given by the transfer function;

• We convert the transfer function in the following format by substituting s = jω

​

​

​

• We call ω = 1/2 , the break point or corner frequency. So for

​

• So when ω << 1 , (i.e., for small values of ω), then G( jω ) ≈ 1

​

• Therefore taking the log magnitude of the transfer function for very small values of ω, we get

​

• Hence below the break point, the magnitude curve is approximately a constant.

​

• So when ω >> 1, (i.e., for very large values of ω), then

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If u=1,

The magnitude asymptotes intersect at the 0dB line when u=ω/ω_n=1.

The difference between the actual magnitude curve and the asymptotic approximation is a function of ζ. The max value of the frequency response occurs at the resonant frequency, ω_r.

When ζ→0, ω_r→ ω_n.

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301

The peak will have an amplitude of 1/(2·ζ)=5.00 or 14 dB.

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303

Draw a Bode diagram for

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305

Sum of all plots except the gain

306

Add the gain to obtained the total plot

307

308

Nyquist Plot or Polar Plot

• Nyquist Plots were invented by Nyquist - who worked at Bell Laboratories, the premiere technical organization in the U.S. at the time.

​

• Nyquist Plots are a way of showing frequency responses of linear systems. 

​

• There are several ways of displaying frequency response data, including Bode' plots and Nyquist plots.

​

• Bode' plots use frequency as the horizontal axis and use two separate plots to display amplitude and phase of the frequency response.

​

• Nyquist plots display both amplitude and phase angle on a single plot, using frequency as a parameter in the plot. 

​

• Nyquist plots have properties that allow you to see whether a system is stable or unstable.

309

• A Nyquist plot is a polar plot of the frequency response function of a linear system.

​

• That means a Nyquist plot is a plot of the transfer function, G(s) with s = jω.  That means you want to plot G(j ω).

​

• G(j ω) is a complex number for any angular frequency, ω, so the plot is a plot of complex numbers.

​

• The complex number, G(j ω), depends upon frequency, so frequency will be a parameter if you plot the imaginary part of G(j ω) against the real part of G(j ω).

Sketch the Polar plot of Frequency Response

To sketch the polar plot of G(jω) for the entire range of frequency ω, i.e., from 0 to infinity, there are four key points that usually need to be known:

​

1. The start of plot where ω = 0,
2. The end of plot where ω = ∞,
3. Where the plot crosses the real axis, i.e., Im(G(jω)) = 0, and
4. Where the plot crosses the imaginary axis, i.e., Re(G(jω)) = 0.

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Problem-1: Polar Plot of Integrator

Representing G(s) in the frequency response form G( jω ) by replacing s = jω:

The magnitude of G( jω ), i.e., | G( jω) |, is obtained as;

The phase of G( jω ), denoted by, φ , is obtained as;

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Problem-2: Polar Plot of First Order System

Consider a first order system

Representing G(s) in the frequency response form G( jω ) by replacing s = jω:

The magnitude of G( jω ), i.e., | G( jω) |, is obtained as;

The phase of G( jω ), denoted by, φ , is obtained as;

The start of plot where ω = 0

The end of plot where ω = ∞

The mid part of plot where ω = 1/T

312

Problem-3: Polar Plot of Second Order System

Consider a second order system

313

Representing G(s) in the frequency response form G( jω ) by replacing s = jω:

The magnitude of G( jω ), i.e., | G( jω) |, is obtained as;

The phase of G( jω ), denoted by, φ , is obtained as;

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Example :Plot the Nyquist plot for the open loop system

315

Four Important Points for Derivation the Nyquist Stability Criterion

1. The concept of mapping points;

​

• The concept of mapping contours.

​

3. The relationship between the poles of 1 + G(s)H(s) and the poles of G(s)H(s);

4. The relationship between the zeros of 1 + G(s)H(s) and the poles of the closed-loop transfer function, T(s);

​

​

316

The Concept of Mapping Points

• If we take a complex number s = x+jy on the s-plane and substitute it into a function, F(s), another complex number results. This process is called mapping.

OR

• The term mapping is defined as the substitution of a complex number into a function, F(s), to get another complex number.

​

• For example, substituting s = 4 + j3 into the function F(s) = (s² +2s + 1) yields 16+j30. We say that 4 + j3 maps into 16+j30 through the function s² +2s + 1.

317

The Concept of Mapping Contours

The collection of points, called a contour.

318

Examples of Contour Mapping

The contour B maps in a counter clockwise direction if F(s) has just poles that are encircled by the contour, Also, you should verify that, if the pole or zero of F(s) is enclosed by contour A, the mapping encircles the origin .

319

Examples of Contour Mapping

In this last case, the pole and zero rotation cancel, and the mapping does not encircle the origin.

The number of encirclements of the origin of F(s) contour is N=Z-P;

Where Z number of zeros and P number of poles of F(s) inside s contour

320

​

1. The poles of 1 + G(s)H(s) are the same as the poles of G(s)H(s), the open-loop system.

​

• The zeros of 1 + G(s)H(s) are the same as the poles of T(s), the closed-loop system.

F(s) combine the characteristic equations of the open loop and closed loop systems

The Function F(s) for Stability

321

THE NYQUIST STABILITY CRITERION

• A linear closed-loop continuous control system is absolutely stable if the roots of the characteristic equation have negative real parts.

​

• Equivalently, the poles of the closed-loop transfer function, or the zeros of the denominator, 1 + GH(s), of the closed-loop transfer function, must lie in the left-half plane.

​

• The Nyquist Stability Criterion establishes the number of zeros of 1 + GH(s) in the right-half plane directly from the Nyquist Stability Plot of GH(s).

322

We use the Nyquist plot of the open loop system and the number of

encirclement of (-1,j0) point

Nyquist Path

323

Nyquist Stability Criterion

324

325

326

327

Nyquist or Polar Plot

Nyquist Diagram

B-The open loop system is critically stable

328

• The region to the right of the contour has been shaded.

​

• Clearly, the (-1,0) point is not in the shaded region; therefore it is not enclosed by the contour and so N ≤ 0.

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• The poles of GH(s) are at s = 0 and s = -1, neither of which are in the right-hand-plane RHP; hence P_o = 0. Thus N = -P_o = 0, and the system is absolutely stable.

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The Nyquist Stability Plot for GH(s) = 1/s(s-1) is given in the figure below.

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• The region to the right of the contour has been shaded and the (-1,0) point is enclosed; then N > 0. (It is clear that N = 1).
• The poles of GH are at s = 0 and s = +1, the latter pole being in the RHP. Hence P_o = 1.

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• N ≠ P_o indicates that the system is unstable.

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• Z = N + P_o = 1 + 1 = 2, therefore the poles of the closed-loop transfer function, or the zeros of the characteristic equation 1+GH in the right-hand plane RHP.

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Relative Stability

• Phase Crossover Frequency (ω_p) is the frequency at which the phase angle of the open-loop transfer function equals –180°.

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• Gain Crossover Frequency (ω_g) is the frequency at which the magnitude of the open loop transfer function, is unity.

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• Gain margin (K_g) is the reciprocal of the magnitude of G(jω) at the phase cross over frequency.

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Phase margin (γ) is that amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability.

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Figure 14.10 Gain and phase margins on a Nyquist plot.

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