1 Figure. a. Block diagram repreentation o a ytem; b. block diagram repreentation o an interconnection o ubytem
2 REVIEW OF THE LAPLACE TRANSFORM Table. Laplace tranorm table
3 Table. Laplace tranorm theorem
4 Figure. Block diagram o a traner unction
5 MECHANICAL SYSTEMS TRANSFER FUNTIONS Table.4 Force-velocity, orcediplacement, and impedance tranlational relationhip or pring, vicou damper, and ma
6 Example: Find the traner unction X / F or the ytem given below Figure.5 a. Ma, pring, and damper ytem; b. block diagram
7 Figure.6 a. Free-body diagram o ma, pring, and damper ytem; b. tranormed ree-body diagram
8 t t x dt t dx dt t x d M v We now write the dierential equation o motion uing Newton law Taking the Laplace tranorm, auming zero initial condition F X M F X X X M v v
9 Solving or traner unction yield G X F M v [Sum o impedance]x [Sum o applied orce] Note that the number o equation o motion required i equal to number o linearly independent motion. Linear independence implie that a point o motion in a ytem can till move i all other point o motion are held till. Another name o the number o linearly independent motion i the number o degree o reedom.
10 Example : Find the traner unction X /F Figure.7 a. Two-degree-oreedom tranlational mechanical ytem 8 ; b. block diagram
11 Figure.8 a. Force on M due only to motion o M b. orce on M due only to motion o M c. all orce on M
12 [M v v3 ] X v3 X F - v3 X [M v v3 3 ]X 0 Figure.9 a. Force on M due only to motion o M ; b. orce on M due only to motion o M ; c. all orce on M
13 The traner unction X /F i [M v v3 ] v3 v3 [M v v3 3 ]
14 Note that Sum o impedance connected to the motion at X X - Sum o impedance between X and X X Sum o applied orce at X Sum o impedance connected to the motion at X X - Sum o impedance between X and X X Sum o applied orce at X
15 Equation o Motion by Inpection Problem : Write, but not olve, the equation o motion or the mechanical ytem given below. Figure.0 Three-degree-o-reedom tranlational mechanical ytem
16 Uing ame logic, or M Sum o impedance connected to the motion at X X - Sum o impedance between X and X X - Sum o impedance between X and X 3 X 3 Sum o applied orce at X and or M - Sum o impedance between X and X X Sum o impedance connected to the motion at X X - Sum o impedance between X and X 3 X 3 Sum o applied orce at X
17 Similarly, or M 3 - Sum o impedance between X and X 3 X - Sum o impedance between X and X 3 X Sum o impedance connected to the motion at X 3 X 3 Sum o applied orce at X 3 Anymore we can write the equation or M, M and M 3 Note that M ha two pring, two vicou damper and ma aociated with it motion. There i one pring between M and M and one vicou damper between M and M 3.
18 0 ] [ X X X M v v v Equation or M or M ] [ F X X M X v v v and or M 3 0 ] [ X M X X v v v v Thee equation are the equation o motion. We can olve them or any diplacement X, X or X 3, or traner unction.
19 ROTATIONAL MECHANICAL SYSTEM TRANSFER FUNTIONS Table.5 Torque-angular velocity, torque-angular diplacement, and impedance rotational relationhip or pring, vicou damper, and inertia
20 Rotational mechanical ytem are handled the ame way a tranlational mechanical ytem, except that torque replace orce and angular diplacement replace tranlational diplacement. Alo notice that the term aociated with the ma i replaced by inertia. The value o, D and J are called pring contant, coeicient o vicou riction and moment o inertia, repectively. Writing the equation o motion or rotational ytem i imilar to writing them or tranlational ytem; the only dierence i that the ree body diagram conit o torque rather than orce.
21 Example : Find the traner unctionθ / T or the rotational ytem given below. Figure. a. Phyical ytem; b. chematic; c. block diagram
22 Firt, obtain the chematic rom the phyical ytem. Draw ree-body diagram or J and J uing uperpoition. Let tart with J : Figure.3 a. Torque on J due only to the motion o J b. torque on J due only to the motion o J c. inal ree-body diagram or J
23 For J : Figure.4 a. Torque on J due only to the motion o J ; b. torque on J due only to the motion o J c. inal ree-body diagram or J
24 Now, write the equation o motion by umming the torque on J and J 0 D J T D J θ θ θ θ
25 0 D J T D J θ θ θ θ The required traner unction i ound to be T θ J D - - J D
26 Thee equation have that now well-known orm Sum o impedance connected to the motion at θ - Sum o impedance between θ andθ θ Sum o applied torque at θ θ Sum o impedance connected to the motion at θ - Sum o impedance between θ andθ θ Sum o applied torque at θ θ
27 Equation o Motion by Inpection Problem : Write, but not olve, the equation o motion or the mechanical ytem given below. Figure.5 Three-degree-oreedom rotational ytem
28 - Sum o impedance connected to the motion at θ Sum o impedance between θ andθ θ - θ Sum o impedance between θ andθ Sum o impedance connected to the motion at θ θ - θ - Sum o impedance between θ andθ 3 Sum o impedance between θ andθ 3 θ 3 θ 3 Sum o applied torque at θ Sum o applied torque at θ - Sum o impedance between θ andθ 3 θ - Sum o impedance between θ andθ 3 θ Sum o impedance connected to the motion at θ 3 θ 3 Sum o applied toque at θ 3
29 D D J D D D J T D J θ θ θ θ θ θ θ Hence,
30 Traner Function or Sytem with Gear In indutrial application, generally gear aociate to a motor which drive the load. Gear are ued to obtain more peed and le torque or le peed and more torque. The interaction between two gear i depicted in the igure below. An input gear with radiu r and N teeth i rotated through angleθ t due to a torque, T t. An output gear with radiu r and N teeth repond by rotating through angleθ t and delivering a torque, T t. Figure.7 A gear ytem
31 Let u ind the relationhip between the rotation o Gear, θ t and Gear, θ t. A the gear turn, the ditance traveled along each each gear circumerence i the ame. Thu, r θ r θ or θ θ r r N N
32 What i the relationhip between the input torque, T and the delivered torque, T? I we aume the gear do not aborb or tore energy into Gear equal the energy out o Gear. T θ T θ and we get T T θ θ N N All reult are ummarized in the igure below
33 Let u ee what happen to mechanical impedance that are driven by gear. The igure above how gear driving a rotational inertia, pring and vicou damper. For clarity, the gear are hown by an end-on view. We want to repreent Figure.9a a an equivalent ytem at θ without the gear. In other word, can the mechanical impedance be relected rom the output to the input, thereby eliminating the gear? We know T can be relected to the output by multiplying by N /N. the reult i hown in Figure.9b. We write the equation o motion a J D θ T N N
34 Now convertθ into an equivalentθ, N N T N N D J θ Ater impliication, T N N N N D N N J θ we get the equivalent ytem hown in Figure.9c
35 Generalizing the reult, we can make the ollowing tatement : Rotational mechanical impedance can be relected through gear train by multiplying the mechanical impedance by the ratio Number o teeth o gear on detination hat Number o teeth o gear on ource hat The next example demontrate the application o the concept o a rotational mechanical ytem with gear.
36 Example : Find the traner unction θ / T or the ytem o Figure.30a Figure.30 a. Rotational mechanical ytem with gear; b. ytem ater relection o torque and impedance to the output hat; c. block diagram
37 Let u irt relect the impedance J and D and the torque T on the input hat to the output a hown in Figure.30b, where the impedance are relected by N /N and the torque i relected by N /N. The equation o motion can now be written a J e De e θ T where N N N J J N D e D D N J e N and Solving orθ / T, the traner unction i ound to be G a hown in Figure.30c N θ / T J e e N D e e
38 In order to eliminate gear with the large radii, a gear train i ued to implement large gear ratio by cacading maller gear ratio. A chematic diagram o a gear train i hown in Figure.3. Figure.3 Gear train Next to each rotation, the angular diplacement relative toθ ha been calculated a N N 3 5 θ 4 θ N N 4 N 6 For gear train, we conclude that the equivalent gear ratio i the product o individual gear ratio. We now apply thi reult to olve or the traner unction o a ytem that doe not have lole gear. N
39 Example : Find the traner unction θ / T or the ytem o Figure.3a Figure.3 a. Sytem uing a gear train; b. equivalent ytem at the input; c. block diagram Thi ytem, which ue a gear train, doe not have lole gear. All o the gear have inertia and or ome hat there i vicou riction. To olve the problem, we want to relect all o the impedance to the input hat, θ. The gear ratio i not ame or all impedance. For example, D i relected only through one gear ratio a D N /N, wherea J 4 plu J 5 i relected through two gear ratio a J 4 J 5 [N 3 /N 4 N /N ]. The reult o relected all impedance toθ i hown in Figure.3b
40 The equation o motion i J e D e θ T The traner unction i G θ T J e D e a hown in Figure.3c
41 ELECTROMECHANICAL SYSTEM TRANSFER FUNCTIONS We talked about mechanical ytem by now. You have already known the electrical ytem. Now, we move to ytem that are hybrid o electrical and mechanical variable, the electromechanical ytem. An application o electromechanical ytem i robot control. A robot have both electrical and mechanical parameter. A robot arm a an example o control ytem that ue electromechanical component i hown in Figure below. Figure.34 NASA light imulator robot arm with electromechanical control ytem component
42 A motor i an electromechanical component that yield a diplacement output or a voltage input, that i, a mechanical output generated by an electrical input. We will derive the traner unction or one particular kind o electromechanical ytem, the armature-controlled dc ervomotor. Schematic o motor i hown in Figure below. In thi igure, a magnetic ield developed by tationary permanent magnet or a tationary electromagnet called the ixed ield. A rotating circuit called the armature, through which current i a t low, pae through thi magnetic ield at right angle and eel a orce, FBli a t, where B i the magnetic ield trength and l i the length o the conductor. The reulting torque turn the rotor, the rotating member o the motor.
43 Modeling o the Permanent Magnet Dc Motor dθ m t V b t b dt We call V b t the back electromotive orce back em; b i a contant o proportionality called the back em contant ; and dθ m t/dtω m t i the angular velocity o the motor. Taking the Laplace tranorm, we get V b θ The relationhip between the armature current i a t, the applied armature voltage e a t and the back em V b t i ound by writing a loop equation around the Laplace tranormed armature circuit b m R a I a L a I a V b E a The torque developed by the motor i proportional to the armature current; thu, T m t I a where T m i the torque developed by the motor, and t i a contant o proportionality, called the motor torque contant, which depend on the motor and magnetic ield characteritic.
44 To ind the traner unction o the motor, we ue the equation Rearranging thi equation yield R L T R a I a L a I a V b E a a a m b θ t Now we mut ind T m in term o θ m i we are to eperate the input and output variable and obtain the traner unctionθ m / E a. Following igure how typical eqivalent mechanical loading on a motor. J m i the equivalent inertia at the armature and include both the armature inertia and, a we will ee later, the load inertia relected to the armature. D m i the equivalent vicou damping at the armature and include armature vicou damping and, a we will ee later, the load vicou damping relected to the armature. m E a
45 D J T m m m m θ E T L R a m b t m a a θ Subtituting the econd Eq. into the irt one yield E D J L R a m b t m m m a a θ θ I we aume that the armature inductance, L a, i mall compared to the armature reitance, R a, which i uual or a dc motor, the lat equation become E D J R a m b m m t a θ
46 Ater impliication, the deired traner unction, θ m / E a, i ound to be a b t m m m a t a m R D J J R E / θ Thi orm i relatively imple; α θ E a m
47 Let u irt dicu the mechanical contant, J m and D m. Conider the Figure.37, which how a motor which inertia J a and damping D a at the armature driving a load coniting o inertia J L and damping D L. Auming that all inertia and damping value hown are known. J L and D L can be relected back to the armature a ome iquivalent inertia and damping to be added to J a and D a repectively. Thu the iquivalent inertia J m and equivalent damping D m at the armature are J m J a J L N N D m D a D L N N Now that how we evaluated the mechanical contant, J m and D m, what about the electrical contant in the traner unction? We will how that thi cotant can be obtained through a dynamometer tet o the motor, where a dynamometer meaure the torque and peed o a motor under condition o a contant applied voltage.
48 Let u irt develop the relationhip that dictate the ue o a dynamometer. Taking L a 0, invere Laplace tr. Ra Tm bθ m Ea t Ra Tm t bω m t ea t I dc voltage e a i applied, the motor will turn at a contant angular velocity ω m with a contant torque T m. Hence, dropping the unctional relationhip baed on the time, the ollowing relationhip exit when the motor i operating at teady-tate with a dc voltage input: R a t T m bω m e b t t a, Solving or T m yield; T m ω m ea Ra Ra From thi equation, we get the torque-peed curve, T m veruω m which i hown below. T tall T m e a e a ω no-load The torque axi intercept occur when the angular velocity reache the zero. That value o torque i called the tall torque, T tall. The angular velocity occuring when the torque i zero i called the no-load peed, ω no,load. Thu, e t a T tall e a and ωno load ω m R a t The electrical contant o the motor can now be ound rom the equation o t T tall and R a e a b e ω The electrical contant, t /R a and b, can be ound rom a dynamaeter tet o the motor, which would yield T tall andω no-load or a given e a. a b no load
49 Example : Given the ytem and torque-peed curve o the ollowing igure, ind the traner unctionθ L / E a. Solution : Begin by inding the mechanical contant J m and D m. The total inertia and the total damping at the armature o the motor i N N J m Ja J L D N 0 m Da DL N 0 Now we will ind the electrical contant t / R a and b uing torque-peed curve. T tall 500 Nm, ω no-load 50 rad/n, e a 00 V. Hence the electrical contant are R t a T tall ea b e a ω no load 00 50
50 We know the traner unction ormulation orθ m / E a a b t m m m a t a m R D J J R E / θ / E a θ m In order to indθ L / E a, we ue the gear ratio, N / N / 0, and ind E a L θ a hown in igure below.
51 ELECTRIC CIRCUIT ANALOGS In thi ection, we how the commonality o ytem rom the variou dicipline by demontrating that the mechanical ytem with which we worked can be repreented by equivalent electric circuit. An electric circuit that i analogou to a ytem rom the another dicipline i called an electric circuit analog. Serie Analog : Conider the tranlational mechanical ytem hown in Figurea whoe the equation o motion M v X F. irchho meh equation or the imple RLC network hown in Figureb i L R I C E Figure.4 Development o erie analog: a. mechanical ytem; b. deired electrical repreentation; c. erie analog; d. parameter or erie analog
52 A we previouly pointed out, thee two mechanical and electrical equation i not directly analogou becaue diplacement and current are not analogou. We can create a direct analogy by operating on the mechanical equation to convert diplacement to velocity by dividing and multiplying the let-hand ide by, yielding M X M v v V F When we have more than one degree o reedom, the impedance aociated with the motion appear a the erial electrical element in a meh, but the impedance between adjacent motion are drawn a a erie electrical impedance between the correponding mehe. We demontrate with an example. Example : Draw a erie analog or the mechanical ytem o igure below.
53 V M V F V V M v v v v v v
54 Paralel Analog : A ytem can be converted an equivalent paralel analog. Conider the tranlational mechanical ytem in the Figure a, whoe the equation o motion i given by M v X M v V F Figure.43 Development o parallel analog: a. mechanical ytem; b. deired electrical repreentation; c. parallel analog; d. parameter or parallel analog irchho nodal equation or the imple paralel RLC network hown in Figureb i C E I R L Clearly, Figurec i equivalent to Figurea in the ene o analogy.
55 Example : Draw a parallel analog or the ame mechanical ytem with previou example which i V M V F V V M v v v v v v
56 NONLINEARITIES In thi ection, we ormally deine the term linear and nonlinear and how to ditinguih between the two. We will how how to approximate a nonlinear ytem a a linear ytem. A linear ytem poee two propertie : Superpoition and Homogeneity. The property o uperpoition mean that the output repone o a ytem to the um o input i the um o repone to the individual input. Thu, i a input o r t yield an output o c t and an input o r t yield an output o c t, then an input o r tr t yield an output o c tc t. The property o homogeneity decribe the repone o the ytem to multiplication o the ytem by a calar. Speciially, in a linear ytem, the property o homogeneity i demantrated i or an input o r t that yield an output o c t, an input Ar t yield an output o Ac t; that i, multiplication o an input by a calar yield a repone that i multiplied by the ame calar. We can viualize the linearity a hown in Figure.45. Figure.45 a. Linear ytem; b. Nonlinear ytem
57 Figure.46 how ome example o phyical nonlinearitie. Figure.46 Some phyical nonlinearitie A deigner can oten make a linear approximation to a nonlinear ytem. Linear approximation imply the analyi and deign o a ytem and are ued a long a the reult yield a good approximation to reality. For example, a linear relationhip can be etablihed a a point on the nonlinear curve i the range o input value about that point i mall and the origin i tranlated to that point. Electronic ampliier are an example o phyical device that perorm linear ampliication with mall excurion about a point.
58 LINEARIZATION In thi ection, we how how to obtain linear approximation to nonlinear ytem in order to obtain traner unction. The irt tep i to recognize the nonlinear component and write the nonlinear dierential equation, we linearize it or mall-ignal input about the teady-tate olution when the mall-ignal input i equal to zero. Thi teadytate olution i called equilibrium and i elcted a the econd tep in the linearization proce. Next we linearize the nonlinear dierential equation, and then we take the Laplace tranorm o the linearized dierential equaiton, auming all zero initial condition. Finally we eperate input and output variable and orm the traner unction. Let u irt ee how to linearize a unction; later, we will apply the method to the linearization o dierential equation. I we aume a nonlinear ytem operating at point A, [x 0,x 0 ] in Figure.47. Small change in the input can be related to change in tje output about the point by way o the lope o the curve at a point A. Thu, i the lope o the curve at point A i m a, then mall excurion o the input about point A, δx, yield mall change in the output, δx, related by the lope at point A.
59 Figure.47 Linearization about a point A Thu, [ x xo ] ma x x0 rom which δ x m aδx and o a 0 0 x x ] m x x x m δx a Thi relationhip i hown graphically in Figure.47, where a new et o axe, δx andδx, i created at the point A, and x i approximately equal to x 0, the ordinate o the new origin, plu mall excurion, m a δx, away rom point A. Let u look an example.
60 Example : Linearize x 5cox about xπ/ Solution : We irt ind that the derivative o x i d/dx -5inx. At x π/, the derivative i -5. Alo x 0 π/5coπ/0. Thu, rom the equation o a 0 0 x x ] m x x x m δx The ytem can be repreented a x -5δx or mall excurion o x about x π/. The proce i hown graphically in the igure, where the coine curve doe not indeed look like a traight line o lope -5 nearπ/. a Figure.48 Linearization o 5 co x about x π/
61 Taylor Serie Expanion : Another approach to linearization i Taylor erie. The previou dicuion can be ormalized uing the Taylor erie expanion, which expree the value o a unction in term o value o thi unction at a particular point. The Taylor erie o x i For mall excurion o x rom x 0, we can neglect high order term. The reulting approximation yield a traight-line relationhip between the change in x and the excurion away rom x 0. Neglecting the high order term in the equation above, we get d x x0 x x0 dx x x 0 or δ x maδx Which i a linear relationhip between δx and δx or mall excurion away rom x 0. The ollowing example demontrate linearization.the irt example demontrate linearization o a dierential equation and the econd example applie linearization to inding a traner unction.
62 Example : Linearize the ollowing dierential equation or mall excurion about xπ/4 d dt x dx co x dt Solution : The preence o the term cox make thi equation nonlinear. Since we want to linearize the equation about xπ/ 4, we let xδ π/ 4, whereδx i the mall excurion aboutπ/ 4, and ubtitue x into given equation, d But d π δx 4 dt π π δx d δx 4 4 π co 0 δx dt dt 4 d δx dt and 0 π d δ x 4 dt Finally, the term coδx π/ 4 can be linearized with the truncated Taylor erie. Subtituting xcoδx π/ 4, x 0 π/ 4co π/ 4, and x-x 0 δx into the equation d x x0 x x0 yield dx x x 0 d δ x dt π π co δx co 4 4 d cox dx π δx in δx π x 4 4 Solving thi equation or coδx π/ 4 yield
63 x x x δ δ π π π δ 4 in 4 co 4 co Note that we have obtained by now three equation which are 4 dt x d dt x d δ π δ dt x d dt x d δ π δ 4 x x x δ δ π π π δ 4 in 4 co 4 co Subtituting thee three equation into the irt equation we wrote which i 0 4 co 4 4 π δ π δ π δ x dt x d dt x d yield the ollowing linearized dierential equation x dt x d dt x d δ δ δ
64 Example : Nonlinear Electrical Network Find the traner unction V L /V or electrical network hown in the igure which contain a nonlinear reitor whoe voltage-current relationhip i deined by i r 0.e 0. V r Where i r and V r are the reitor current and voltage, repectively. Alo vt in the igure i mallignal ource. Figure.49 Nonlinear electrical network
65 We will ue the irchho voltage law to um the voltage in the loop to obtain the nonlinear dierential equation, but irt we mut olve or the voltage acro the nonlinear reitor. Taking the natural log o reitor current-voltage relationhip, we get V r 0ln i r Applying the irchho voltage law around the loop, where i r i, yield di L 0ln i 0 v t * dt Next, let u evaluate the equilibrium olution. Firt, et the mall-ignal ource, vt, equal to zero. Now evaluate the teady-tate current. With vt0, the circuit conit o 0 V battery in erie with the inductor will be zero, ince v L Ldi/dt and di/dt i zero in the teady,tate, given acontant battrey ource. Hence the reitor voltage v r i 0 V. Uing the characteritic o the reitor, i r e 0. we ind that i r i4.78 amp. Thi current, i 0, i the equilibrium value o the network current. Hence ii 0 δi. Subtituting thi current into * equation yield V r
66 t v i i dt i i d L δ δ Uing the linearizing equation x x dx d x x x x to linearize ln 0 i i δ, we get i i i i i di i d i i i i i i i δ δ δ δ ln ln ln 0 0 or i i i i i δ δ ln ln 0 0ln 0 0 t v i i i dt i d L δ δ Subtituting thi equation into **, the linearized equation become **
67 d δ i Letting L and i , the inal linearized dierential equation i i v t dt Taking the Laplace tranorm with zero initial condition and olving orδi yield V δi *** d dδi But the voltage acro the inductor about the equilibrium point i V L t L i0 δ i L dt dt Taking the Laplace tranorm V L Lδi δi Subtituting the *** equation into the lat equation yield V L rom which the inal traner unction i V V L V or mall excurion about i4.78 or, equivalently, about vt0
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