Control Principles 2. 1 System Elements and Analogies. Learning Outcomes

Transcription

1 Cotrol Priciples 2 Learig Outcomes This chapter buils upo a extes the basic priciples of cotrol systems that are ealt with i Cotrol Priciples 1. O completio you shoul be able to: 1 Appreciate the ifferet forms of system elemet, a aalogous elemets. 2 Ietify a rage of first-orer systems, a escribe their respose to step isturbace iputs. 3 Distiguish betwee first-orer a seco-orer systems. 4 Describe the respose of seco-orer systems to both step iputs a forcig fuctios, particularly with referece to system ampig. 5 Explai the effect of mechaical resoace, a methos of reucig its effects. 6 Uersta the cocepts of proportioal, ifferetial a itegral cotrol strategies. 1 System Elemets a Aalogies All types of system cosist of elemets or compoets that are itercoecte i such a way that the behaviour of each elemet has a effect o the behaviour of the whole system. Some system elemets are capable of storig eergy whilst others ca oly issipate eergy. Of the eergy storage elemets, some are associate with kietic eergy (or its equivalet); the others beig associate with potetial eergy (or its equivalet). I orer to put these cocepts ito perspective cosier Table 1, which shows the basic Table 1 Eergy storage a issipative elemets System KE storage PE storage Dissipative Mechaical 1 mass mv sprig ks 2 2 amper s 122 Electrical 1 iuctor LI capacitor CV 2 2 resistor q

2 Cotrol Priciples mechaical a electrical elemets classifie ito the three types metioe above. Cosierig Table 1, the followig aalogies ca be mae: Both the mass a the iuctor store eergy i a form associate with movemet. I the case of the mass, this will be kietic eergy by virtue of its velocity. For the iuctor, the eergy is store i the magetic fiel prouce by the curret flowig. emember, v s / metre per seco; a I q / coulomb/seco (amp), i.e. the curret is the rate at which charge is isplace. Both the sprig a the capacitor store eergy by virtue of positio or potetial that is, potetial eergy. For example, the further the sprig is isplace the more eergy it will store. Similarly, the more charge place o the plates of a capacitor the larger its p.., hece more eergy store. Both the amper a the resistor have a slowig ow effect i their respective systems. As a result they cause heat eergy to be issipate. The aalogies ca ow logically be extee to the system variables a costats such that: voltage, V force, F charge, Q isplacemet, s curret, I velocity, v resistace, amper coefficiet, iuctace, L mass, m capacitace, C reciprocal of sprig stiffess, 1/ k 2 First-orer Systems A first-orer system is oe which cotais oly oe of the two possible types of eergy storage elemet, i combiatio with the associate eergy issipative elemet. For the elemets so far cosiere, the followig combiatios form first-orer systems: Mass-amper; sprig-amper; iuctor-resistor; a capacitorresistor. The last two combiatios we have alreay stuie i etail whe cosierig.c. trasiets i the book, Further Electrical a Electroic Priciples, ISBN (Chapter 8). I this situatio the systems (circuits) were subjecte to a step iput, by coectig to or iscoectig from a.c. supply. I each case it was fou that the system respose followe a expoetial law. Let us review the saliet poits regarig the C- a L- series circuits, a see how these results ca be traslate i terms of the respose of aalogous mechaical systems.

3 124 Cotrol Priciples 2 C- series circuit a sprig-amper system Figure 1 shows a C- circuit that is subjecte to a sue step iput isturbace whe the switch is move. I this case the capacitor will charge up util the voltage betwee its plates is equal to the applie voltage. The relevat equatios for this system are as follows. C q i V V C E Fig. 1 E v vc volt q sice v i ; a vc q q the, E C q C volt...[ 1] The above equatio is kow as the ifferetial or system equatio for the C- circuit. We have see that its solutio shows that the circuit p..s, charge, a curret vary expoetially with time util the capacitor reaches its fully-charge state, such that: vc E( 1e t / τ) volt; i Ioe t/ τ amp qq( 1e t/ τ ) coulomb; a τ C secos S k F 2 F F 1 Fig. 2 The sprig-amper arragemet show i Fig. 2 is subjecte to a step iput by suely applyig a force F to it. The sprig has a stiffess of k ewto/metre, a the system will isplace util the reactio force of the sprig ( F 2 ) is equal to the applie force. The amper has a ampig coefficiet of ewto per metre/seco, a will exert a reactio force ( F 1 ) oly whilst the system is i motio. This may be compare to a resistor proucig a p.. oly whilst charge is movig through it,

4 Cotrol Priciples i.e. whilst carryig curret. Combiig the forces i this mechaical system results i the followig ifferetial equatio: F F1F2 ewto s where F1 v ; a F2 ks s so, F ks ewto...[ 2] Comparig eq [1] a [2] we ca coclue that the isplacemet, s, of the system will icrease, a the velocity, s / t, will ecrease expoetially with time. Thus: where where a system time costat, s F k v F k F ( 1e t/ τ ) metre, fial isplacemet F e t/ τ me iitial velocity τ k secos tre/seco, Note : C 1 or k 1 as state i Sectio 1 k C If the switch i the C- circuit is ow reture to its origial positio the the capacitor will ischarge expoetially. Similarly, if the applie force is ow remove from the sprig, it will retur to its ormal legth. This isplacemet will also be expoetial i form. L- series circuit a mass-amper system From Chapter 8 we kow that whe a L- circuit is coecte /iscoecte to/from a.c. supply, that the curret will icrease/ ecrease expoetially. The aalogous mechaical system is the massamper subjecte to a suely applie force, as show i Fig. 3. F 1 m F F 2 Fig. 3

5 126 Cotrol Priciples 2 Sice we have a aalogous system we ca coclue that the mass will accelerate util the reactio force of the amper causes the spee to stabilise at some costat value. The reaso for this particular coclusio is that electrical curret is aalogous to velocity, a i the L- circuit the curret buils up to some steay value limite by the circuit resistace. The system ifferetial equatio for the massamper will be as follows: F F1F2 ewto where, F1 mass acceleratio ma ewto m v ewto a F2 v ewto therefore, F m v v ewto F Hece, v ( 1e t/( τ) ) metre/seco; m a τ secos where F / is the fial velocity. The system equatio for the L- circuit is: V L i i volt The cocept of first-orer systems is ot cofie to the electrical a liear mechaical examples so far cosiere. otary mechaical systems, ivolvig agular acceleratio ( α ), agular velocity ( ω ), agular isplacemet ( θ ) a applie torque ( T ), will react i a similar maer. So too will the equivalet hyraulic a thermal systems. Thus the behaviour of ay first-orer system i respose to a step iput isturbace may be summarise as follows. System equatio of the form Y A x t Bx the system variable, x, will respo expoetially such that Y x ( 1e t/ τ ) for expoetial growth B Y x e t/ τ for expoetial ecay; B a τ A secos B

6 Cotrol Priciples Worke Example 1 Q A sprig-amper system as show i Fig. 4 is subjecte to a force of 25 N. The sprig stiffess is 200 N/ m a the amper coefficiet is 600 Ns/m. If this force is suely remove, etermie (a) the system time costat, (b) the time take for the system to move 50 mm from its iitial startig poit, a (c) the time take for the sprig to retur to its ormal ucompresse legth. A F 25 N; k 200 N/m; 600 Ns/m F 25 N 600 Ns/m k 200 N/m Fig. 4 (a) τ k secos therefore τ 3 s As (b) I orer to etermie etails of how the sprig will retur to its ustresse legth, we firstly have to etermie the istace the applie force will iitially isplace it. Iitial isplacemet, F si k metre mm Comparig this system with its electrical aalogue, we ca say that this isplacemet correspos to the iitial charge place o a capacitor. Whe the force is release it will be equivalet to startig the ischarge process for the capacitor. Thus the isplacemet for which the time is require to be calculate ivolves a ecayig expoetial fuctio. s si e t/ τ metre s e t/ τ si t s τ si t hece t 275. s As (c) The system will take approximately five time costats to reach its ew steay state, so time take for sprig to ustretch 15 s As

7 128 Cotrol Priciples 2 3 Ieal Seco-orer Systems A seco-orer system is oe which cotais both forms of eergy storage elemet (oe for KE a oe for PE ). Let us firstly cosier a ieal seco-orer system. This is oe i which there is o loss or issipatio of eergy, hece the eergy issipative elemet is abset. Such a ieal is impossible to achieve i practice sice it woul result i perpetual motio. However, we ee to stuy the behaviour of such a system i orer to appreciate the behaviour of a real secoorer system. Cosier a mass-sprig system as i Fig. 5, which is free from all ampig effects, such as frictio etc. Figure 5(a) illustrates the settig of the iitial coitios, whereby the system is give a iitial isplacemet by the applicatio of a force. (a) s m k PEmax KEzero (b) m k PEzero KEmax s k (c) m PEmax KEzero atum Fig. 5 Uer these iitial coitios the system isplacemet will be at its maximum value, a the eergy iput to the system will be store as PE i the sprig (0.5 ks 2 joule). No eergy will be store i the mass at this stage sice it will be statioary. At some time t 0, the system is release by removig the applie force. The sprig will assert itself a rive the mass back towars its ormal atum positio. I so oig the sprig gives up eergy to the system. Sice the mass is beig accelerate, the it is gaiig KE (0.5 mv 2 joule). This eergy ca oly have bee obtaie from the sprig. Thus, the PE i the sprig is beig trasferre ito KE i the mass. Figure 5(b) shows the istat i time where the system has reache its ormal atum poit (i.e. zero isplacemet). At this time the sprig will have trasferre all of its eergy to the mass, so the mass will have reache its maximum velocity. The iertia of the mass will ow cause it to cotiue beyo the atum poit. As it oes so the sprig

8 Cotrol Priciples will begi to stretch. Thus the mass will ow be trasferrig eergy back to the sprig. Figure 5(c) shows the istat that the mass comes to rest, at the system maximum isplacemet o the other sie of the atum. At this poit all of the eergy will oce more be store i the sprig. The whole sequece will the be repeate, a the system will oscillate back a forth with a costat amplitue of s metre. Sice o eergy is lost from the system, the these oscillatios woul cotiue iefiitely perpetual motio! I exactly the same way as was use i the first-orer systems, we ca euce the system ifferetial equatio for this system by equatig the forces ivolve, as follows. Iertia force ue to the mass ma m v ewto a stiffess force ue to sprig ks ewto therefore, F m v ks ewto s a sice v ; the v/ 2s/2 t 2 so, F m s ks ewto 2 a at time t 0, F 0, so equatio becomes: 2 0 m s ks 2 The above equatio escribes simple harmoic motio, a the solutio to such a equatio is a cosie waveform. This meas that a graph of isplacemet will be a cosie waveform, as show i Fig. 6. If the mass is large a the sprig soft, the the frequecy of oscillatio will be relatively low. O the other ha, a stiff sprig a small mass woul result i a relatively high frequecy of oscillatio. I geeral we ca coclue that the frequecy of oscillatio is i some way irectly proportioal to the sprig stiffess k, a iversely proportioal to the mass m. This frequecy of oscillatio is kow as the atural frequecy, f, for the system, a it may be calculate from: s(m) S 0 t(s) S Fig. 6

9 130 Cotrol Priciples 2 ω k ra/s m (1) k so, f 1 hertz 2π m (2) 1 2 E C L Fig. 7 It may occur to you that these equatios seem vaguely familiar. That is as it shoul be, sice the equivalet electrical system is the L-C circuit show i Fig. 7. I this circuit, the capacitor woul iitially be charge with the switch i positio 1. Whe the switch is move to positio 2 the eergy store i C will rive a curret through L, thus trasferrig eergy to the magetic fiel of the iuctor. The result will be a cotiuous oscillatio of curret a eergy betwee the capacitor a iuctor (assumig zero resistace). The frequecy of oscillatio woul be: ω 1 ra/s; or f 1 LC 2π LC hertz You shoul recogise the above equatios as those for the resoat frequecy for a series L-C circuit, or, for a parallel resoat circuit where the resistace is zero. Bearig i mi that accorig to our aalogies C 1/ k, the it may be see that the resoat or atural frequecy equatio for the electrical circuit correspos irectly with that for the mechaical system show. 4 Practical Seco-orer Systems Eve if a amper elemet is ot iclue i a mechaical system, there will always be some uavoiable ampig effect ue to frictio. Similarly, i a electrical circuit, the coectig wires a iuctors will have some resistace. Sice ay form of ampig issipates eergy, the the oscillatios iuce ito a practical seco-orer system must ie away with time. Cosier a practical mechaical seco-orer system that is give a iitial isplacemet, a is the release. The system is represete i Fig. 8. The reactio forces of

10 Cotrol Priciples the three elemets are show as F 1 to F 3, a the system ifferetial equatio will be: F F1F2 F3 ewto 2 m s s F 2 step iput isturbace iertia term ampig term ks stiffess term The solutio of seco-orer ifferetial equatios is beyo the scope of the mathematics require for the course of stuy beig uertake, so the geeral solutio is show below. s Ae αt si ω t metre k F 3 m F 1 F F 2 S i Fig. 8 Aalysig this equatio shows that there is a oscillatory compoet (si ω t ), the amplitue of which ( Ae α t ) is a ecayig expoetial. Thus the system will oscillate at a frequecy of ω ra/s ( f ω /2π hertz), a these oscillatios will ecay expoetially with time. The respose of the system, whe the force F is remove, is show i Fig. 9. S(m) S i expoetial evelope 0 t(s) Fig. 9 The term α i the ecay compoet of the above equatio is a irect fuctio of the system ampig. Thus the greater the ampig, the

11 132 Cotrol Priciples 2 more rapily will the oscillatios ie away. Iee, if the ampig is icrease sufficietly, a poit will be reache where the oscillatios are just prevete from occurrig. This egree of ampig is kow as critical ampig crit. Uer this coitio the respose of the system will be approximately expoetial. Although the system will have a specific value of ampig coefficiet, the overall ampig is ormally referre to i terms of the system ampig ratio ζ (Greek letter zeta). Dampig ratio is efie as the ratio of the ampig preset to that require to just prevet oscillatios. I the form of a equatio this is: ζ crit (3) The egree of ampig preset also affects the frequecy of ay oscillatios. Sice the system cotais both mass a sprig stiffess, the the system will have a atural frequecy of oscillatio, f. As the ampig is icrease so the frequecy of oscillatio is ecrease, so the ampe frequecy f is always less tha the atural frequecy. The relatioship is give by: f f 1ζ 2 hertz (4) The equivalet electrical circuit is show i Fig. 10, a the relevat system equatios are liste below. E v v v L C volt 2q E L q q so, 2 C volt L C i V V L V C 2 1 E Fig. 10 ω 1 1 ra/s; f LC 2π LC ω ω 1ζ f f 2 ra/s; 1ζ2 hertz hertz

12 Cotrol Priciples I this circuit, whe the switch is move from positio 1 to 2 the charge i the system will icrease from zero to its fial steay value, whilst the circuit curret will ecrease from its iitial value, to zero. The graphs of these variatios are illustrate i Fig. 11. eturig the switch to positio 1 woul be equivalet to releasig the mass-sprigamper as previously escribe. q (C) Q i (A) o 0 t (s) 0 t (s) Fig. 11 The effect of ampig ratio ζ o the respose of ay seco-orer system ca be etermie from a set of geeralise respose curves, a example of which is show i Fig. 12. A alterative approach is to use a geeralise form of the system ifferetial equatio. Cosier the above electrical system: E Lq q q/ C where q 2q/ t2 a q q/ a iviig through by the coefficiet L : E L q q L q LC The coefficiet of the last term is 1/ LC ω 2. It is fou that the coefficiet of the seco term, /L 2ζ ω. Thus, if ay seco-orer system has a equatio of the form Y ax bx cx Y bx cx the x a a a Y a x2ζωx ω2 x a (5)

13 134 Cotrol Priciples ζ 0.3 ζ ζ θ o /θ i ζ 1 ζ ζω t Fig. 12 Worke Example 2 Q A liear mechaical system has a total mass of 15 kg, a effective sprig stiffess of 10 N/m, a a ampig coefficiet of 20 Ns/m. If this system is subjecte to a sue applie force of 25 N, write ow the system ifferetial equatio a hece etermie (a) the atural frequecy, a (b) the system ampig ratio. A m 15 kg; k 10 N/m; 20 Ns/m; F 25 N F ms s ks ewto so 2515 s20s 10s As (a) hece, 25 20s 10s s a comparig coefficiets with the geeral equatio (5) so ω f ra/s Hz As 2π

14 Cotrol Priciples (b) 2ζω therefore ζ As Drive Dampe Systems A rivig or forcig fuctio iput may have a waveshape that is siusoial, rectagular, triagular, or iee ay other waveshape. For mechaical systems, such a forcig fuctio may be the result of vibratio. I this case the waveshape will be approximately siusoial. Figure 13 represets a mechaical system subjecte to a siusoial forcig fuctio. k ω s.h.m m Fig. 13 This system will have a atural frequecy of oscillatio f hertz, or ω ra/s. Provie that the forcig iput has a frequecy ω that is very much less tha the system ω, the the mass i the system will closely follow the forcig iput motio. This may easily be cofirme by the followig simple test. Hol oe e of a reasoably soft sprig which has a mass suspee from the other e. Now move your ha up a ow very slowly, a you shoul fi that the mass a sprig move i sympathy with the iput isplacemet. As the iput forcig frequecy is icrease (graually icrease the spee of the up-aow movemet of your ha) it will be fou that the isplacemet of the mass starts to become out-of-step with the iput. I aitio it will be fou that the amplitue of mass isplacemet also icreases. At a particular iput frequecy it will be fou that the isplacemet of the mass is 90 out-of-step with the iput, a its amplitue will be cosierably greater tha the iput isplacemet. Take care with your test rig, sice uer this coitio the mass may well etach itself from the sprig a o some amage! This effect occurs at the resoat frequecy f o for the system. If the iput frequecy is icrease further, the the amplitue of the oscillatios of the mass will begi to ecrease, a its agle of lag will cotiue to icrease. There will come a poit whe the iertia of the mass is such that the mass caot

15 136 Cotrol Priciples 2 keep up with the rivig fuctio, a its isplacemet becomes virtually zero. The relatioship betwee the system isplacemet to the frequecy of the forcig fuctio is show i Fig. 14, a the agular relatioship betwee iput a mass isplacemet is show i Fig. 15. Note that both of these iagrams are geeralise (apply to ay system) by makig the frequecy axes the ratio f /f i each case, a the vertical axis of Fig. 14 the amplitue ratio s o /s i. I the latter case, s i, represets the iput isplacemet, a s o represets the system isplacemet. Thus at very low iput frequecies, s o s i (mass i step with iput), so s o /s i ζ 0.1 S o /S i 2.5 ζ ζ 0.3 ζ 0.4 ζ ζ 0.7 ζ 1 ζ f/f Fig. 14 From Fig. 14 it is apparet that if the system is oly lightly ampe the the isplacemet a velocity of the mass ca be very large at the resoat frequecy. This coul well lea to structural amage, a for this reaso the resoat coitio is ormally avoie i mechaical systems. If iputs at or ear the resoat frequecy are uavoiable,

16 Cotrol Priciples f/f ξ 0.1 ξ 0.2 ξ 0.5 agle φ (egrees) 90 ξ 1 ξ Fig. 15 the the system ampig must be icrease. This will have the effect of reucig the amplitue of oscillatios. From the amplitue/frequecy respose curve it may be see that if ζ 0.7, the the isplacemet of the mass will ot excee that of the iput. Also, at frequecies i excess of 1.5 f o the isplacemet of the mass becomes progressively less a less. A classic example of these effects may be experiece whe rivig a car i which the roawheels are out of balace. I this case it will be fou that at oe particular spee the vibratios ue to the imbalace become very oticeable. If the spee is either icrease or ecrease from this value it will be fou that the vibratios become much weaker, or eve isappear altogether. 6 Cotrol System Strategies I Cotrol Priciples 1 we briefly cosiere the problem of cotrollig the positio of a large raio telescope atea. The iitial solutio to this problem ivolve the use of a very simple close-loop cotrol system, the iagram for which is show i Fig. 16. It will be helpful at this stage to review the maer i which this system operates. This is summarise by escribig the fuctio of each of the system elemets, as follows. Proportioal cotrol This cotrol strategy is calle proportioal cotrol because the feeback sigal is irectly proportioal to the agular positio of the output.

17 138 Cotrol Priciples 2 θ i V i A 1 error sigal A 2 eferece supply V o P2 P 1 eferece supply Costat fiel supply M 1 LOAD θ o Positioal (proportioal) feeback A 3 V o Fig. 16 P 1 a P 2 These are the iput a output potetiometers (trasucers) which provie the electrical sigals ( V i a V o ) that are proportioal to the iput or emae isplacemet a the actual output isplacemet, θ i a θ o respectively. A 1 This is the error etector, i the form of a summig operatioal amplifier. The output of this evice will be the error sigal, equal to V i V o. A 2 This is a power or servo amplifier which provies a much amplifie error sigal to the rive motor. M This is a separately excite.c. rivig motor. This is a step-ow gearbox to provie the large torque require by the loa. A 3 This is a ivertig operatioal amplifier to esure that the feeback sigal V o is i oppositio to the iput V i. This esures that egative feeback is applie. I aitio to the mass of the loa are the masses of motor armature, rive shafts, gearbox etc. Therefore the total mass, a hece iertia, of the system will be large. The stiffess of the system is cotribute

18 Cotrol Priciples by the rive shafts a gearbox, a the amplificatio i the system will have a similar effect to mechaical stiffess. I orer to miimise ay steay state error the system gai (amplificatio) ees to be high. O the other ha, the frictio i the system will be miimal ue to the use of high quality bearigs, lubricatio etc. The whole system will therefore be a severely uerampe seco-orer system which, ue to the high gai, will be very resposive to emae chages. Thus, whe a step ema iput is applie, the system will respo rapily with cosequet violet oscillatios of the atea about its ew positio before fially settlig. This oscillatory respose is obviously uesirable, especially sice large masses are ivolve. The solutio is to slow ow the system by icreasig the ampig ratio. This effect coul be achieve by icreasig the frictio i the system, perhaps by the use of some form of brakig. However, icreasig viscous frictio is to be avoie because it wastes eergy, it icreases steay state error, a it was iitially esige to be miimal ayway! Thus the slowig ow effect of viscous frictio ees to be simulate, without the other averse effects. This may be achieve by meas of erivative feeback as follows. Derivative cotrol The origial positioal (proportioal) feeback must be retaie sice this forms the basic close-loop cotrol which esures that the system is error-actuate. The erivative feeback may be achieve by several meas, oe of which is by the use of a tachogeerator moute o the motor riveshaft as show i Fig. 17. The output of the tachogeerator is irectly proportioal to the spee at which it rotates, so it is proportioal to the rate of chage of output positio, i.e. V θ o /, so the tacho voltage is the erivative of the output positio. This feeback sigal is also of the opposite polarity θ i V o V i A 1 A 2 eferece supply P 2 P 1 eferece supply V Costat fiel supply M T.G. 1 LOAD θ o Derivative (velocity) feeback P 3 Proportioal (positio) feeback A 3 V o Fig. 17

19 140 Cotrol Priciples 2 to the ema sigal, so the total error sigal will ow be give by V i (V o V ) volt. The major avatage of tachogeerator feeback is that it will be at its greatest whe the output shaft is travellig at its fastest (through the oscillatory cross-over poits o the respose graph), a will be zero whe the shaft is statioary. This meas that the ampig effect of this feeback is preset oly whe it is require a will, ulike frictio, ot affect the steay state error i the shaft positio. I aitio, the egree of ampig prouce ca very simply be ajuste by meas of the associate potetiometer, P 3. Although the use of a tachogeerator woul seem to be the ieal solutio to the problem of oscillatory istability i a positioal cotrol system, it ca have a averse effect i a velocity cotrol system, or a positioal cotrol system subjecte to a ramp iput ema. Velocity lag A ramp iput is simply a step velocity iput, as illustrate i Figs. 18(a) a (b). This type of iput woul apply to a positioal cotrol system whe it is require to rotate at some costat velocity. For example, this woul apply to a raar atea whe it is require for the atea to lock o to a follow a target. Sice the system has to be erroractuate, the to keep the motor rivig there has to be some ifferece betwee θ o a θ i, to provie the ecessary error sigal. Cosier the atea rotatig at the appropriate spee to match that of the target ( ω o ω i ). I orer to provie the error sigal to maitai this velocity there will be a agular misaligmet betwee the actual istataeous positio of the atea a the esire positio. This meas that the atea agular positio will be laggig behi the target. This agular ifferece is calle the velocity lag, a from Figs. 19(a) a (b) it may be see that velocity lag (steay state error) icreases with ampig. Hece if a tachogeerator is use, the ampig sigal that it prouces whilst rotatig will icrease the velocity lag. This problem may be avoie if the velocity feeback is prouce by passig the error sigal through either a passive C- ifferetiatig etwork, or a ifferetiatig operatioal amplifier. The ifferetiator a the error etector usually form parts of a elemet calle a cotroller. Thus a ema θ i ema θ i ω i t (s) ramp isplacemet (a) Fig. 18 step velocity (b) t (s)

20 Cotrol Priciples θ θ i θo θ θ i θ o θ ss θ ss t (s) (a) overampe t (s) (b) uerampe Fig. 19 system that employs both proportioal a erivative feeback is sai to have a two-term cotroller. Itegral cotrol The ifferetiatig etwork metioe above is satisfactory for lowpower systems i which the loa o the motor is relatively small a costat. However, where there is sigificat stictio (static or coulomb frictio, iepeet of velocity) i the system, the steay state errors ca still be sigificat, sice stictio icreases the ea-ba. As the loa approaches the esire aligmet positio ( θ o θ i ), the error voltage becomes progressively smaller. Figure 20 illustrates this poit a shows the ea-ba. I this cotext, the otte lies represet the miimum error voltage require to overcome the stictio. Thus, ue to its iertia, the loa coul come to rest at ay positio betwee poits (a) a (b). The system ea-ba coul be reuce by icreasig the system gai, but too much gai tes to prouce istability, i the form of oscillatios. The solutio to the problem is the use of itegral feeback, either i the form of a passive C- itegratig etwork or a itegratig operatioal amplifier. I either case, a p.. is built up across the capacitor whilst the loa is approachig aligmet. This p.. is sufficiet to supplemet the error sigal to the extet that sufficiet error voltage will ow be preset to overcome the ea-ba. With careful ajustmet it is possible to elimiate the steay state error. The arragemet for a three-term or PID cotroller is show i Fig. 21. error 0 ea-ba (a) (b) Fig. 20

21 142 Cotrol Priciples 2 V i error V i V o V o P to servo amplifier D Fig Computer Cotrol of Systems iput ema θ i Icreasigly the cotrol of moer systems is achieve by microprocessors a microcomputers. Oe example is the microprocessor i a washig machie, which etermies the sequece a timig of the various operatios. I orer to achieve this the processor acts o iformatio provie by sesors (trasucers) that moitor such variables as water level, temperature, rum spee, programme selectio, etc. A example of the use of a computer woul be the cotrol of a maufacturig process, which coul ivolve the cotrol of, a respose to, a large umber of variables, icluig material positio a availability, temperature, flowrate, pressure, tool positio, velocity, safety features, etc. The iformatio regarig all the variables will also be supplie by appropriate trasucers. A block iagram for a typical system is show i Fig. 22. Computer/ µprocesser igital error DAC aalogue error Cotroller System output θ o igital feeback ADC aalogue feeback V o Trasucer Fig. 22

22 Cotrol Priciples The variables to be sese a cotrolle are aalogue quatities, whereas the computer or microprocessor is a igital evice. For this reaso the iformatio fe back from the trasucers has to be coverte ito biary ata which ca be processe by the computer. This is achieve by a evice kow as a aalogue to igital coverter (ADC). Similarly, the cotrol sigals geerate by the computer have to be coverte from igital to aalogue sigals. This is achieve by a igital to aalogue coverter (DAC). The cotrol strategies escribe i the previous sectio may be achieve by a computer or microprocessor. The maer i which specific cotrol actios are achieve will ot be ealt with here, sice such etails form part of the moules (uits) coverig computer applicatios a microelectroic systems. Summary of Equatios First-orer systems: Geeral system equatio, Y A x t Bx Time costat, τ A B seco Y Istataeous value for growth, x ( 1e t / τ ) B Istataeous value for ecay, x Y e t/τ B Ieal seco-orer: Geeral equatio (SHM), Y ax bx Natural frequecy, f 1 2π Practical seco-orer: b a hertz Geeral equatio, Y bx cx x x2ζvx v2 x a a a Natural frequecy, Dampig ratio, ζ v 1 2 π crit c a ra/seco Dampig frequecy, f f 1z 2 hertz

23 144 Cotrol Priciples 2 Assigmet Questios 1 Deuce the electrical aalogues for torque, iertia, agular velocity, agular isplacemet, a a torsio bar. 2 A sprig of stiffess 1.5 kn/m is attache to a mass of 20 kg. What will be the atural frequecy of oscillatio of this system whe isturbe? 3 A mass-sprig system has a atural frequecy of 1.5 Hz. If the mass is 4 kg etermie the value of sprig stiffess. 4 A mass-sprig system has a atural frequecy of 0.5 Hz. If it is the ampe by a evice havig a ampig ratio of 0.65, etermie the ampe frequecy of oscillatio. 5 Explai the ifferece betwee first-orer a seco-orer systems, quotig examples of each. 6 A mass of 63 kg is attache to a amper of coefficiet 30 Ns/m. This system is subjecte to a suely applie force of 100 N. Write ow the system ifferetial equatio a hece etermie (a) the iitial acceleratio, (b) the time costat, a (c) the fial velocity. 7 A system comprises a 2.4 kg mass, a sprig of stiffess 9.6 N/m, a amper of coefficiet 3.2 Ns/m, a is subjecte to a step iput force of 0.5 N. Determie the system ifferetial equatio a hece, or otherwise, calculate (a) the atural frequecy, (b) the ampig ratio, (c) the ampe frequecy, () the fial isplacemet, a (e) the value of amper coefficiet that will just prevet oscillatios. 8 Explai why resoace i mechaical systems is ormally avoie, a if it caot be avoie, how the averse effects are miimise. 9 Explai what a tachogeerator is a why it may be employe i a positioal cotrol system. 10 (a) Explai what is meat by the term velocity lag, as applie to cotrol systems, a how it may be reuce, (b) What is a three-term cotroller?

24 Cotrol Priciples Supplemetary Worke Examples Supplemetary Worke Example 1 Q A 4.5 kg mass moves with a velocity of 8 m/s. If all of the kietic eergy store by this mass is trasferre to a sprig of stiffess coefficiet 55 kn/m, calculate (a) the total isplacemet of the sprig, a (b) the value of capacitor that woul be equivalet to this sprig. A m 4.5 kg; v 8 m/s; k N/m For the mass, KE 05. mv joule KE 1 44 J 2 2 (a) Whe this eergy is trasferre to the sprig it will be store as potetial eergy because the sprig will be isplace (compresse or stretche), a PE 05. ks2 joule, where s is the resultig isplacemet s2 144 s metre s mm As (b) Sice capacitace is equivalet to reciprocal of sprig stiffess the C 1 fara 1 k so C 182. µ F As 3 Supplemetary Worke Example 2 Q The buffer o a trai carriage has a sprig of stiffess coefficiet 0.2 MN/m a a amper coefficiet of 300 knm/s. Whe it is subjecte to a impact force of 20 kn, (a) write ow the system ifferetial equatio, a hece, or otherwise, calculate (b) the total isplacemet of the system, a (c) the isplacemet achieve 0.5 s after the impact force is applie. A F N ; k Ns/m; Nm/s; t 0.5 s (a) F s ks ewto s s s so, s As

25 146 Cotrol Priciples knm/s F 20 kn k 0.2 MN/m Fig. 23 (b) Total isplacemet, S, occurs whe the system comes to rest, i.e. s / t 0, so puttig this coitio ito the ifferetial equatio we have: 2020s 2 so, S 10 cm As 20 (c) System time costat, τ k seco τ 1. 5 s s S ( 1e t/ τ ) metre 10 ( 1e 05. / 1. 5) cm s 283. cm As Supplemetary Worke Example 3 Q A costat force of 120 N is applie to a 15 kg mass which is at rest at time t 0. The motio of the mass is resiste by viscous frictio of 10 Ns/m. For this system, (a) erive the system ifferetial equatio, (b) calculate the steay-state velocity achieve, a (c) sketch the equivalet electrical circuit, showig the values of the compoets use. A F 120 N; m 15 kg; 10 Ns/m F F 1 F 2 m Fig (a) The iertia force ue to the mass, F m s 1 ms ewto 2

26 Cotrol Priciples The ampig force ue to frictio, F s 2 s ewto a F F1 F2 ewto so, F ms s ewto s10s s s a, 8 s0. 67s ewto As (b) Steay-state velocity occurs whe the acceleratio ceases, i.e. ṡ 0, so from the ifferetial equatio: s 8 s metre/seco12 m/s As 067. (c) The electrical equivalets of mass, ampig a force are iuctace, resistace a emf, respectively, so: L 15 H; 10 Ω ; E 12 0 V 10 Ω L 15 H Fig. 25 E 120 V Also, sice a series electrical system is aalogous to a parallel mechaical system, the electrical equivalet will be as show i Fig. 25. Whe the switch is close at time t 0, the curret will grow to some steay value, which is limite oly by the resistace of the circuit. q I q amp 120 so, q 12 A 10 Supplemetary Worke Example 4 Q The respose of a seco-orer system to a step-iput isturbace is escribe by the followig ifferetial equatio 105θ 6θ 4θ Usig this equatio, etermie: (a) the system s atural uampe frequecy;

27 148 Cotrol Priciples 2 (b) the system ampig ratio; a (c) the ampe frequecy of oscillatio. A (a) 105 θ6θ 4θ θ θ θ θ1. 2θ 0. 8θ The staarise geeral form of ifferetial equatio for such a system is Y x2ζω x ω 2 x, a comparig coefficiets we ca say that a ω2 08. ω f hertz 2π f Hz As ra/s (b) 2ζω ζ 2ω ζ As (c) f f f Hz As ζ 2 2 Supplemetary Worke Example 5 Q The switch i the circuit of Fig. 26 is left i positio 1 util the capacitor becomes fully ischarge. Some time later, at time t 0, the switch is move to positio 2. (a) Write ow the system ifferetial equatio a hece, or otherwise, calculate (b) the system atural frequecy, (c) the ampig ratio, a () the actual ampe frequecy of oscillatio. L C 59.6 mh 1.5 kω 68 F 1 E 24 V Fig. 26

28 Cotrol Priciples A 1500 Ω ; L H; C F; E 24 V (a) This is a seco-orer system i which the charge ( q ) will icrease from zero to its fial steay-state value, with a oscillatory compoet. 2q E L q q volt 2 C E 2q q q L 2 L LC 24 2q 1500 q q q q q As 2 (b) The geeral form of equatio for a ampe seco-orer system is: Y a 2 x x 2ζω ω 2 x 2 a comparig coefficiets with the system equatio: ω ω ra/s ω f hertz 2π 2π f 2. 5 khz As (c) Agai comparig coefficiets: 2ζω ζ ζ 08. As () f f f 1 ζ 2 hertz khz 1. 5 khz As Supplemetary Worke Example 6 Q The seco-orer system illustrate i Fig. 27 has a mass of 25 kg, a amper coefficiet of 22 Ns/ m, a a sprig stiffess coefficiet of 12 N/m. Calculate (a) the system atural frequecy, (b) the icrease of amper coefficiet require to just prevet oscillatios (critical ampig), a (c) the actual ampe frequecy of oscillatio.

29 150 Cotrol Priciples 2 k m Fig. 27 A m 25 kg; 22 Ns/m; k 12 N/m (a) F ms s ks ewto 25 s22s 12s F s s s F s 088. s 048. s 25 Y x 2ζωx ω 2 x a comparig coefficiets a ω ω f 2π f 0.11 Hz As (b) 2ζω ζ but ζ so crit crit crit 22 ewto seco/metre ζ Ns/m a icrease of ampig require Ns/m As (c) f f 1 ζ 2 hertz f Hz As