Diploma | b-Teh- E letvical vy ZB (246 Vecants Block Diagram and Signal Flow ihm Graph - Part Il oa es 4 oR -6y [75 Sue + 25 Mon Tech] v Cech.) on Aptitde ) 120 minutes —_Rahath Agrawal « (No Nagadive wonakieg) Gotied oni Cantronsysbarns='S3Modelling of Control Systems Mathematical modeling Mechanical system Electrical system a Translational Rotational SeriesRLC Parallel RLCMechanical Systems 1, Mechanical Translational System: Cau) Input = Force (F), ? tr Output = Linear displacement (x) or Linear velocity (V) F ™ > may FaF,+F,+F, Fo dompiny cr dx Fe ON ky r Mata ko = a at : 2 Mechanical Rotational System Input = Torque (r), >; Output = —_ Canin ERY Tomoment & inuttia [=e T eo] yr F rdampur ety fe Ka Spring Congtost-Electrical Systems no Sed fiat = Yn 1 Series RLC Circult: R “ at @= Magnetic fluxee otra ae dt © Rotstens |T = T d+ fede + 0 ate ot ® Electrol @ Sess ke |V= 4M +R 4 47 fe ann a de. © foulld kc [T= CAs +(t)4d + 6 ae get (AorElectrical System Charge -C Current —I Voltage -V Resistance -Q. Capacitance -F Thermal System Heat flow —J Heat flow rate — temperature °C Sresmoh °C Resistance -——— W Capacitance — we *G > enoyy Pang XK time “4 (Heer Aw)(=c av at CS txdt dv J ahormod Joule , 4€& we °C eer as fae ahetm< C |e Servomechanism Servo ‘Transfer function Application ‘Armature as) Ky |Jo)__K, o. Og vloxt controlled DC |1'E,(s) __s(st, +1)'IE.(9)_ Gt, +D] djs placement servomotor - K,, = motor gain constant tm = motor time constant Field 65) _ K, wo yO * an pulort Sonoma! E,(s) s(5t;+1)(Stacs + te Fates ‘ fom gonerek, +, ~Field winding time constant dow i Timah = Mechanical time constant AC os) _ Ka servomotor Ve (s) = s+) Potentiometer Te. Gnstant) ‘otentiometer gain constant. ‘Tachometer lex K,, = Tacho-generator constant.‘Statoe rem apeseenet One rang mapped seca owns gran ree Wincng Coreen Be weg cuuaemegrapect Bearng ‘Shaft ‘ead en oop hetnae rears ud Pace = Encoder reese ar Sap neds Bee etreaners te prten ee Encoder CableAn actuator is a device that produces a motion by converting energy and signals going into the system. The motion it produces can be either rotary or linear. ewes ‘Siena cneesySexvomotos - «© Te motors which otapond ty the enwor signah ob and [accelerate tre dood quickls] axe ald “the Yomoloss + Thue motors have {adlowiny {cahora - © They one capable Auveloriny high fo¥qae and hold a Stoke position . © thoy axe able ty accelenk | decelerate” and weveise it's ¥ ditection of; YoRROn = quick Sewomebts one wed for frecise peed Control ond fiecise position Control + Such mofory ote wid in Peedbacl, Conttol fg stensMenawsrement ¢ £ gts ule veloc Lise _| —_ ce Cj © De _Genesabor ‘oe “Tachoratey =] — +A tachogeneraroy works on = a same > pif , =e ns body wohwe Spetd is fp be “ie coup led te Shayt oh qenenatoy + As coi] could t shyt volts ble permanent manus , emy is tna actoss “Gil which ig megwred by a valtmeter. a) 3 for a dc ye. E= ¢¢ Nt a =Kt + (5) Galt Re slachy- TE a Ni | cspeed) ad Congteitet) 2 et) w= do e(t) = ko(t) at e(t) = k La! {ro(=2| wlS) = ¢.0lS) “dt dt ‘ By taking Laplace transform- E(s) s0(s)© Advantages and function disadvantage of transfer ‘Advantages Disadvantages Output of the system can be easily determined. Applicable only for Tine time invariant system. Poles and zeros of the system are directly identified. Initial condition must be taken as zero. Stability can be easily determined. Applicable only for single input and single output. Gives simple mathematical algebraic equation. Controllability and observability cannot be determined, as in state variable approach. Valid only for LTI | system Internal states of system cannot be determined. Valid only for SISO It does not describe al state, This method gives only zero state response (ZSR) Tyany Functor State Vo viele method * Advantages and disadvantage of state variable model. Valid for All System (i.e. LTI_& Non LTI System) Gives both zero state response (ZSR) zero input response (ZIR) Disadvantages State explosion problem Output (Ry pons) d 4 Zero State Re forse Zen input RepoteeYi c Taially wally Relaxed Tes Vols) Gig Vols)> 20 State $e PE Lritial entaeyy is prsent in Gpacitos lp S 2IR 21K No inal pis mde Zot ont B olp — onte. pouent, ° ath (nit i 4 iosBlock Diagrams Iny ransfer Output (8) ren 6s) as) Figure-3.1: Elementofa block diagram Qs) = A{s)- G(s) S) olTF= G(s) =Block Diagram of a Closed-loop System A(s) = reference input signal C(s) = output signal or controlled variable B(s) = feedback signal E(s) = actuating signal or error signal G(s) = a = forward path transfer functio H(s) = feedback path transfer functions G(s) G(s) H(s) = open-loop transfer function Fe) 1+ G(s) H(s) - Ts) = ae = closed-loop transfer functionBlock Diagram Transformation Theorems 1. Blocks in Cascade: Baa] am} = ememam pe (@) (e) 2 Blocks in Parallel: lel: Two blocks aye in yoxalleh only ath hey the diction ap is came FS) Fe a)+ Gta) + Gyo) LO. 6 (6)3. | Consecutive summing points can be interchanged Ros) © Ris) + Xs) 7 Ris) + X{s)- Vis) Ros) © Ris)- Ys) Ris)= Yis)+ Xie) xis) ys) ys) (@) (0) Xs) 4& Shifting of a take off point after the block es) =A(5)-()5. Shifting of a take off point b. fore the block ra) a of ROA Gy LS os) =his¥ GIS) ° {as | as) =R iS) Risa) £ oe (0) (a) 6 Shifting of a summing point taffor the block. Re) A San }2 is)= 1R(s)# M81 Gle)__AS)_A_ Gey] ~~ @7. Shifting of a summing point befote tne block. ie) = (Rls) He) #189) A de anni B_ O16)= (Rs) Xs) tes }K8) = IRL) Gs) + Xs]4) 4 Take - off point com ater Sernenity point then it ig Wider os Sey le node.(a) Elimination of a summing point in a closed-loop system: E(s) = [A(s) + B(s)] Ris) + () &s) { aw) | cs) : a—i Ris) G(s) a oy [#8 | | TFG(s) Hi) . (a) ® for (vy Flbs [Che ai) [+ Quy) fos Go Fly: [CKTA = Gs) 1 Alo HIS)transfer function relating As) and A(s) for a closed-loop control system Els) | 1 Rls) ~ 1¥G(s)H(s)‘Moving a takeott point | ¥= Px ‘beyond a blockpd STR —> sey Si Right x) + TIL > Take-off - LyFor the block diagram shown in figure, determine the overall trasnfer function a ane (x) ; efter R— & | «.t43} ——| |» c 5, ce —— R "Eo2H,Reduce the block diagram shown in figure into single block representation form.Simplify the block diagram shown in figure and obtain the transfer function £5) edaEffect of feedback on overall gain is (a) Increases gain (b) Decrease gain (LL Increases or decreases gain pote (d) No effect on gain Negabive Alps Te= O 4 bu gio Poritve Elo» ChTF =Non-touching loops are said to be non- touching, if- (a) none of these (b) they do not pass any common node (c) they do not pass any common loop (d) when they have multiple inputsFor the block diagram shown in figure, the Atl transfer function is equal to - moe &© ces)Determine the closed loop transfer function C(s)/R(s) for the block diagram shown in Fig. a 1 [+ BH, — bith, 0. 1-G.H, +G,H, aH On, C(s) A G, R(s) 14G,H,+G,H,H, (d)Which one of the following effects in the system is NOT caused by negative feedback? Qain x B40 = CongfefUr (a) Reduction in gain (b) Increase in bandwidth F Increase in distortion (d) Reduction in output impedance Negative Flb - (Reduces goin 1 Incseaes Pee ® Reduce noise §& def bition @® te Up impedante , 4 ofp impedance. ® ty stability [ We Aa decrease SbbiliThe system of open loop and closed loop system are respectively we aad. (b) ee andl 1+GH 1+GH (c) wml (d) 1 andConsider the following three cases of block diagram a sae and C 2=(vag- 1) A= Karty - Voy = (WG) -Y) Ga = ¥lig, ~ VG, Block diagram B x+@ 2. .X Se} Zz + val ll ’ Block diagram C ‘hich of the above relations are correct? fa) A and B and C N@ Aand A, Band CApplications of negative feedback to a certain amplifier reduced its gain Trom 200 to QOoIF (Ge) the gain with the same feedback is to be raised to (50) in the case of another such appliance, the mo the amplifier without feedback must (5° Ee have been (a 400 wyaso CASE 2019) = Gs) (c) 500 600 Tpalox Cuse(D? OLTF = 200- OS) 200 CHIE = AC) = 100 150+ 3 u'(n=His) EOI) HIS) q 2 s 1 ee = WO > |Hos)= = | \50= et IF 200h18) bo Civ '/5) = 600For the block diagram as shown in figure, the overall transfer function & is G,G2H; (1-G\H, -G2H>) Gi be a GG, (thi) 2He, (1-G\H, +G,n,) G,G,K, © 56H, +G,R) @ GG Wan, GyThe characteristic polynomial of a system can be defined as (a) Denominator polynomial of given transfer function. eee (b) Numerator polynomial of given transfer function (c) Numerator polynomial of a closed-loop transfer function. jenominator polynomial of a closed-loop transfer function. Choyac. 4”For the block diagram shown in FigCORG) ) Ri(s)=0 GG, +G, 1+G, +G,H, 2G, G, ay en,For what value of K, are the two block diagrams as shown below equivalent? tl C(s) (a) 1 (b) 2 (c) (s +1) (d) (s+ 2)Consider the following statements in connection with the feedback of control systems: 1. Feedback can improve stability or be harm{al to stability if is not properly applied. 2. Feedback can always improve stability. In many situations the feedback can reduce the effect of noise and disturbance on system performance. 4. In general the sensitivity of the system gain of a fecdback system to a parameter variation depends on where the parameter is located. Which of these statements are correct? (a) 1,2 and 3 only (b) 1, 3 and donly (c) 1,2 and 4 only (d) 1,2,3 and 4The open-loop transfer function of a unity feedback control system is G(s)= (s+2)° The closed-loop transfer function will have poles at (a) -2,-2 (b) -2,-1 (©) -2, +2 (d) 2 41The transfer function of the system C/R is R () c (a) G/(1+GH) (b) G/(1-GH) (c) GH(1+GH) (d) GH(1-GH)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.