hapter MATHEMATIAL MODELS OF PHYSIAL SYSTEMS.. INTODUTION For the analyi and deign of control ytem, we need to formulate a mathematical decription of the ytem. The proce of obtaining the deired mathematical decription of the ytem i known a modeling. The baic model of dynamic phyical ytem are differential equation obtained by application of the appropriate law of nature. Thee equation may be linear or nonlinear depending on the phenomena being modeled. The differential equation are inconvenient for the analyi and deign manipulation and o the ue of Laplace Tranformation which convert the differential equation into algebraic equation i made ue of. The algebraic equation may be put in tranfer function form, and the ytem modeled graphically a a tranfer function block diagram. Alternatively, a ignal flow graph may be ued. Thi chapter i concerned with differential equation, tranfer function, block diagram, ignal flow graph, etc., of different phyical ytem namely, mechanical, electrical, hydraulic, pneumatic and thermal ytem. Analyi of a dynamic ytem require the ability to predict it performance. Thi ability and the preciion of the reult depend on how well the characteritic of each component can be expreed mathematically. One of the mot important tak in the analyi and deign of control ytem i mathematical modeling of the ytem. The two mot common method are the tranfer function approach and the tate equation approach. The tranfer function method i valid only for linear time-invariant ytem, wherea the tate equation are firt-order to ue tranfer function and linear tate equation the ytem mut firt be linearized, or it range of operation mut be confined to a linear range. Although the analyi and deign of linear control ytem have been well developed, their counterpart for nonlinear ytem are uually quite complex. Therefore, the control ytem engineer often ha the tak of determining not only how to accurately decribe a ytem mathematically, but alo, more important, how to make proper aumption and approximation, whenever neceary, o that the ytem may be adequately characterized by a linear mathematical model.
8 Analyi of Linear ontrol Sytem x(t) Table.. Some Laplace tranform pair Unit impule δ(t) Unit tep u(t) x() t t n n n! e λt λ n! t n e λt n ( λ) ω in ωt ω co ωt ω ω e λt in ωt ( λ ) ω λ e λt co ωt ( λ ) ω We can thu conclude that {e λt } (.3) S λ The reult of Example. i ueful in many way. Firt, the Laplace tranform of the exponential function i of interet in it own right. Second, ince the exponential i a generating function of many other intereting function, we can ue Eq. (.3) to provide the Laplace tranform of thee other function. To tart, note that a tep function u(t) i defined a u(t) t 0 0 t < 0 Note that u(t) e 0t (.4)
Mathematical Model of Phyical Sytem 3... x (n ) (0) x n (0) (.) where x(0), x (0), x ( 0),..., x n, and x n denote the value of x(t), x () t, x () t,..., (d n /dt n ) x(t), (d n /dt n ) x(t) evaluated at t 0. Note that if all initial value of x(t) and it derivative are zero, the Laplace tranform of the nth derivative of x(t) i n X() and the differentiation operation i equivalent in the Laplace domain to the operator. We can arrive at the integration theorem through ue of Eq. (.0) a follow. Let y(t) x () t ; thu x(t) y () t dt A a reult of Eq. (.0), we have Thu, { y () t dt } y() t dt y() t dt t {y(t)} { } 0 Y( ) y (0) (.) where y (0) y(t) dt evaluated at t 0. The hift theorem deal with the Laplace tranorm of x(t τ), where τ i a delay given that x(t) i defined and x(t) 0, t < 0. Applying the fundamental expreion (.), we have X() 0 σ x( σ) e dσ ( ) ( ) t τ τ 0 x t e dt τ e x( t τ) e dt Thu, {x(t τ)} e τ X() (.3) The final-value theorem can be obtained from the differentiation theorem (.0) by taking the limit a 0, with the reult lim xt ( ) t 0 The initial-value theorem i tated a t 0 0 t lim X ( ) (.4) lim xt ( ) lim X ( ) (.5) We will conider now an important cae involving the convolution integral defined for the two function h(t) and x(t) by the relation y(t) h(t) * x(t) (.6) or y(t) ht ( τ)() x τ d τ (.7) 0
3 Analyi of Linear ontrol Sytem The aterik denote the convolution operation. The Laplace tranform of y(t) i obtained uing Eq. (.) a Y() 0 t y() t e dt (.8) Let u note that both h(t) and x(t) are aumed to be zero for t < 0. We can thu conclude that h(t τ) i zero for τ > t. A a reult, we can write y(t) A a reult, Eq. (.8) can be written a 0 ht ( τ)() xτ d τ (.9) Y() ( )() t τ τ τ 0 h t x d e dt 0 Interchanging the order of integration, we get Y() t e h( t τ )() x τ dtd τ 0 0 () t ( ) x τ τ τ 0 e h t dt d 0 onider the integral with repect to t and let σ t τ; thu t e h( t τ) dt 0 ( στ) σ e h ( σ) d σ t Thu, we have τ e e h( σ) dσ 0 σ H () e τ Y() τ e x() τ H() dτ 0 Since H() i independent of τ we get Y() H() τ e x() τ dτ 0 A a reult, Y() H() X() (.0) Our concluion i that the Laplace tranform of the convolution of h(t) and x(t) i the product of the Laplace tranform H() and X(). The Invere Laplace Tranform onider the imple differential equation d x dx a a ax 0 u(t) dt dt Application of the Laplace tranform to both ide, auming zero initial condition, give u (a a a 0 ) X() U()
Mathematical Model of Phyical Sytem 33 Thi i an algebraic equation which can be written a U() X() a a a 0 Suppoe now that the input function u(t) i a unit tep; thu U() A a reult, the Laplace tranform of x(t) i given by X() a ( a a ) 0 Finding the function x(t) whoe tranform i a given above i ymbolized by the invere tranform operator; thu (t) {X()} a ( a a0) A formal definition of the invere Laplace tranform i given by x(t) {X()} π a j a j t X () e d (.) where a i a real contant. It i quite poible (although omewhat involved) to obtain the invere Laplace tranform by performing the integration indicated in Eq. (.). A much more effective way i commonly employed in control ytem engineering, which relie on performing a partial fraction expanion that reult in an expreion of X() a the um of the function X (), X (),..., X n (). X() X () X ()... X n () (.) The function X (), X (),... can then be looked up in a table of Laplace tranform pair and hence we can obtain the correponding invere X (t), X (t),...,x n (t). The final reult i thu X(t) X (t) X (t)... X n (t) (.3) Partial Fraction Expanion In mot control ytem engineering application it i deired to find the invere Laplace tranform of a function F() expreed a the ratio of two function N() and D() according to N () F() (.4) D () The numerator and denominator function are commonly obtained a polynomial in of the form N() m j b j (.5) j0
34 Analyi of Linear ontrol Sytem D() i0 a i i (.6) The denominator i an nth-degree polynomial, while the numerator i of mth degree, with n > m. For example, a typical function F() i 4 F() 3 Thu N() 4 D() 3 In order to carry out the partial fraction expanion procedure it i neceary to obtain N() and D() in the following factored form: N() ( z ) ( z )... ( z m ) (.7) D() ( p ) ( p )... ( p n ) (.8) In Eq. (.7), the z, z,..., z m are called the zero of the function F(), while in Eq. (.8), the p, p,..., p n are called the pole of the function F(). Obtaining the pole and zero from definition of Eq. (.5) and (.6) may or may not be traightforward. In the preceding example it i eay to ee that z 4 p p Thi follow ince D() i a econd-order polynomial which can be eaily factored. For higher-order polynomial it i often neceary to reort to an iterative procedure to obtain the neceary factor. onider, for example, the expreion D() 4 9 3 6 4 It i eay to ee that i a common factor. D() D () with D () 3 9 6 4 Now D () i a third-order polynomial and we have to find it root. Althouth there i a well-defined procedure for doing jut that, we illutrate the ue of an iterative method uch a Newton method. We tart with an etimate (0) if the olution, and obtain () uing D () (0) ( ) (0) D ( ) where D () i the derivative of D (). In our example we have D () 3 8 6 Thu take (0) to obtain () 9 6 4 3 8 6.55 (0)
Put Put A a reult, Mathematical Model of Phyical Sytem 37 5; thu A 8 ; thu A 3 54 3 F() 54 5 The invere Laplace tranform i thu f(t) ( t 3 5t t ) e e e 54 onider now the cae when a pole i repeated in the F() of Eq. (.9). Aume that p p... p n in Eq. (.9), o that we have N () F() n ( p ) ( p )...( p ) n n (.37) learly, we need an alternative expreion to Eq. (.30). The required expreion i F() A A... p ( p ) ( ) A A A n n... n n p pn pn (.38) To obtain the coefficient A, A,..., A n we adopt the procedure outlined previouly with a light modification, a hown in the next two example. Example.4. onider the function f(t) defined by it Laplace tranform F() ( 3) Find f(t) uing partial fraction expanion. Solution: We write Thu, Put F() A A A3 3 A ( 3) A ( 3) A 3 0; thu A 3
The invere Laplace tranform i thu given by Mathematical Model of Phyical Sytem 39 f(t) 9 ( e 3t 3te 3t ) There are ituation where the preence of complex-conjugate pole make it impoible to find a real root uing Newton method. The following example involve uch a ituation and propoe a method for treatment. Example.6. Find the invere Laplace tranform of F() 4 3 6 7 4 8 Solution: The denominator function i given by D() 4 6 3 7 4 8 We attempt to factor D() into the product of two econd-order polynomial D() D ()D () where, D () a b D () a b Performing the multiplication and equating coeffcient of equal power, we conclude that a a 6 b b a a 7 a b a b 4 b b 8 The equation above can be combined to yield one equation in b given by b 6 7b5 6b4 6b3 68b 5508b 583 0 An iterative olution yield b 3 Thu, 8 b 6 b Hence, 6a 3a 4 Alo, a a 6 A a reult, we find that a a 4 We can thu conclude that D() ( 3)( 4 6) Intead of performing the partial fraction expanion in term of imple pole, we do it in term of the econd-order term a A B A B F() 3 4 6
40 Analyi of Linear ontrol Sytem Since F() ( 3)( 4 6) we thu have (A B )( 4 6) (A B )( 3) Expanding, we obtain (B B ) 3 (A A 4B B ) (4A 6B A 3B ) (6A 3A ) Equating the coefficient of equal power in on both ide, we get B B 0 A A 4B B 0 4A A 6B 3B 0 6A 3A Solving the equation above, we obtain A 9 A 5 9 B 9 B 9 Thu, the partial fraction of F() i 5 F() 9 4 6 3 To get the invere Laplace tranform, we write F() a ( ) ( ) F() 9 ( ) ( ) ( ) ( ) Thu, the invere Laplace tranform i obtained a t t t t ƒ(t) e in t e co t e in e co t 9.3. MEHANIAL SYSTEM ELEMENTS Mot feedback control ytem contain mechanical a well a electrical component. From a mathematical viewpoint, the decription of electrical and mechanical element are analogou. In fact, we can how that given an electrical device, there i uually an analogou mechanical counterpart, and vice vera.
Mathematical Model of Phyical Sytem 4 The analogy, of coure, i a mathematical one; that i, two ytem are analogou to each other if they are decribed mathematically by imilar equation. The motion of mechanical element can be decribed in variou dimenion a tranlational, rotational, or a combination of both. The equation governing the motion of mechanical ytem are often directly or indirectly formulated from Newton law of motion..3.. Tranlational motion Tranlational motion take place along a traight line and the variable involved in decribing a traight-line motion are diplacement, velocity and acceleration. Newton law of motion govern the linear motion. According to thi law, the product of ma and acceleration i equal to the algebraic um of force acting on it. Newton law of motion tate that the algebraic um of force acting on a rigid body in a given direction i equal to the product of the ma of the body and it acceleration in the ame direction. The law can be expreed a Σforce Ma (.39) where M denote the ma and a i the acceleration in the direction conidered..3.. Ma The function of ma in linear motion i to trore kinetic energy. Ma cannot tore potential energy. Suppoe a force i applied to ma M a hown in Fig.., the ma tart moving in x direction a hown. For the time being, we will aume other force uch a friction, etc. to be zero. Hence, according to Newton law, d x M dt f(t) (.40) x () t M f () t Fig...3.3. Linear pring A pring can tore potential energy. In a ytem, there may be a pring where ome component uch a elatic tring, cable, etc. may work a a pring. Strictly peaking, a pring i a non-linear element. However, we can aume it to be linear for mall deformation. Let u aume a pring ha negligible ma and connected to a rigid upport a hown in Fig...
46 Analyi of Linear ontrol Sytem ytem and vice vera. It i alway advantageou to obtain electrical analogou of the given mechanical ytem a we are well familiar with the method of analying electrical network than mechanical ytem. There are two method of obtaining electrical analogou network, namely,. Force-voltage Analogy, i.e. Direct Analogy. (Table.3). Force-current Analogy, i.e. Invere Analogy. (Table.4) Mechanical Sytem onider imple mechanical ytem a hown in the Fig..7. Due to the applied force, ma M will diplace by an amount x(t) in the direction of the force F(t) a hown in the Fig..7. K... M X(t) B F(t) Fig..7 According to Newton law of motion, applied force will caue diplacement x(t) in pring, acceleration to ma M againt frictional force having contant B. F(t) Ma Bv K x(t) (.45) where, a acceleration, v velocity d xt () dxt () F(t) M B Kx() t dt dt (.46) Taking Laplace, F() M X() B X() K X () (.47) Thi i equilibrium equation for the given ytem. Now we will try to derive analogou electrical network. Force-voltage Analogy (Loop Analyi) In thi method, to the force in mechanical ytem, voltage i aumed to be analogou one. Accordingly we will try to derive other analogou term. onider electric network a hown in the Fig..8. The equation according to Kirchhoff law can be written a, L V(t) i(t) Fig..8
48 Analyi of Linear ontrol Sytem The equation according to Kirchhoff current law for above ytem i, I I L I I (.53) Let node voltage be V, I Vdt V dv L (.54) dt Taking Laplace, V() V() I() V () (.55) L But to get thi equation in the imilar form a that of F() we will ue, V(t) d φ dt Subtituting in equation for I() where φ flux (.56) V() φ () i.e. φ () V() (.57) I() φ() φ () φ () (.58) L omparing equation for F() and I() it i clear that, (i) apacitor i analogou to ma M. (ii) eciprocal of reitance i analogou to frictional contant B. (iii) eciprocal of inductance L i analogou to pring contant K. Table.6. Force-urrent Analogy Tranlational otational Electrical F Force T urrent I M Ma J B Friction B / K Spring K /L x Diplacement θ φ x Velocity 0 θ ω Voltage e Note: The element which are in erie in F V analogy, get connected in parallel in F I analogou network and which are in parallel in F V analogy, get connected in erie in F I analogou network. Stepwie Procedure to Solve Problem on Analogou Sytem : Identify all the diplacement due to the applied force. The element pring and friction between two moving urface caue change in diplacement. φ
50 Analyi of Linear ontrol Sytem Subtituting value of θ () from equation () in equation (), we get J K)( J f K) K θ() K T() θ Tranfer function () T() ( J K )( J f K ) K 4 3 ( ) K J J J f KJ KJ Kf K An. Example.8. Obtain the differential equation decribing the complete dynamic of the mechanical ytem hown in Fig... K B M K B X M f (t) Fig.. Solution: The mechanical network diagram i hown in Fig..3. K X X X B f (t) M M K B or or Node x Node x Fig..3 ƒ (t) x x F() (M B K ) X (t) (B K ) X () Mx K (x x ) B ( ) K (x x ) B ( ) x x M x Bx K x [M (B B ) (K K )] X () [B K ] X () 0 The electrical analog baed on force-voltage analogy i hown in Fig..4.
Mathematical Model of Phyical Sytem 5 The electrical analog circuit i drawn with the help of electrical analog equation which are obtained from nodal equation in Laplace domain. The electrical analog equation are E() L Q() Q() L e _(t) i i L and Fig..4 L Q() Q() 0 where Q charge. onverting the above equation into differential equation form di e () L idt i i idt dt and di L i i idt idt dt i idt 0 Note: Above equation are alo obtained, if we apply Kirchhoff law to the electrical network hown in Fig..4. Example.9. Obtain the nodal equation for the ytem hown in Fig..5 and draw it analogou electrical network. K X (t) X M K X M Fig..5
Mathematical Model of Phyical Sytem 53 or d q q q L 0 dt L Q Q () () Note: Thee equation are imilar to the nodal equation. Example.0. Obtain the nodal equation for the ytem hown in Fig..8 and draw it electrical analog baed on force-current analogy. 0 x x f (t) k f M M f f k Fig..8 Solution: The mechanical network i hown in Fig..9. x x f f (t) M k f M k f Node x Node x Fig..9 ( ) M x fx f x x Kx ƒ (t) Mx fx Kx f ( x x ) or L /k t i (t) f (t) M f M L K f Fig..0
54 Analyi of Linear ontrol Sytem Electrical analog circuit baed on force-current analogy i hown by Fig..0. Example.. Obtain tranfer function for the ytem hown in Fig... B X M X 0 B Fig.. Solution: Writing equation for the given ytem baed on Newton law, we get ( ) B x x Mx 0 Bx (0) 0 B X () (M B B ) X 0 () X0() X () B B M B B M B B An. Example.. Obtain tranfer function for the ytem hown in Fig... X B k k X 0 Fig.. Solution: Writing Newton law equation K (x x 0 ) ( ) B x x K x 0 0 0 [B K ] X () (B K K ) X 0 () 0 X0() X () B K B K K An..6. TANSFE FUNTION It ha been hown already that the input and output of a linear ytem in general, i related by a linear or a et of linear differential equation. Such relationhip are capable of completely decribing the ytem behaviour in the preence of a particular input excitation and known initial condition.
58 Analyi of Linear ontrol Sytem differential equation of Eq. (.63) i eldom ued in it original form for the analyi and deign of control ytem. To obtain the tranfer function of the linear ytem that i repreented by Eq. (.63), we imply take the Laplace tranform on both ide of the equation, and aume zero initial condition. The reult i ( n a n n-... a a )() (b m m b m m-... b b )() (.64) The tranfer function between r(t) and c(t) i given by m m () G() () bm bm... b b n n (.65) a n... a a We can ummarize the propertie of the tranfer function a follow:. Tranfer function i defined only for a linear time-invariant ytem. It i meaningle for nonlinear ytem.. The tranfer function between an input variable and an output variable of a ytem i defined a the Laplace tranform of the impule repone. Alternately, the tranfer function between a pair of input and output variable i the ratio of the Laplace tranform of the output to the Laplace tranform of the input. 3. When defining the tranfer function, all initial condition of the ytem are et to zero. 4. The tranfer function i independent of the input of the ytem. 5. Tranfer function i expreed only a a function of the complex variable. It i not a function of the real variable, time, or any other variable that i ued a the independent variable. Tranfer Function (Multivariable Sytem) The definition of tranfer function i eaily extended to a ytem with a multiple number of input and output. A ytem of thi type i often referred to a the multivariable ytem. In a multivariable ytem, a differential equation of the form of Eq. (.63) may be ued to decribe the relationhip between a pair of input and output variable. When dealing with the relationhip between one input and one output, it i aumed that all other input are et to zero. Since the principle of uperpoition i valid for linear ytem, the total effect on any output variable due to all the input acting imultaneouly i obtained by adding up the output due to each input acting alone. A number of example i appropriate to illutrate the concept of tranfer function. Example.3. onider the integrating configured network hown in Figure.3. The current at the output terminal in zero, and we can write the input voltage a V i () I()
Mathematical Model of Phyical Sytem 59 V i () I() I() V 0 () Fig..3. Integrating onfiguration Network The output voltage i given by V 0 () I() The tranfer function i thu obtained a V0 () V () i Example.4. For the differentiating configured network hown in Figure.4, we can write the tranfer function a V0 () V () i / V i () V 0 () Fig..4. Differentiating onfiguration Network Example.5. For the pring-dahpot ytem hown in Figure.5, we can write a force balance equation a or Bx 0 Kx 0 Kx i Employing the Laplace tranform, we thu have (B K) X 0 () KX i () A a reult, the tranfer function i given by X0 () X () i X0 () X () i K B K ( B/ K ) Note the imilarity of thi tranfer function and that of Example.3.
6 Analyi of Linear ontrol Sytem Thu, the tranfer function i given by V0 () Vi () L Thi i clearly imilar to the tranfer function of Example.8. L V i () V 0 () Fig..9 L Network Example.0. For the lead-lag network hown in Figure.30, we can write the following impedance function: Let Thu, Alo, Let Thu, Z / / τ a Z τa Z τ b τb Z
Mathematical Model of Phyical Sytem 63 V i () V 0 () Fig..30. Lead-Lag Network The tranfer function i thu given by V0 () V () i Z Z Z Thi i written in term of the network element a Let τ ab ; thu We can rewrite thi a V0 () V () i V0 () V () i V0 () V () i ( τ )( τ ) a ( τ )( τ ) a b b ( τ )( τ ) a b a b a b ab τ τ ( τ τ τ ) τ aτb τ τ where, τ τ τ a τ b τ a τ b τ ab τ τ The example above help to illutrate the concept of a tranfer function..7. BLOK DIAGAM ALGEBA Introduction If a given ytem i complicated, it i very difficult to analye it a a whole, with the help of tranfer function approach, we can find tranfer function of each and every element of the complicated ytem. And by howing connection between the element, complete ytem can be plitted into different block and can be analyed conveniently. Thi i the baic concept of block diagram repreentation. Baically block diagram i a pictorial repreentation of the given ytem. It i very imple way of repreenting the given complicated practical ytem. In block diagram, the interconnection of ytem component to form a ytem can
Mathematical Model of Phyical Sytem 65 A pictorial repreentation of the relationhip between ytem variable i offered by the block diagram. In a block diagram, three ingredient are commonly preent.. Functional block. Thi i a ymbol repreenting the tranfer between the input U() to an element and the output X() of the element. The block contain the tranfer function G(), a hown in Figure.3. The arrow directed into the block U() X() G() Fig..3. Functional Block repreent the input U(), while that directed out of the block repreent the output X(). The block hown repreent the algebraic relationhip X() G()U() (.66). Summing point. Thi i a ymbol denoted by a circle, the output of which i the algebraic um of the ignal entering into it. A minu ign cloe to an input ignal arrow denote that thi ignal i revered in ign in the output expreion. Figure.7 how the relationhip E() () () (.67) () E() () Fig..3. Summing Point () () () Fig..33. Takeoff Point 3. Takeoff point. A takeoff point on a branch in a block diagram ignifie that the ame variable i being utilized elewhere, a hown in Figure.33. A fundamental block diagram configuration i the ingle-loop feedback ytem hown in Figure.34a. The output variable () i modified by the feedback element with tranfer function H() to produce the ignal B(): B() () H() (.68) The ignal B() i compared to a reference ignal () to produce the error.
Mathematical Model of Phyical Sytem 67 () G() G() H() () Fig..35. educed form of Fig..34 Equation (.7) i valid for negative feedback ytem. Hence, for a poitive feedback ytem we have M() () G () () (.7) GH () () In general, for a poitive/negative feedback ytem, the control ratio i given by M() () G () () ± (.73) GH () () a the cae may be. Let u now dicu about the block diagram eduction Technique ueful in the analyi of complex control ytem. ule (): ombining block in cacade G G G G G G G G G ule (): ombining block in Parallel G G ± G ± G ± G ule (3): Moving a pick-off point after a block G G G G /G ule (4): Moving a take off point ahead of a block G G G G G G G
68 Analyi of Linear ontrol Sytem ule (5): Moving a umming point after a block ± G[ ± ] G G G ± ± G[ ± ] ule (6): Moving a umming point ahead of a block G G G G ± G ± ( ± ) G ± G G ± ule (7): Eliminating a feedback loop /G ± G G GH H Uing the rule given above, let u reduce the block diagram to arrive at the tranfer function of the ytem. Example.. Simplify the block diagram hown in Fig..36 and obtain the cloed loop tranfer function ()/(). () G () Fig..36. Block diagram of Example. Solution: The given block diagram of Fig..37 can be reduced a follow. Uing ule. (combining block in parallel)
Mathematical Model of Phyical Sytem 69 () () G Fig..37. Block diagram of Example. Fig..37 i in the canonical form of control ytem and o the tranfer function i obtained uing rule 7 a () () which i hown in Fig..38. ( G G ) 3 ( G G )( G G ) 3 4 () (G ) (G ) ( ) () Fig..38. educed form of Fig..37 () Example.. Evaluate () from the block diagram hown in Fig..39. () () X() G () 3 () () () H () Fig..39. Block diagram of Example. Solution: Let u define the output of () a X(). At point (), we have (). G () At point (), we have G XH and for X(). X [G XH ] GG X G HG
Mathematical Model of Phyical Sytem 7 Step 3: Uing rule, (combining block in parallel), we have Fig..4 reduced to Fig..43 G H Fig..43. educed form of Fig..4 Step 4: Shifting the umming point ahead of the block by uing rule 4, we have the block diagram of Fig..43 reduced to Fig..44. G G H Fig..44. educed form of Fig..43 Step 5: educing Fig..44 uing rule, we have Fig..44 reduced to Fig..45. (G ) H G Fig..45. educed form of Fig..44 Step 6: educing Fig..45 uing rule, we have Fig..45 reduced to Fig..46. (G ) ( ) Fig..46. educed form of Fig..45 Therefore, the tranfer function of the ytem of Fig..46 i given by G( G) G( H H) Example.4. Uing block diagram reduction technique, find the cloedloop tranfer function of the ytem whoe block diagram i given in Fig..47.
Mathematical Model of Phyical Sytem 75 G H H / Fig..57. educed form of Fig..56 Step 4: Eliminating the feedback loop uing rule 7, we have Fig..57 reduced to Fig..58. G G H G Fig..58. educed form of Fig..57 Step 5: Uing rule of combining two block in parallel, we have Fig..59 reduced to Fig..60. G G 3 G H G Fig..59. educed form of Fig..58 Hence, the tranfer function i given by GG G3 H G G G G HGG 3 Example.6: Simplify the block diagram hown in Fig..60 G b a H Fig..60. Block diagram of Example.6
76 Analyi of Linear ontrol Sytem Solution: Firt, move the takeoff b to a a hown in Fig..6a. Now we can ee that G and are in parallel and the block diagram reduce to that hown in Figure.6b. The feedback loop with a forward gain of and feedback element H can be reduced a hown in Fig..6c. Finally, Figure.6d how that the overall tranfer between and. G H (a) G H (b) H G (c) G H (d) Fig..6. Simplified Block Diagram Example.7. Ue block diagram reduction technique to obtain the ratio / for the ytem hown in the block diagram of Fig..6. G H Fig..6. Block diagram of Example.7
78 Analyi of Linear ontrol Sytem Solution: Evaluation of / Aume 0. Therefore, umming point No. 5 can be removed. Shift take off point No. 4 beyond block. H 3 / G G G 3 3 6 H Fig..65. Eliminate the feedback loop between point 3 and 6. H 3 / G 6 H Fig..66. Eliminating the feedback loop again. G 6 H 3 H Fig..67. GG G3 GH HG GGGH 3 3 3 An. Evaluation of / Aume 0. Thu, umming point No. can be removed.
Mathematical Model of Phyical Sytem 79 H 3 3 5 G 4 6 H Fig..68. Shifting the umming point No. and rearranging beyond. H 3 3 G H Fig..69. earranging, we get H 3 G H Fig..70. earranging and eliminating the feedback loop
80 Analyi of Linear ontrol Sytem H 3 G H earranging, Fig..7. G H H 3 Fig..7. Eliminating the feedback loop, we get G3( ) HG G( GGH H ) 3 3 An. Exmaple.9. Find the tranfer function for the block diagram hown in Fig..73. G O H Solution: Fig..73. G H Fig..74.
Mathematical Model of Phyical Sytem 8 G ( ) H Fig..75. G ( ) H ( ) Fig..76. G ( ) H G ( ) Fig..77. GG 3( G) H GG 3( G) H H G G3( G) GG 3( G) ( H ) HG( G ) G 3 Example.30: Find cloed-loop tranfer function of ytem hown in Fig..78. G H Fig..78.
8 Analyi of Linear ontrol Sytem Solution: G H Fig..79. G G H Fig..80. G G G H G H Fig..8. or ( G 4H)( GG GG 3) GG H H GG H H 3 G ( )( GH 4 ) GHH ( G G) 3 An..8. SIGNAL FLOW GAPHS A ignal flow graph may be regarded a a implified notation for a block diagram, although it wa originally introduced by S. J. Maon a a caue-andeffect repreentation of linear ytem. In general, beide the difference in the phyical appearance of the ignal flow graph and the block diagram, the ignal flow graph to be contrained by more rigid mathematical relationhip, wherea the rule of uing the block diagram notation are far more flexible and le tringent. A ignal flow graph may be defined a a graphical mean of portraying the input-output relationhip between the variable of a et of linear algebraic equation. onider that a linear ytem i decribed by the et of N algebraic equation y j Σ N a k kj y k j,,..., N (.74)
Mathematical Model of Phyical Sytem 83 It hould be pointed out that thee N equation are written in the form of caue-and-effect relation: N jth effect Σ (gain from k to j) (kth caue) (.75) k or imply output Σ (gain)(input) (.76) Thi i the ingle mot important axiom in the contruction of the et of algebraic equation from which a ignal flow graph i drawn. Thi method i believed to provide a fater mean for determining the repone of multiloop ytem than do the block diagram reduction technique dicued in the previou ection. onider a et of linear equation having the form n y i Σ a j ij y j i,,..., n A node i aigned to each variable of interet a hown in Fig..8a. A branch between two node relate the variable at both end. In a fahion imilar to block diagram the gain between the variable i indicated on the branch of y y y 3 y a y (a) (b) y a 4 a 4 y 4 y a 43 y 3 (c) Fig..8. Defining Signal Flow Graph (a) Node (b) Directed Branch (c) Summation ule an aociated arrow. Thu, in Figure.8b we have y a y The value of a variable at a node i equal to the um of all incoming ignal. Thu, in Fig..8c, y 4 a 4 y a 4 y a 43 y 3 A number of definition i appropriate at thi time and will be dicued in next part..8.. Singal Flow Graph (SFG) Algebra (a) Addition ule: The value of the variable deignated by a node equal the um of all ignal entering the node. For the equation given by
84 Analyi of Linear ontrol Sytem X i the SFG i repreented by Fig..83. Σ n a j ij X j (.77) x a i a i x i x x n a i n Fig..83. SFG of equation (.77) (b) Tranmiion ule: The value of the variable deignated by a node i tranmitted on every branch leaving that node. The equation given by X a ij X j, i,,..., n, j-fixed (.78) i repreented by SFG a in Fig..84 x a j x x j a nj a j x n Fig..84. SFG of equation (.78) (c) Multiplication ule: A erie (cacade) connection of branche with tranmittance a, a 3, a 43,... etc., can be replaced by a ingle branch with a new tranmittance equal to the product of the individual tranmittance. X n [a, a 3, a 43,...] X (.79) The SFG of equation (.79) i repreented in Fig..70. X a a 3 a 43 X a,a 3,a 43... X n (a) X X n Fig..85. SFG of equation (.79)
Mathematical Model of Phyical Sytem 85 The variou term involved in the SFG are defined a follow..8.. Definition in SFG (a) Node: A ytem variable that equal the um of all the incoming ignal i defined a Node. A uch the variable X i and X j are repreented by a mall dot which i called Node. a 4 a a 3 a 33 a43 a 54 X X X 3 X 4 X 5 a 3 Fig..86. A ignal flow graph. (b) Branch: A ignal travel along a branch from one node to another in the direction indicated by the branch arrow and the ignal get multiplied by the tranmittance (tranmiion function) of the branch. X, X,... X 5 are different node which have been connected by branche a hown in Fig..86. (c) Path: A path i a continuou, unidirectional ucceion of branche along which no node i travered more than once. In Fig..86, X to X, X to X 3 etc., are path. (d) Input node: (ource node): It i a node with only outgoing branche. In Fig..86, X i an input node. (e) Output node: (ink node): It i a node with only incoming branche. In Fig..86, X 5 i an output (ink) node. (f) Forward Path: It i a path from the input node to the output node. In Fig..86, X to X to X 3 to X 4 to X 5 i a forward path. X to X to X 4 to X 5 i another forward path. (g) Feedback loop: It i a path which originate and terminate on the ame node. In Fig..87, X to X 3 and back to X i a feedback path. (h) Self loop: It i a feedback loop coniting of only one branch. In Fig..86, a 33 i a elf loop. (i) Gain: The gain of a branch i the tranmittance of that branch when the tranmittance i a multiplicative operator. For example, a 33 i the gain of the elf-loop. (j) Non-touching loop: Loop which do not have a common node are aid to be non-touching. (k) Path gain: It i the product of the branch gain encountered in travering a path. For example, the path gain of the forward path from X to X to X 3 to X 4 to X 5 i given by a. a 3. a 43. a 54. (l) Loop gain: It i the product of the branch gain of the loop. For example, the loop gain of the feedback loop from X to X 3 and back to X i a 3. a 3.
86 Analyi of Linear ontrol Sytem.8.3. ontruction of Signal Flow Graph The ignal flow graph of a liner feedback control ytem can be contructed by direct reference to the block diagram of the ytem. Each variable of the block diagram become a node and each block become a branch. Let u conider the block diagram of a canonical feedback control ytem which i hown in Fig..87 () E() G() () B() H() Fig..87. anonical form of control ytem. The ignal flow graph (SFG) i eaily contructed from Fig..88. G E H Fig..88. SFG of Fig..87 We ee from Fig..88 that the or ign of the umming point i aociated with H. The SFG of a ytem can be contructed from it decribing equation. To explain the procedure, let u conider a ytem decribed by the following et of imultaneou equation: x a x a 3 x 3 x 3 a 3 x a 3 x a 33 x 3 (.80) x 4 a 4 x a 43 x 3 There are four variable namely, x, x, x 3 and x 4 and o four node are required. We arrange them from left to right and connect them with appropriate branche by which we obtain the ignal flow graph of Fig..89. a 4 a a 3 a 3 a 33 a 43 x x x 3 x 4 a 3 Fig..89. SFG of Equation (.80)
Mathematical Model of Phyical Sytem 87 x a a 4 x x 4 a 3 a 3 a 3 a 43 x 3 a 33 Fig..90. Modified SFG of Fig..89 The overall ytem gain from input to output may be obtained by Maon gain formula..8.4. Maon Gain Formula It i poible to reduce complicated block diagram to canonical form, from which the control ratio i written a () G () () ± (.8) G () H () It i poible to write the input-output relationhip and hence the control ratio of SFG by inpection; which i accomplihed by uing the Maon gain formula given by where, ΣPi i i M (.8) P i path gain of the ith forward path. determinant of the graph (SFG) [um of loop gain of all individual loop] [um of all gain product of two non-touching loop] [um of all gain product of three non-touching loop]... (.83) P P P... j j j3 j j j i the value of for that part of the graph not touching the ith forward path. P jk jth poible product of K non-touching loop gain. M overall gain of the ytem Let u illutrate the Maon gain rule by finding the overall gain of the ignal flow graph given in Example.3.
88 Analyi of Linear ontrol Sytem Example.3. Determine the overall tranfer function / from the ignal flow graph hown in Fig..9 H G 8 G 6 G G G 7 5 Fig..9. SFG of Example.3 Solution:. There are ix forward path with path gain P G 6 P G 5 G 7 P 3 G. G 7 P 4 G 8 G 6 P 5 G. G 8. G 6 P 6 G 8 H G G 7. There are three individual loop with loop gain P H P G 5 P 3 G G 8 H 3. There i only one poible combination of two non-touching loop with loop gain product. P H G 5 4. There are no combination of three non-touching loop, four non-touching loop, etc. Therefore, we have P j3 P j4... 0 Hence, from the equation (.76), we have [ H G 5 G G 8 H ] [H G 5 ] G 8 H G 5 G G 8 H H G 5 5. The firt forward path i not in touch with one loop (with gain G 5 ). Therefore, ( G 5 ) G 5 [written from the value of ] The econd forward path i not in touch with one loop (with gain H ). Therefore ( H ) H [written from ] The other forward path, namely, third, fourth, fifth and ixth are in touch with all loop individually. Hence, we have
Mathematical Model of Phyical Sytem 89 3 4 5 6 [written from ] From equation (.8), the overall gain [ GGG M 4 6( GH 5 ) GG 5 7( HG 4) GGG 7 GG 6 8 GG G6G8H GGGG 3 8 7H] GHGH HG HG HHGG 8 5 4 4 5 The value of,,, etc. have to be found out very carefully. Example.3. For the ytem repreented by the following equation, find the tranfer function X()/U() by ignal flow graph technique. x x α 3 u x β x x α u x β x α u Solution: In order to repreent differentiated variable, we need to Laplace Tranform the given et of equation. Hence, we have the tranformed equation a x x α 3 u (.84) or x β x x α u x x α β β u (.85) x β x α u or x β α x u (.86) Making ue of the equation (.84), (.85) and (.86) we have the SFG a hown in Fig..9. α / β u α / x β x x α 3 β / Fig..9. SFG of Example.3 Maon gain rule of equation i given by (.8) i given by P i i Σ M i. There are three forward path with path gain α P ( β )
90 Analyi of Linear ontrol Sytem P α β P 3 α 3. There i only one loop with loop gain β β P β ( β) 3. There i no poibility of having non-touching loop. Hence, we have β β ( β ) ( β) 4. The firt forward path with gain P touche the loop with gain P. Therefore, (written from ) The econd forward path with gain P touche the loop at node X. Therefore, (written from ) The third forward path with gain P 3 doe not touch loop. Therefore, β 3 (written from ) ( β) The overall gain (tranfer function) of the ytem i given by M () P P P 3 3 () α α β α 3 ( β ) β ( β) β ( β ) X() U() 3 β β α α α [ β β ] Example.33. Obtain the tranfer function Y()/X() of the SFG hown in Fig..93 / / b X() Y() a a Fig..93. SFG of Example.33
Mathematical Model of Phyical Sytem 9 Solution:. There i only one forward path with path gain. b P b. There are two feedback loop with loop gain P a a P 3. There i no poible combination of two non-touching loop. Therefore, a a a a 4. The forward path touche both the loop. Therefore. The overall gain i given by Y() X() M Y() b [ P ] X() a a 6 a a which i the required tranfer function of the ytem (SFG). Example.34. Obtain the tranfer function Y()/X() of the SFG hown in Fig..94. b X() / / b a Y() a Fig..94. SFG of Example.34 Solution:. There are two forward path with path gain P b P. There are two feedback loop with loop gain P a / P a / b
9 Analyi of Linear ontrol Sytem 3. There are no non-touching loop. Therefore, we have a a a a 4. Both the forward path touch the individual loop. Therefore, The overall gain i given by Y M () X() [P P ] b b a a Y() X() b b a a which i the tranfer function of the SFG hown in Fig..94. Example.35. onider the block diagram of Example.6; a ignal flow diagram of the ytem i hown in Figure.95. There are two forward path with gain G and and one loop with gain H L G H Fig..95. SFG of Example.35 The individual determinant are The diagram determinant i H In accordance with Maon gain formula, we get G G G G Thi agree with our concluion in Section.8..8.5. Signal Flow Graph from Block Diagam The eaiet method of determining the control ratio of a complicated block diagram i to draw the ignal flow graph of the block diagram and then to ue
Mathematical Model of Phyical Sytem 93 Maon gain rule to obtain the control ratio. Takeoff point and umming point are eparated by a unity-gain branch in the ignal flow graph when uing Maon gain rule. Example.36. Draw the SFG and find / for the ytem hown in Fig..96 G H Fig..96. SFG of Example.36 Solution: The ignal flow graph for the ytem of Fig..96 i hown in Fig..97. G H Fig..97. SFG of Fig..96. There are two forward path with path gain P G P G. There are four individual loop with loop gain P G P H P 3 G P 4 G 3. There i only one poible combination of two non-touching loop with loop gain product P G H 4. Therefore, the value of in Maon gain rule i [P P P 3 P 4 ] P G H G G G H 5. The firt forward path i in touch with all the loop. Therefore,
94 Analyi of Linear ontrol Sytem The econd forward path i in touch with all the loop. Therefore, The overall ytem gain i given by M P P GG G GGG 4 3 4 GG H G H GG G GGG GG G H H 4 4 3 4 4 Example.37. Draw the ingal flow graph and determine / for the block diagram hown in Fig..98. G H Fig..98. SFG of Example.37 Solution: The ignal flow graph of Fig..98 i hown in Fig..99. G H Fig..99. SFG of Fig..98. There are two forward path with path gain P G P G. There are five individual loop with loop gain P G P G H P 3 P 4 G P 5
Mathematical Model of Phyical Sytem 95 3. There i no poible combination of two or more non-touching loop. 4. The value of in Maon gain rule i [P P P 3 P 4 P 5 ] G G H G 5. The firt forward path i in touch with all the loop. Therefore, The econd forward path i in touch with all the loop. Therefore, The overall gain i given by P P GG G3 GG 4 GG G3 GG H GG3H GG 4 G4H Example.38. Find the overall T.F. by uing Maon gain formula for the ignal flow graph given in the Fig..00. G G 5 G 6 H Fig..00. Solution: Two forward path, K, T G G 5 G 6 T G G 6 Loop are, L H L G 5 L 3 Out of thee, L and L 3 i combination of two non-touching loop [L L L 3 ] [L L 3 ] L L 3 H Fig..0. Non-touching loop
96 Analyi of Linear ontrol Sytem Eliminate L, L, L 3 a all are touching to T from Eliminate L and L 3, a they are touching to T, from But L i non-touching hence keep it a it i in [L ] T G G 6 L H Fig..0. L non-touching to T Subtitute in Maon Gain formula, T T T.F. GGG 3 4GG 5 6[] GG G6[ G4H] T.F. GH GGGH GH GGHH 4 3 4 5 4 Example.39. alculate given below: Y Y 7 of the ytem, whoe ignal flow graph i G 5 H 4 Y G Y 3 Y 4 G Y 3 Y 5 Y 7 H Y6 H 3 Fig..03. SFG of Example.39 Solution: Forward path for Y to Y 7 are two T G T G G 5 Individual feedback loop are: G G H H 3 (a) (b) (c) Fig..04. Different Loop of Fig..03
Mathematical Model of Phyical Sytem 97 L G H, L, L 3 G H 3, Self-loop L 4 H 4 ombination of two non-touching loop, L L G H L L 4 G H H 4 L L 4 H 4 L 3 L 4 G H 3 H 4 One combination of three non-touching, L L L 4 G H H 4 [L L L 3 L 4 ] [L L L L 4 L L 4 L 3 L 4 ] [L L L 4 ] all loop are touching L a L i non-touching to forward path. Y7 T T GG GG G5( G3H Y Y Now to find the ratio Y. Forward path for Y to Y i one. T Now i ame and L L 4 L L 4 a L and L 4 are nontouching to T. H 4 Y T Y GH 3 H4 GH 3 H4 Y Y 7 Y Y Y Y 7 GGGG 3 4 GGG 5( GH 3 ) GH 3 H4 GH 3 H4 GG GG GG ( G H ) 3 4 5 3 GH H GH H 3 4 3 4 Example.40. Find () () by uing Maon gain formula.
98 Analyi of Linear ontrol Sytem G 5 () G () H Fig..05. SFG of Example.40 Solution: Number of forward path K Maon gain formula, T.F. Σ T T T k k k T G, T G 5 Individual feedback loop are: (a) L H H G (b) L G G 5 (c) L 3 G 5 Fig..06. Different loop of Fig..05 L and L 3 are two non-touching loop. [L L L 3 L 4 ] [L L 3 ] [ H G G 5 ] [ H G 5 ] H G G 5 G 5 H Now conider different forward path, T G All loop are touching to thi forward path.
Mathematical Model of Phyical Sytem 99 G 5 L i non-touching to T, H Fig..07. Non-touching Loop of Fig..06 onider T G 5 [L ] ( H ) H () () () () T T GG GG 5 4( GH) GGGG GG( GH) 3 4 4 5 GH GGGGH GGH GGGHH 3 4 5 4 5 4 Table.7. omparion of Block Diagram and Signal Flow Graph Method Sr. No... 3. 4. 5. 6. 7. Block Diagram Baic importance i given to the element and their tranfer function. Each element i repreented by a block. Tranfer function of the element i hown inide the correponding block. Summing point and takeoff point are eparate. Feedback path i preent from output to input. For a minor feedback loop preent, the formula ± can be ued. Block diagram reduction rule can be ued to obtain the reultant tranfer function. Signal Flow Graph Baic importance i given to the variable of the ytem. Each variable i repreented by a eparated node. The tranfer function i hown along the branche connecting the node. Summing and takeoff point are abent. Any node can have any number of incoming and outgoing branche. Intead of feedback path, variou feedback loop are conidered for the analyi. Gain of variou forward path and feedback loop are jut the product of aociative branch gain. No uch formula ± i neceary. The Maon gain formula i available which can be ued directly to get reultant tranfer function without reduction of ignal flow graph.
00 Analyi of Linear ontrol Sytem 8. 9. 0. Method i lightly complicated and time-conuming a block diagram i required to be drawn time to time after each tep of reduction. oncept of elf loop i not exiting in block diagram approach. Applicable only to linear time invariant ytem. No need to draw the ignal flow graph again and again. Once drawn, ue of Maon Gain Formula give the reultant tranfer function. Self loop can exit in ignal flow graph approach. Applicable to linear time invariant ytem..9. UNSOLVED POBLEMS. Find out the tranfer function of the following block diagram a hown in Fig..08. () G () H Fig..08. The block diagram of a control ytem i hown in Fig..09 below. Obtain tranfer function (a) ()/()/ N0 and (b) ()/N()/ 0. G () N() () () () H () () Fig..09 3. For the ytem hown in Fig..0, obtain the cloed-loop tranfer function by block reduction method. () G () H Fig..0 4. Draw ignal flow diagram for the ytem hown in Fig... Alo find overall tranfer function uing Maon gain formula.
Mathematical Model of Phyical Sytem 0 3 N() () S 4 0 () 0.5 () Fig.. 5. Draw the ignal flow graph and obtain the tranfer function of the ytem hown in Fig... () G () H H 3 Fig.. 6. Find ()/() in the following Fig..3. () B 3 F F F 3 () B B Fig..3 7. Draw ignal flow graph for the following block diagram Fig..4. () G () H Fig..4 8. Draw the ignal flow graph for the following et of equation. x x 4x 3 6x 4 0 x x 3 5x 4 0 7x 3x 3 x 4 0
0 Analyi of Linear ontrol Sytem x x x 3 3 x 4 4 5 7 6 Fig..5 9. Find the tranfer function of the ignal flow graph hown in Fig..6 uing Maon gain formula. G 5 () H G G 6 3 () Fig..6 0. educe the following block diagram into ignal flow graph and then determine the tranfer function uing Maon gain formula. () G () H Fig..7. Find the tranfer function of the ignal flow graph hown in Fig..8. H () G () Fig..8. Find tranfer function for the ignal flow graph given below in Fig..9.
04 Analyi of Linear ontrol Sytem 5. Draw ignal flow graph for the following et of equation. d x 5 00x 0y 0 dt d y 5 00y 0x 0 dt 0 dt dt d x dt 0.5 0. d y dt d y dt 0 0.5 Fig...0. MULTIPLE HOIE QUESTIONS. Match Lit I (ignal) with Lit II (Laplace tranform) and elect the correct anwer. Lit I Lit II (A) e -t u (t). (B) u (t). ( ) () tu (t) 3. (D) te -t u (t) 4. u(t) denote the unit tep function. A B D (a) 4 3 (b) 3 4 (c) 4 3 (d) 3 4. The Laplace tranform of current in an L erie circuit with ohm, L H and / F i I(). The voltage acro the inductor L will be (a) e -t in tu (t) (b) e -t in tu (t) (c) e -t (in t cot t) u(t) (d) e -t (cot t in t) u(t)