ELEN 30123 Feedback Control System PDF

ELEN 30123 FEEDBACK CONTROL SYSTEM POLYTECHNIC UNIVERSITY OF THE PHILIPPINES - MANILA, COLLEGE OF ENGINEERING, ELECTRICAL ENGINEERING DEPARTMENT http://www.free-powerpoint-templates-d...

ELEN 30123 FEEDBACK CONTROL SYSTEM POLYTECHNIC UNIVERSITY OF THE PHILIPPINES - MANILA, COLLEGE OF ENGINEERING, ELECTRICAL ENGINEERING DEPARTMENT http://www.free-powerpoint-templates-design.com CLASS INTRODUCTION Week 1 Subject Agenda 01 Introduction Basic classroom rules, subject grading system, and other important and Learning reminders. Objectives Review and Preview of the Principle 02 To familiarize commonly used electrical systems, flow charts, mathematical methods like Differential Equations, Laplace Transforms, and Introduction to Control Systems Control System Analysis 03 To obtain mathematical model application to a system, time domain and frequency domain analysis, block diagrams, and system behavior analysis. Design and Implementation 04 To obtain control system applications, design of the system with preferable system output.  Subject Timeline W4 W3 W2 W1 Modeling in the Time Domain Modeling in the Frequency Domain Control Systems Introduction Introduction Introduction, State Space Introduction, Laplace History, Role of Control System Classroom Interaction, Representation, System Transform, Transfer Functions, Engineer, Systems Instructors profile, rules and Conversion, Linearization Electrical Network configuration and Analysis, regulations and grading system Applications Design Process W5 W6 W7 Time Response Reduction of Multiple Subsystems Steady State Errors Design Introduction, System Introduction, Block diagrams, Introduction, Error Frequency Response Responses, State Equations, Analysis and design of specifications, disturbances, Techniques, state space, digital Order Systems Feedback Systems, Signal flow sensitivity Control System Graphs Meet your Instructor BENSON G. PULGA Class Instructor, REE PUP Graduate Batch 2015 Former Student Assistant, 2014-2015 Worked in Property Management, 2015- Present PUP EE Instructor, 2016 to Present Adviser, PUP EE WIRED Specialization; Industrial Motor Controls, Wiring Installations, Electrical Equipment Repair and Maintenance bensonpulga/facebook (+63) 0965 199 0283 [email protected]  Grading System ATTENDANCE RESEARCH ASSIGNMENTS MAJOR EXAMS QUIZZES 10% 10% 10% 40% 30%  3C’s in Classroom Principle Communication Consideration Cooperation Cooperation PREVIEW OF SOME ELECTRICAL SYSTEMS WITH CONTROL SYSTEM APPLICATIONS  Water System Elevated Tank 1 Potable Water Tank – 25,000 Gal 1 Flushing Water Tank - 33,000Gal Booster Pump (Flushing) Capacity: 5.5HP Water Supply Cut in: 20psi 2 - 5.5HP Booster Pump for Flushing Water BSOCC Cut out: 40psi 2 - 4HP Booster Pump for Potable Water Quantity: 2 units Both units supplying water for 16th to 23rd floor Booster Pump (Potable) running at the same time. Capacity: 4HP Cut in: 20psi Cut out: 40psi Transfer Pumps Quantity: 2 units Total of 4 units transfer pumps, 2 units 20 HP Function: Cistern Tank transfer pump for potable water, 2 units 25 Increases water pressure HP for flushing water , functioning 2 at a Water from BWC flows time, distributes water from cistern to fill for the higher floors (16th – 23rd Floor) separately to cistern tank. each overhead tank. Cistern Tank Cistern Tank (Potable) (Flushing) 66,000 gallons 99,000 gallons  Fire Alarm and Detection System Fire Alarm / Manual Push Station FLOW SWITCH Activates evacuation alarm for relevant area or Detects water flowing in the zone. (1 unit per floor) system and causes the FDAS Annunciator to sound the alarm. Indicates the affected area through lighted LED. (1 unit per floor) SUPERVISORY SWITCH Detects the unauthorized Smoke Light Indicator Indicates the affected room through operation of main control lighted LED. (1 unit per room) valve and causes the FDAS to sound the alarm. Alarm Sequence: Pre-alarm – FACP will indicate Fire Alarm Control Panel Smoke Detectors activated initiating device. The controlling component of the Sends signal to FACP once Positive Alarm – affected floor will fire alarm system. Activates alarm the effects of fire (smoke) sound the alarm after 15 sec. sequence once it received signal have been detected. that pre-alarm was not from initiating devices. Addressable Quantity per area: acknowledged. Loop 1 – BSO retail Common Area: General Alarm – FACP will sound Loop 2 – Basement 2- 6th Floor EE Rooms 1 pc. the general alarm after 5 Loop 3 – 7th Floor- 10th Floor Telco Rooms 1 pc. minutes. that positive alarm was Loop 4 – 11th Floor- 15th Floor AHU Rooms 1 pc. not acknowledged. Loop 5 – 16th Floor- 19th Floor Hallways 1 pc. Loop 6 – 20th Floor- Roofdeck Roof deck/GF 1 pc.  Fire Protection System Supervisory / Tamper Switch Fire Hose Cabinet Detects the unauthorized High pressure water outlets located in different areas. (2 cabinet per floor) operation of main control valve and causes the FDAS to sound the alarm. Flow Switch Detects water flowing in the system and causes the FDAS to sound the alarm. Pressure Gauge Indicates system pressure. Wet Stand Pipe Serves as water outlet Sprinkler Head for firefighters Discharges water when the effects of fire have been (Pressurized) detected. 68°C for hallway/rooms, 80°C for kitchen. @175 psi JOCKEY PUMP No. of Sprinkler Heads Capacity: 5 HP FIRE PUMP Fire Exits 1 unit/floor cut in: 230 psi Capacity: cut out: 240 psi 100hp Type: Submersible Cut in: 215psi Dry Stand Pipe Serves as water inlet for AFSS Tank firefighters. 4500 gallons  Building Electrical System To major equipment and To EE Room per Floor (Tenant) common area MVSG (Medium Voltage Switch Pad Mounted Transformer Low Voltage Switch Gear Gear) 34,500V 34,500V - 400V 400VAC, 3P, 4W, 60 Hz  HVAC System Condenser Pump COOLING TOWER is used to pump warm water A cooling tower is a heat outside to the cooling tower. rejection device, which extracts Chiller waste heat to the atmosphere It is the main source that though the cooling of a water provides chilled water for Air- stream to a lower temperature. conditioning system. Secondary Pump is a pump designated for each chiller used to return the chilled water from AHU to be chilled again. AIR HANDLING UNIT Primary Pump (AHU) Equipment are the pumps used to deliver the installed to condition chilled water to the AHU units air to the building going to the tenants. thru the supply air duct and returns air thru Return Air Duct. Cistern Tank 20,000 gallons  Assignment and Reminders 1. Review Advance Mathematics. (Differential Equations, Laplace Transforms, Transfer Functions) 2. Review Circuit Analysis 1,2 and 3. 3. Short Quiz before our class next week. THANK YOU  ELEN 30123 FEEDBACK CONTROL SYSTEM POLYTECHNIC UNIVERSITY OF THE PHILIPPINES - MANILA, COLLEGE OF ENGINEERING, ELECTRICAL ENGINEERING DEPARTMENT http://www.free-powerpoint-templates-design.com  Summary Quiz 1. Give other name for input. 2. Give other name for output. 3-4. Give atleast 2 advantage of control system. 5. Person credited for Root Locus Technique. 6-7. Classification of control system based on feedback path 8-10. Give atleast 3 difference of Open and closed Loop control system. Principles of Control System Analysis Week 2  Control System History of Control Benefits in Applications System studying Control Systems Basic Features and Analysis and Design Configurations of Design Process Objectives Control System TOPIC AGENDA AND LEARNING OBJECTIVES  Principles of Control System Control System Definition A control system consists of subsystems and processes (or plants) assembled for the purpose of controlling the outputs of the processes. A control system provides an output or response for a given input or stimulus. Other used names for Input; Some Applications; Stimulus Desired Response Body system – Pancreas (blood sugar control), Eyes (follows moving Reference object), Hands (Grasp the object), etc. i/p x (t) Space Explorations (Guidance, Navigation and Control) Personal and Public Mobility (Land, Air, and Sea Transportation) Other used names for Output; Response Actual response Manufacturing Plant (Equipment Operation) Controlled variable o/p Building System Automation y (t) Home Appliances and Devices  Principles of Control System Control System Sample Application  Principles of Control System Control System Advantages We build control systems for four primary reasons; 1. Power Amplification (Gain) – A system that requires large amount of power to its output.  Principles of Control System Control System Advantages We build control systems for four primary reasons; 2. Remote Control – A system that can compensate for human disabilities. It is also useful in remote or dangerous locations.  Principles of Control System Control System Advantages We build control systems for four primary reasons; 3. Convenience of Input form – A system used to provide convenience by changing the form of the input.  Principles of Control System Control System Advantages We build control systems for four primary reasons; 4. Compensation for disturbances – A system that able to yield the correct output even with a disturbance or a system that measures the amount of disturbance then return the process commanded by input. * disturbances such as Noise, Errors, Outside forces (wind, dusts, power fluctuations etc.)  Principles of Control System Control System Background Stability criterion for a third-order system based on the coefficients of the Differential Equations. “A Treatise on the Stability of a Given State of Motion” The General Problem of Stability of Motion. Routh-Hurwitz Criterion for Stability. Analysis of feedback amplifiers, sinusoidal frequency analysis and design techniques. Walter R Evans - Graphical technique to plot the roots of a characteristic equation of a feedback system whose parameters changed over a particular range of values..  Principles of Control System Control System Engineer Project development from Space shuttle control systems concept through design and testing. Control engineer can be working with sensors and Performs hardware selection, motors as well as electronic, design, and interface, including pneumatic and hydraulic total subsystem design. circuits.  Principles of Control System Response Characteristics and System Configuration 1. Transient Response – Instantaneous change of the input against the gradual change of the output. 2. Steady State Response – Approximation to the commanded or desired response. 3. Steady State Error – The accuracy of the output that could make it different from the input. 4. Stability – Characteristics of input with less or zero natural responses.  Principles of Control System Classification of Control System 1 Based on type of signal used 1. Continuous time control system – signals are continuous 2. Discrete-time control system – there exists one or more in time discrete (intervals) time signals.  Principles of Control System Classification of Control System 2 Based on number of inputs and outputs 1. SISO (Single Input and Single Output) – control system with 2. MIMO (Multiple Inputs and Multiple Outputs) – control one input and one output. systems with more than one input and outputs.  Principles of Control System Classification of Control System 2 Based on number of inputs and outputs  Principles of Control System Classification of Control System 3 Based on feedback path 1. Open-Loop Control System – output is not fed-back to the input. The control action is independent of the desired output.  Principles of Control System Classification of Control System 2. Closed Loop Control System – output is fed-back to the input. The control action is dependent on the desired output. *The error detector produces an error signal, which is the difference between the input and the feedback signal. This feedback signal is obtained from the block (feedback elements) by considering the output of the overall system as an input to this block. Instead of the direct input, the error signal is applied as an input to a controller. So, the controller produces an actuating signal which controls the plant. In this combination, the output of the control system is adjusted automatically till we get the desired response. Hence, the closed loop control systems are also called the automatic control systems.  Principles of Control System Classification of Control System Difference of Open Loop and Close Loop Control system Open Loop Control System Closed Loop Control System Control action is independent of Control action is dependent of the desired output the desired output. Feedback path is not present Feedback path is present. These are also called as non- These are also called feedback control systems. as feedback control systems Easy to design Difficult to design These are economical. Expensive Inaccurate Accurate  Principles of Control System Control System Analysis and Design Objective Control system is dynamic: It responds to an input by undergoing a transient response before reaching a steady-state response that generally resembles the input. 1. Producing the desired transient response – Analyzing the system for its existing transient response and adjust the parameters or design components to yield a desired transient response. 2. Reducing of steady-state errors – Analyzing a system’s steady state error, then design corrective action to reduce the steady state error. 3. Achieving system stability – Analyzing a system for its “useful and effective operation” where, the natural response must approach to zero (leaving only the forced response) or oscillate. *Total response = Natural Response + Forced Response Natural Response – describes the way the system dissipates or acquires energy(dependent only to the system, not in input). Forced Response - disturbances/error (dependent on the input). Transient response is the sum of the natural and forced responses, while the natural response is large. Steady state response is the sum of the natural and forced responses, but the natural response is small.  Principles of Control System The Design Process Create a Schematic Develop a Mathematical Model (Block 3 4 Diagram) Approximations about the system and neglecting certain phenomena. Mathematically modelling a system using differential equation (transfer function) via Laplace Transform. Reduce the Block Diagram Draw a Functional Block Diagram 5 Reduction of large system’s block Component parts of the system (Function and 2 diagram to a single block with a Hardware), showing their interconnection. mathematical description that represents the system from its input to its output. Analyze and Design Determine a physical system and Evaluate the time response of a system specifications from the 6 1 START for a given output. requirements. Transient response, Steady state, Steady Overall concept/description of the system. state errors, Stability.  Summary Quiz 1. Give other name for input. 2. Give other name for output. 3-4. Give atleast 2 advantage of control system. 5. Person credited for Root Locus Technique. 6-7. Classification of control system based on feedback path 8-10. Give atleast 3 difference of Open and closed Loop control system. THANK YOU! Modeling in Frequency Domain Week 3 TOPIC AGENDA AND L E A R N I N G O B J E C T I V E S Discuss the Linearity in Control System Introduction to the Transfer Review the Laplace and Inverse Function in Control System Laplace Transform Formulas and Theorems.  Linearity in Control System Block Diagram Representation of a System Input Output System r (t) c (t) Block diagram where the input, output and system are distinct and separate parts Input Output System System System r (t) c (t) Block diagram representation of an interconnection of subsystem. Note: The input, r(t), stands for reference input. The output, c(t), stands for controlled variable.  Linearity in Control System Linearity of the Control System r₁(t) G₁ c₁(t) ; c₁(t) = G₁r₁(t) r₂(t) G₂ c₂(t) ; c₂(t) = G₂r₂(t) r₁(t) G₁ c₃(t) ; c₃(t) = G₁r₁(t) + G₂r₂(t) r₂(t) G₂ c₃(t) = c₁(t) + c₂(t)  Linearity in Control System Types of Feedback Feedback plays an important role to improve the performance of the control system. Positive Feedback The positive feedback adds the reference input, R(s) and feedback output. C(s) = [ R(s) + C(s)H ] G = G R(s) + G C(s) H C(s) (1 – GH) = GR(s) T= Output = C(s) = __G___ Input R(s) (1 - GH)  Linearity in Control System Types of Feedback Negative Feedback The negative feedback adds the reference input, R(s) and feedback output. C(s) = [ R(s) - C(s)H ] G = G R(s) – G C(s) H C(s) (1 + GH) = GR(s) T= Output = C(s) = __G___ Input R(s) (1 + GH)  Transfer Function Transfer Function Transfer function of a control system is the ratio of Laplace transform of output to Laplace transform of input. Transfer Function = L { C ( t ) } L{R(t)} with zero initial conditions Transfer Function with Positive Feedback Transfer Function with Negative Feedback T= Output = C(s) = __G___ T= Output = C(s) = __G___ Input R(s) (1 - GH) Input R(s) (1 + GH)  Transfer Function The Laplace Transform The Laplace transform, named after its inventor Pierre-Simon Laplace (/ləˈplɑːs/), is an integral transform that converts a function of a real variable (t) (often time) to a function of a complex variable (s) (complex frequency). The transform has many applications in science and engineering because it is a tool for solving differential equations.In particular, it transforms differential equations into algebraic equations and convolution into multiplication. The Laplace Transform Equation L {c₁F₁(t) + c₂F₂(t)} = c₁ L {F₁(t)} + c₂ L {F₂(t)} The Laplace Transform Equation dealing with Constants, C.  Transfer Function The Laplace Transform  Transfer Function Problem Exercises I. Find the Laplace Transform of the following; 1. Find L {eᵅᵗ} 6. Find L {t² + 4t - 5} 2. Find L {1} 7. Find L {t³ - t² + 4t} 3. Find L {t} 8. Find L {e¯²ᵗ+ 4e¯³ᵗ} 4. Find L {sin at} 9. Find L {3e⁴ᵗ- e¯²ᵗ} 5. Find L {cos at} 10. Find L {t¯½} II. Find the Inverse Laplace Transform of the following; 15 3𝑠 1. Find L¯¹ { } 5. Find L¯¹ { } 2 𝑠 +4𝑠+13 𝑠 2 +4𝑠+13 𝑠+1 𝑠 2. Find L¯¹ { } 6. Find L¯¹ { } 2 𝑠 +6𝑠+25 𝑠 2 +6𝑠+13 1 1 3. Find L¯¹ {2 } 7. Find L¯¹ { } 𝑠 +2𝑠+10 𝑠 2 +4𝑠+4 2𝑠 −3 1 8. Find L¯¹ { } 4. Find L¯¹ {2 } 𝑠 2 −4𝑠+8 𝑠 −4𝑠+8  Transfer Function Electric Network Transfer Functions Components and the relationships between voltage and current and between voltage and charge under zero initial conditions Note; v(t) = V(volts), i(t) = A(amps), q(t) = Q(coulombs), C = F(Farads), R = Ω(ohms),G = mhos, L = H(Henries)  Transfer Function Sample Electrical Network THANK YOU! Modeling in Frequency Domain Week 4 Topic Agenda and Learning Objectives Inverse Laplace Review using Partial 01 Fraction Expansion Laplace Transform of a Differential 02 Equation System Response from the Transfer 03 Function 04 Electric Network transfer Functions  Transfer Function Partial Fraction Expansion Algebraic conversion of complicated function to a sum of simpler terms which we can determine the Laplace transform of each term. Conditions: F(s) = N(s) ; where N(s) < D(s), direct partial fraction expansion can be made. D(s) ; where N(s) > D(s), N(s) must be divided by D(s) successively until the result has a remainder whose numerator is of order less than its denominator. I. Sample problems 𝑠 3 +𝑠 2 +6𝑠+7 15𝑠 3 +2𝑠 2 +7 1. Find L¯¹ { } 4. Find L¯¹ { } 𝑠 2 +𝑠+5 𝑠+1 3𝑠 3 +6𝑠 2 +2𝑠+1 4𝑠 3 +𝑠+2 2. Find L¯¹ { } 5. Find L¯¹ { } 𝑠 2 +2𝑠−15 (𝑠+2)² 7𝑠 3 +4𝑠 2 +2𝑠+12 3. Find L¯¹ { 2 } 𝑠 +4𝑠−18  Transfer Function Partial Fraction Expansion CASE 1. Roots of the denominator of F(s) are Real and Distinct. Each factor of the denominator is raised to unity power. 𝑎 𝐾₁ 𝐾₂ = (𝑠+𝑎) + (𝑠+𝑏) ; where K is called residues. (𝑠+𝑎)(𝑠+𝑏) CASE 2. Roots of the denominator of F(s) are Real and Repeated. One or same factor of the denominator is/are raised to an integer power higher than 1. 𝑎 𝐾₁ 𝐾₂ 𝐾₃ = (𝑠+𝑎) + + (𝑠+𝑎)(𝑠+𝑏)² 𝑠+𝑏 (𝑠+𝑏)² CASE 3. Roots of the denominator of F(s) are Complex or Imaginary. One or same factor of the denominator have complex roots. 𝑎 𝐾₁ 𝐾₂𝑠+𝐾₃ = + 𝑠(𝑠 2 +𝑎𝑠+𝑏) 𝑠 (𝑠 2 +𝑎𝑠+𝑏)  Transfer Function Partial Fraction Expansion I. Sample problems 𝑠 2 −7 2𝑠 2 +5𝑠−4 1. Find L¯¹ {3 2 } 6. Find L¯¹ { } 𝑠 +4𝑠 +3𝑠 𝑠 3 +𝑠 2 −𝑠 5𝑠 3 − 6𝑠−3 2𝑠 2 +1 2. Find L¯¹ { 3 } 7. Find L¯¹ { } 𝑠 (𝑠+1)² 𝑠(𝑠+1)² 16 4𝑠+4 3. Find L¯¹ { } 8. Find L¯¹ { } 𝑠 (𝑠²+4)² 𝑠 2 (𝑠−2) 1 1 4. Find L¯¹ { } 9. Find L¯¹ { } 𝑠 2 +𝑎𝑠 𝑠³(𝑠 2 +1) 𝑠+2 5𝑠−2 5. Find L¯¹ { } 10. Find L¯¹ { } 2 𝑠 −6𝑠+8 𝑠 2 (𝑠+2)(𝑠−1)  Transfer Function Laplace Transform Solution of a Differential Equation 𝑛 where, 𝑓 denotes the nth derivative of the function f. For first-order derivative: For first-order derivative: 𝑑𝑦 𝑑𝑐(𝑡) 𝑑𝑟(𝑡) L{f′(t)} = sL{f(t)} − f(0) 𝑓′(𝑡) = 𝑑𝑥 = 𝑑𝑡 = 𝑑𝑡 For second-order derivative: For second-order derivative: L{f′′(t)} = s²L{f(t)} − sf(0) − f′(0) 𝑑²𝑦 𝑑²𝑐(𝑡) 𝑑²𝑟(𝑡) 𝑓′′(𝑡) = = = 𝑑𝑥² 𝑑𝑡² 𝑑𝑡² For third-order derivative: For third-order derivative: 𝑑³𝑦 𝑑³𝑐(𝑡) 𝑑³𝑟(𝑡) L{f′′′(t)}=s³L{f(t)} −s²f(0) − sf′(0) − f′′(0) 𝑓′′′(𝑡) = = = 𝑑𝑥³ 𝑑𝑡³ 𝑑𝑡³ 𝑡 𝐹 (𝑠) L[‫׬‬−0 𝑓 𝑡 𝑑𝑡 ] = Transform of an integral equation 𝑠  Transfer Function The Transfer Function and System Response 1. Formulation of system representation by establishing a viable definition for a function that algebraically relates a system’s output to its input. 2. This function will allow separation of the input, system and output into three separate and distinct parts. 3. The function will allow us to algebraically combine mathematical representations of subsystems to yield a total system representation. 𝑎𝑛 𝑠 𝑛 𝐶 𝑠 + 𝑎𝑛−1 𝑠 𝑛−1 𝐶 𝑠 +.. +𝑎0 𝐶 𝑠 = 𝑏𝑚 𝑠 𝑚 𝑅 𝑠 + 𝑏𝑚−1 𝑠 𝑚−1 𝑅(𝑠)+.. +𝑏0 𝑅(𝑠) 𝐶 𝑠 𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 +.. +𝑎0 = 𝑅 𝑠 (𝑏𝑚 𝑠 𝑚 + 𝑏𝑚−1 𝑠 𝑚−1 +.. +𝑏0 ) Simplified general nth order, linear, time invariant differential equation Input 𝑏𝑚 𝑠 𝑚 + 𝑏𝑚−1 𝑠 𝑚−1 +.. +𝑏0 Output R (s) 𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 +.. +𝑎0 C (s) Block diagram where the input, output and system are distinct and separate parts 𝐶(𝑠) 𝑏𝑚 𝑠 𝑚 + 𝑏𝑚−1 𝑠 𝑚−1 +.. +𝑏0 =𝐺 𝑠 = 𝑅(𝑠) 𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 +.. +𝑎0 C(s)= 𝑅(𝑠)𝐺 𝑠  Transfer Function The Transfer Function and System Response Sample Problem 𝑑𝑐(𝑡) 1. Find the transfer function represented by; +2c(t) = r(t) 𝑑𝑡 Solution; 𝑑𝑐(𝑡) C(s) = R(s) G(s) +2c(t) = r(t) 𝑑𝑡 1 1 s C(s) +2C(s) = R(s) C(s) = 𝑠 ( 𝑠+2 ) C(s) [s + 2] = R(s) 1 𝐴 𝐵 C(s) = = 𝑠 + 𝑠+2 𝑠(𝑠+2) 𝐶(𝑠) 1 = G(s) = Transfer Function G(s) 𝑅(𝑠) 𝑠+2 1 1/2 1/2 C(s) = = - 𝑠+2 𝑠(𝑠+2) 𝑠 1 1 c(t) = − 𝑒 −2𝑡 System Response c(t) 2 2  Transfer Function The Transfer Function and System Response Sample Problem 𝑑³𝑐(𝑡) 𝑑²𝑐(𝑡) 𝑑𝑐(𝑡) 𝑑2 𝑟(𝑡) 𝑑𝑟(𝑡) 2. Find the transfer function represented by; +3 +7 +5c(t) = +4 + 3r(t) 𝑑𝑡³ 𝑑𝑡² 𝑑𝑡 𝑑𝑡² 𝑑𝑡 𝑑²𝑐(𝑡) 𝑑𝑐(𝑡) 𝑑𝑟(𝑡) 3. Find the transfer function represented by; +6 +2c(t) = 2 + r(t) 𝑑𝑡² 𝑑𝑡 𝑑𝑡 𝑠 4. Find the ramp response for a system whose transfer function is G(s) = (𝑠+4)(𝑠+8) 𝑠 −13 5. Find the ramp response for a system whose transfer function is G(s) = 𝑠²(𝑠+12)(𝑠+7)  Transfer Function Electric Network Transfer Functions Application of the transfer function to the mathematical modeling of electric circuits including passive networks and operational amplifier circuits. Components and the relationships between voltage and current and between voltage and charge under zero initial conditions Note; v(t) = V(volts), i(t) = A(amps), q(t) = Q(coulombs), C = F(Farads), R = Ω(ohms),G = mhos, L = H(Henries)  Transfer Function Simple Circuits via Mesh Analysis 1. Solution via Differential Equation Using Kirchhoff’s voltage law by writing the differential equation of the components first then taking Laplace transform of the transformed circuit. Sample Problem. Find the transfer function relating the capacitor voltage, Vc(s) to the input voltage, V(s).  Transfer Function Simple Circuits via Mesh Analysis 2. Solution via Impedance Transfer Function Using Kirchhoff's voltage Law and obtaining direct Laplace transform to the transformed circuit. Taking Laplace transform of the equations in the voltage-current column in previous table, assuming zero initial conditions. For Capacitor (C) ; For Resistor (R) ; For Inductor (L) ; 1 V(s) = 𝐶𝑠 𝐼(s) V(s) = 𝑅𝐼(s) V(s) = 𝐿𝑠 𝐼(s) For Impedance (Z) ; 1 𝑉(𝑠) 1 ( ) I(s) = V(s) Z(s) = (𝐶𝑠 + R + Ls ) I(s) = V(s) 1 1 𝐼(𝑠) For parallel circuit For series circuit 𝐶𝑠+R+Ls [sum of the impedances] I(s) = [sum of applied voltages] Summarized steps for getting Impedance Transfer Function 1. Redraw the original network showing all time variables such as v(t), i(t), and vc(t), as laplace transforms V(s), I(s), Vc(s), respectively. 2. Replace the component values with their impedance values. This replacement is similar to the case of DC circuits, where we represent resistors with their resistance values.  Transfer Function Simple Circuits via Mesh Analysis 1. Solution via Impedance Transfer Function Sample Problem. Find the transfer function relating the capacitor voltage, Vc(s) to the input voltage, V(s). 1 V(s) = 𝐿𝑠 𝐼 s , V(s) = 𝑅𝐼 s , V(s) = 1 𝐼 s Vc(s) = I(s) 𝐶𝑠 𝐶𝑠 1 Z(s) = (Ls + 𝑅 + 1 ) V(s) = Cs Vc(s)(Ls + 𝑅 + 𝐶𝑠), V(s) = Vc(s)(s²LC + 𝑠𝑅𝐶 + 1), 𝐶𝑠 1 1 𝐼(𝑠) 1 𝑉𝑐(𝑠) 1 𝑉𝑐(𝑠) V(s) = I(s) (Ls + 𝑅 + ), = = , = 𝐿𝐶 𝐶𝑠 𝑉(𝑠) Ls+𝑅+ 1 𝑉(𝑠) s²LC+𝑠𝑅𝐶+1 𝑅 1 𝐶𝑠 𝑉(𝑠) s²+ 𝐿 𝑠+𝐿𝐶 THANK YOU!  ELEN 30123 FEEDBACK CONTROL SYSTEM POLYTECHNIC UNIVERSITY OF THE PHILIPPINES - MANILA, COLLEGE OF ENGINEERING, ELECTRICAL ENGINEERING DEPARTMENT http://www.free-powerpoint-templates-design.com Modeling in Frequency Domain Week 5 Topic Agenda and Learning Objectives Basic AC/DC Circuits Computation 01 Review Identification of Current and Voltage in 02 each junction of a Circuit  Transfer Function Review of DC Circuits Computation Solution via Loop Rule (Junction) -I +I c. Voltage typically flows from “-” to “+”, indication of Voltage lift, V will have “+” sign.(Gain) d. If voltage flows from “+” to “-”, it indicates of Voltage drop (voltage consumption, V will have “-” sign.(Drop) +V -V Steps in Solving a simple DC Circuits using Loop Rule. 1. Identify the Current and Voltage Sign Rule in a Circuit. a. Current typically flows from “+” to “-”, indication of Current lift, I will have “+” sign. b. If current flows from “-” to “+”, it indicates Current drop, I will have “-” sign.  Transfer Function Review of DC Circuits Computation Steps in Solving a simple DC Circuits using Loop Rule. 2. Choose the direction of current flow in the circuit. If there’s 3. Properly identify the sign of every component once you two or more loops, it is recommended to have same directions. chose the direction of current flow.. Wrong directions will only indicate “-” sign in current. 4. Solve the circuit by using Ohm’s Law. Note: Sum of Iin = Sum of Iout Sum of Voltages in the circuit = 0  Transfer Function Review of DC Circuits Computation Sample Circuit Problems 1. Find the Current I1 in the circuit and voltage drop in each components Solution; I1 = 12 volts / 60 ohms Voltage Drop = I x R I1 = 0.2 amps 𝑉𝐷 = 0.2 amps x 60 ohms 𝑉𝐷 = 12 V  Transfer Function Review of DC Circuits Computation Sample Circuit Problems 2. Find the Current I2, I4, I5 in the circuit , voltage drop in each components and voltage in each nodes 1-8  Transfer Function Review of DC Circuits Computation Sample Circuit Problems 3. Find the Current I1 in the circuit , voltage drop in each components and voltage in each nodes 1-0  Transfer Function Review of DC Circuits Computation Sample Circuit Problems 3. Find the Current I1 in the circuit , voltage drop in each components and voltage in each nodes 1-8 2 3 5 1 4 8 6 7 THANK YOU!