Question paper solution. 1. Compare linear and nonlinear control system. ( 4 marks, Dec 2012)


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1 Question paper solution UNIT. Compare linear and nonlinear control system. ( 4 marks, Dec 0) Linearcontrol system: obey super position theorem, stability depends only on root location, do not exhibit limit cycle, can be analyzed by Laplace, Fourier and Z transforms Nonlinear control system: do not obey super position theorem, stability depends on root location and initial condition of input, exhibit limit cycle, cannot be analyzed by Laplace, Fourier and Z transforms.. Distinguish between open loop and closed loop control system. Describe two examples for each. (0 Marks), Jan 009, June, Dec,July 08, July 009, Dec 00 Control systems are classified into two general categories based upon the control action which is responsible to activate the system to produce the output viz. ) Open loop control system in which the control action is independent of the out put. ) Closed loop control system in which the control action is some how dependent upon the output and are generally called as feedback control systems. Open Loop System is a system in which control action is independent of output. To each reference input there is a corresponding output which depends upon the system and its operating conditions. The accuracy of the system depends on the calibration of the system. In the presence of noise or disturbances open loop control will not perform satisfactorily. input Actuating signal output Controller System EXAMPLE  Rotational Generator The input to rotational generator is the speed of the prime mover ( e.g steam turbine) in r.p.m. Assuming the generator is on no load the output may be induced voltage at the output terminals. Dept. of ECE/SJBIT Page
2 Speed of the Prime mover Inputs Rotational Generator Fig  Rotational Generator Induced Voltage Output EXAMPLE Washing machine Most ( but not all ) washing machines are operated in the following manner. After the clothes to be washed have been put into the machine, the soap or detergent, bleach and water are entered in proper amounts as specified by the manufacturer. The washing time is then set on a timer and the washer is energized. When the cycle is completed, the machine shuts itself off. In this example washing time forms input and cleanliness of the clothes is identified as output. Time Washing Machine Fig 3 Washing Machine Cleanliness of clothes EXAMPLE 3 WATER TANK LEVEL CONTROL To understand the concept further it is useful to consider an example let it be desired to maintain the actual water level 'c ' in the tank as close as possible to a desired level ' r '. The desired level will be called the system input, and the actual level the controlled variable or system output. Water flows from the tank via a valve V o, and enters the tank from a supply via a control valve V c. The control Valve Vvalve C is adjustable manually. Waterin Valve V O Desired Water level r WATER TANK Actual Water level c C Water out Fig 4 b) Open loop control Fig.4 a) Water level control A closed loop control system is one in which the control action depends on the output. In closed loop control system the actuating error signal, which is the difference between the input signal and the feedback signal (output signal or its function) is fed to the controller. Dept. of ECE/SJBIT Page
3 Reference input Error detector Actuating/ error controller Control elements Forward path System / Plant Controlled output Feed back signal Feed back elements Fig.5: Closed loop control system EXAMPLE THERMAL SYSTEM To illustrate the concept of closed loop control system, consider the thermal system shown in fig 6 here human being acts as a controller. He wants to maintain the temperature of the hot water at a given value r o C. the thermometer installed in the hot water outlet measures the actual temperature C0C. This temperature is the output of the system. If the operator watches the thermometer and finds that the temperature is higher than the desired value, then he reduce the amount of steam supply in order to lower the temperature. It is quite possible that that if the temperature becomes lower than the desired value it becomes necessary to increase the amount of steam supply. This control action is based on closed loop operation which involves human being, hand muscle, eyes, thermometer such a system may be called manual feedback system. Human operator Steam Steam Thermometer Hot water Desired hot water.tempr o c + + Brain of operator (rc) Muscles and Valve Actual Water temp C o C C old water Thermometer Drain Fig 6 a) Manual feedback thermal system EXAMPLE HOME HEATING SYSTEM The thermostatic temperature control in hour homes and public buildings is a familiar example. An electronic thermostat or temperature sensor is placed in a central location usually on inside wall about 5 feet from the floor. A person selects and adjusts the desired room temperature ( r ) b) B Dept. of ECE/SJBIT Page 3
4 say 5 0 C and adjusts the temperature setting on the thermostat. A bimetallic coil in the thermostat is affected by the actual room temperature ( c ). If the room temperature is lower than the desired temperature the coil strip alters the shape and causes a mercury switch to operate a relay, which in turn activates the furnace fire when the temperature in the furnace air duct system reaches reference level ' r ' a blower fan is activated by another relay to force the warm air throughout the building. When the room temperature ' C ' reaches the desired temperature ' r ' the shape of the coil strip in the thermostat alters so that Mercury switch opens. This deactivates the relay and in turn turns off furnace fire, which in turn the blower. Desired temp.r o c + Relay switch Furnace Outdoor temp change (disturbance) Blower House Actual Temp. C o C Fig 7 Block diagram of Home Heating system. A change in out door temperature is a disturbance to the home heating system. If the out side temperature falls, the room temperature will likewise tend to decrease. 3. For the two port network shown, obtain the transfer functions, i) V (s) / V (s) and ii) V (s) / I (s) ( 8 marks, Dec 0) Ans. Taking laplace transform of the given network Dept. of ECE/SJBIT Page 4
5 Applying KVL Dept. of ECE/SJBIT Page 5
6 4. For the system shown in fig. i) draw the mechanical network ii) write the differential equations iii) obtain torque to voltage analogy. ( 8 Marks, Dec 0) Dept. of ECE/SJBIT Page 6
7 5. Write the differential equations for the mechanical system shown in fig. (b) and obtain FV and FI analogous electrical circuits. ( 0 Marks), Jan Mention merits and demerits of open loop and closed loop control systems and give an example for each. ( 6 Marks), July 009 An advantage of the closed loop control system is the fact that the use of feedback makes the system response relatively insensitive to external disturbances and internal variations in systems parameters. It is thus possible to use relatively inaccurate and inexpensive components to obtain the accurate control of the given plant, whereas doing so is impossible in the openloop case. From the point of view of stability, the open loop control system is easier to build because system stability is not a major problem. On the other hand, stability is a major problem in the closed loop control system, which may tend to overcorrect errors that can cause oscillations of constant or changing amplitude. It should be emphasized that for systems in which the inputs are known ahead of time and in which there are no disturbances it is advisable to use openloop control. closed loop control systems have advantages only when unpredictable disturbances it is advisable to use openloop control. Closed loop control systems have advantages only when unpredictable disturbances and / or unpredictable variations in system components used in a closed loop control system is more than that for a corresponding open loop control system. Thus the closed loop control system is generally higher in cost. 7. Obtain the transfer function of an armature controlled DC servomotor. (6 Marks), July 009, Dec 00 A DC motor is used in a control system where an appreciable amount of shaft power is required. The DCmotors are either armaturecontrolled with fixed field, or fieldcontrolled with fixed armature current.dc motors used in instrument employ a fixed permanentmagnet field, and the control signal is appliedto the armature terminals. Dept. of ECE/SJBIT Page 7
8 Fig.. (a) Schematic diagram of an armaturecontrolled DC servo motor, (b) Block diagram In order to model the DC servo motor shown in Fig., we define parameters and variables as follows. Ra = armaturewinding resistance, ohms La = armaturewinding inductance, henrys ia = armaturewinding current, amperes if = field current, amperes ea = applied armature voltage, volts eb = back emf, volts θ = angular displacement of the motor shaft, radians T = torque delivered by the motor, lbft J = moment of inertia of the motor and load referred to the motor shaft, slugft.f = viscousfriction coefficient of the motor and load referred to the motor shaft, lbft/rad/sec The torque T is delivered by the motor is proportional to the product of the armature current iaandthe air gap flux Ψ, which is in turn is proportional to the field current Ψ = Kf if Dept. of ECE/SJBIT Page 8
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11 UNIT. Illustrate how to perform the following in connection with block diagram reduction rules: i) Shifting a take  off point after a summing point. ii) Shifting a take  off point before a summing point. ( 4 marks, Dec 0) Shifting a take  off point before a summing point. The performance equations of a controlled system are given by the following set of linear algebraic equations: i) Draw the block diagram. ii) Find the overall transfer function C(s)/ R(s) using block diagram reduction technique. ( 8 marks, Dec 0) Dept. of ECE/SJBIT Page
12 3. Draw the corresponding signal flow graph for the given block diagram is shown in fig. and obtain the overall transfer function by Mason s gain formula. ( 8 marks, Dec 0) ; ;  G G H; L =  G G3 H; L3 =  G H; Dept. of ECE/SJBIT Page
13 4. Obtain the C/R ratio for the block diagram shown using block diagram reduction technique ( 06 Marks, July 009) 5. Find C(S)/ R(S) by Mason s gain formula ( 06 Marks, July 009) 6. Explain briefly the following termsi) Forward path ii) path gain iii) loop gain iv) canonical form (08 Marks, July 009) Path: A path is a traversal of connected branches in the direction of branch arrow. Loop: A loop is a closed path. Self loop: It is a feedback loop consisting of single branch. Loop gain: The loop gain is the product of branch transmittances of the loop. Nontouching loops: Loops that do not posses a common node. Dept. of ECE/SJBIT Page 3
14 Forward path: A path from source to sink without traversing an node more than once. Feedback path: A path which originates and terminates at the same node. Forward path gain: Product of branch transmittances of a forward path. 7. Define the transfer function. Explain Mason s gain formula for determining the transfer function from signal flow graphs. ( 06 marks, Dec 00) The relationship between an input variable and an output variable of a signal flow graph is given by the net gain between input and output nodes and is known as overall gain of the system. Masons gain formula is used to obtain the over all gain (transfer function) of signal flow graphs. Masons Gain Formula Gain P is given by P k P k k Where, P k is gain of k th forward path, is determinant of graph =(sum of all individual loop gains)+(sum of gain products of all possible combinations of two nontouching loops sum of gain products of all possible combination of three nontouching loops) + k is cofactor of k th forward path determinant of graph with loops touching k th forward path. It is obtained from by removing the loops touching the path P k. 8. For the block diagram shown in fig.q (b), determine the transfer function Q (s)/q(s) using block diagram reduction algebra. (8 Marks, Dec 00) Dept. of ECE/SJBIT Page 4
15 8. For the system described by the signal flow graph shown in fig. Q(c), obtain the closed loop transfer function C(s) / R(s), using Mason s gain formula. ( 6 Marks, Dec 00) h c X a X b e X 5 f X 3 X 4 X 3 i d g There are two forward paths. The gain of the forward path are: P =abcd, P =ac There are five loops with loop gains: L =eh, L =cg, L 3 = bei, L 4 =edf, L 5 =befg There is one combination of Loops L and L which are nontouching with loop gain product L L =ehcg = +eh+cg+bei+efd+befg+ehcg Forward path touches all the five loops. Therefore =. Forward path does not touch loop L.Hence, = + eh The transfer function T = X X 3 P P eh abef ac cg bei efd eh befg ehcg 9. Obtain the closed loop transfer function C(s)/ R(s) for the signal flow graph of a system shown in fig. Q(b) using Mason s gain formula. ( 0 Marks, June 0) Dept. of ECE/SJBIT Page 5
16 P 3 =G G G 7 There are three forward paths. The gain of the forward path are: P =G G G 3 G 4 G 5 P =G G 6 G 4 G 5 There are four loops with loop gains: L =G 4 H, L =G G 7 H, L 3 = G 6 G 4 G 5 H, L 4 =G G 3 G 4 G 5 H There is one combination of Loops L and L which are nontouching with loop gain product L L =G G 7 H G 4 H = +G 4 H +G G 7 H +G 6 G 4 G 5 H +G G 3 G 4 G 5 H + G G 7 H G 4 H Forward path and touch all the four loops. Therefore =, =. Forward path 3 is not in touch with loop. Hence, 3 = +G 4 H. The transfer function T= C s R s P P P 3 3 G 4 H G G G G G G G 7 3 H G G G G G G G H 6 G G G G G G G 5 H G 4 H G G G 4 7 H H 0. Derive an expression for the closed loop transfer function of a negative feedback system. ( 4 Marks, Dec 0) B(s) R(s) + C(s) C(s) G(s)  B(s) H(s) Fig.4 Dept. of ECE/SJBIT Page 6
17 The ratio of the feed back signal B(s) to the actuating error signal E(s) is called the open loop transfer function. open loop transfer function = B(s)/E(s) = G(s)H(s) The ratio of the output C(s) to the actuating error signal E(s) is called the feed forward transfer function. Feed forward transfer function = C(s)/E(s) = G(s) If the feed back transfer function is unity, then the open loop and feed forward transfer function are the same. For the system shown in Fig.4, the output C(s) and input R(s) are related as follows. C(s) = G(s) E(s) E(s) = R(s)  B(s) = R(s)  H(s)C(s) but B(s) = H(s)C(s) Eliminating E(s) from these equations C(s) = G(s)[R(s)  H(s)C(s)] C(s) + G(s)[H(s)C(s)] = G(s)R(s) C(s)[ + G(s)H(s)] = G(s)R(s) C(s) G(s) = R(s) + G(s)H(s) C(s)/R(s) is called the closed loop transfer function. Dept. of ECE/SJBIT Page 7
18 UNIT3. Derive the expression for peak time. ( 4 marks, Dec 0). The loop transfer function of transfer function is given by sec i) Determine the static error coefficients ii) Determine steady state error coefficients for the input r(t) = t + 5t +0 ( 8 marks, Dec 0) Dept. of ECE/SJBIT Page 8
19 3. ( 8 marks, Dec 0) Comparing with standard nd order system equation Dept. of ECE/SJBIT Page 9
20 T s = Sec 4. Derive expressions for peak response time t p and maximum overshoot M p of an under damped second order control system subjected to step input ( 6 Marks, June 0, Dec 00, Dec 0) W. K.T. for the second order system, T.F. = C(S) = W n () R(S) S + W n S+ W n Assuming the system is to be underdamped ( < ) Rise time(tr):  W. K.T. C(t r ) =  e  Wnt Cos Wdt r +.sinwdt r Let C(t r ) =, i.e., substituting t r for t in the above Eq:  Then, C(t r ) = =  e  Wntr Coswdt r +.sinwdt r Cos wdt r + sinwdt r = tan wdt r =   = wd.  Dept. of ECE/SJBIT Page 0
21 jw Thus, the rise time t r is, jwd t r =. tan   w d =  secswn  Wn Wd wd When must be in radians.  Wn Peak time : (tp) S Plane Peak time can be obtained by differentiating C(t) W.r.t. t and equating that derivative to zero. dc = O = Sin Wdtp Wn. e  Wntp dt t = t p  Since the peak time corresponds to the st peak over shoot. Wdtp = = tp =. Wd The peak time tp corresponds to one half cycle of the frequency of damped oscillation. Maximum overshoot : (MP) The max over shoot occurs at the peak time. i.e. At t = tp =. Wd Mp = e ( / Wd) ( /  ) or e Settling time : (ts) An approximate value of ts can be obtained for the system O < < envelope of the damped sinusoidal waveform. by using the Time constant of a system = T =. Wn Setting time ts= 4x Time constant. = 4x.for a tolerance band of +/ % steady state. Wn Dept. of ECE/SJBIT Page
22 5. For a unity feedback control system with G(s) = 0(S+) / S (S+). Find i) The static error coefficients ii) Steady state error when the input transform is ( 6 Marks, June 0) i) G(s) H(s) = 0(S+) / S (S+) it is a type II system ii) Kp = G(s) H(s) Kp = 0 Kv = S G(s) H(s) Kv = 0 Ka = S G(s) H(s) Ka = 0 iii) E ss = Dept. of ECE/SJBIT Page
23 UNIT4. Explain RouthHurwitz s criterion for determining the stability of a system and mention any three limitations of RH criterion. ( 0 Marks, June 0) Routh s Stability Criterion: It is also called Routh s array method or RouthHurwitz s methodrouth suggested a method of tabulating the coefficients of characteristic equationin a particular way. Tabulation of coefficients gives an array called Routh s array. Consider the general characteristic equation as, F(s) = a 0 s n + a s n + a s n a n = 0. Method of forming an array : S n a 0 a a 4 a 6. S n a a 3 a 5 a 7 S n b b b 3 S n3 c c c S 0 a n Coefficients of first two rows are written directly from characteristics equation. Dept. of ECE/SJBIT Page 3
24 From these two rows next rows can be obtained as follows. a a a 0 a 3 a a 4 a 0 a 5 a a 6 a 0 a 7 b =, b =, b 3 = a a a From nd & 3 rd row, 4 th row can be obtained as b a 3 a b b a 5 a b 3 C =, C = b b This process is to be continued till the coefficient for s 0 is obtained which will be a n. From this array stability of system can be predicted. Routh scriterion : The necessary & sufficient condition for system to be stable is All the terms in the first column of Routh s array must have same sign. There should not be any sign change in first column of Routh s array. If there are sign changes existing then,. System is unstable.. The number of sign changes equals the number of roots lying in the right half of the splane. Limitation of Routh scriterion :. It is valid only for real coefficients of the characteristic equation.. It does not provide exact locations of the closed loop poles in left or right half of splane. 3. It does not suggest methods of stabilizing an unstable system. 4. Applicable only to linear system. Dept. of ECE/SJBIT Page 4
25 . A unity feedback control system is characterized by the open loop transfer function 3. Define: i) Marginally stable systems; ii) absolutely stable system; iii) conditionally stable systems. ( 06 Marks, June 0) Definition of stable system: A linear time invariant system is said to be stable if following conditions are satisfied.. When system is excited by a bounded input, output is also bounded & controllable.. In the absence of input, output must tend to zero irrespective of the initial conditions. Unstable system: A linear time invariant system is said to be unstable if,. for a bounded input it produces unbounded output.. In the absence of input, output may not be returning to zero. It shows certain output without input. Besides these two cases, if one or more pairs simple non repeated roots are located on the imaginary axis of the splane, but there are no roots in the right half of splane, the output response will be undamped sinusoidal oscillations of constant frequency & amplitude. Such systems are said to be critically or marginally stable systems. Critically or Marginally stable systems: A linear time invariant system is said to be critically or marginally stable if for a bounded input its output oscillates with constant frequency & Amplitude. Such oscillation of output are called Undamped or Sustained oscillations. Dept. of ECE/SJBIT Page 5
26 C(t) For such system one or more pairs of non repeated roots are located on the imaginary axis as shown in fig6(a). Output response of such systems is as shown in fig6(b). Constant Amplitude &frequency oscillations Fig 6(a) C(t) X J steady state output X J X  J X  J non repeated poles on J axis. If there are repeated poles located purely on imaginary axis system is said to be unstable. t J x x J x x steady state output splane t Conditionally Stable: A linear time invariant system is said to be conditionally stable, if for a certain condition if a particular parameter of the system, its output is bounded one. Otherwise if that condition is violated output becomes unbounded system becomes unstable. i.e. Stability of the system depends the on condition of the parameter of the system. Such system is called conditionally stable system. Splane can be divided into three zones from stability point of view. Dept. of ECE/SJBIT Page 6
27 J axis Left half of splane Right half of splane Stable UnstableReal SPlane (repeated) unstable (non repeated roots) Marginally stable 4. Define the term stability applied to control system sand what is the difference between absolute stability and relative stability. ( 4 Marks, Dec 0) A linear time invariant system is stable if following conditions are satisfied. For bounded input, the response of the system should be bounded and controllable. In absence of input the output should be zero. The location of roots of the characteristic equation for a stable system is in left half of an splane. Absolute stability: same above 3 conditions. Relative stability: The response decays to zero much faster as compare to the poles close to jw axis. When poles are located far away from jwaxis or imaginary axis in LHS of Splane. 4. Using Routh s stability criterion determine the stability of the following systems: i) Its open loop transfer function has poles at s = 0, s= , s = 3 and s =  5. Gain K = 0. Dept. of ECE/SJBIT Page 7
28 ii) It s a type system with an error constant of 0/sec and poles at s = 3 and s = 6 ( 8 Marks, Dec 0) i) Poles at s = 0, s= , s = 3 and s = 5. Gain K = s+50 = 0 No sign change in first column therefore it is a stable system. ii) It is a type system There is a pole at s =0 poles at s = 3 and s = 6 and K v = 0/sec Type  Thus the characteristic equation K = 80 Dept. of ECE/SJBIT Page 8
29 S 3 +9S +8S+80 = 0 There are two sign changes so two roots are located in RHS of splane. Hence the system is unstable. 5. Using RH Criterion determine the stability of the system having the characteristic equation, S 4 + 0S S + 70S + 75 = 0 has roots more negative than S = . ( 8 Marks, Dec 0) Dept. of ECE/SJBIT Page 9
30 UNIT 5. Sketch the root locus for a unity feedback control system with open loop transfer function: ( Marks, june 0) K G(s) s(s 6s 5) ) Root locus is symmetrical about real axis. ) There are no open loop zeros (m=0). There are three open loop poles at s=0, 3 j4(n=3). Three branches of root loci start from the open loop poles when K= 0 and all the three branches go asymptotically when K. 3) Entire negative real axis lies on the root locus as there is a single pole at s=0 on the real axis. 80 (q ) 4) The asymptote angle is A =, q 0,, n m 0,,. n m Angle of asymptote are A = 60, 80, ) Centroid of the asymptote is ( 3 3).0 3 σ A (sum of poles) n (sum of m zeros) p 6) The brake points are given by dk/ds =0. 3 K s(s 6s 5) (s 6s 5s) dk ds s K,, 3s 34 s j.087and j For a point to be break point, the corresponding value of K is a real number greater than or equal to zero. Hence, S, are not break points. 7) Angle of departure from the open loop pole at s=0 is 80. Angle of departure from complex pole s= 3+j4 is 80 (sum of the angles of (sum of the angles of vectors to a complex pole in question from other poles) vectors to a complex pole inquestion from zeros) Dept. of ECE/SJBIT Page 30
31 p 80 (80 tan ) Similarly, Angle of departure from complex pole s= 3j4 is φ p 80 ( ) 33.3 or ) The root locus does cross the imaginary axis. The cross over point and the gain at the cross over can be obtained by Rouths criterion 3 The characteristic equation is s 6s 5s K 0. The Routh s array is s s s s K 5 K For the system to be stable K < 50. At K=50 the auxillary equation is 6s +50=0. s = ±j5.or substitute s= j in the characteristic equation. Equate real and imaginary parts to zero. Solve for and K. s ( 3 jω 6ω 6s jω K) 5 jω jω ω 5 ω 0, j5 K 0,50 The plot of root locus is shown in Fig s K 0 K 0 0 Dept. of ECE/SJBIT Page 3
32 . The open loop transfer function of a feedback control system in Check whether the following points are on the root locus. If so, find the value of K at these points, i) S = .5 ii) S = j ( 6 Marks, Dec 0) Dept. of ECE/SJBIT Page 3
33 3. Sketch the root locus plot for a negative feedback control system characterized by an open loop transfer function, Comment on stability. (4 Marks, Dec 0) 4. Define brake away / in point on a root locus. Explain any one method of determining the same. (6 Marks, Dec 00) Breakaway and breakin points.the breakaway and breakin points either lie on the real axis or occur in complex conjugate pairs. On real axis, breakaway points exist between two adjacent poles and breakin in points exist between two adjacent zeros. To calculate dk these polynomial 0 must be solved. The resulting roots are the breakaway / breakin ds points. The characteristic equation given by Eqn.(7), can be rearranged as B(s) KA(s) 0 (4) where, B(s) A(s) (s K(s p )(s z )(s p ) (s z ) (s p ) and n z m ) Dept. of ECE/SJBIT Page 33
34 The breakaway and breakin points are given by dk ds d ds A B A d ds B 0 (5) Note that the breakaway points and breakin points must be the roots of Eqn.(5), but not all roots of Eqn.(5) are breakaway or breakin points. If the root is not on the root locus portion of the real axis, then this root neither corresponds to breakaway or breakin point. If the roots of Eqn.(5) are complex conjugate pair, to ascertain that they lie on root loci, check the corresponding K value. If K is positive, then root is a breakaway or breakin point. 5. For the following characteristic polynomial K G(s) draw the s(s 4)(s 4s 0) root locus for 0 K. (8 Marks, Dec 00) ) Root locus is symmetrical about real axis. ) There are no open loop zeros (m=0). There are three open loop poles (n=3) at s = 0, 4,  j4. Three branches of root loci start from the three open loop poles when K= 0 and to infinity asymptotically when K. 3) Root locus lies on negative real axis between s = 0 to s = 4.0 as there is one pole to the right of any point s on the real axis in this range. 80 (q ) 4) The asymptote angle is A =, q n m 0,,, 3 n m Angle of asymptote are A = 45, 35, 5, 35. 5) Centroid of the asymptote is ( ).0 4 σ A (sum of poles) n (sum of m zeros) 6) The root locus does branch out, which are given by dk/ds =0. K dk Break point is given by ds 3 4s 4s 7s s (s s s 3 s(s (s 8s 4 )(4s.0, K.0 4)(s 8s 3 6s 6s 64; 4s 36s 3s j.45, K 40) 0) 80s) 40s Dept. of ECE/SJBIT Page 34 0
35 The root loci brakeout at the open loop poles at s = .0, when K = 64 and breakin and breakout at s=+j.45, when K=00 7) The angle of departure at real pole at s=0 is 80 and at s=4 is 0. The angle of departure at the complex pole at s =  + j4 is p (sum of the angles of (sum of the angles of vectors to a complex pole in question from other poles) vectors to a complex pole inquestion from zeros) θ θ θ p 4 tan 63.4 or6.6  atan(4,) tan 63.4, θ ( tan 0 90 ) The angle of departure at the complex pole at s =  j4 is 90 p 80  ( ) The root locus does cross the imaginary axis, The cross over point and gain at cross over is obtained by either Routh s array or substitute s= j in the characteristic equation and solve for and gain K by equating the real and imaginary parts to zero. Routh s array The characteristic equation is 4 s 3 8s 36s 80s K 0 70 ) The Rouths array is s s s s s K 080 8K 6 K K Dept. of ECE/SJBIT Page 35
36 For the system to be stable K > 0 and 0808K > 0. The imaginary crossover is given by 0808K=0 or K = 60. At K = 60, the auxiliary equation is 6s +60 = 0. The imaginary cross over occurs at s= j 0. or s 4 put ω jω ω Equate real and 8ω 4 4 8s 4 3 s ω 36ω jω 80ω 36s jω 3 K K 0 80s 36 jω imaginary parts to zero 0 j ω K 8ω K 3 0, 0 80 jω 80ω j 60 0;s K 0 j 0 0 The root locus plot is shown below Dept. of ECE/SJBIT Page 36
37 UNIT 6. State the advantages and limitations of frequency domain approach. ( 6 Marks, Dec 0) Advantages:. Determine the transfer function, of a system whose asyptotic gain plot is shown in fig. ( 0 Marks, Dec 0) Starting slope at 0dB = no pole or zero at the origin. Wc = factor is / + T S = / +S Shift at W = is 0dB. 0 log K = 0dB Therefore K = Next change in slope occur at W = W c. Dept. of ECE/SJBIT Page 37
38 3. List the effects of lead compensation. ( 4 Marks, Dec 0) The lead compensation adds a dominant zero and a pole, this increases the damping of the closed loop system. Increasing damping means less overshoot, less rise time, less settling time. It improves phase margin. Since it improves phase and gain margin result in improving the relative stability. 4. Explain Nyquist s stability criterion. ( 4 Marks, Dec 00, June 00) It states that the number of unstable closedloop poles is equal to the number of unstable openloop poles plus the number of encirclements of the origin of the Nyquist plot of the complex function D(S). This can be easily justified by applying Cauchy s principle of argument to the function D(S)with the plane contour. Note that Z and P represent the numbers of zeros and poles, respectively, of D(S) in the unstable part of the complex plane. At the same time, the zeros of D(S) are the closedloop system poles, and the poles of D(S) arethe openloop system poles (closedloop zeros). Dept. of ECE/SJBIT Page 38
39 The above criterion can be slightly simplified if instead of plotting the function, D(s) = + G(S) H(S) we plot only the function G(S) H(S) and count encirclement of the Nyquist plot of G(S) H(S) around the point ( , j0) so that the modified Nyquist criterion has the following form. The number of unstable closedloop poles (Z) is equal to the number of unstable openloop poles (P) plus the number of encirclements (N) of the point ( , j0) of the Nyquist plot of G(S) H(S), that is Z = P + N Dept. of ECE/SJBIT Page 39
40 UNIT 7. a) (Dec 0) GH(j ) GH(j ) Im Dept. of ECE/SJBIT Page 40
41 c) Dept. of ECE/SJBIT Page 4
42 UNIT 8. June 0 statetransition matrix is a matrix whose product with the state vector at an initial time gives at a later time. The statetransition matrix can be used to obtain the general solution of linear dynamical systems. It is also known as the matrix exponential. Properties of state transition matrix Let A Rn n. Then the following hold: (a) e At e At = e A(t+t) for all t, t R; (b) Ae At = e At A for all t R; (c) (e At ) = e At for all t R; (d) det e At = e (tra)t for all t R; (e) If AB = BA, then e At e Bt = e (A+B)t for all t R. Dec 0 State: A state of a system is the current value of internal elements of the system, that change separately (but not completely unrelated) to the output of the system. In essence, the state of a system is an explicit account of the values of the internal system components. Dept. of ECE/SJBIT Page 4
43 State Variables The state variables represent values from inside the system, that can change over time. In an electric circuit, for instance, the node voltages or the mesh currents can be state variables. In a mechanical system, the forces applied by springs, gravity, and dashpots can be state variables. We denote the input variables with u, the output variables with y, and the state variables with x. In essence, we have the following relationship: Where f(x, u) is our system. Also, the state variables can change with respect to the current state and the system input: Where x' is the rate of change of the state variables. We will define f(u, x) and g(u, x) in the next chapter. State space: In a state space system, the internal state of the system is explicitly accounted for by an equation known as the state equation. The system output is given in terms of a combination of the current system state, and the current system input, through the output equation. These two equations form a system of equations known collectively as statespace equations. The statespace is the vector space that consists of all the possible internal states of the system. Because the statespace must be finite, a system can only be described by statespace equations if the system is lumped. b) advantages:. it can be applicable for for nonlinear and time varying system.. It can be applied to MIMO 3. Any type of of input can be considered for designing the system 4. Method involves matrix algeebra can be made conviniently adapted for the digital computers. Dept. of ECE/SJBIT Page 43
44 c) Dept. of ECE/SJBIT Page 44
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