Selamat Kembali Ke Parit Raja.. BDA30703 Kejuruteraan Kawalan Seksyen S5

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1 PROFESOR MADYA DR. ZAMRI BIN OMAR Timbalan Dekan (Hal Ehwal Pelajar & Alumni) Fakulti Kejuruteraan Mekanikal & Pembuatan Selamat Kembali Ke Parit Raja.. BDA30703 Kejuruteraan Kawalan Seksyen S5 zamri@uthm.edu.my Phone : Room: A41 st Floor Mobile : (text only) Assoc. Professor at Aeronautical Eng. Dept. B.Eng(MechAero), UTM M.Eng (Mech), UTM PhD (Aerospace), RMIT University Private Pilot License (DCA) Semester I 2015/2016 BDA ~ Control Engineering Section 5 Meeting: 3 hrs/week Venue: G3BKB 1 Day : Sunday (10am11am) Thursday (8am10am) 1

2 Important Reminder.. 1. Class participation 3. Communication 4. ELearning: Author 5. Some rules; punctuality, dress code, don'ts.. BDA Control Engineering Sem I 15/16 Chapter 1 Introduction & Block Diagram Chapter 2 Mathematical Modeling Chapter 3 Time Response Analysis Chapter 4 Stability & Root Locus Method Chapter 5 Frequency Response Analysis Chapter 6 Control System Design RPP; a. Syllabus b. Assesment c. References Total lecture weeks : 14 1st Part : 7 Weeks 2st Part : 7 Weeks MidSem Break : Study Week : Exam : 2

3 Chap 1 : Intro & Block Diagram How to control? Chap 1 : Intro & Block Diagram What is Control Engineering? deals with dailylife physical systems. dynamic systems, contain several variables need to be controlled control approach: develop math. model, time & frequency response analysis, stability. U3 Systems? an assemblage of interrelated components. governed by physical laws. also has diverse meaning. Engineering? Engineer?? 3

4 Slide 6 U3 stop here UTHM, 10/09/2015

5 C1 : Block Diagram Representation Systems Representation represented as block diagram disturbances, n(t) input, u(t) System components output, y(t) a block has input, output. arrow represents signal (arrow in is i/p, arrow out is o/p) Single input/output(siso), multiple input/output (MIMO) disturbance may occur. input; controlled signal output; dependent signal disturbance; uncontrolled i/p signal C1 : Block Diagram Example; Power Plant System Fuel rate, f Boiler Steam pressure, p Turbine Mechanical rotational power, T Electrical generator Electrical power, W disturbance disturbance disturbance input; fuel rate, f (m 3 /s) output; electrical power, W (KW) System components; represent by blocks output of a block = input to another block disturbance; system inefficiency, weather 4

6 C1 : Control System & Open Loop Knob setting Control System a device to control the power source to cause the output is a depenndent of the input it has objectives!! to design a startegy; the o/p is as desired. 2 control strategies; open loop, closed loop Open Loop Control o/p has no effect on the control action. contains one signal path Switch Controller 1.Low (27 o C ) 2. Med (20 o C ) 3. High (16 o C ) R 1 R 2 R3 Aircond system Process Room C1 : Open Loop Control Switch Controller Process Knob setting 1.Low (27 o C ) 2. Med (20 o C ) R 1 R 2 Room 3. High (16 o C ) R3 Open Loop Control room T is controlled by knob settings each setting causes a diferrent electrical current (power) syst. Performance; depends on the controller accuracy user can t control; ambient temperature, efficiency of cooling element if any disturbance occurred; a/s system connnot correct itself 5

7 C 1 : Open Loop Block Diagram The aircond system can be represented by the following block diagram Cooling level ( switch ) Controller Current Cooling element Cooling rate Room Room temperature components; controller, system a one way signals Might not get the desired room temperature; controller inefficiency ambient temperature C1 : Closed Loop Control Closed Loop Control the performance of a/c system can be improved by introducing; an operator (Ali) dedicated to control the knob settings a room temperature sensor desired T = medium, set knob to 2 a hot day, T 20 o C; change the knob setting to 3 after a while, T = 16 o C (too cold); change knob setting to 2 please!! the T can be regulated/set at 20 o C, but may be oscillated needs a large effort from the operator this system is known as a closed loop (feedback system); 6

8 C1 : Closed Loop Block Diagram Closed Loop Control extra components; actuator & sensor if the operator is replaced by a mechancial/electrical device; an automatic feedback system C1 : Block Diagram Elements Block Diaram Elements a block / blocks input signal, u System component output signal, v summing point V 1 V V 2 V = V 1 V 2 takeoff point V V V 7

9 C1 : Exercise; Block Diagram Exercise 1 a car driver uses a control system to maintain the speed of the car at a prescribed level. Sketch a block diagram to illustrate the feedback system. C1 : Exercise; Block Diagram Exercise 2 an operator is to maintain the liquid level in the reservoir. The operator compares the actual level with the desired level and opens and closes the valve, adjusting the fluid flow out to maintain the desired level. Sketch the block diagram for this system. 8

10 Block Diagram Examples Block Diagram Examples 9

11 C1 : Transfer Function Concept Transfer Function (TF) a ratio of output to input. Symbol/letter in the block represents TF, G ratio of Laplace transformation of o/p to Laplace transformation of i/p for an open loop system Input, X Transfer function, G Output, Y G ( s ) = Y( s ) X( s ) for a feedback system, C1 : Closed Loop Transfer Function the forward path TF, the feedback path TF, C( s ) G ( s ) = E( s ) F( s ) H ( s ) = C( s ) the open loop TF, the closed loop TF, F( s ) G ( s )H( s ) = R( s ) C( s ) G( s ) = R( s ) 1 G( s )H( s ) 10

12 C1 : Block Diagram Reduction To derive TF Transfer Function in Series consider a system with 2 blocks in series, θ 1 θ2 3 F 2 θ It can be reduced/simplified to, θ3 F 2 TF in series is, θ3 = F1( s )F2 ( s ) C1 : Blocks in Series TF Derivation θ 1 θ2 3 F 2 θ θ 1 and θ 1 can be written as, θ2 = F1 ( s ) θ2 = F1 ( s ) (1) θ3 = F2 ( s ) θ3 = F2 ( s ) θ2 (2) θ2 eliminate θ 2, θ 3 = F2 ( s ) θ2 = F2 ( s )F1 ( s ) θ 3 = F2 ( s )F1 ( s ) TF θ 1 in series proved! For blocks in series, multply all individuals TF to obtain total TF 11

13 C1 : Blocks in Series Example The TF for this power plant system is, W = G( s )M( s )R( s ) f C1 : Blocks in Parellel TF Derivation Transfer Function in Parellel parellel; arrows are in the same direction, not necessarily same signs consider a system with 2 blocks in parellel, θ 2 θ4 F 2 θ 3 It can be reduced/simplifed to, θ4 F 2 12

14 C1 : Blocks in Parellel TF Derivation θ 2 θ4 F 2 θ 3 Tf for each block can be written as, θ F ( s ) 2 θ 1 = θ2 = F1 ( s ) (1) F ( s ) 3 2 = θ3 = F2 ( s ) (2) at the summing point; θ4 = θ2 θ3 (3) thus the overall TF (θ 4 /θ 1 ) is, [ F ( s ) F ( s )] θ4 = θ2 θ3 = F1 ( s ) F2 ( s ) = 1 2 θ 4 = F1 ( s ) F2 ( s ) TF in parellel proved! θ 1 For blocks in parellel, add up all individuals TF to obtain total TF C1 : Blocks in Parellel Example θ 2 θ4 the TF is, θ4 = F1 ( s ) F2 ( s ) Prove it!! F 2 θ 3 C D F 2 F 3 the TF is, D( s ) C( s ) = F1 ( s ) F2 ( s ) F3 ( s ) F4 ( s ) F 4 13

15 C1 : Feedback System (Closed Loop); TF Derivation Transfer Function for Feedback feedback, positive feedback consider a feedback systems, θ 2 θ4 θ 3 F 2 It can be reduced/simplifed to, F θ F F 1 2 carefully examine!! C1 : FB Loop TF Derivation θ 2 θ4 θ 3 F 2 The overall TF (θ 4 /θ 1 ) we can write the following, θ θ 2 = θ3 (1) 4 = F1( s ) θ4 = F1 ( s ) θ2 (2) θ2 θ3 = F2 ( s ) θ4 (3) Eliminate θ 2 and θ 3 θ4 = F1 ( s ) [ θ θ ] [ θ F ( s θ ] θ4 = F1( s ) 1 2 ) 4 1 θ4 = F1 ( s ) θ4f1 ( s )F2 ( s ) 3 θ 1F1 ( s ) = θ4 θ4f1( s )F2 ( s ) [ 1 F ( s )F ( s )] F1( s ) = θ4 1 2 θ F ( s ) 4 = 1 θ 1 1 F1( s )F2 ( s ) 14

16 C1 : FB Loop TF Derivation θ 2 θ4 θ 3 F 2 The overall TF for a FB loop, Forward Blocks TF = 1 Forward Blocks X Feedback Blocks or, Forward Blocks TF = 1 OL Blocks C1 : Block Diagram (BD) Reduction Rules A summary of BD reduction rules is here. Assignment 1 is here. 15

17 C1 : Block Diagram Reduction Example 1 obtain the TF for the following system, R G 1 G 2 G 3 C H 1 H 2 C1 : Block Diagram Reduction Example 2 obtain the TF for the following system, X G 1 G 2 G 3 Y H 2 H 3 H 1 16