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Cs Unit 1

This is control system unit 1

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30 views21 pages

Cs Unit 1

This is control system unit 1

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BEC602 : Control System UNIT-1 : INTRODUCTION TO CONTROL SYSTEMS (1-1 C to 1-38 C) Basic Components of a control system, Feedback and its effect, types of feedback control systems. Block diagrams Reduction and signal flow graphs, Modeling of Physical systems: electrical networks, mechanical systems elements, free body diagram, analogous Systems, sensors and encoders in control systems, modeling of armature controlled and field controlled DC servomotor. UNIT-2 : STATE-VARIABLE ANALYSIS (2-1 C to 2-31 C) Introduction, vector matrix representation of state equation, state transition matrix, state-transition equation, relationship between state equations and high- order differential equations, relationship between state equations and transfer functions, Decomposition of transfer functions, Controllability and observability, Eigen Value and Eigen Vector, Diagonalization. UNIT-3 : TIME DOMAIN ANALYSIS OF CONTROL SYSTEMS . (3-1 Cto 3-30 C) Time response of continuous data systems, typical test signals for the time response of control systems, unit step response and timedomain specifications, time response of afirst order system, transient response of a prototypesecond order system, Steady-State error, Staticand dynamic error coefficients, error analysis for different types of systems. : STABILITY OF LINEAR CONTROL SYSTEM unr: (4-1Ct04-32.q) Bounded-inputbounded output stability continuous data systems, zero-input and asymptotic stability of continuous Gata systems, Routh Hurwitz criterion, Root-Locus i ‘Technique: Introduction, Properties of he Root Loc, Design Introduction to aspects of the Root Loci Control Systems Resonant peak and Resonant the prototype Second ord zero to the forward ps CONTENTS + Basic Compsnents of a Control the Bode plot gain margin and phase margin. 3, el = 1-2€ to 1-70 SHORT QUESTIONS (6Q-1 C to $Q-14¢) vi SOLVED PAPERS (2018-19 TO 2023-24) (SP-1 C to SP-17 Block Diagrams Reduction Signal Flow Graph" PC fo Fave sos BTC to 1-890 # Sensors and Bae0ders i rcnno IBS ae St IC to 1-870. of Armature Controlled aad : Field Controlled DC Servomotor 1-1€ EC-Sem-6) a __—_—sss=S<(} recess [+ output i sof a control system ? GueTA.| What are the basie component a control sy Examples: Fig: LL, ‘Automatic washing machine c ‘Answer sre rious components of lose loop ystem are show 2. Immersion rod 3. field controlled DC motor, B. 1 2 . Closed loop control system : Closed loop control system is also known as feedback control system. In closed loop control system the control action is desired output. dependent on the Command Tapot [_element feedback paths forms a closed loop * en sind apc ret reference input. The difference o is regulates the output according to the signal Reference, & input Ge Gene [Controlled process] from erro 5 Controlled syste: fea feedback Joo feedback loop. a Feedback element : This element feedback the output of the error detector for comparison with the reference input. GueiS | Explain open loop and closed loop control system with the help of suitable examples. pean] A. Open loop control system : {L_‘The open loop control ystem is also known as control system without feedback or non-feedback control system. is independent of the desired Examples: 1. Air conditioners 2 Autopilot aeroplane 3. Electric iron ‘Compare the open loop control system and closed loop control system, also give few examples for each system. 2 Inopen loop systems the control output. system the output is nat compared with the reference input. Introduction to Control Systems 1-4€ (EC-Sem-6) Taw | aloo No] Ope control system the gain Gts) reduces by a factor } as compared to open loop control aystom. ‘The open loop pole: If unity negative fee transfer funetion of a 2-7. introduced in'the system, the overall op system becomes, non-linearity. B_| Ieisnot much accurate, Tis very accurate. ST ies very sensitive t errors | Loss sensitive to errors and and disturba dintorbances. 4) Tehas small bandwidth. Tehas large bandwidth. [| Simple in construction and is | Complicated in design and costly. - cheap. 6. | Highly affected by ‘Less affected by no. + Blectrie iron, 7 |Examples + Washing | Examples + e machine, traffic signal. automatic gear. Que id. | Discuss the effect of feedback on: i. Overall gain Stability iii, Noise and Disturbance. i. Overall Gain : For open loop + Rs) oo ce) Fig. 14. cw) = £2 Gis) Te) = a) For closed loop : Ts) = C9). __ Ge) Rs) 1+ GOH) RG) aa Ga] cis) He) | Fig. 142) oe VF K RO "7, Kea) aT Sg 3. Thus the closed loop pole is now located at s =—(K + 7) Fig. 144) 4. Thus, the feedback controls the time Fésponse, the output of the system adversely. i Some external disturbance signals also make the aystem output inaccurate. fe pene a 3. The examples of uch external disturbances ar high fea h are high fequency noise in electronic applications thermal oie in amplifier tes mind pute on antenna of radar systom ete 4. The disturbance may be in the forward path, feedback path or output of asystem. Introduction to Control Systems 4 Control System Liedesems) 1-6 (EC-Sem6) aback on the following: LI i. Linear versus non-linear systems : 1. Linear systems do not exist in practice, sine all physical systems are nonlinear to some extents. Anewersel 2. Linear feedback control systems are idealized models fabricated by the i. Sensitivity: on tule purely forthe split of analysis and design. 00 ; eeu 3. When the magnitude ofsignalsin contol system are limited torangesin 95 - BGuanse nT * Bio ‘which system components exhibit linear characteristics, the ater is called ee c* 100 linear feedback control system." Gan 4. But when the magnitudes of signals ae extended beyond the range of 8-3 the linear operation, the systom is called non-linear feedback control sytem, fe data control systems are subdivided into sampled data and digital control system. [AKTU 2022-25, Marks 10 . Introduction to Control Systems 1-8C (EC-Sem- PART-2 “Block Diagrams Reduction ond Signal Flow Graphs. | Control System 1-90 (EC-Sem-6) 4. Negative feedback connection: GueT | Explain the block diagram reduction rules. al ranch pint: branch print or 8 A Branch pin pocscncurenay oer lsks oF rnming it. erase output of control system is feedback to the B._ Summing porijeeltis compared with the reference input C. Blocks : 1. Ablock diagram of asyste performed by each component 2. Assystem is represented by using block di construct even for the complicated system. function of individual element in the block ‘Take-off point ti Ot Gar cts) Summing post : ake off point is a point from which the smisa pictorial representation of the functions itand ofthe flow of signals. 1s, These are easy to ‘easy to visualize the .gram of a system. cre) ce), Sema] o,f cw, ? Gita) } Bis), Hw lie 5. Moving the summing point ahead of the bloc! 2s (a+ biG . oe 40 ¢ » = c 6 Moving the summing point before the block : +e: ope b ° a) : 1 ttercangng ip a st a a e a4 6 nutty C6 tea a ingstacett pone cnpas x 1} 10. Moving a take-off point before a block : G}ee ) introduction to Control Systems 1-10 EC-Sem-6) iven block diagram obtain the transfor funetion of the given bl era z (Bc Ris) a Gi] 2] ARTO 2025-24, Marks 10 ‘Anawer Given: Rs) (GG [Gr cay Control System 1-11 € EC-Sem-6) Fig 1.95. “( Gi G31, 4G) i) , = ? T+ GiG,Fh, cia) Ti Lh Fig.196. Gain = eal ST TG) Ho G,G,(G 460 1+ G,Gathy~ GyGrG,Ha- GGG bce ‘Que Reduce the block diagram to its canonical form and Fees Introduction to Control Systems ye12c (EC-Sem-6) Control System 1-13 (BC-Sem-6) (Ec-Sem-6) Introduction to Control Systems 4c Control System 1-18C (EC-Sem-6) Surisytem BC C-Sem-6 G; Fig. 1114. GG, | GyGqt Go > Vis) REQ Ta, Gyih [| Tae Gua ‘Answer . A. Shift take off point after block @, 2 65 oe] ma cs) - Giese) al 1+ GG 4H Rr THT Gy Hip (Gye GOST / Figtizs.” Y 3. ake of point after block, . % Fig. 18. G,G,GNO,+ GN L, ea cs) TG HVT + G+ G GHG C= 616205 (6,* GH: Ys) . For the following block diagram, determine over all transfer function using block diagram reduction technique, Gi 176,64, Fig 24) 4. Now reduce the internal block Ly as jecacsa®) Introduction to Control Systems _ Control System 1-17 (EC-Sem-6) ne 2 Tofind the overall transfer funetion we can use the su r sae ; ;perposition rule = Cie 19) by simply applying the one input ata time and other being equal to Hi, + G,03H| a 17 G,6,H + 6160 8 First consider the input as Rs) and put X's) equals to zero. The block diagrams, Re) Fig. 112.5. e oe VF enw two thm renter ection eo ea i 7 Fig. 1.13.4, GGG © tents “ i, low reduce the internal block L, as, FAG, CGH, - C6,00.4i canes = 7 a @ Using block diagram reduction technique, find overall ; ee transfer function forthe following system £ Fig. 113.5, ‘the forward gain which are in cascade HB TE, ow : J Fig. 1.13.8. ‘The above block can be further reduced and give Xi) Rw) f. 6G, + GH, + CGH, OG (118.1) ig. 118.2. c t only X(2) is applied and R(s) equals to zero. The 1 co the Ly, here G,which form a negative feedback and is cascaded _ Gy : Xe 6G i 2 cus RO Gay a y i Fes? 7 tie Fig. 113.7. Fig iss. ‘Then further reduce the block diagram given in lop with -1 feedback, og 1-18 (EC-Sem-6) Introduction to: = CoD! Stag i Xe + = cw) SE Contret System (2C-Som-0) Que 1-14] Discuss the effect of feedback on stability of the system. Determine the overall transfer fun jon of the given system using block diagram reduction technique. ‘Be sven #7" 2966, oe ow x, Li, « 8*T+Gs| Fig: 1198: Fig. aa. ‘9. Reducing the above block diagram using feedback gain formula we get Anewor, A oo ge B, Ritectof feedback on stability Refer 1.4, Page 1-4C, Unit. Xa = & Oo). SE 1. First move summing point before block —[ Ey} Res (ae Le ey Coy (1.132) Fig. 1142, 10. Combining eq. (1.13.1) 2 After separating both loop : Boe ee GH, TR, GG, 0,6, 1+ G,G,,H,-6,6,H, Aaa, 3, From Fig. 1.14.4, we can write the transfer function of the system as Gs) (G,6,-G.G)0+GH,) RG) ~ 1+G,G,H,4, -G.GHH,” Que 1.15. | Briefly explain Mason's gain formula. Ris) ls) = Gis) Xs) Where G(s) = Matrix of transfer function given as Rw) [1 GH GEG, GG, GEG, ———_GG,GG,__ 146, + GH, +G,.G,H, *G.GG, 1+6,+6,H,+6,G,H, +66, ssiasticaeciieditidii ———— introduction to Control Systems, 1-20€ (EC-Sem-6) ae Mason's het ‘overall gain can be Aevermine by M a ae where, Py Forward path gain ah al oan outpot nae javolvesclosed-oop gain and mutual 4 Determinant which inves op gain! d Pelt if all possible pair of gain formula as, 1th froma specified input node gain products Tap gain products ofall possible triplets of ath and Gita acacia with he concerned path and ‘+ Favolve al closed Henps inthe graph whic evalu ofthe graph rasing the hy fom the rar patent pth facia forth hy pth egal othe vais % Totmanto Sigal ox raph which exis irom the ea i jow Graph) Queli6.] Define the following SFG (Signal FI PI nut node is defined as Forward Path : A path from the output nod cd mor ea path in which no node is enco Loom acon be part of lp ss oo mth She branch inte the node and atleast one branch ov Control System 1-21 (8C-Sem-6) Nk Sclt Loop : A focdback loop consisting of only one node i called welt 0p. Non-Touching : If there is no node common in between the two or ‘more loops, such loops are said to be non touching loops, Determination of transfer funct ‘The signal Now graph of Fig. 1.16 Fig. 1.16.2. where, P,= Forward path gain 4, = 1= (Gum of loops touching to P) +(Sum 6f2non-tovching 3. Forward paths: P,= 6, G,G,;P,=6, . Ly= ~O,Gy My Ly=~G,Hy Ly =G, Gy, ay21-0=1 42° 1-1-6, 6, H,-G, 1H, +G, G,H,) 47 146, G,H;+0,1,-G,G,H, 4. ‘There is no two non-touching loop. A= 1-1, +1, +19 B= 1-b-G,Gglhy- GH, + 6,611) © FIAGGyI,+ Gt, -G,6,21, aco (& G,6,+G, 8.8.6, 0.-00 26,1, 0,0, 1) Re) 17G, 6,11, +6,H, +, G, H, Que 1.17,] Using Maso: function for the following gain formula, find over all transfer 1-22€ (BC-Sem-6) (ARTO 2018-19, Marks 07 === fe gain of the forward paths, P,=0,G,6,60P2= G,G,G,65 F sei L,=GHy Ly=Gells LyaGHe ba= CoH 1 non-touching 10 a Teen tye OL abs WGC Liban GsOrHly Cab EH, + Set Oy Gla Ola Gh OTE Oe i ap Aye 1a Whyt Lys Geile GH) aye l=, + L)=1- Git GH) 5. Mason'sgain formula, ‘Que 118; | Find overall transfer’ Rie) j z cw TARTU 2016-19, Marks 07 Introduction to Control Systems Control Syatem 1-23 (EC-Sem-6) Tee GGG 1+G,G,G, +G,H, + C.GH, + HHH, «GHG, HGH, Que For the system shown in Fig. 1.19.1, obtain the transfer ig 1.19.1, funetion by signal flow graph method. [ARTO 2021-22, Marks 10 r Introduction to Control Syston, 1-24 (EC-Sem-6) : Control System “Answer 1-25 € (EC-Sem-6) * Fig. 1192. v=2 No. of forward paths, N 3 oer = first forward path sy 2 9277 92-3 OA ain Bye abode Fig. 1.2022 ws First forvard path gain P, ot Ye-27e where 9915 0,0,1, + 0, 65Hy~ 0, 0,6, + OH, second forward path is y+ 9277 93 saree veonsi=s _ 126.6, H,-6,6.1,+6, 1, Sa ea I, sedh, fy di and ly =P = . 0,6,6,+6,6, * 15GG, 1, +6, 6, 1,76, H, ‘Que 121. ] Find the transfer function ¥¥Y, of the signal flow graph shown in Fig. 121.1, gh sedh odi f+ (hjdi SBS (0 pogh vedh edi fh ob) dio bf Mat towching to fire forward ‘There is no loop whi aye 1 Similarly, s Bat PS, 3 re BAR (adede (ey rahe oll Que 120.] Using Mason's gain form transfer function : Dede bf valuate the overall Fig. 121.1. ‘AKTU 2022-25, Marks 10) ‘Answer Considering the signal flow graph the value of can be obtained as A= 1 ~ (sum of the gains of all individual loops) + (aum of products of gains of all combination of non touching loops) ~ (aum of products of gains of all le combination of thi na tudhlng ipa) 9 Po tak 1+ Oty» Gay + Oy + 040g» Hg Gy Gy C, Qty 6,0, He 6, 0 nite ob, G0k Hecho Mee He Chetan MS OO OOM OE, A, introduction to Control Systems 1-26¢ (EC-Sem-6) ransfor function Yq/ ¥, is given 05+ ee, GG,GG.0+ GH +o he a raph for the following set of Fir] Construct the signal ow coe vtancous equations and obtain the Sse ein Om 4% y= Aa + Aaa * Asks xa haket Ake AR 2008-24, Marks 10) ‘overall transfer function CS ye Ahi Aas sake tAs%s Aas ‘dures equations are RO ia taki edad Aa, Asy Ls Nye di PANatAse%s pe ae Ae 7a x, Sedaka 2 Aw ‘The complete signal flow graph is Control System 1-270 (2C-Sem-6) Forward pathe : Loops: ysAs ‘The loops Land L, cre teaching to each other Se t-i,+L4) th the oop ae touching hence = y= = 1 touching hance i, ot in ota X, _ TA, + TA, + Ta, #76, x” a T= Ay Ay An PART-3 stems : Electrical Networks, Mechanical "ree Body Diagram, Analogous Systems, Que 1.23. | Derive force-voltage analogy. 2 ead anes eas : reat ue aaa —— fetes ‘1 . The equation of motion for the system is obtained by applying D’Alembert's principle, 2 Here, Introduetion to Control s; 1-29 (EC-Sem-6) ‘onal mechanical system : Refer Q. 1.22, Page 1-27C, Equation of spring-mass-damper system jon of voltage equation joring electrical circ ee |. Force-Current analogy : |. The nodal equation as obtained for the analogous to spring-mass-damper syste: 1.24.1) circuit shown in Fig, 1.24.1 is sm deseribed by the eq. (1.24.1). revolt 36% 2,2, Electrical cirevit fr foree sot Fig, 1.24.1. Electrical circuit for force-current analogy. -Thevoluage equation forthe circuits. a8 fl 2 According to nodal analysi rie ch ‘As the curront is the rate of flow of elect 1.24.2) ments of the cireuit is 1.24.3) (1.244) (1.24.5) 4.5) in-eq, (1.24.2), (1.24.6) kes associated with inductance Las 24.7) in eq. (1.24.6), we got Que i.24, | Derive force-current analogy. ay 1db, Cae Rat 7 oo syand (1.24.8), we get 7 Comparing te 1.24.1. Foree-Current Analogy Aye i 7 Capacitance, C Dass. 1-81 C (EC-Sem-6) juslinkage,@ | | Fluxlinkage,¢ 7] Damping coefficient, B Conductance, @ Beifiness, K (compliance, UK) aw the analogous electrical cireuit for the Fig. 125.3. ‘Que 126. | Determine the transfor function ¥,(s)/F(a) of the system FLV and F-l analogy. All the symbols have y Atnode y,:M, 2% + K,y,-y,)=0 Taking Laplace of eq. (1.2 Fig. 1.26.2. wy, dy, #Y sky, +8 % 4 Ky, -. ae * Kt BT + Kyo,-¥.) (1.26. (1.26.2) eget + K,I¥ (8) -¥o)] (1.26.8) +Kzp+ Ky +Bsl-K,Y fs) a. rrovetin to Cot ya aaciecsom®) eee 1-33 C (EC-Sem-6) ee 2. Free body diagram of My: Ku,-yd—-— fy Mid’ a , Modeling of Armature Controlled and Field Controlled DC Servomotor. Figs 127.2. Que 128. | What is the role of sensors and encoders in control system ? Explain the construction and principle of potentiometer. ea A. Role of sensors and encoders : Sensors ‘monitor the performance and feedback in co Introduction to Control S; 1-34 EC-Sem-6) cacpl of4 esruction ana surton a reser ical enereY- sl displacement © entiometer + mat converts mechanical ene, intoel “The mechanic the potentiometer ono tage is appl \e fixed termi When the vol aera For precision cont thas infinite resolution Write a short note on armature-contrel DC servomator. on Derive the transfer function for armature controlled DC servo mot Control System 1.35 C (EC-Sem-6) 2 Let R, = Armature resistance L, = Armature inductance f, = Armature current E-= Induced emfin armature V= Applied armature voltage displacement of the motor shaft jeveloped by motor f inertia of motor shaft 9 Tw) ew at. +R aaa + BY aK, Fig. 1.29.2. (30, | Discuss about field controlled DC servomotor. Introduction to Control g, 1-36c (EC-Sem-6) 3. The torque T developed by the motor Ss Proper tur-gop flux 6 and armature current‘, £0 eRe T+K,¢ current ipare Propo asumed 7. From the above equations the transfer function ofthis «ystem is obtained a) k, Ea) ~ Sark, se ris proportional to product of (Contra! Systara 1-87€ (EC-Sem-6) 8. Block diagram is shown in Fig. 1.90.2. , 1_|Lyn [e a La + Ry [s(J, + B) ay Fig. 1.30.2. VERY IMPORTANT QUESTIONS. Following questions are very important. These questions may be asked in your SESSIONALS as well as UNIVERSITY EXAMINATION. ‘uss the effect of feedback on following: Stability’ ii, Sensitivity @.2. Obtain overall transfer function for the given block diagram shown in Fig. 1 using block reduction method. i] = oH Rie) + + ~ Gee Cy G Ty Fig. An& Refer Q.1.11 Q3. For the following block diagram, determine over all transfer function using block diagram reduction technique. Ris) Sy ss cis) Introduction to Control g, 1-38 C (EC-Sem-6) Ang Refer Q. 1.12. own in Fig. 3, obtain the transfer func, :ph method. Q.4. For the system shi by signal flow grat Ane ReferQ 119 Q.5. Find the transfer function rw, shown in Fig. 4 of the sigmal Mow ra ana Refer Q320 Q.6. Determine the transfer function Ys) / Fla) of the system shown in Fig. 5 asc Refer Q. 1.26.

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