Control System Master File PDF

CP15 SEMESTER 3 CONTROL SYSTEM – MASTER FILE SME : Surendar S Prepared by:Surendar S Verified by: Shankar MG Approved by: Shankar MG Rev No : 0 Released Date : 02/01/2020 ...

CP15 SEMESTER 3 CONTROL SYSTEM – MASTER FILE SME : Surendar S Prepared by:Surendar S Verified by: Shankar MG Approved by: Shankar MG Rev No : 0 Released Date : 02/01/2020  ` Table of Content Contents 1.0 Introduction to Control System.......................................................................................... 2 1.1 Control System................................................................................................................... 2 1.2 Transfer Function:............................................................................................................ 21 1.3 Basic Elements of Electrical based system...................................................................... 29 1.4 Mathematical modeling of a System............................................................................... 36 1.5 Control Systems – Electrical Analogies of Mechanical Systems...................................... 49 1.6 (a) Transfer Function Of An Armature Controlled Dc Motor................................................ 57 1.6 (b) Transfer Function Of A Field Controlled Dc Motor......................................................... 63 1.7 Basic Elements of Block Diagram, Basic Connections for Blocks & Block Diagram Reduction Rules............................................................................................................................................ 68 2.0 COMPONENTS OF CONTROL SYSTEM............................................................................... 78 2.1 Introduction to Control System Components....................................................................... 78 2.2 Potentiometer....................................................................................................................... 83 2.3 Controllers............................................................................................................................. 89 3.0 Analyse- Response of Control System............................................................................ 102 3.1 Time Response Analysis of systems.................................................................................... 102 3.2 Steady State Errors.............................................................................................................. 123 3.3 Stability............................................................................................................................... 129 3.4 Frequency Response Analysis............................................................................................. 136 4.0 Process Control System................................................................................................. 141 4.1 Basics of Process Control and Control methods................................................................. 141 1.4 4.2 Adaptive Control System......................................................................................... 147 4.3 Boilers................................................................................................................................. 150 4.4 Computerized Process Control........................................................................................... 160 5.0 Mechatronic Systems.................................................................................................... 169 5.1 Mechatronics – Concept Design......................................................................................... 169 5.2 Difference between conventional system & Mechatronics System.................................. 172 5.3 Application of Mechatronics in different fields.................................................................. 174 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 1  ` 1.0 Introduction to Control System 1.1 Control System In life we use many machines, tools, equipments etc… to do some work. For example: Fan to circulate air, wachine machimne to wash clothes, vehicle to move from one place to another, traffic signal light to control the traffic. In each of these, we have multiple components and parts. Each of them interact with each other and give us the desired results. System Simply put, a system is an organized collection of parts (or subsystems) that are highly integrated to accomplish an overall goal. Every system takes some inputs, then manipulate, transform or process it to give desired output. A system is arrangement or a combination of different physical components connected or related in such a manner so as to form an entire unit to attain a certain objective. Input The Stimulus or excitation applied to a control system from an external source in order to produce the output is called input. Input is something put into a system or expended in its operation to achieve output or a result. Typed text, mouse clicks, data are inputs for a computer system. Electricity and switches are inputs for a fan. Diesel/petrol and driver’s actions are the inputs for a car. Output The actual response obtained from a system is called output. The output depends on the input and the activity performed on the inputs by the system. The electrical bulb glowing, motion of car, rotating fan are examples of output of a system. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 2  ` Control We use systems to get well defined results. For example: audible sound, air circulation to make us comfortable. So, we give the inputs or configure the system to get the results that we need. This process of regulating the inputs or directing the system so that the desired objective is attained is called control. Combining above definitions Control System It is an arrangement of different physical elements connected in such a manner so as to regulate, direct or command itself to achieve a certain objective. A control system is a system, which provides the desired response by controlling the output. Here, the control system is represented by a single block. Since, the output is controlled by varying input, it is called as the control system. We will vary this input with some mechanism. Difference between System and Control System The output of a system may not be the desired output. For example, an audio amplifier, whose output is so low that we can not hear. Let us take an example to understand the difference between a system and a control system. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 3  ` A Fan: Can’t say system A Fan Without blades cannot be a “ SYSTEM”. Because it cannot provide a desired/proper output i.e airflow A Fan: Can be a system but not control system A Fan with blades but without a regulator can be a System, because it can provide a proper output i.e airflow. But it cannot be a “Control System”, because it cannot provide desired output i.e Controlled airflow A Fan: Control system A Fan with blades and a regulator is a Control System, because it provides a Desired output i.e Controlled airflow. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 4  ` Classification of Control Systems Control Systems can be classified as open loop control systems and closed loop control systems based on the feedback path. a. Open Loop Control System In open loop control systems, output is not fed-back to the input. So, the control action is independent of the desired output. A System in which the control action is totally independent of the output of the system is called as “open loop system” NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 5  ` Examples of Open Loop Control System: NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 6  ` Light Switch: Lamp glows whenever light switch is ON irrespective whether light is required or not. Volume on Stereo System: Volume is adjusted manually irrespective of output volume level. Advantages of Open Loop Control System a. Simple in construction and design. b. Less expensive. c. Easy to maintain. d. Stable operation. e. Convinent to use when the output is difficult to measure. For example: Difficult to measure speed of air circulation and then control speed of fan. Disadvantages of Open Loop Control System a. They are inaccurate. The output may change due to external and internal disturbances. For example: Inaccurate speed of fan due to fluctuation in power supply. b. They are unreliable. The output is inconsistent over time. For example, outpiut drops due to wear and tear of machine parts. c. Any change in output is not corrected automatically. Video Link - Understanding - Open-Loop Control Systems: https://www.youtube.com/watch?v=FurC2unHeXI&t=101s NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 7  ` Closed Loop Control System The goal of any electrical or electronic control system is to measure, monitor and control a process. One way in which we can accurately control the process is by monitoring its output and “feeding” some of it back to compare the actual output with the desired output so as to reduce the error and if disturbed, bring the output of the system back to the original or desired response. The quantity of the output being measured is called the “feedback signal”, and the type of control system which uses feedback signals to both control and adjust itself is called a Close-loop System. “Feedback”, simply means that some portion of the output is returned “back” to the input to form part of the system excitation. Closed-loop systems are designed to automatically achieve and maintain the desired output condition by comparing it with the actual condition. It does this by generating an error signal which is the difference between the output and the reference input. In other words, a “closed-loop system” is a fully automatic control system in which its control action being dependent on the output in some way. For example, consider our electric clothes dryer. Suppose we used a sensor or transducer (input device) to continually monitor the temperature or dryness of the clothes and feed a signal relating to the dryness back to the controller. Block Diagram of Closed Loop Control System NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 8  ` Examples of Closed Loop Control System Automatic Electric Iron – Heating elements are controlled by output temperature of the iron. The sensor reads the output temperature and compares it with the desired temperature. The difference between desired temperature and output temperature generates an error signal (voltage). This is given as input to the timer module. The timer module will accordingly set the on-off time of the heating element. Servo Voltage Stabilizer – Voltage controller operates depending upon output voltage of the system. The output voltage is tapped by the control circuit.This is compared with the desired reference voltage. Depending upon the error the servo motor will rotate in a direction to reduce the error. The servo motor moves the auto transformer tap. The voltage tapped from auto transformer is used to boost or buck the input voltage through the buck boost transformer. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 9  ` Water Level Controller – Input water is controlled by water level of the reservoir. Asensor probe is used to sense the water level in the tank. The output in the form of voltage proportional to water level is received by the probe controller. The controller operates the solenoid valve to rotate the pneumatic valve to let in / cut off the water supply. An Air Conditioner – An air conditioner functions depending upon the temperature of the room. The sensor checks the room temperature and generates an error signal by comparing it with the reference temperature/voltage. The error signal is read by the controller to switch on/off the AC accordingly. Cooling System in Car – It operates depending upon the temperature which it controls. Advantages of Closed Loop Control System a. Closed Loop Control Systems are more accurate even in the presence of non linearity. A nonlinear system is one in which the change of the output is not proportional to the change of the input. b. Highly accurate as any error arising is corrected due to presence of feedback signal. c. Facilitates automation. d. This system is less affected by noise. Noise is unwanted fluctuations in the inputs. Example: Voltage fluctuation, magnetic effect due to presence of transformer near by. e. Decision Making & Initiative Action is very fast (in msec). NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 10  ` Disadvantages of Closed Loop Control System a. They are Costlier. b. They are complicated to design. c. Required more maintenance. d. Stability is the major problem and more care is needed to design a stable closed loop system. Video Link - Understanding – Closed - Loop Control Systems: https://www.youtube.com/watch?v=5NVjIIi9fkY&t=154s Basic Elements of a Closed Loop System Any closed loop system will have well defined desired output and is called the reference value. This output is measured, and is called the controlled variable. Any closed loop system will have the following five elements in it a. Comparison element b. Control element c. Correction element d. Process element e. Measurement element NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 11  ` Let us take an example to understand each of these elements in a system. Various elements of controlling the Room Temperature: Consider a room heater used to maintain 240 centigrade temperature during a cold night. Controlled Variable – The room temperature. Reference Value – The required room temperature. We want to maintain 24 degree temperature in a room. Comparison Element- The person in the room comparing the measured value with required temperature. Error Signal- Difference between measured and required temperature. It may be positive or negative. Control Unit – The person in the room. Correction Unit- The on-off switch present on the on the heating equipment. Process – The heating of the air by the heating equipment. Measuring Device – A thermometer. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 12  ` Various elements of Water Level Controller: Controlled Variable – Water Level in Tank. Reference Value – Initial setting of the float and lever position. Comparison Element- The lever. Error signal- The difference between the actual and initial settings of the lever positions. Control Unit – The Pivoted Lever. Correction Unit- The Flap Opening or Closing the water supply. Process – The water level in the tank. Measuring Device – The floating ball and lever. Video Link - Understanding – Components of a Feedback Control System: https://youtu.be/u1pgaJHiiew NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 13  ` Difference between Open Loop and Closed Loop Control System Basis For Open Loop System Closed Loop System Comparison Definition The system whose control In closed loop, the output action is free from the depends on the control action output is known as the open of the system. loop control system. Other Name Non-feedback System Feedback System Components Controller and Controlled Amplifier, Controller, Process. Controlled Process, Feedback. Construction Simple Complex Examples Traffic light, automatic Air conditioner, temperature washing machine, control system, speed and immersion rod, TV remote pressure control system, etc. refrigerator, and toaster. Video Link - Understanding – Open Loop & Closed Loop Control System: https://www.youtube.com/watch?v=bPyh1F-_kpg Questions 1. What is meant by System? Give one example. 2. What is meant by Control System? Give one example. 3. What are the types of control system? 4. What type of control system is the traffic signal light system? 5. Traffic signal light is implememnted as a closed loop system. Identify various elements of the system. 6. Which among the following is not an advantage of an open loop system? a. Simplicity in construction & design b. Easy maintenance c. Rare problems of stability d. Requirement of system recalibration from time to time NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 14  ` 7. Which of the following is true in an open loop control system? a. Output is independent of control input b. Output is dependent on control input c. Only system parameters have effect on the control output d. Input depends on the output. 8. Which system has the tendency to oscillate? a. Open loop system b. Closed-loop System c. Both 1 & 2 d. None of the above 9. A good control system has all the following features except a. Good stability b. Slow Response c. Good Accuracy d. Sufficient power handling capacity 10. A car is rising at a constant speed of 50 km/h, which of the following is the feedback element for the driver? a. Clutch b. Eyes c. Needle of the speedometer d. Steering wheel 11. What type control system is an automatic toaster? 12. By which of the following the control action is determined when a man walks along a path ? (a) Brain (b) Hands (c) Legs (d) Eyes 13. Which of the following is a closed loop system? (a) Auto-pilot for an aircraft (6) Direct current generator (c) Car starter (d) Electric switch NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 15  ` Not for Examination (Page No 14 – 18) Laplace transform In a control system our interest is to find the relationship between input and output. We call this a transfer function. It is typically written in the frequency domain (S-domain), rather than the time domain (t-domain). In other words we analyse the response of the system for different frequencies rather thatn analysing how output varies with respect to time. The control systems have feedback. Due to this the outpiut oscillates with varying amplitude. The relationship between input and output contains complex differential equations. It is easier to solve these equations in frequency domain. The time domain function is converted into frequency domain function by multiplying the equation by e-jwt. By this multiplication the magnitude does not change. This process is called laplace transform. In Laplace Transform: s = jw Laplace transform also implies mathematically that the derivatives are converted in to complex number algebraic equation, which areeasier to solve. To evaluate the performance of an automatic control system commonly used mathematical tool is Laplace transform. Laplace transform converts the differential equation into an algebraic equation in ‘s’. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 16  ` Why Laplace Transform? In order to compute the time response of a dynamic system, it is necessary to solve the differential equations (system mathematical model) for given inputs. Laplace transform is one of the favored ways by control engineers to do this. This technique transforms the problem from the time (or t) domain to the Laplace (or s) domain. The advantage in doing this is that complex time domain differential equations become relatively simple s domain algebraic equations. When a suitable solution is arrived at, it is inverse transformed back to the time domain. The laplace transform of a time function f(t) is defined as: Where 𝑠 is complex variable 𝜎 ± 𝑗𝜔 and is called the Laplace operator. Example: Find Laplace transform for f(t) = 1 Laplace transform for f(t) = e-at NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 17  ` Derivatives: the Laplace transform of a time derivative is Where (0), 𝑓 ′ (0) are the initial conditions, or the values of Example: Find the Laplace transform of the following differential equation given: NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 18  ` NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 19  ` Table 1: Common Laplace Transform Pairs: NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 20  ` 1.2 Transfer Function: In any system we are interested in knowing how output varies when we vary the input. A transfer function models the device's output for each possible input. In its simplest form, this function is a two-dimensional graph of an independent input (voltage) versus the dependent output (voltage), called a transfer or characteristic curve. Consider the RL circuit. The input is Vin and the output is VL. 𝐿𝑠 VL(s) = Vin(s). 𝑅+𝐿𝑠 Here the output depends on the resistance of the inductance L, which depends on the frequency. We therefore us suffix(s). The transfer function, HL for the inductor is: VL(s) 𝐿𝑠 HL(s)= =. 𝑉𝑖𝑛(𝑠) 𝑅+𝐿𝑠 Notice that the voltages are frequency dependent. Therefore transfer functions are defined using Laplace transforms. The Transfer functions for components are used to design and analyze systems assembled from components, particularly using the block diagram technique, in electronics and control theory. The unit of the transfer function models the output response of the device. For example, the transfer function of an amplifier is the ratio of voltage at the output as a to the voltage applied to the input. The transfer function of an electromechanical actuator might be the mechanical displacement of the movable arm as a function of electrical current applied to the device. The transfer function of a photodetector might be the output voltage as a function of the luminous intensity of incident light of a given wavelength. A Transfer function is defined as the ratio of the Laplace Transform of the output to the Laplace transform of the input under the assumption that all the initial conditions are ZERO. Transfer function characterizes the input-output relationship of components or systems that are described by linear time-invariant differential equations. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 21  ` Linear system: Outpt is directly proposrtional to input. Time-invariant: system has the property that a certain input will always give the same output , without regard to when the input was applied to the system. Linear time-invariant controllers are very popular in the electrical, mechanical and aerospace. In linear time-invariant systems, the system response is always predictable no matter when it occurs or at what frequency it occurs. Every response is the result of the current impulse plus the sum of past impulses. 𝑪(𝒔) 𝑮(𝒔) = 𝒂𝒍𝒍 𝒕𝒉𝒆 𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏𝒔 = 𝟎 𝑹(𝑺) Feedback In Systems In any system we are always interested in the output. The output should be with in the required range, reliable and stable. So there has to be a method how to ensure output is as we require. Hence output (part of output or measure of output) should be one of the inputs to the system. A feedback control system is a system whose output is controlled using its measurement as a feedback signal. This feedback signal is compared with a reference signal to generate an error signal which is filtered by a controller to produce the system's control input. We will concentrate on continuous-time linear time-invariant (LTI) feedback systems. The Laplace transform will be our main tools for analysis and design. Need for Feedback a. To counteract disturbance signals affecting the output. For example, we would like to maintain the level of liquid in a process tank close to 2m. But an uncontrolled drain valve produces unpredictable disturbances in NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 22  ` the output flow rate. Then, we need to use a level sensor to measure the level of liquid and to control the input feed valve to counteract the effect of the disturbance created by the drain valve. b. To improve system performance in the presence of model uncertainty. Consider traffic signal light system. It uses fixed time slot for changing lights. Due certain reasons, if traffic increases more than what was expected or if timer circuit starts malfunctioning, then system will be inefficient. If traffic congestion is taken as the feedback to operate the lights, the system will work satisfactorily. c. To linearize nonlinear systems. Hi-fi audio equipment makes extensive use of feedback control to produce high-quality audio signals with low distortion in spite of its components (transistors), which have highly nonlinear characteristics. Types of Feedback There are two types of feedback systems − a. Positive feedback b. Negative feedback Positive Feedback Consider the example of number of cattle running in a village. It will lead to panic. Panic will cause more number of cattle to run. The output is contributing to the input in such a way that output further increases. Positive feedback is in phase with the input, in the sense that it adds to make the input larger. Positive feedback tends to cause system instability. When the loop gain is above 1, the growth will be exponential, increasing oscillations, chaotic behavior. System parameters will accelerate towards extreme values, which may damage or destroy the system. Positive feedback is used in digital electronics to force voltages away from intermediate voltages into '0' and '1' states. Thermal runaway is a positive feedback that can destroy semiconductor junctions. A familiar example of positive feedback NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 23  ` is the loud squealing sound produced by audio feedback in public address systems: the microphone picks up sound from its own loudspeakers, amplifies it, and sends it through the speakers again. Positive feedback does not necessarily imply instability of an equilibrium, for example stable on and off states may exist in positive-feedback architectures. The positive feedback adds the reference input, R(s) and feedback output. The following figure shows the block diagram of positive feedback control system. C(S) = [R(S) + C(S) H] G = GR(S) + G C(S) H C(S) (1 - GH) = G R(S) 𝑶𝒖𝒕𝒑𝒖𝒕 𝑪(𝑺) 𝑮 Transfer function 𝑻 = = = 𝑰𝒏𝒑𝒖𝒕 𝑹(𝑺) 𝟏−𝑮𝑯 When GH = 1, the system is unstable, so does not have a well-defined gain; the gain may be called infinite. Where,  T is the transfer function or overall gain of positive feedback control system.  G is the open loop gain, which is function of frequency.  H is the gain of feedback path, which is function of frequency The Schmitt trigger circuit uses positive feedback to ensure that if the input voltage creeps gently above the threshold, the output is forced smartly and rapidly from one logic state to the other. R-S ('reset-set') flip-flop made from two digital nor gates uses positive feedback. Red and black mean logical '1' and '0', respectively. A bistable multivibrator", is a circuit that due to high positive feedback is not stable in intermediate state. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 24  ` Negative Feedback Consider the example of our own body temperature system. When we feel cold we shiver. Shivering increases body temperature. If body temperature increases then we sweat and decrease the body temparature. The positive and negative changes are reduced by adding or subtracting the feedback. Negative feedback reduces the error between the reference input, R(s) and system output. Negative feedback is preferred in control system. The negative feedback results in better stability in steady state and rejects any disturbance signals. It also has low sensitivity to parameter variations. Hence negative feedback is preferred in closed loop control systems. The following figure shows the block diagram of the negative feedback control system. C(S) = [R(S) - C(S) H] G = GR(S) - G C(S) H C(S) (1+ GH) = G R(S) 𝑶𝒖𝒕𝒑𝒖𝒕 𝑪(𝑺) 𝑮 Transfer function 𝑻 = = = 𝑰𝒏𝒑𝒖𝒕 𝑹(𝑺) 𝟏+𝑮𝑯 Where,  T is the transfer function or overall gain of negative feedback control system.  G is the open loop gain, which is function of frequency.  H is the gain of feedback path, which is function of frequency. Thermostats - A thermostat senses the temperature and either turns the heater on or off. The negative feedback loop is used to control or decrease the excess output of the system, by subtracting some amount of signal from input. An electronic amplifier is the example of negative feedback system. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 25  ` Example: Find the closed loop gain and circuit behavior of a negative feedback loop amplifier, when the feedback resistors are R1= 1K and R2 = 10K. The feedback loop has resistors 1K and 10K connected in parallel. The feedback loop gain can be calculated as β = R1 / (R1+R2) = 0.0909 = 1 / 11 The closed loop gain of the system can be calculated as A_c = 1/β = 11 Analyzing the closed loop behavior: Assume the open loop gain as 1000, by comparing the closed loop gain with the transfer function formula , we get T = (G (s))/(1+ G(s).H(s) ) = A/(1+ A.β) = 1000/ (1 + 1000/11) = 10.88. So the actual gain of the closed loop amplifier is 10.88, but the gain is 11, in practical. When negative feedback is applied to the signal, how would an open loop gain will affect the closed loop gain of a system? Let’s say the open loop gain of the system as 5000, then finding the closed loop gain we get T = A/(1+ A.β) = 5000 / (1 + 5000/11) = 10.97. This means, if we increase in the open loop gain 400%, the closed loop gain will change 0.8%. This means the gain of a system doesn’t depends upon the temperature and other variables. It only depends on the feedback loop gain only. Video Link - Understanding - Transfer Function of System & Types of Feed Back: https://www.youtube.com/watch?v=WrVk_lT60dk https://www.youtube.com/watch?v=7aIcsFiUigQ&list=PLWPirh4EWFpGpH_Rb6 Q4iQ6vGGRA6MORZ&index=3 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 26  ` Questions 1. The output of the feedback control system must be a function of _____. a) Reference input b) Reference output c) Output and feedback signal d) Input and feedback signal 2. Which of the following is applicable when deriving the transfer function of a linear element? a) Both initial conditions and loading are taken into account b) Initial conditions are taken into account but the element is assumed to be not loaded c) Initial conditions are assumed to be zero but loading is taken into account d) Initial conditions are assumed to be zero and the element is assumed to be not loaded 3. If the initial conditions for a system are inherently zero, what does it physically mean? a) The system is at rest but stores energy b) The system is working but does not store energy c) The system is at rest or no energy is stored in any of its part d) The system is working with zero reference input 4. What is the transfer function of a system having the input as X(s) and output as Y(s) is? 5. In the circuit shown below, if current is defined as the response signal of the circuit, then determine the transfer function. a) H(s)=C/(S2 LC+RCS+1) b) H(s)=SC/(S2 LC-RCS+1) c) H(s)=SC/(S2 LC+RCS+1) ---- (Ans) d) H(s)=SC/(S2 LC+RCS-1) 6. In the circuit shown above question 5, if voltage across the capacitor is defined as the output signal of the circuit, then the transfer function is? a) H(s)=1/(S2 LC-RCS+1) b) H(s)=1/(S2 LC+RCS+1) --- (Ans) c) H(s)=1/(S2 LC+RCS-1) d) H(s)=1/(S2 LC-RCS-1) 7. Which one has tendency to oscillate, Open loop or closed loop? NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 27  ` 8. The transfer function is applicable to which of the following ? a) Linear and time-in variant systems b) Linear and time-variant systems c) Linear systems d) Non-linear systems 9. The DC gain of system represented by the transfer function 12/( s + 2 )( s + 3 ) is a. 1 b. 2 c. 6 d. 12 10. The DC gain of system represented by transfer function 12/( s + 2 )( s + 3 ) is a. 1 volt b. 2 volts c. 2.5 volts 1.5 volts 11. Why do we feedback in control systems? Explain with the help of an example. 12. Explain positive feedback with the help of an example. 13. In the bock diagram given below, identify negative and positive feedback control systems. 14. What type of feedback is used in the circuit given below? NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 28  ` 1.3 Basic Elements of Electrical based system Basic Elements of Electrical based system are a. Resistor b. Inductor c. Capacitor Resistor Consider an electrical circuit having the resistance R and the voltage applied across this circuit is e and the current flowing through resistor is i. The resistor has resistance R and the unit of resistance is ohm. The current flowing through the resistance is i and the voltage is e. The relationship between the voltage drop across the resistor and the current is g given by Ohm's law. Ohm’s Law: e=iR so according to Ohm's law, V is the voltage drop across the resistor R is equals to current I multiplied with the resistance R. It is the relationship between the voltage drop and the current flowing through the resistor. Inductor Inductors are spiral coil of wire that creates magnetic field when current passes through it. The current, i flowing through the coil produces a magnetic flux, NΦ around it that is proportional to this flow of electrical current. Changing in magnetic field gives interesting properties known as inductance.  Inductance is the property of the coil due to which it resists any variation in the current passing through it.  The current passing through the coil generates the field about it, the magnitude of the field depends on the strength of the current. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 29  `  The variation in the magnitude of the field, due to the variation in the current, produce emf in the conductive element, this emf resists the voltage that is varying the current. V = -L(di/dt) The current, i that flows through an inductor produces a magnetic flux that is proportional to it. But unlike a Capacitor which oppose a change of voltage across their plates, an inductor opposes the rate of change of current flowing through it due to the build up of self-induced energy within its magnetic field. In other words, inductors resist or oppose changes of current but will easily pass a steady state DC current. Consider the electrical circuit having the having the inductance L and i is the current flowing through it and e is the voltage applied across its terminals, So if we write the voltage drop across the inductance, 𝑑𝑖 𝑒=𝐿 𝑑𝑡 Relationship in terms of i 1 𝑖 = ∫ 𝑒. 𝑑𝑡 𝐿 So it is the relationship between the current and the voltage drop across the inductance and in this way we can represent an inductor in an electrical system. Example: A steady state direct current of 4 ampere passes through a solenoid coil of 0.5H. What would be the back emf voltage induced in the coil if the switch in the above circuit was opened for 10mS and the current flowing through the coil dropped to zero ampere. The current of 4 amps drops to 0 in 10ms. Hence di = 4 amps and dt = 10ms So with a decreasing current the voltage polarity will be acting as a source and with an increasing current the voltage polarity will be acting as a load. So for the same rate of current change through the coil, either increasing or decreasing the magnitude of the induced emf will be the same. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 30  ` Capacitor Capacitors are passive device that can store an electrical charge on their plates when connected to a voltage source. A capacitor consists of two or more parallel conductive plates which are not touching each other, but are electrically separated either by air or by some form of a insulating material such as waxed paper, mica, ceramic, plastic or a liquid gel. The insulating layer between a capacitors plates is called the Dielectric. When used in a DC circuit, a capacitor charges up to its supply voltage but blocks the flow of current through it because the dielectric of a capacitor is non- conductive and basically an insulator. However, when a capacitor is connected to an AC circuit, the flow of the current appears to pass through the capacitor with little or no resistance. Consider the electrical circuit having the capacitance C and e is the voltage applied across its terminals and I is the current. Capacitance is defined as the ratio of the charge on the capacitance to the voltage across the capacitor terminal. Unit of capacitance is farad. Charge on a capacitor at any instant is : 𝑞 Charge 𝑐 = 𝑒 𝑞 = 𝑐𝑒 Assume that the capacitor is fully discharged and the switch connected to the capacitor has just been moved to position A. The voltage across the 100uf capacitor is zero at this point and a charging current i begins to flow charging up the capacitor until the voltage across the plates is equal to the 12v supply voltage. The charging current stops flowing and the capacitor is said to be “fully-charged”. Then, Vc = Vs = 12v. Current through a Capacitor Electrical current can not actually flow through a capacitor as it does a resistor or inductor due to the insulating properties of the dielectric material between the two plates. However, the charging and discharging of the two plates gives the effect that current is flowing. The current that flows through a capacitor is directly related to the charge on the plates as current is the rate of flow of charge with respect to time. As the NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 31  ` capacitors ability to store charge q between its plates is proportional to the applied voltage e, the relationship between the current and the voltage that is applied to the plates of a capacitor becomes: 𝑑𝑒 i= 𝑐 𝑑𝑡 1 𝑒 = ∫ 𝑖. 𝑑𝑡 𝑐 As the voltage across the plates increases (or decreases) over time, the current flowing through the capacitance deposits (or removes) charge from its plates with the amount of charge being proportional to the applied voltage. Then both the current and voltage applied to a capacitance are functions of time. 𝑑𝑞 𝑖= 𝑑𝑡 𝑞 = ∫ 𝑖𝑑𝑡 So these are the two equations, one is for the voltage across the capacitor and second is for the current across the capacitor so in this way we can represent our capacitor in an electrical system. Series combination of the resistor (R), inductor (L) and capacitor(C) Consider the electrical circuit having the series combination of resistor having the resistance R and inductor having the inductance as L and a capacitor having the capacitance as C. These three elements they are connected to a voltage supply giving the voltage e and the current flowing through the network is i(t). So, let us write the equation for electrical system. We will apply KVL, so we will get the sum of the voltage in this loop is equal to zero. 𝑑𝑖 1 𝑒 − 𝑅𝑖 − 𝐿 − ∫ 𝑖𝑑𝑡 = 0 𝑑𝑡 𝑐 These are instantaneous values. So if we shift these three terms to the right hand sides, 𝑑𝑖 1 𝑒 = 𝑅𝑖 + 𝐿 + ∫ 𝑖𝑑𝑡 𝑑𝑡 𝑐 𝑑𝑞 𝑖= 𝑑𝑡 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 32  ` 𝑑𝑞 𝑑2𝑞 𝑞 𝑒=𝑅 +𝐿 2 + 𝑑𝑡 𝑑𝑡 𝑐 Parallel combination of the resistor (R), inductor (L) and capacitor(C) Consider the electrical circuit having the parallel combination of the three elements resistor, inductor and capacitor these three elements they are connected to a current source i (t) in the circuit. The current will be divided in these three elements and we will have iR as the current flowing across the resistor R and iL is the current flowing across the inductor L and iC is the current flowing across the capacitor C. According to the Kirchhoff’s current law which states that the algebraic sum of the currents at a node it is equal to zero. 𝑖(𝑡) = 𝑖𝑅 + 𝑖𝐿 + 𝑖𝐶 𝑒 𝑖𝑅 = 𝑅 1 𝑖𝐿 = ∫ 𝑒𝑑𝑡 𝐿 𝑑𝑒 𝑖𝐶 = 𝑐 𝑑𝑡 Relationship in terms of current, 𝑒 1 𝑑𝑒 𝑖(𝑡) = + ∫ 𝑒𝑑𝑡 + 𝑐 𝑅 𝐿 𝑑𝑡 Relationship in terms of current and Flux, 𝑑∅ e= 𝑑𝑡 ∅ = 𝑓𝑙𝑢𝑥 1 𝑑∅ 1 𝑑2∅ 𝑖(𝑡) = + ∅+𝐶 𝑅 𝑑𝑡 𝐿 𝑑𝑡 2 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 33  ` Questions 1. What possible values of resultant resistance can one get by combining two resistances, one of value 2 ohm and the other 6 ohm? 2. How will you connect three resistors of 2 ohm, 3 ohm and 5 ohm respectively so as to obtain a resultant resistance of 2.5 ohm? Draw the diagram to show the arrangement. 3. How many milliamperes of current flow through a circuit with a 40 V source and 6.8 k of resistance? 4. How much resistance is required to limit the current from a 12 V battery to 3.6 mA? 5. How the inductance of an inductance calculated? 6. Which formula is used to find the capacitance C? a) Q/v b) Qv c) Q-v d) Q + v 7. In which of the following, the capacitor doesn’t allow sudden changes? a) Voltage b) Current c) Resistance d) Capacitance 8.. In which of the following, the inductor doesn’t allow sudden changes? a) Voltage b) Current c) Resistance d) Inductance 9. The voltage applied to a pure capacitor of 50*10-6 F is as shown in figure. Calculate the current for 0-1msec. a) 5A ---- (Ans) b) 1A c) -5A d) -1A 10. The voltage applied to the 212mH inductor is given by v(t)= 15e-5tv. Calculate NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 34  ` the current. a) 16.782e-10t b) 15.75e-5t c) 11.27e-10t d) 14.15e-5t --- (Ans) 11. If the voltage across a capacitor is constant, then what is the current passing through it? 12. An Inductor works as a ___________ (open/close) circuit for DC supply. 13. The circuit shown in the figure consists of resistance, capacitance and inductance in series with a 100V source when the switch is closed at t = 0. Find the equation obtained from the circuit in terms of current. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 35  ` 1.4 Mathematical modeling of a System When a microprocessor switches on a motor. How will the rotation of the motor shaft vary with time? The speed will not immediately assume the full-speed value but will only attain that speed after some time. When a hydraulic system is used to open a valve which allows water into a tank to restore the water level to required level. The water level will not immediately assume the required level but will only attain that level after some time. To understand the behaviour of systems, mathematical models are used. These mathematical models are equations which describe the relationship between the input and output of a system. They are used to forecast the behaviour of a system under specific conditions. The basis for any mathematical model is provided by the fundamental physical laws that govern the behaviour of the system. Advantages of Mathematical modeling a. Mathematics - very precise language. Helps to formulate ideas and identify underlying assumptions. b. Mathematics - concise language, well-defined rules for manipulations. c. Computers can be used to perform numerical calculations Example:  Motion of mechanical system – Newton’s law of motion  Electrical system – Kirchhoff’s law Motion of mechanical system – Newton’s law of motion Newton’s Three Laws of Motion 1. Every object in a state of uniform motion will remain in that state of motion unless an external force act on it. 2. Force equals mass times acceleration [f= Ma]. 3. For every action there is an equal and opposite reaction. Mechanical System and basic Elements Mechanical systems can be either translational or rotational. Although the fundamental relationships for both types are derived from Newton’s law, they are different enough to warrant separate consideration. Three elements typically encountered in mechanical systems are mass, the linear damper and the linear spring. The linear damper produces a force proportional to the applied velocity, and the linear spring produces a force proportional to the applied displacement. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 36  ` In the analysis of mechanical systems, three essential basic elements are mass, spring and dashpot which occurs in various ways. Mass The mass of a body though distributed, we can assume that the entire mass is concentrated at one point called the CG of the body. The ideal mass element represents a particle of mass of a body concentrated at the centre of the mass and it has inertia. In mathematical modelling of mechanical system mass is represented by the symbol: Spring The elastic deformation of a body is represented by the ideal element known as spring. It stores energy during the variation of its shape due to elastic deformation resulting from the application of the force. Spring is represented by, Damper/Dash-pot Friction exists in physical systems whenever mechanical surfaces are operated in sliding contact. Dampers are used to minimize the vibrations to improve the dynamics of the system. Damper/Dash-pot is represented by, Most of the mechanical stsyems can be modelled in terms of these three systems. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 37  ` There are two types of mechanical systems based on the type of motion. a. Translational mechanical systems b. Rotational mechanical systems Translational motion in mechanical System A mechanical system in which motion is taking place along a straight line is known as Translational motion. These systems are characterized by displacement, linear velocity and linear acceleration. Let us now subject each of the three basic elements to translator motion and see how we can model them. Mass When a force ‘F’ is applied to a mechanical body of mass M displacement takes place then it is opposed by an opposing force 𝒇𝒎 due to mass. This opposing force is proportional to the acceleration of the body. Fm α a 𝑑2𝑥  Fm = M a = M 𝑑𝑡 2 𝑑2𝑥  F = Fm = M 𝑑𝑡 2 Where,  F is the applied force  Fm is the opposing force due to mass  M is mass  a is acceleration  x is displacement Spring If a force (f) is applied on spring K, then it is opposed by an opposing force due to elasticity of spring 𝒇𝒌. This opposing force is proportional to the displacement of the spring (x). NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 38  ` The force 𝒇𝒌 is given by, F α x  Fk = K x  F = Fk = K x Where,  F is the applied force  Fk is the opposing force due to elasticity of spring (force in newton)  K is spring constant  x is displacement in meter When spring has displacement on both ends then, According to Newton’s third law of motion, F = fk Hence F = k(x1 – x2) Damper/Dash-pot Damper absorbs the velocity of a body. If a force (f) is applied on dashpot B, then it is opposed by an opposing force due to friction of the dashpot 𝒇𝒃. This opposing force is proportional to the velocity of the body. Where,  Fb is the opposing force due to friction of dashpot  B is the frictional coefficient  v is velocity  x is displacement NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 39  ` When the Dash-pot has displacement at both ends then, Mathematical modeling of a Translational System Now let us mathematically model these systems. Free Body Diagram A free body diagram (FBD) is a graphical illustration used to visualize the applied forces, moments, and resulting reactions on a body in a given condition. To obtain the mathematical model of a mechanical system, it is necessary to draw a free bogy diagram including the various forces acting on it. Procedure to draw free body diagram a. Assume the direction of displacement of the mass as positive direction. b. Find all the forces with directions acting on the mass. c. Using Newton’s law of motion, express all the forces in terms of displacement or velocity of the mass. Example Draw free body diagram for given mechanical translational system. It has all the three mechanical elements. Steps: a. Consider the mass alone b. Draw the free body with forces Fb is the opposing force due to friction of dashpot. Fk is the opposing force due to elasticity of spring (force in newton) Fm is the opposing force due to mass. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 40  ` c. Represent all the forces in terms of displacement using Newton’s Law of motion. Total force F will sum of all forces acting. Then consider the doisplacment x. An example of a system that is modeled using the based-excited mass-spring-damper is a class of motion sensors sometimes called seismic sensors. Accelerometers belong to this class of sensors. The spring and damper elements are in mechanical parallel and support the ‘seismic mass’ within the case. Video Link - Understanding - Mathematical Modelling of Mechanical Translational System: https://www.youtube.com/watch?v=D3h55h2B-KQ&t=331s NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 41  ` Rotational motion in mechanical System Rotational Mechanical System involves rotational motion (motion of an object about its own axis) Mechanical System and basic Elements In the analysis of rotational mechanical system, three essential basic elements are Moment of inertia (J) of mass, stiffness constant (k) of the spring and rotational friction coefficient (B) of dash-pot which occurs in various ways. Mass (Rotation) In translational mechanical system, mass stores kinetic energy. Similarly, in rotational mechanical system, moment of inertia stores kinetic energy. Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn that where the force is applied and how the force is applied is just as important as how much force is applied when we want to make something rotate. The rotational equivalence of mass is moment of inertia, I. It accounts for how the mass of an extended object is distributed relative to the axis of rotation. For a point mass m connected to the axis of rotation by a massless rod with length r, I = mr2. The moment of inertia of the system is assumed to be concentrated at the centre of gravity of the body. Torque is the rotational equivalence of force. So, a net torque will cause an object to rotate with an angular acceleration. Because all rotational motions have an axis of rotation, a torque must be defined about a rotational axis. A torque is a force applied to a point on an object about the axis of rotation. The angular displacement Ɵ is equivalence of displacenet. The moment of inertia is represented by (J), angular displacement (Ɵ) and torque T. 𝑑2𝑥 Comparing with F = ma=m in linear motion, for rotational motion we get 𝑑𝑡 2 𝑑2Ɵ T =TJ= J 𝑑𝑡 2  T is the applied torque  Tj is the opposing torque due to moment of inertia  J is moment of inertia  θ is angular displacement NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 42  ` Spring In translational mechanical system, spring stores potential energy. Similarly, in rotational mechanical system, torsional spring stores potential energy. The elastic deformation of a body is represented by the ideal element known as spring. It stores energy during the variation of its shape due to elastic deformation resulting from the application of the torque. If a torque is applied on torsional spring K, then it is opposed by an opposing torque due to the elasticity of torsional spring. This opposing torque is proportional to the angular displacement of the torsional spring. Assume that the moment of inertia and friction are negligible. Spring represented in rotational motion, Comparing with F = Fk = K x in linear motion, for rotational motion we get T =Tk= KƟ (Here K is torsional spring constant)  T is the applied torque  Tk is the opposing torque due to elasticity of torsional spring  K is the torsional spring constant  θ is angular displacement When the spring has angular displacement on both ends then, Damper/Dash-pot Damping occurs whenever a body moves through a fluid. Dampers are used to minimize the vibrations to improve the dynamics of the system. These are different rotary dampers can be found in rotational motion of mechanical system. If a torque is applied on dashpot B, then it is opposed by an opposing torque due to the rotational friction of the dashpot. This opposing torque is proportional to the angular velocity of the body. Assume the moment of inertia NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 43  ` and elasticity are negligible. Damper/Dash-pot is represented in rotational motion by, Tb ∝ ω 𝑑𝜃 ⇒Tb = Bω =B 𝑑𝑡 𝑑𝜃 T=Tb=B 𝑑𝑡 Where,  Tb is the opposing torque due to the rotational friction of the dashpot  B is the rotational friction coefficient  ω is the angular velocity  θ is the angular displacement When the dashpot has angular displacement on both ends then, NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 44  ` Mathematical modeling of a Rotational System Free body diagram To obtain the mathematical model of a mechanical system, it is necessary to draw a free bogy diagram including the various torques acting on it. Procedure to draw free body diagram a. Assume the direction of rotation of the mass as positive direction. b. Find all the torques with directions acting on the mass. c. Using Newton’s law of motion, express all the torques in terms of angular displacement or angular velocity of the inertia element. Example: Draw free body diagram for given mechanical rotational system. Steps: a. Consider the toque alone b. Draw the free body with torques  Tb is the opposing torque due to the rotational friction of the dashpot.  Tk is the opposing torque due to elasticity of spring.  TJ is the opposing torque due to moment of inertia. c. Represent all the torques in terms of angular displacement using Newton’s Law of motion. Video Link - Understanding - Mathematical Modelling of Mechanical Rotational System: https://www.youtube.com/watch?v=VH_YdzycRHs NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 45  ` Example of Modeling Cruise Control of a Car  Cruise control refers to the control of a velocity of a car on a road.  The input comes in the form of some kind of an accelerating force(F)  The measured output is usually the car velocity (V).  We usually measure the position of the car how much distance it has travelled; typically, from reference point.  A simple model is developed to control car velocity. So, how do we capture these dynamics? There is a certain force which generates a displacement (x) and in turn a velocity(V). Modeling of Cruise control Step: 1 what is the purpose of the model? The purpose of the model is to understand the velocity behaviour of a car, and therefore to control the velocity. Step: 2 How do we define the boundaries? We look at the car as a system to be controlled and then it travels on a highway or a road that forms a part of the environment. Step:3 What could be a typical structure here? In this system the thermal energy which comes from the engine structure is converted into kinetic energy or the velocity. Some part of the energy is lost due to friction between the road and car. So, the car would have a certain mass(M) and then since we already said that there is some energy loss, there should be some kind of a damping element or damper (B) introduced in the model, and before we write down the relevant energy NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 46  ` conservation or the force conservation loss. For simplicity, we will neglect the rotational inertia of the wheels. Step: 4 Select basic variables of interest. So, the variables of interest are the displacement x which again gets the velocity V, all via this force F. So, have this car modelled as a mass M and the friction between the car and the surface on which it is moving the road is modelled as a damper with the B. Step: 5 Mathematical description of each model elements: There are two elements here: a. One is the inertia element (accelerating force Fa) which is modelled as Fa=Ma b. The frictional force(𝐹𝑓 ) would be related to the velocity via a damping coefficient 𝐹𝑓 = 𝐵𝑣 Step: 6 Apply Relavant Physical laws: Frictional force 𝐹𝑓 is to be subtract from the input force (F) to get the accelerating force(𝐹𝑎 ) 𝑭 − 𝑭𝒇 = 𝑭𝒂 Fa=Ma 𝐹𝑓 =Bv 𝑭 − 𝑩𝒗 = 𝑴𝒂 Step: 7 Final mathematical Model: 𝑩 𝑭 𝒂+ 𝒗= 𝑴 𝑴 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 47  ` Questions 1. What are the advantages of mathametical modeling? 2. What are the three basic mechanical elements used in mathematical modeling? 3. Write mathematical representation for a mass under translatory motion. 4. The spring is modelled as F=kx, where x is the spring extension. Hiw will you model if two springs are connected in series? 5. Write mathematical representation for a spring under rotational motion. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 48  ` 1.5 Control Systems – Electrical Analogies of Mechanical Systems We can represent mechanical systems in terms of equivalent electrical systems. We camn then apply electrical laws and equations to analyse. There are two types of electrical analogies of translational mechanical systems. Those are force voltage analogy and force current analogy. There are two types of electrical analogies of Rotational mechanical systems. Those are Torque voltage analogy and Torque current analogy. Force Voltage Analogy In force voltage analogy, the mathematical equations of translational mechanical system are compared with mesh equations of the electrical system. Consider the following translational mechanical system as shown in the following figure. Consider the following electrical system as shown in the following figure. This circuit consists of a resistor, an inductor and a capacitor. All these electrical elements are connected in a series. The input voltage applied to this circuit is 𝑉 volts and the current flowing through the circuit is 𝑖 Amps. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 49  ` By comparing Equation 1 and Equation 3, we will get the analogous quantities of the translational mechanical system and electrical system. The following table shows these analogous quantities. Voltage (v) is expressed in terms of charge (q). Torque Voltage Analogy In this analogy, the mathematical equations of rotational mechanical system are compared with mesh equations of the electrical system. Rotational mechanical system is shown in the following figure. The torque balanced equation is – NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 50  ` By comparing Equation 4 and Equation 3, we will get the analogous quantities of rotational mechanical system and electrical system. The following table shows these analogous quantities. Voltage (v) is expressed in terms of charge (q). Force Current Analogy In force current analogy, the mathematical equations of the translational mechanical system are compared with the nodal equations of the electrical system. Consider the following electrical system as shown in the following figure. This circuit consists of current source, resistor, inductor and capacitor. All these electrical elements are connected in parallel. Notice the following: The RLC are in parallel. Current source is being used. The current is distributed between the three branches. Flow of current creates magnetic flux in inductor. This will induce the voltage. The voltage developed across inductor due to magnetic flux is the same as the voltage across capacitor and resistance. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 51  ` According to Faraday’s law, 𝒅∅ Voltage induced v(t) = -N where∅ is flux due to current. 𝒅𝒕 In the circuit number of turns N is constant therefore we can replace -N∅ with Ψ Minus (-) sign indicates the direction of volyage. It opposes the direction of current. 𝒅∅ 𝒅𝜳 Then v(t) = -N = 𝒅𝒕 𝒅𝒕 By comparing Equation 1 and Equation 6, we will get the analogous quantities of the translational mechanical system and electrical system. The following table shows these analogous quantities. Current (i) is expressed in terms of magnetic flux (Ψ). NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 52  ` Similarly, there is a torque current analogy for rotational mechanical systems. Let us now discuss this analogy. Torque Current Analogy In this analogy, the mathematical equations of the rotational mechanical system are compared with the nodal mesh equations of the electrical system. The torque equation we derived earlier is, By comparing Equation 4 and Equation 6, we will get the analogous quantities of rotational mechanical system and electrical system. The following table shows these analogous quantities. Current (i) is expressed in terms of magnetic flux (Ψ). NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 53  ` Gears - Transformer Analogy Gears are used to increase or decrease the RMP of shafts. In electrical engineering ransformers are used to step or step down the voltages. We can draw an analogy between these two systems. Transformer Gears Transmits electrical energy from one A rotating machine to transmit torque circuit to another through some electromagnetic induction. Changes Voltage Level Changes speed and direction of motion NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 54  ` Video Link - Understanding - Force Voltage Analogy & Force Current Analogy: https://www.youtube.com/watch?v=BwOPhQ0YL3Q&t=206s https://www.youtube.com/watch?v=SOugjtdDBL8 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 55  ` Questions 1. In force-voltage analogy, what is velocity analogous to? a. current b. charge c. inductance d. capacitance 2. What is the mathematical equation for elastic (spring) element (K) is (where, x = displacement)? 𝑑2𝑥 𝑑𝑥 𝑘 a. k 2 b. k c. kx d. 𝑑𝑡 𝑑𝑡 𝑥 3. Which of the following is the analogous quantity for mass element in force- voltage analogy? a. Resistaance b. Capacitance c. Induction d. Transformer 4. What is electrical analogous quantity for spring element (K) in force-voltage analogy? a. L b. R c. 1/c d. c 5. In force-current analogy, what is electrical analogous quantity for displacement (x)? a. current b. flux c. inductance d. capacitance 6. What is electrical analogous quantity for dash-pot in force-current analogy? a. capacitance b. flux c. inverse of inductance d. conductance (1/R) 7. In force-voltage analogy, what is electrical analogous quantity for displacement (x)? a. voltage b. flux c. inductance d. charge 8. In mathemativcal representation equation, how is a mass element (where, x = displacement) 𝑑2𝑥 𝑑𝑥 𝑀 a. M 2 b. M c. M.x d. 𝑑𝑡 𝑑𝑡 𝑥 9. In mathemativcal representation equation, how is a dosh-pot element (where, x = displacement) 𝑑2𝑥 𝑑𝑥 𝐵 a. B b. B c. B.x d. 𝑑𝑡 2 𝑑𝑡 𝑥 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 56  ` 1.6 (a) Transfer Function Of An Armature Controlled Dc Motor An electric motor converts electrical energy into mechanical energy. The basic working principle of a DC motor is: "whenever a current carrying conductor is placed in a magnetic field, it experiences a mechanical force". The direction of this force is given by Fleming's left-hand rule and its magnitude is given by F = BIL. Where, B = magnetic flux density, I = current and L = length of the conductor within the magnetic field. When armature windings are connected to a DC voltage, an electric current sets up in the winding. Magnetic field may be provided by field winding (electromagnetism) or by using permanent magnets. The Speed of the DC Motor depends on Armature Voltage (current) and Flux of the magnetic field. Speed is Directly Proportional to Armature Voltage and Inversely Proportional to Flux of the poles. Speed ∝ Armature Voltage 𝟏 Speed ∝ 𝑭𝒍𝒖𝒙 Let us assume magnetic flux to be Constant. Then the speed will be controlled by armature voltage. Armature Circuit NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 57  ` Terms & Parameters Related to Armature Circuit Related to input: 𝑽𝒂=𝑨𝒓𝒎𝒂𝒕𝒖𝒓𝒆 𝑽𝒐𝒍𝒕𝒂𝒈𝒆 𝒊𝒂=𝑨𝒓𝒎𝒂𝒕𝒖𝒓𝒆 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 Related to motor: 𝑳𝒂=𝑨𝒓𝒎𝒂𝒕𝒖𝒓𝒆 𝑰𝒏𝒅𝒖𝒄𝒕𝒂𝒏𝒄𝒆 𝑹𝒂=𝑨𝒓𝒎𝒂𝒕𝒖𝒓𝒆 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 According to fundamental laws of nature, no energy conversion is possible until there is something to oppose the conversion. In case dc motors this opposition is provided by back emf. When the armature of a motor is rotating, the conductors are also cutting the magnetic flux lines and hence according to the Faraday's law of electromagnetic induction, an emf induces in the armature conductors. The direction of this induced emf is such that it opposes the armature current. The induced emf is in opposite direction to the armature voltage. The induced emf is called back emf. If the load on a dc motor is suddenly reduced. Then torque required will be treduced. Speed of the motor will start increasing due to the excess torque. Hence, being proportional to the speed, magnitude of the back emf will also increase. With increasing back emf armature current will start decreasing. Torque being proportional to the armature current, it will also decrease until it becomes sufficient for the load. Thus, speed of the motor will regulate. Hence, presence of the back emf makes a dc motor ‘self-regulating’. Magnitude of back emf is directly proportional to speed of the motor. Back EMF 𝒆𝒃 = 𝒃𝒂𝒄𝒌 𝒆𝒎𝒇,𝑽 Back emf ∝ Angular velocity = Kb ω ω = Angular velocity 𝒓𝒂𝒅 Kb = Back emf constant 𝒔𝒆𝒄 T = Torque developed by the motor Torque ∝ Armature current = Kt ia NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 58  ` 𝐾𝑡 = Torque constant ia = Armature current Related to output Ɵ=𝑨𝒏𝒈𝒖𝒍𝒂𝒓 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕,𝒓𝒂𝒅 Connected Load 𝑱=𝒎𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒊𝒏𝒆𝒓𝒕𝒊𝒂 𝒎𝒐𝒕𝒐𝒓 & 𝒍𝒐𝒂𝒅,𝑲𝒈𝒎𝟐/𝒓𝒂𝒅 𝑩=𝑭𝒓𝒊𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒎𝒐𝒕𝒐𝒓 & 𝒍𝒐𝒂𝒅,𝑵𝒎/(𝒓𝒂𝒅) 𝒔𝒆𝒄 Transfer Function of Armature controlled DC Motor in terms of Identified Terms & Parameters In motor, Input – Electrical Signal Output – Mechanical Energy So the Transfer Function of Motor under ZERO initial condition is, 𝐋𝐚𝐩𝐥𝐚𝐜𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦 (𝐌𝐞𝐜𝐡𝐚𝐧𝐢𝐜𝐚𝐥 𝐎𝐮𝐭𝐩𝐮𝐭) Transfer function = 𝐋𝐚𝐩𝐥𝐚𝐜𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦 (𝐄𝐥𝐞𝐜𝐭𝐫𝐢𝐜𝐚𝐥 𝐈𝐧𝐩𝐮𝐭) So as per diagram, Electrical input = Armature Voltage 𝑉𝑎 Mechanical Output = Angular displacement Ɵ 𝐋𝐚𝐩𝐥𝐚𝐜𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦 (𝐀𝐧𝐠𝐮𝐥𝐚𝐫 𝐃𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭) Transfer function = 𝐋𝐚𝐩𝐥𝐚𝐜𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦 (𝐀𝐫𝐦𝐚𝐭𝐮𝐫𝐞 𝐕𝐨𝐥𝐭𝐚𝐠𝐞) NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 59  ` Derive Transfer Function of Armature Controlled DC motor Equivalent Circuit of Armature The current ia is due to Armature voltage and resistance of the armature winding. Hence it is represented using a resistance. The back emf is due to voltage induced in the armature coil hence represented as inductance. By KVL, 𝑑𝑖𝑎 𝑖𝑎 𝑅𝑎 + 𝐿𝑎 + 𝑒𝑏 = 𝑉𝑎 --------------------------------------------------- (1) 𝑑𝑡 𝑇 ∝ 𝑖𝑎 𝑇 = 𝐾𝑡 𝑖𝑎 ------------------------------------------------------- (2) 𝑑Ɵ 𝑒𝑏 ∝ 𝑑𝑡 𝑑Ɵ 𝑒𝑏 = 𝐾𝑏 ---------------------------------------------------------------------- (3) 𝑑𝑡 Mechanical Rotational System of Motor Converting a Load in to Mechanical System, so  Rotational force of motor = Torque T  The output angle of rotation =Ɵ  The Load will generate the opposing torque as per applied torque due to moment of inertia = J Opposite torque due to moment of inertial ∝ angular acceleration J = moment of inertia 𝑑2Ɵ = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑑𝑡 2 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 60  ` 𝑑2Ɵ Due to Inertia Tj = 𝐽 𝑑𝑡 2  The opposing torque due to friction or dashpot = B Opposite torque due to dashpot or friction ∝ angular velocity To remove the proportionality we multiplied with frictional coefficient B, so B=frictional coefficient 𝑑Ɵ = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑑𝑡 𝑑Ɵ Due to friction Tf = 𝐵 𝑑𝑡 So, Sum of all the Opposing torque is equal to sum of all the applied torque. 𝑑2Ɵ 𝑑Ɵ Torque applied = Tj + Tf = 𝐽 +𝐵 = T---------------------------------- (4) 𝑑𝑡 2 𝑑𝑡 Taking Laplace Transform for equation 1, 2, 3&4 respectively 𝐼𝑎 (𝑠)𝑅𝑎 + 𝐿𝑎 𝑆𝐼𝑎 (𝑠)+ 𝐸𝑏 (𝑠) = 𝑉𝑎 (s) ---------------------------------------------- (5) 𝑇(𝑠) = 𝐾𝑡 𝐼𝑎 (𝑠)--------------------------------------------------------------------- (6) 𝐸𝑏 (𝑠) = 𝐾𝑏 (𝑠)Ɵ(𝑠)-------------------------------------------------------------- (7) 𝐽𝑠 2 Ɵ(𝑠)+𝐵𝑠Ɵ(𝑠) = 𝑇(𝑠)-------------------------------------------------------- (8) Substitute equation (6) in equation (8) 𝐽𝑠 2 Ɵ(𝑠)+𝐵𝑠Ɵ(𝑠) = 𝐾𝑡 𝐼𝑎 (𝑠) 𝐽𝑠 2 𝜃(𝑠)+𝐵𝑠 𝜃(𝑠) 𝐼𝑎 (𝑠) = ------------------------------------------------------------------ (9) 𝐾𝑡 Substitute equation (7) & (9) in equation (5) 𝐽𝑆 2 𝜃(𝑠) + 𝐵𝑠𝜃(𝑠) 𝐽𝑆 2 𝜃(𝑠) + 𝐵𝑠𝜃(𝑠) 𝑅𝑎 [ ] + 𝐿𝑎𝑆 [ ] + 𝐾𝑏 𝑠𝜃(𝑠) = 𝑉𝑎 (𝑠) 𝐾𝑡 𝐾𝑡 𝐽𝑆 2 𝜃(𝑠) + 𝐵𝑠𝜃(𝑠) [ ] (𝑅𝑎 + 𝐿𝑎𝑆) + 𝐾𝑏 𝑠𝜃(𝑠) = 𝑉𝑎 (𝑠) 𝐾𝑡 𝐽𝑆 2 + 𝐵𝑠 𝜃(𝑠) [ ] (𝑅𝑎 + 𝐿𝑎𝑆) + 𝐾𝑏 𝑠 = 𝑉𝑎 (𝑠) 𝐾𝑡 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 61  ` (𝐽𝑆 2 + 𝐵𝑠)(𝑅𝑎 + 𝐿𝑎𝑆) + 𝐾𝑡 𝐾𝑏 𝑠 𝜃(𝑠) [ ] = 𝑉𝑎 (𝑠) 𝐾𝑡 𝜃(𝑠) 𝐾𝑡 = 𝑉𝑎 (𝑠) (𝐽𝑆 2 + 𝐵𝑠)(𝑅𝑎 + 𝐿𝑎𝑆) + 𝐾𝑡 𝐾𝑏 𝑠 The Transfer Function of the armature Controlled DC Motor is identified. Video Link - Understanding - How to find out the Transfer Function of Armature Controlled DC Motor?: https://www.youtube.com/watch?v=53IQg-lepQ8 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 62  ` 1.6 (b) Transfer Function Of A Field Controlled Dc Motor The Speed of the DC Motor depends on Armature Voltage and Magnetic Flux. Speed ∝ Armature Voltage 𝟏 Speed ∝ 𝑭𝒍𝒖𝒙 Field Controlled DC Motor: Speed is Inversely Proportional to Flux. 𝟏 Speed ∝ 𝑭𝒍𝒖𝒙 In this the armature voltage (𝑽𝒂 ) is Kept Constant and the field current is varied. Steps: 𝑉𝑎 − 𝐾𝑒𝑝𝑡 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 The voltage applied to the field coil is varied 𝑉𝑓 Varies, 𝑖𝑓 𝑣𝑎𝑟𝑖𝑒𝑠 𝑠𝑜 𝑓𝑙𝑢𝑥 𝑎𝑙𝑠𝑜 𝑣𝑎𝑟𝑖𝑒𝑠, 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑙𝑢𝑥 𝑣𝑎𝑟𝑖𝑒𝑠 𝐷𝐶 𝑚𝑜𝑡𝑜𝑟 𝑠𝑝𝑒𝑒𝑑 𝑎𝑙𝑠𝑜 𝑉𝑎𝑟𝑖𝑒𝑠. Terms & Parameters Related to Field Circuit Related to input 𝑽𝒇=𝒇𝒊𝒆𝒍𝒅 𝑽𝒐𝒍𝒕𝒂𝒈𝒆 , V 𝒊𝒇=𝒇𝒊𝒆𝒍𝒅 𝑪𝒖𝒓𝒓𝒆𝒏𝒕 , A 𝑹𝒇=𝒇𝒊𝒆𝒍𝒅 𝑹𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 ,Ω 𝑳𝒇=𝒇𝒊𝒆𝒍𝒅 𝑰𝒏𝒅𝒖𝒄𝒕𝒂𝒏𝒄𝒆 , H NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 63  ` 𝑻=𝑻𝒐𝒓𝒒𝒖𝒆 ,𝑵−𝒎 Torque ∝ Field current T = Kt if 𝑲𝒕=𝑻𝒐𝒓𝒒𝒖𝒆 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 ,𝑵𝒎/𝑨 Related to output Ɵ=𝑨𝒏𝒈𝒖𝒍𝒂𝒓 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕,𝒓𝒂𝒅 Related to connected Load 𝑱=𝒎𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒊𝒏𝒆𝒓𝒕𝒊𝒂 𝒎𝒐𝒕𝒐𝒓 & 𝒍𝒐𝒂𝒅,𝑲𝒈𝒎𝟐 /𝒓𝒂𝒅 𝑩 𝒓𝒂𝒅 =𝑭𝒓𝒊𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝑵−𝒎/( ) 𝒔𝒆𝒄 Transfer Function of Field controlled DC Motor in terms of Identified Terms & Parameters In motor, Input – Electrical Energy Output – Mechanical Energy So the Transfer Function of Motor under ZERO initial condition is, 𝐋𝐚𝐩𝐥𝐚𝐜𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦 (𝐌𝐞𝐜𝐡𝐚𝐧𝐢𝐜𝐚𝐥 𝐎𝐮𝐭𝐩𝐮𝐭) Transfer function = 𝐋𝐚𝐩𝐥𝐚𝐜𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦 (𝐄𝐥𝐞𝐜𝐭𝐫𝐢𝐜𝐚𝐥 𝐈𝐧𝐩𝐮𝐭) So as per diagram, Electrical input = Field Voltage 𝑉𝑓 Mechanical Output = Angular displacement Ɵ 𝐋𝐚𝐩𝐥𝐚𝐜𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦 (𝐀𝐧𝐠𝐮𝐥𝐚𝐫 𝐃𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭) Transfer function = 𝐋𝐚𝐩𝐥𝐚𝐜𝐞 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦 (𝐅𝐢𝐞𝐥𝐝 𝐕𝐨𝐥𝐭𝐚𝐠𝐞) NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 64  ` Derive the Transfer Function of Field Controlled DC motor with help of Equation The current if is due to field voltage and resistance of the field winding. Hence it is represented using a resistance. The magnetic flux is generated by the filed coil, hence represented as inductance. 𝑑𝑖𝑓 𝑖𝑓 𝑅𝑓 + 𝐿𝑓 = 𝑉𝑓 ------------------------------------------------------------------- (1) 𝑑𝑡 𝑇 ∝ 𝑖𝑓 𝑇 = 𝐾𝑡𝑓 𝑖𝑓 ---------------------------------------------------(2) 𝑑2Ɵ 𝑑Ɵ 𝐽 2 + 𝐵 = 𝑇---------------------------------------------(3) 𝑑𝑡 𝑑𝑡 𝐼𝑓 (𝑠)𝑅𝑓 + 𝐿𝑓 𝑆𝐼𝑓 (𝑠) = 𝑉𝑎 (s) ------------------------------------------------------------ (4) 𝑇(𝑠) = 𝐾𝑡𝑓 𝐼𝑓 (𝑠)---------------------------------------------------------------------- (5) 𝐽𝑠 2 Ɵ(𝑠)+𝐵𝑠Ɵ(𝑠) = 𝑇(𝑠)--------------------------------------------------------------(6) Substitute equation (5) in equation (6) 𝐽𝑠 2 Ɵ(𝑠)+𝐵𝑠Ɵ(𝑠) = 𝐾𝑡𝑓 𝐼𝑓 (𝑠) 𝐽𝑠 2 𝜃(𝑠)+𝐵𝑠 𝜃(𝑠) 𝐼𝑓 (𝑠) = --------------------------------------------------------------------- (7) 𝐾𝑡𝑓 NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 65  ` Substitute equation (7) in equation (4) 𝐽𝑆 2 𝜃(𝑠) + 𝐵𝑠𝜃(𝑠) 𝐽𝑆 2 𝜃(𝑠) + 𝐵𝑠𝜃(𝑠) 𝑅𝑓 [ ] + 𝐿𝑓𝑆 [ ] = 𝑉𝑓 (𝑠) 𝐾𝑡𝑓 𝐾𝑡𝑓 𝐽𝑆 2 𝜃(𝑠) + 𝐵𝑠𝜃(𝑠) [ ] (𝑅𝑓 + 𝐿𝑓𝑆) = 𝑉𝑓 (𝑠) 𝐾𝑡𝑓 (𝐽𝑆 2 + 𝐵𝑠)(𝑅𝑓 + 𝐿𝑓𝑆) 𝜃 (𝑠 ) [ ] = 𝑉𝑓 (𝑠) 𝐾𝑡𝑓 𝜃(𝑠) 𝐾𝑡𝑓 = 𝑉𝑓 (𝑠) 𝑆 𝐽 𝐵(1+ 𝑠 ) 𝑅 𝑆𝐿𝑓 𝐵 𝑓(1+ 𝑅𝑓 ) The Transfer Function of the Field Controlled DC Motor is identified. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 66  ` Questions 1. How can we control the speed of a DC motor. 2. The speed of a dc motor is a) Always constant b) Directly proportional to back emf c) Directly proportional to flux d) Inversely proportional to the product of back emf and flux. 3. Back emf depends on which one of the following: a. Armature speed b. Armarure current c. Armature voltage d. Armature resistance 4. What are the inputs and outputs of a Armature controlled DC motor? 5. What are the inputs and outputs of a field current controlled DC motor? 6. Draw equivalent circuit of armature. 7. Draw equivalent circuit of filed coil. 8. What do the following represent in the circuit diagram given below: Ia, Va, Te, ωm, ∅f, Ea, Ra, La NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 67  ` 1.7 Basic Elements of Block Diagram, Basic Connections for Blocks & Block Diagram Reduction Rules A block diagram is a diagram of a system in which the principal parts or functions are represented by blocks connected by lines that show the relationships of the blocks. They are used in engineering in hardware design, electronic design, software design, and process flow diagrams. Block diagrams are typically used for higher level, less detailed descriptions that are intended to clarify overall concepts without concern for the details of implementation. Block diagrams consist of a single block or a combination of blocks. These are used to represent the control systems in pictorial form. Basic Elements of Block Diagram The basic elements of a block diagram are a block, the summing point and the take-off point. Let us consider the block diagram of a closed loop control system as shown in the following figure to identify these elements. The above block diagram consists of two blocks having transfer functions G(s) and H(s). It is also having one summing point and one take-off point. Arrows indicate the direction of the flow of signals. Block The transfer function of a component is represented by a block. Block has single input and single output. The following figure shows a block having input X(s), output Y(s) and the transfer function G(s). NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 68  ` 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. The following figure shows the summing point with two inputs (A, B) and one output (Y). Here, the inputs A and B have a positive sign. So, the summing point produces the output, Y as sum of A and B. I.e. Y = A + B. Take-off Point The take-off point is a point from which the an input signal can be passed through one or more branches. That means with the help of take-off point, we can apply the same input to one or more blocks, summing points. In the following figure, the take-off point is used to connect the same input, R(s) to two more blocks. In the following figure, the take-off point is used to connect the output C(s), as one of the inputs to the summing point. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 69  ` Basic Connections for Blocks There are three basic types of connections between two blocks. Series Connection Series connection is also called cascade connection. In the following figure, two blocks having transfer functions G1(s) and G2(s) are connected in series. The transfer function of this single block is the product of the transfer functions of those two blocks. The equivalent block diagram is shown below. Similarly, you can represent series connection of ‘n’ blocks with a single block. The transfer function of this single block is the product of the transfer functions of all those ‘n’ blocks. Parallel Connection The blocks which are connected in parallel will have the same input. In the following figure, two blocks having transfer functions G1(s) and G2(s) are connected in parallel. The outputs of these two blocks are connected to the summing point. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 70  ` The transfer function of this single block is the sum of the transfer functions of those two blocks. The equivalent block diagram is shown below. Feedback Connection The following figure shows negative feedback control system. Here, two blocks having transfer functions G(s) and H(s) form a closed loop. The transfer function of this single block is the closed loop transfer function of the negative feedback. The equivalent block diagram is shown below. Similarly, you can represent the positive feedback connection of two blocks with a single block. The transfer function of this single block is the closed loop transfer function of the positive feedback, Block Diagram Reduction Rules Follow these rules for simplifying (reducing) the block diagram, which is having many blocks, summing points and take-off points. Rule 1 − Check for the blocks connected in series and simplify. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 71  ` Rule 2 − Check for the blocks connected in parallel and simplify. Rule 3 − Check for the blocks connected in feedback loop and simplify. Rule 4 − If there is difficulty with take-off point while simplifying, shift it towards right. Rule 5 − If there is difficulty with summing point while simplifying, shift it towards left. Rule 6 − Repeat the above steps till you get the simplified form, i.e., single block. Note − The transfer function present in this single block (after reducing) is the transfer function of the overall block diagram. Example Consider the block diagram shown in the following figure. Let us simplify (reduce) this block diagram using the block diagram reduction rules. Step 1 − Use Rule 1 for blocks G1 and G2. Use Rule 2 for blocks G3 and G4. The modified block diagram is shown in the following figure. Step 2 − Use Rule 3 for blocks G1G2 and H1. Use Rule 4 for shifting take-off point after the block G5. The modified block diagram is shown in the following figure. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 72  ` Step 3 − Use Rule 1 for blocks (G3+G4) and G5. The modified block diagram is shown in the following figure. Step 4 − Use Rule 3 for blocks (G3+G4) G5 and H3. The modified block diagram is shown in the following figure. Step 5 − Use Rule 1 for blocks connected in series. The modified block diagram is shown in the following figure. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 73  ` Step 6 − Use Rule 3 for blocks connected in feedback loop. The modified block diagram is shown in the following figure. This is the simplified block diagram. Therefore, the transfer function of the system is Block Diagram Representation of Electrical Systems Let us represent an electrical system with a block diagram. Electrical systems contain mainly three basic elements — resistor, inductor and capacitor. Consider a series of RLC circuit as shown in the following figure. Where, Vi(t) and Vo(t) are the input and output voltages. Let i(t) be the current passing through the circuit. This circuit is in time domain. NTTF_DIPLOMA IN MECHATRONICS ENGINEERING & SMART FACTORY_SEMESTER 3_CONTROL SYSTEM 74  ` By applying the Laplace transform to this circuit, will get the circuit in s- domain. The circuit is as shown in the following figure. Induct