"y fo UHUUgEY HioKuooee ne 3.2_Concept of Transfer Function : In any system, first the system parameters are gned and their values Selection of system parameters Selected System parameters | —»—— Output input Fig. 3.1 Now performance of system can be expressed in terms of its output as, desi; are selected as per requirement. The input is selected next to see the performance of the system designed. This is shown in the Fig. 3.1 Effect of system parameters on the selected input Output = Output = Input x Effect of system parameters, Output +. Effect of system parameters = This effect of system parameters, role of system parameters in the performance o cystems can be expressed as ratio of output (0 input. Mathematically such a functi x the effect of system parameters on input to produce output is called n. Due to the own characteristics of the system parameters, the inp! applied to the system. This is the concept of transt* given in the next section. explaining transfer func nsferred into output, once {inition of the transfer function is gets tran function. The exact de! (36) Scanned with CamScanner3 transfer Function 3.3.1 Definition thematically it is defined 9s the ratio of Laplace transform of output (response) system to Laplace Transform of input (excitation of driviny; function), under the ption that all initial conditions are zero, of th assump symbolically system can be represented as shown in Fig, 3.2(a), While the transfer ction of system can be shown as in the Fig 3.2(b) fu Fig. 3.2(a) Fig. 3.2(b) Cs Transfer function of this system, te where C(s) is Laplace of c(t) and R¢s) is RG Laplace of r(t). If T(s) is the transfer function of the system then, Tye) = Laplace transform of output _ C(s) Laplace transform of input R(s) 3.3, calculate its transfer function where Vi()) is output stem. —x1 For a system shown in and V(b is input to the Fig. 3.3 Sol. : We can write for this system, equations by applying KVL as, vi) = Rid Ef ite at () 1, and V,(t) = c i(t) dt we 2) We are interested in Ye where V,(s) is Laplace of Vo(t) and Vy(s) is Laplace of ils) y, os lt) and initial conditions are to be neglected. a $9 taking Laplace of above two equations and assuming initial conditions zero we ” write, ~ Scanned with CamScanneransfer Functioy Control Engineering ___ 38 TF 1 & Imputsn ‘ atis)+ EH Vis) = RIB SE 1 sc Iis) = sCV¥o (6) K(s) Substituting in equation (3), Vi0) = sCVols) ret] sc. sCR V,(s) Vols) = Vols) [1+ sCR] Vi(s) = Vo) 1 Vis) 1+8CR We can represent above system as in Fig. 34 Fig. 3.4 Ex. 20 Find out the iven network, . | Fig. 3. Sol.: Applying KVL we get the equatione = = ineidi,t = Retry le. ae’ os it i O/p=E ve ° Laplace transform of f raya ~ Neglecting initial condition 5 and Laplace transform of 4i(t) Fi = skis at (s) ~ Neglecting initial conditio® Take Laplace transform, Ey) = ae 1 | Scanned with CamScannerControl! Engineering 39 Transfer Function & Impulse Response = a 7 e Ei) [Rese 4] ° sc , il Now Bo = Lfiae ® E,() = 4b of) = 10) : Ks) = sC E,(s) @ Substituting value of I(s) in equation (2) = 1 _ [rs sL+ 1 1 [Rests] sc Fig. 3.6 3.3.2 Advantages and Features of Transfer Function : i) It gives mathematical models of all system components and hence of the overall system. Individual analysis of various components is also possible by the transfer function approach ii) As it uses a Laplace approach, it converts integro-differential time domain equations to simple algebraic equations iii) It suggests operational method of expressing integro-differential equations which relate output to input. iv) The transfer function is expressed only as a function of the complex variable ‘s’, It is not a function of the real variable, time or any other variable that is used as the independent variable. v) It is the property and characteristics of the system itself. Its value is dependent on the parameters of the system and independent of the values of inputs. In the example 1, if the output ie, focus of interest is selected as voltage across resistance R rather than the voltage across capacitor C, the transfer function will be different. So transfer function is to be obtained for a pair of input and output and then it remains constant for any selection of input as long as output variable Scanned with CamScannerction & Impul Control Engineering __40_ Transfer Functio mpulse Respons, ame. It helps in calculating the output for any type of Input applied yy stem, vi) Once transfer function is known, output response for any type of reference jn. can be calculated. vii) It helps in determining the important information about the system ie Poles: zeros, », characteristic equation etc.. Vili) It helps in the stability analysis of the system. ix) The system differential _equation_can be obtained by replacing variable ‘y d/dt. 3.3.3 Disadvantages : i) Only applicable to linear time invariant systems. —" ii) It does not provide any information concerning the physical structure of 4, system. iii) Effects arisis ig due to initial_conditions are totally neglected. Hence init conditions loose their importance. 3.3.4 Procedure to Determine the Transfer Function of a Control System : ‘The procedure used in Ex. 1 and Ex.2 can be generalised as below : 1) Write down the time domain equations for the system by introducing differet variables in the system. 2) Take the Laplace transform of the system equations assuming. all. init conditions to be zero. 3) Identify system input and output variables, 4) Eliminating introduced varial output variables, 5) Take the ratio of Lapl input variable bles, get the resultant equation in terms of input an! lace transform of output variable to Laplace transform 0 to get the transfer function of the system, Ex. 3. Find out the T.F. of the given network Scanned with CamScanner41. Transfer Function & Impulse Response Controt Engineering 1 KVL we can write, sol. : Apply Bw = Lol, f i(t) dt + A(R 0) dtc while E,Q) = GR . 2) where 1) = input and E(t) = output Taking Laplace of equations (1) and (2), neglecting, the initial conditions E\(s) = sts) + 4 4 RI (3) Ej(s) sLl(s) + & + RI(s) E(s) = Is)R - (4) 1 Ey) ty [s+ Le] from (3) Substituting Kis) fe) from (4) in the above equation ¥ [FP 1+ =] x |store scr E\G s(5) xe Els) _ skC E\(s) ss? LC+sRC+1 This is the required transfer function. Note ; The network in Ex. 2 and Ex. 3 is same but as focus of interest i changed, the transfer function is changed. For a fixed output, transfer function is constant and independent of any type of input applied to the system. 3.4 Impulse Response and Transfer Function : output is The impulse function is defined as, rt) = A fort=0 =0 fort# 0 A unit impulse function 6(t) can be considered a narrow pulse (of any shape) occuring at zero time such that area under the pulse is unity and the time for which the pulse occurs tends to zero. In the limit t 0, the pulse reduces to a unit impulse S(t), Consider a narrow rectangular pulse of 4A width A and height \\ units, so that the area under the pulse = 1, as shown in the + a ——_—_+| 1 Fig. 3.8(a). Fig. 3.8 (a) Scanned with CamScannerFunction & Impulse p, Control Engineering 42__Transfer Ft "SPOn gy Now if we go on reducing width A and maintain the a as unity then the height ba Will go on increasing. Ultimately when A >> 0, \y > @ and the pulse is of infinite magnitude may then be called an impulse of Magnitude unity and it is denoted by 6(t). It is ot possible to draw an impulse functi Paper, hence it is represented by a_ vertical arrow at t=0 as shown in the Fig. 3.8(b) on on 0 Symbol of S(t) Fig. 3.8 (b) So mathematically unit impulse is defined as, 8(t) = 1, t=0 =0, t#0 If in the ebove example the area under the narrow pulse is maintained at K Units while the period of pulse is reduced, it. called to be an impulse of magnitude 4 and is denoted by KA(t), as shown in thy Fig. 3.8 (0). An important property of impulse function is that if it is multiplied by any 0 c : y Symbol of Ks function and integrated then the result is yenbol ct S(t the value of the function at t = 0 Fig. 3.8 (c) +0 t o Thus J (5 at = J f(t) 8 (t) dt = J {QS (dt = {|= 9. = o 0 This is called ‘sampling’ property of impulse. transform of 8(t) as, Hence if we define Laplace Mem) = [sme ta by definition 0 = est =e ; ho ~» by sampling Property. =e. Lis(y) = Thus Laplace Trans! Now So response C(s) ca T(s) is determined. Scanned with CamScannerControl Engineering 43° Transfer Function & Impulse Response Note : The equation [C(s) = Ris). T(s)| is applicable only in Laplace domain and cannot be used in time domain, The equation [e(l) = rfl)» t)] is not at all valid in time domain, Now consider that input be unit impulse ie. rt) = 6(t) = impulse input Is) L(Y) = 1 Substituting in above, Cis) = 1- Ts) = Th) Now e) = LMC) = LT) = TH Thus we can sa’ y that for impulse input, impulse response C(s) equals the transfer function T(s). So impulse response is ¢(t) = T(t) as C(s) = T(s) hence we can conclude that, Laplace transform of impulse response of a linear time invariant system is its + function with all the initial conditions asssumed to be zero. Ex. 4 The unit impulse response of a certain system is found to bee. Deterinine its transfer function. Sol. : Laplace of unit impulse response is the transfer function. L{e#} = Ts) 1 Ts) = © a Ex. 5 The Laplace inverse of the transfer function in time domain of a certain system ise while its input is 1(t) = 3. Determine its output c(t). Sol. : Let T(s) be the transfer function L1{T(s)] = TW) =e given r(t) = 2 But c(t) # r(t) x T(t), It is mentioned earlier that a = T(t) is not at all valid in time domain, so r c(t) % 20%, Hence the equation valid according to the definition of transfer function is, Cs) TS) = a 1 so T(s) = LT()} = L{e s "os Rw) = 2 Scanned with CamScannerCs) = C{s) Taking Laplace inverse of this equation, ct) = 04-040" This is the required output expression. 3.5 Some Important Terminologies Related to T.F. : As transfer function is a ratio of Laplace of output to input it can be expressed as. @ ratio of polynomials in ‘s’, us Pos) | QW)! This can be further expressed as, . TR = ags™ +ays™ 1 bays™-2 4. +a bos" + bys"=1 4 bash? tootby The numerator and denomina\ lor can be factorised to get the factorised form of the transfer finction as, TR, = KS 59)(5~ 5p) vel Sym) (51) (653)... (6 Sn) where K is called system gain factor. Now if in are substituted as s,, 5. infinity. a the transfer function, values of 's! 2/83 -~Sq in the denominator then value of TR. will become 3.5.1 Poles of a Transfer Function : Definition : The values of ’s’, which m, ake the TE. infinite af denominator of a T.R. are calle weak = er si itution in the cd Poles! of that" - er substitution in So values of $1 52,53 Sn are called poles of the TR. These poles are nothing but the Toots of the denominator of a T.F. to zero, equation obtained by equating For example, let the transfer function of a system be, Tis) = 20842) s(s44) The equation obtained by equating denominator to is s(s +4) = 0 2270 is, Scanned with CamScannerControl Engineeri ee Sinecring 45 Transfer Function & Impulse Response s 0 and sey I these valtes are used inv the des infinity. Hence poles of th Nominator, the value of transfer function becomes: Ms transfer function are s If the poles are tike s = they are called simple pote called repeated pote, conjugate poles. = Oand - 4 0, 4, #5, sm ie real and without repeated values, A pole having same value or more than that * pair of poles with complex conjugate values is called complex e.g. For Ts) = +2) The poles are the roots of the e uation (s+ 4)? (s? +254 2)(s+1) = 0. . Poles are & sj so T(s) has simple pole at s Repeated pole at S = ~4, (two poles) Complex conjugate poles at s = -1 + jl Poles are indicated by °X’ (cross) in s-plane. 3.5.2 Characteristic Equation of a Transfer Function : Definition : The equation obtained by equating denominator of a TF. to zero, whose roots are the poles of that T.F. is called characteristic equation of that system. F(s) = bos" + bys" + b98"-2 4. + by = 0 is called the characteristic equation. 3.5.3 Zeros of a Transfer Function : Similarly now if the values of ‘s’ are substituted as sy ,5 numerator of a T.F., its value becomes zero. ~ Definition : The values of ’s’ which make the TF. zero after substituting, in the numerator are called ‘zeros’ of that T.F. vet Sm in the Such zeros are the roots of the equation obtained by equating numerator of a TF. to zero. Such zeros are indicated by ‘0’ (zero) in s-plane. Poles and zeros may be real or complex-conjugates or combination of both types. Poles and zeros may be located at the origin in s-plane. Similar to the poles, the zeros also are called simple zeros, repeated zeros and complex conjugate zeros depending upon their nature. 2(s +1)? (s +2) (s? +25 s? (s+ 4) (s? + 654.25) g. Tis) This transfer function has zeros which are roots of the equation, As+1)? (5+ 2)(s? +28+2) = 0 Scanned with CamScannerControl Engineering 46 Transfer Function & Impulse Responey Simple zero at Repeated zero at Comple =1 (twice) X conjugate zeros ats = -1 + jl. The zeros are indicated by small circle ‘0’ in the s-plane 3.5.4 Pole-Zero Plot : eros of a TF. in splane Definition : Plot obtained by locating all poles and zeros of a TF. in soplane s called pole-zero plot of a system 3.5.5 Order of a Transfer Function : Definition : The highest power of ‘s’ present in the characteristic equation ie in er function of a system is call the denominator polynomial of a closed loop transfer function alle ‘Order’ of a tem. For example, consider Example 1 discussed earlier. The system TF, is —! 1+sKC Fe. 1 + SRC = 0 is its characteristic equation and system is first order system Then s =~ 1/RC is a pole of that sy and T.F. has no zeros. Imaginary tem The corresponding pole-zero plot can be shown as in Fig, 3.9. Similarly for Example 2, the TF. calculated is, 1 hE = S*LC+sRC+1 The characteristic equation is, Imaginary . TF. has no zeros, Pole2 Pole 1 Now if Selected are real, c alues of R, Land such that both poles unequal and negative SeFesponding pole-zero plot can *OWN as in Fig. 3. 10 Scanned with CamScannerControl Engineering 47 Transfer Function & Impulse Response For a system having TLE, as, Cis) (s+ 2) Re) os Parse 2][s? 97 The characteristic equation is, 2s+2)(s? 475412) = 0 ie. System is Sth order and there are 5 poles. Poles are 0, 144, -3, A while ere is located at ‘2’ The corresponding pole-zero plot can be drawn as shown in Fig. 3.11. | | Fig. 3.11 After getting familiar with introductory ‘remarks about control system, now it is necessary to see how overall systems are represented and the methods to represent the given system, based on the transfer function approach. Scanned with CamScannerEx.6 The transfer function of a system is given by, K(s+6) Ts) = ——_8* 9) $(5+2)(s+5)(s? +7412) Determine i) Poles ti) Zeros ii) Characteristic equation and de ”) Pole-zero plot Sol. : i) Poles are the roots of the equation obt aine roots of, by equating denominator to zero ie. Scanned with CamScannerControl Engineering _ 49 Transfer Function & Impulse Response s(s+2(S+5)(8? #75412) = 0 ie. s(s + 2) (s + 5) (s +3) (8 + 4) =0 So there are 5 poles located at s = 0, — 2 nd —4 ii) Zeros are the roots of the equation obtained by equating numerator to zero ie roots of K (s + 6) = 0 ° ies =-6 There is only one zero iii), Characteristic equation is one. whose roots are the poles of the transfer function. ie So + 14s? +7159 +1545? +1208=0 iv) Pole-zero plot This is shown in the Fig, 3.14 Fig. 3.14 "(1 -cos 2t) Ex.7 The unit impulse response of a system is given by T(t) = Determine its transfer function Sol. : Laplace transform of the impulse response is the transfer function. T(s) L (T(t) = Lle™' (cos 2t)) = L le} = L fe * cos 2} Se) (s+1) 4 (s+ 1)(s? +2545) Scanned with CamScannerEx. 10. The dynamic behaviour of the system is described by the equation AC 4 10C = 40 e where e is the input and C is the output. Determine the fer function of the system. I. : Take Laplace of the given differential equation and assume all initial Conditions zero, SC(s)+10C() = 40 E(s) (s +10) C(s) = 40 Ef) cs) _ _40 Es) ~ s#10 Ex. 11 Find the transfer function a of a system having differential equation 5 siven below, 2d cH), ac) “ae dt Sel. : Taking Laplace transform of the given equation and assuming all init conditions zero we get, 25°C(s) + 25C(s)+ClS) = Ris)+2e7* RG) Laplace transform of delayed function is, Lif(t-T)) = eS? Fey Lir(t-1} = eR) Combining terms of C(s) and Ris) we get, (2s? +2s+1)C(s) = Ris) (14207 5) Ss) 1+#2e7s RO” Peay + C(t) = r(t) + 2r (t - 1) (Refer Table 2.1) Scanned with CamScannerControl Engineerin contro! ‘9 57 Transfer Function & Impulse Response Ex..14° Determine the transfer Ey, re the transfer function if the dc, gain is equal to 10 for the system whose pole-zero plot is shown below Imaginary Fig. 3.27 Sol. : Fi ol en the transfer function has 3 poles at s = 0, s and j. And it has one zero at s = Ts) = ——KO+3) (s+1)(s+2+)j)(s+2-j) - K(s+3) (+ [is+2* ~( K(s+3) (+1)[s° +4s ] 0 which is given as 10. Now d.c. gain is value of, T(s) at dic. gain = at 1 = S83 1x5 16.667 16.667 (s+ 3) (5 +1) (5? #48 +5) Ts This is the required transfer function. Ex. 15 For a certain system c(t) is the output andr) is the input. It is repres | d? elt det) id rt differential equation #58 + MO 4 8 et = i ern) Determine its transfer function. inding Laplace of the given equation, neglecting the initial conditions, $?C(s) + S8C(s) + 8C(s) = 2sR(s) + Ris) Sol. Scanned with CamScannerFunction & Impulse Re. Control Engineering 58 Transfer S C(s) [s? + 5s + 8] = R(s) 25 + U Cs) _ 2841 R(s) 5? +58+8 This is the required transfer function. Ex. 16 If the system transfer function is Y(s) s+4 X(s)'57 +2545 Obtain the differential equation representing the system. 4 Sol. : MG). 2 _st4 X(s)_ 5? 42845 " (s? +2s+5) Y(s) <. s? Y(s) + 2s Y(s) +5Y(s) (s + 4) X(s) 5X(s) + 4X(s) Replacing variable s by <4 and Y(s) by y(t) and X(s) by x(t) we get, 2 Ss y(t)+ 2s y(t) +5y(t) = xy +4x(t) wy dy a5 = = — +4x dt? dt dt This is the required differential equation. a Scanned with CamScanner7.2 Definition and Classification of Time Response : Ti time. So it can be defined as below. ne response of a control system means, how output behaves with respect to Definition : Time Response : The response given by the system which is function of the time, to the applied excitation is called Time Response of a control system. In any practical system, output of the system takes some finite time to reach to its final value. This time varies from system to system and is dependent on different factors. Similarly final value achieved by the output also depends on the different factors like friction, mass or inertia of moving elements, some nonlinearities present ete. For example consider a simple ammeter as a system. It is connected in a system sO as to measure current of magnitude 5A. Ammeter Pointer hence must deflect to show us 5A reading on it. So 5A is its ideal value that it must show. Now pointer will eo. Scanned with CamScannercontrol Engineering __Time Response Analysis take some finite time to stabilise depends on Various factors like Securate SA oF not to indicate son reading, and aft friction, pointer in abilising, alse, it tia ete. whether it will show us Based on this example, we can eh first is the part of ontpat during the second is the final value att sify the total output time, it takes to reach to its final value. And ained by output which will be near te asires gystem is stable and accurate. P il ‘ar to its desired 1 d by consideri “ d by considering, another practical example, Suppose we want to travel from city A to city B. So our final desired position is city B. But it will take some finite time to reach to city B. Now this time depends on whether we travel by a bus or a train or a plane, Similarly whether we will reach to city Bor not depends on number of factors like vehicle condition, road condition, weather condition ete, So in short we can classify the output as, nse into two parts, hue if can be further explaine i) Where to reach ? ii) How to reach ? Successfulness and accuracy of system depends on the final value reached by the system output which should be very close to what is desired from that system. While reaching to its final value, in the mean time, output should increase smoothly. Thus final state achieved by the output is called steady state while output variations within the time it takes to acl response of the system. ieve the steady state is called transient Definition : Transient Response The output variation during the time, it takes to achieve its final value is called as Transient Response. The time required to achieve the final value is called transient period. : This can also be defined as that part of the time response which decays to zero after some time as system output reaches to its final value. ‘The transient response may be exponential or oscillatory in nature. Symbolically it is denoted as C, (1). To get the desired output, system must pass satisfactorily through transient period. Transient response must vanish after some time to get the final value closer to the desired value, Such systems in which trar are called Stable Systems. ent response dies out after some time Mathematically for stable operating systems, Lim Gy eo tow From transient response we can get following: information about the system, i) When the system has started showing its response to the applied excitation 2 Scanned with CamScannerControl Engineering 242 Time Re! ii) What is the rate of rise of output. From this, parameters of system can be designed which can withstand such rate of tis tem is fast behaving or not ponse Analysis . [also gives indication whether iii) Whether output is increasing exponentially or it is oscillating. iv) If output is oscillating, whether it is over shootting its final value, v) When it is settling down to its final value ? All this information matters much at the time of designing the systems. Definition : Steady State Response It is that part of the time response which remains after complete transient response vanishes from the system output This also can be defined as response of the system as time approaches infinity from the time at which transient response completely dies out. The steady state response is generally the final value achieved by the system output. Its significance that it tells us how far away the actual output is from its desired value, This response indicates the accuracy of the system. The symbol for steady state output is Cys (0 From steady state response we can get following information about the system : i) How much away the system output is from ils desired value which indicates error. ii) Whether this error is constant or vary’ about response. g with time. So the entire information stem performance can be obtained from transient and steady state Hence total time response C(t) we can write as, Ut) = Cys (H+ Cet) The difference between the desired output and the actual output of the system is called steady state error which is denoted ase... This error indicates the accuracy and plays an important role in designing the system. ‘The above definitions can be shown in the waveform as in Fig. 7-1 (a), (b) where input applied to the system is step type of input. ett) a(t) Steady "Pr? Transient state of Transient time system Fig. 7.1 (a) C; (t) is exponential Fig. 7.1 (b) G (t) Is 08 Scanned with CamScannerControl Engineering ——___243 7.3 Standard Test Inputs ; In practice, man: sed as referenc, als are ay ailable be inputs for the ramp, sawtooth type, square wave, highly impossible to consider Hence from the analysis point of view as reference inputs are defined as St dard Test mn be done on the basis of the responce pig inputs. Once system behaves satisl input is assumed to be upto the whi Uiangular ete, each one of it mark, These standard test signals are, i) Step Input (Position function) : Weis the sudden application of the input at a spec Mathematically it can be Ri) = A 0 If A = 1, then it is called unit step ion and denoted by u(t). described as, for t20 for t<0 fun Laplace transform of such input is“, $ ii) Ramp Input (Velocity function) : It is constant rate of change in input ic. gradual application of input as shown in the Fig. 7, Magnitude of Ramp input is nothing but its slope. Mathematically it is defined as, Rit) = At for t>0 = 0 fort <0 If A = 1, itis called Unit Ramp input. Its | iii) Parabolic Input (Acceleration Function) : This is the input which is one degree Various control systems. These sig, those signals which are __Time Response Analysis ich are the functions of time and can are step, analysing the systems it is, 's an input and study the response most commonly used aluation of the system But while Inputs. The ev: given by the system to the. standard test actorily to a te: st input, its time response to actual ified time as shown in the Fig, 7.2. i) Al Fig. 7.2 Fig. 7.3 A Laplace transform is s? +— Slope =At : vn in faster than a ramp type of input as shown the Fj 1 Fig. 7.4 Scanned with CamScanner__Time Response An Control Engineering Mathematically this function is described as, Ry = Av, for t 20 2 =0, for t<0 where A is called tude of the parabolic input. Function is expressed as 4 t so that in Laplace transforms of different standard inputs similarity will get maintained. If A = 1, ie. R(t) = z itis called unit parabolic A input. Its Laplace transform is ie) Impulse Input : It is the input applied instantaneously (for short duration of time) of very high amplitude as shown in Fig. 7.5 It is the pulse whose magnitude is infinite while its width tends to zero ie. t > 0, applied momentarily . Area of the impulse is nothing but i magnitude. If its area is unity it is called Unit Impulse Input. Mathematically it can be expressed as, Ri) = A, for t=0 =0 for t#0 Its Laplace transform is always 1 if A = 1. ie. for unit impulse response. (Refer Chapter-2). The unit impulse is denoted as (1). - RO _| Unit ramp Unit parabolic Unit impulse Table 7.1 Scanned with CamScannerControl Engineering 245 Time Response Anal 7.4 Steady State Analysis : As discussed earlier steady state is that part of the output which remains after transients compleiely vanish from the output. Mainly the steady state response has following two specifications, i) How much time system takes to reach its steady state which is called settling time which is discussed later in connection with transient response. It is related to transient response also because same time will be required by the transients to die out completely from the fem output ii) How far away actual output is reached from its desired value which is called steady state error (e,.). Out of the two specifications, the steady state error is the most important specification which is related only to the steady state. So let us see on which factor it depends, how to calculate it and how to reduce it. the difference between the actual output and the Definition : desired output Steady State Error : It Now reference input tells us the level of desired output and actual output is fed back through feedback element to compare it with the reference input. Hence to be precise it can be defined as the difference between reference input and the feedback signal (actual output). Mathematically it is defined in Laplace domain as, L [ e(t) E(s) = R(s)-C(s)-H(s), for non unity feedback systems and L { e(t) | = E(s) = R(s) ~ C(s), for unity feedback systems. 7.5 Derivation of Steady State Error : nple closed loop system using negative feedback as shown in Fig. 7.6 B(s) = Feedback signal Consider a F(s) = Error signal, and where Fig. 7.6 Now, E(s) = ~ Bis) But Bis) = C(s)- H(s) E(s) = R(s) - C(s) H(s) 4 Scanned with CamScannerTime Response Analysis. Control Engineering __246 and ls) = ES) Gos) Es) = R(s) = E@) Gls) Hts) E(s) + Els) G(s) H(s) = R(s) £(6) = —8___ for nonunity feedback 1+ G(s) HS) Ee) = SO for unity feedback 1+ G(s) This E(s) is the error in Laplace domain and is expression in *s'. We want to calculate the error value. In time domain corresponding error will be e(t). Now steady state of the system is that state which remains as t > 7 Lim Steady state error, es = 5 e(t) >n Now we can relate this in Laplace domain by using final value theorem which states that, Lim Fe Lim s vhere.F(s) = L { F oe a where F(s) (F(t) | Li oy = E™ 5%) where E(S) is L [ e(t) |. Therefore, = e(t) = too s>0 Substituting E(s) from the expression derived, we can write Lim sR(s) s>0 1+G() His) For negative feedback systems use positive sign in denominator while use negative sign in denominator if system uses positive feedback. ess From the above expression it can be concluded that steady state error depends on, i) R(s) ie. reference input, its type and magnitude. ii) G(s) H(s) ie. open loop transfer function, iii) Dominant nonlinearities present if any. Now we will study the effect of change in input and product G(s) H(s) on the value of steady state error. As transfer function approach is applicable to only linear systems the effect of nonlinearities is not discussed. 7.6 Effect of Input (Type and Magnitude) on Steady State Error : (Static Error Coefficient Method) : Consider a system having open loop T.F. G(s) H(s) and excited by, a) Reference input is step of magnitude A : Rw) = A s Scanned with CamScannerControl Engineering ar ____Time Response Analysis, Lim sR) so0 14 G(s) Hos) Lim sAus * s01G Lim A 530 14+G(s) HG) oo = A . Wo ae Vee yoo) H{(s) Fig. 7.7 Lim For a system selected," G(s) H(s) is constant and called Positional Error so 7 Coefficient of the system denoted as Ky, Lim Ky = 1) 69) HIG) = positional error coefficient and corresponding error is, A Css = 1+K, So whenever step input is selected as a reference input, positional error coefficient Kj, will control the error in the system alongwith the magnitude of the uit applied. b) Reference input is ramp of magnitude “A‘ : Ris) Lim __ sR(s) 5301+G(5) Hs) Css Lim 0 S[l+ GG) I Fig. 7.8 Lim s—0 s#sG(s) H(s) A Tim : 866) HE) Css so Scanned with CamScannerControl Engineering __Time Response Analysis, So whene coefficient K applied. So sta er parabolic input is selected as a reference input, acceleration error will control the error in the nitude of input error coefficients are given in Tabl Static Error Coetficionts ny 8) HIS) ae | | Ky of 8 G(s) H(s) Mee oo? Ges) He) Table 7.2 7.7 Effect of Change in G(s) H(s) on Steady State Error (TYPE of a System) : This can be studied by focusing on to the dominant elements of G(s) H(s) from error point of view. Such elements are constant of system °K’ and poles of G(s) H(s) at origin if G(s) H(s) is expressed in a particular form called time constant form. This is as shown below, _ _Kd+T d+ Gs) H(s) at E28he sl (14 Ty §)(14Tps).... where K = Resultant system gain and j = TYPE of the s\ “TYPE! of the system means number of poles at origin of open loop TF. G(s) HG) of the system So stem = 0, TYPE zero system J j = 1, TYPE one system j = 2, TYPE two system j =n, TYPE‘n’ system So ‘TYPE’ is the property of open loop TF, G(s)H(s) while ‘Order’ is the property ( of closed loop T.F. This is because, as defined earlier, order is the highest 5) 14 G(s) His) power of s present in the denominator polynomial of closed loop T.F. of the system. 7.8 Analysis of TYPE 0, 1 and 2 Systems : Note : A popular method to assess steady state performance of servomechanisms or unity feedback systems is to find their error co-efficients Kp, Ky and Ky. Scanned with CamScanner— - + A Kontrol Engineering as Time Response Analysis. where, K p = Position error constant, x i = Velocity error constant and © K, = Acceleration error constant. Obviously in order to find these error constants the system must be stable, t because for an unstable system there is no steady state and K,, Ky and K, are undefined Hence before we proceed to find Ky, Ky and Ky we must ensure (either by pole location or by Routh table of the closed loop system) that it is stable Thus the concept of Ky, Ky and K, is applicable only if, | ') System is unity feedback ie. only if His) = 1 ii) Only if the ystem is stable, Consider the input selected as step of magnitude “A | 1) Let us assume say system is of TYPE ‘0’, ie. Gis) Hig) = KOFTIA + Ths). | (1+T, s)(1+T, s) Li For step input Ky = my GES) His) = K ess ie. TYPE °0' systems follow the step type of input with finite error ; A which + can be reduced by change in A’ or °K’ or both as per requirement, Now °K’ can be increased by introducing a variable gain amplifier in the forward Path and error can be reduced. But there is limitation on the increase in value of °K’ form stability point of view that will be discussed afterwards. But increase in *K' is one way to reduce the error. Corresponding response will be as shown in Fig. 7.10 . (a) or (b). Fig. 7.10 (a) Fig. 7.10 (b) Scanned with CamScannerControl Engineerin: 251 e Response Analysis, sontrol Engineering | Response Analy: ii) If for the same input now TYP in Gis) His). increased to ‘one’ by adding, pole at origin TYPE 1: ie. Gis) Hig) = KOL+ Ts + Tas). As input is step iii) Similarly if now TY! TYPE is further increased to “two! ie. KG + Ts +T (1+ T,s)(1+ Tyo) Gs) Hos) = Lin Asinputsstep, Ky = EP Gay Hig)= In general for any TYPE of system more than zero, Ky, will be infinite (r) and error will be zero, Though mathematically answer for errar is zero, practically small error will be present but it will be ne ly small. Such type of responses may take ‘one of the forms shown in Fig, 7.11 (a) and (b). tt) ay A eno a Time a Tine | Fig. 7.11 (a) Fig. 7.11 (by Thus TYPE 1 and above systems follow a step type of refer nce input of any magnitude, successfully, with negligibly small error. Let us now char will control the error the selected input from step to ramp of magnitude °A’ so Ky Scanned with CamScannerThis is true for any system of TYPE more than one . Hence all systems of TYPE 2 and more than two follow ramp type of input with negligibl take the form as shown in Fig, 7.14 (a) and (b). \ small error and may CO} ow) Fig. 7.14 (a) Fig. 7.14 (b) Let us now change the selected input from ramp to parabolic input of magnitude A hence coefficient K, will control the error, vii) Consider TYPE 0 system : Gs) He) = Kot (+ Lim > K, = 50 s~ G(s) H(s) = 0 Ts) Ue s) (+ Ty 9) - viii) Consider TYPE 1 system : K(1+Ty s) (14 Ty 8) vee GS) HG) = GTT dey) Li ! kK, = goo 8 OU) His) = 0 : ara For both TYPE *0’ and “I’ systems, error will be very large and uncontrollable if parabolic input is used. Hence parabolic input should not be used as a reference to excite TYPE ‘0’ and TYPE *1’ systems. The output may take the form as shown in Fig. 7.15 (a) and (b) if excited by such input. Fig. 7.15 (a) Fig. 7.15 (b) Scanned with CamScannerAnalysis. Control Engineering 254 Time Response Analysis. ix) If TYPE 2 system is used K+Ts) (+7; Gis) His) = = s*(+T, s) G+T) 8) A im 5 - = A finite Then x, - | ") 82 Gls) His) = K % Amie error A/K which nce TYPE 2 syste: aw Parabolic input with finite Hence TYPE 2 systems will follow Parabolic inpu Cee shown in can be controlled by change in A or K or both and output may take Fig.7.16 (a) and (b). ) Fig. 7.16 (a) Fig. 7.16 (b) And for any system of TYPE ibly_ small. 3 or more if parabolic input is used error will be Type of Error Coefficients Errore, for System " Ky Ky K,_| Step input__| Ramp input | Parabolic input 0 kK 0 0 » a 1 » kK 0 ° A » « 2 & 2 kK 0 0 A K Table 7.3 7.9 Disadvantages of Static Error Coefficient Method : The disadvantages of Static Error Coefficient Method are 1) Method cannot give the error if inputs are other than the three standard test inputs. 2) Most of the times, method gives mathematical answer of the error as *0’ or ‘infinite’ and hence does not provide precise value of the error. 3) Method does not provide variation of error with respect to time, f ‘4 which will be otherwise very useful from design point of view. will be Scanned with CamScannerGontrol Engineering —255__________ Time Response Analysis. } 1) Brror constants ane defined for the loap transfer function Gls) Ht plicable to only the system contig Re) ®)“ 5] rT | Fig. 7.17 striely, hence the method is ap mu ration shown below 5) As final value theorem is u error so before applying His necessary to check if sE(s) has any poles on the jo axis or in the right half of the s-plane. This means before applying this method, the system’ must be checked for its stability. The method can not be applied to unstable systems. ed to calculate steady stat 6) When the system configutation is different than as shown above then it is Recessary to establish the expression for the error signal and apply the final ¢ theorem directly, without the use of error coefficients. The method plicable only to stable systems, Bee: muse of these disadvantages Dynamic error coeffici method) is dev ‘eloped which eliminates above said disadyant: nt method (Error series 7.10 Generalised Error Coefficient Method : (or Dynamic Error Coefficients) As explained earlier, E(s) = 1+ Gs) H{s) Let us assume that this is the product of two polynomials of © E(s) = Fy(s)- Fy(s) 1 Where F,(s) = TrGe He , Fy(s) = Ris) Now If Fis) = Fl): F s) then using convolution integral, ‘ LRG) = FQ) = [FD Fa(t-9) dr 0 ‘ t Similarly e() = fFG) Fy (2) de = JF) RQHa) ds 0 a R (t ~ 2) can be expanded by using Taylor series form as, Scanned with CamScannerime Response Analysis. Control Engineering 5G 2 De Rt = RO - RW +L RW a ROW be 3 . ‘ r 2 oe “(4 de Substituting et) = fF, (9) [ro-« ROH OF Rr ! x 2 0 1 ' = PROF) de =f RW) Fy) de + 3 Lim Now e tn 1 ! = 2 pw F()de =f RQ) FC) det pone Lo 0 = RW PFs) de- RW fey) de ROL SRO dea 0 a 0 Where Ky = fF (2) ds Ky = fr? FG) dr... 0 7 Substituting these values we have, ess = Ky ROK, RO+ Ko Re) + Where Ko , Ky, Ky « To calculate these coefficients use the following method; According to definition of Laplace Transform, are called dynamic error coefficients. F(s) = Tre ee dr 0 Now K ~ fre dr 0 Multiplying by e ~** to both sides, iplying by Kye *° = f() dre 0 Taking limit as s > 0 of both sides, Lim Ky e787 = Lim F,(s) 3-90 s-¥0 Ky = Lim Fis) Where F,(s) 1 Ste G(s) HG) Scanned with CamScannerControl Engineering 2s? Time Response Analy Taking derivative of Ky eo" wort =1Ky es = Gils) ds Substituting Ky = frye 0 -+ [Fi dre = aks) 4 ds ie Kye Fis) ds Taking limit a 50 of both sides, Lim 4 Fu) ~ sin ds ~ (s) , dm Lin — soo ds In general, XK, error coefficient method. This method eliminates the disadvantages of stat i Transient Response Analy: Transient response is the part of the total output which can not be separated out and hence to analyse it, it is necessary to calculate total output C(t) for the applied input. This contains steady state value alongwith transient response. From the expression of C(t) only transient response C,(t) can be analysed. 7.11.1 Method to determine total output C(t) : i) Determine the closed loap transfer function of the system a s Co) __ GIs) R(s) 1+ G(s) H(s Find the expression for K(s) from the information of reference input R(@®) to be ii) applied to the system. ili) Substitute R(s) in the closed loop T.F. to obtain expre: Take Laplace inverse of C(s) by using partial fraction method to obtain total C(t) of the system to the applied input R(\). As CH = CoD + CO The transient output can be studied from the above expression. ‘Transient response may be exponential or oscillatory in nature. ion for C(s) iv) Scanned with CamScannerResponse Analysis. Control Engineering nt response depen In general traw: it is independent of the i) Poles of the closed loop TF. . 4 re the magnitude of the selected i) Location of the poles of the closed loop TE 1 magnitude of the input applied. Any change i INE Meet tom input will not have any effect on the transient respons : 7.12.1 Closed loop poles of first order system = n by, The closed loop transfer function of a system is give! Ce) _ RS : The equation which gives poles of system is called characteristic equation which is, 1 + G(s) His) = 0 For first order system this equation is also first order in general of the form. 1+Ts = 0 - + Closed loop poles are the roots of characteristic equation. So for first order system there is only one closed loop pole ie. =—2 7 The pole-zero plot is as shown in the The value of closed loop pole appears output. Imaginary eg. in the system considered above as_an exponential index in the transient the closed loop pole is s = z and in output Ci) = 1 - e RC that is the . 1 - . exponential index is ~ =F which is nothing Ri but the closed loop pole. Fig. 7.20 Consider two systems, one with closed loop pole at 5) = -2 while other having ~ 4. In the output of two systems we will get the terms e* closed loop pole at in first system and e-* in second system. Now as t > x, e* will vanish earlier than * Hence second system transients will vanish quickly and Will stabilise to its final 7 rst system, value earlier than the From this we can conclude that locations of closed loop poles affects the transient behaviour. As closed loop pole moves in the left half away fon the imaginary axis in the s-plane, ‘ransients die out more quickly making system more stable. Scanned with CamScannerControl Engineoring _261 Time Response Analysis 7.13 Analysis of Second Order System : Every practical system takes finite time to reach to ils steady state and during this period, it oscillat decided by type of clo or increases exponentially. The behaviour of system gets sed loop poles and locations of closed loop poles in s-plane. T closed loop poles are dependent on selection of the parameters of the system. Every system has tendency to oppose the oscillatory behaviour of the system which is called damping.) Now this tendency controls the type of closed loop poles and hence the nature of the response. This damping is measmed by a factor or a ratio called damping ratio of the system. This factor explains us, how much dominant the apposition is to the oscillations in the output, In oscillate but slowly ome systems it will be low in which case system will : h damped frequency, In some systems it may be so high that system output will not oscillate at all and not only that it will be exponential, $0 slow that it will take very long time to reach the steady state. This damping tatio explaining such behaviour is denoted by a greek symbol (Zeta) & Now as this measures the opposition by the system to the oscillatory behaviour, if it is made zero tem will oscillate with maximum frequency. As there is no opposition from em, system naturally and freely oscillates under such condition. Hence. this frequency of oscillations under & = 0 condition is called Natural frequency of oscillations of the system and denoted by the symbol o, rad / see For a second order system the denominator of closed loopT.F. s quadratic and the coefficients of this equation are directly related to & and wp as shown The C.L-TF. (Closed loop transfer function) for a standard second order system takes the form as, cls) _ on S+Oq Where characteristic equation is, s*+2&0, s+o2 = 0 The standard second order system is that where in C.L.T.F. numerator is or. In practice it is not necessary thal numerator_must be always w2. It may be other constant or polynomial of ‘s' but denominator middle term coefficient and last term coefficient always reflect ‘26 wq! and “w?' of the system. Hence always denominator of TE, must be compared with the standard form 7 42a, S+OR = 0 to decide the values of € and op of the system. The numerator should not be used to compare to obtain the values of £ and op ~ Scanned with CamScannerResponse Analysis. Control Engineering eg. RG) 1 G@) HE) 10 Mord rm is not 2 but denominator This C.L.T.F. is not standard as numerator te i gz the denominator always reflects € and a). The values can_be decided by comparin} y xy ce with the standard characteristic equation 5? +25, $+ On on = 12 ie. @, = VI2 rad / sec While Bio, = 3 Lb 7.14 Effect of £ on Second Order System Performance : Consider input applied to the standard second order system is unit step. Ris) = 1/s ‘ 2 While C&L = RG) 52420, stod 2 Cs) = On s(s7 +2%o, s+) Finding the roots of the equation s? +2E wy s+02 = 2 wy £46? w2 — don3 je, REVS Bn Tn 2 ie. 51,2 = ~b0,£ oy Ye? -1 We can write, Cls) = SM, + 1 FF] Now nature of these rools is dependent on damping ratio &, Consider following i 2 ler followin; cases, Scanned with CamScannerControl Engineering 263 Time Response Analysis. i) 1<8 <«, the roots are, S12 = - 80, to, ie. real, unequal and negative, say ~K, and-K> Cs) = — 8(s+Kj) (6+ K3) Taking Laplace inverse, C(t) will take the following form, Cit) = Cy + Be Kt Cen KR, Where Cy = Steady The output is purely exponential, This means damping is so high that there are no oscillations in the output and is purely exponential. Hence such systems are called “Overdamped. te output = A Hence nature of response will be as shown in Fig. 7.22. As & increases, output will take more time to reach its stead state and hence become sluggish and slow. ii) When & = 1, the roots are, 83,2 = ~My Op ie. real, equal and negative. oo c Cs) = ——22_ —_ + (9) = eTaa)Gtoq) e+e) (sto) Tal ng Laplace Inverse, C(t) will take the following form ...(Refer Chapter-2 for Laplace Inverse) Whe! Co = This is purely exponential. But in comparison with overdamped case, settling time required for this case 1s less and because of repetitive occurrence of roots the system is called“ Hy Damped’. This is critical value of damping ratio because if it is decreased further roots will become complex conjugates and this is the least value of damping ratio for which roots are real and negative. Cit) = Cyt Bre et+Ce ont st eady state output value = A Scanned with CamScannerTime Response Analysis Control Engineering 264 e corresponding response jg amped and th P ly di So exponential and as show 1, systems are critica in the Fig. 7.22. ols are, Hi) When range of & is 0.< & < 1, the i e that real part will be ic. complex conjugates with negative real part. Note always negative, he yl S+OR Taking Laplace Inverse, C(t) will take the following form, t+ °) CW) = C+ Ke “bent sin(og 41 (Detailed proof discussed in next section). Where = Steady state output value = A This response is but decreasing associated with exponential term with negative index e 75". Such quency of such oscillations is called oscillatory, with oscillating frequency wm amplitude as it oscillations are called Damped oscillations and fre Damped frequency of oscillations «3 which is nothing but ay @q = Y1-E? rad/sec As damping is reduced, it is not sufficient to damp the oscillations completely hence oscillations are of damped type. As damping is not sufficient, systems are called Underdamped Systems. - In such response, real part of complex roots controls the amplitude while imaginary part controls the frequency of damped oscillations, The nature of the output response for such systems when excited by unit step input will be as shown in Fig. 7.23. Fig. 7.23 Scanned with CamScannerControl Engineering 265 Time Response Analysis. iv) When & = 0, then roots are, 54, = 4) ie. complex conjugates with zero real part, ic. purely imaginary. A, Bic Peon) 8 * Cis) = S(S+]O_) Sj) sl But instead of finding out partial fractions, the corresponding C(t) can be obtained by substituting & = 0 in the expression for C(t) for underdamped condition which is, C(t) = Cy +K" sin(oy t+0) K" = constant Cyo = Steady state output value = A The response is purely oscillatory, oscillating with constant frequency and amplitude. The frequency of such oscillations is the maximum frequency with which output can oscillate. As this frequency is under the condition § = 0 Le. no oppe condition, system oscillates freely and naturally. Hence this frequency is called natural frequency of oscillations denoted by (oq tad/sec. The systems are classified as Undamped Systems. The nature of the response will be as) shown in Fig.7.24. Fig. 7.24 Summarising all cases as in Table 7.4. Sr. Range of § Type of elosed loop | Nature of response | System Classification No. poles 1 5 Purely imaginary Oscillations with Undamped constant frequency and amplitude 2 O) “Sh (148T)) (14 STp) one as (1 + sT, ) (1+sT,) hence value of */ for given system is zero. Hence it is Type t is clear that quadratic denominator can be expressed Zero system. Error coefficients : Lim Lim 250 Ky = 2 Gis) His) = a P= 549 SOMO = Sys +2148 440 Lim Li 5 } K, = sGis) Hg) = os =0 sO 330 + 214s +40 Lim > K, = 2 ct Hie) = Lim 250 =0 soo soo 214s +40 Now for first input, 10 ) rts 10 ~ RG)=—, so A=10 Scanned with CamScannerTime Response Analy: Sentrol Engineering 282 eu = ii) (ty = 5t, ii) ra) = 14st Se? 2 ie A, = step, Az A 4 = 6 parabolic 5, 6_) ey = + 5 s 00 te 4 Thus as system is type 0 system, error is finite only for step input. K s? (s+ 2) (843) Ex. 8 For a system Gis) His) = Find the value of K to limit steady state error to 10 iwhen input to system is 1 + 101 + Sol. : The input is combination of three inputs. So let us calculate their magnitudes. rit) = Ay Az= tmp = 10, 45 = parabolic = 40 Now K, Gis) Hy = HK Po sso 50 st 4 2543) Lim Lim Ky = SGis) Hs) = - 599 GE) His) sO, “2 Lim © Sais) His) = bm i >0 so0g G Pea + €. Ay ot O52 +53 = Ly Oss 2 KK, Now ey, is to be limited to 10, 1 = 10 = K= 4 Value of K'is 24 to limit the steady state error to 10 Scanned with CamScannerControl Engineering 283 Time Response Analysis D1 t and it is desired that es, < 0.005, find the range of values of K vithin specified limit for Ex. 6 Assuming r(t for error to be a m1 system, K (s) sisi) Sol. : From the system shown we can write, Gis) = j H(s) = 1 The input is r(t) = 0.1 the error, i.e. ramp of magnitude 0.1. For ramp input Ky controls Lim sk sO s(s+1) Lim sG(s)H(s) = s>0 0.005 = For any value of K greater than 20, eg, will be less than 0.003. Hence the range of value of K for e.. < 0.005 is, : 205 K This error is to be reduced by 5% of existing value, with new gain of G(s) as Kz ‘ad of 200, &sst = — “Scanned with CamScannerTime Response Analy: 284 Control Engineering" ons 100 0.076 New error is 0.076. same input and H(s) = 1 with New 7 Ky = east = 0.076 = So new gain is 210.52 Ex. 8 Inv tie system shown below the error allowed after achieving steady state is 0.01 for unit step input applied to it, Find the value of K. Roe) + cs) TOs +1 Sol.: For step input, K, controls the error, Lim 5 GO) HS) lim) Kk 0 (0s+1) And . ee where A = magnitude of step = 1 0.01 = | 1+K 1+K 100 K = 99, Scanned with CamScannerControl Engineerin: 5 Kontrol engineering 285, ___Time Response Analysis Ex. 9 Find the steady state error E, if Tis unit slep input and R= 0. eo; : — 0.18 Ris) E(s) ~ — 10 }—(X)-] 10 Trot 8) Sol. : When system error coefficients, ‘Snot in the simple closed loop form then we can not apply the In such ease, we have to use final value theorem Lim Lim Le. Cs, = e(t) = sE(s) tow ) s>0 {s) In the system given F(s) = =.C(s) when Ris) = 0 “(s) Now let us find out so that for unit step disturbance we can calculate Cs) and hence E(s). When R = 0, summing point at R(s) can be removed and block of ~ is to be added to consider sign of the signal at that summing, point. cs) Scanned with CamScanner ~Time Response Analysis i 206 Control Engineering - + diagram. drawing, the ding Combining the two summing points and redrawing cls) Negative sign ot { =100 ) can be taken out to change sign of the signal at the 1401s) summing point from positive to negative. Ts) 700 CIs) [co] Larorsi| Now the two blocks are in parallel, in the feedback path, Cis) ; A (1s? +s?+ 10054 10000) 1 For T(s) x w+ 01s) (1s? +s? Gone +374 100s5 + 10000) Scanned with CamScannerControl Engineerin: _ Now 207 F() = bs Steady state error = Cs Lim s Es) 0 s0 = 0.01 Lim =100(1-4 0.15) _ 100 s(0.159 +8? + 100s-+ 10000) \ Time Response Analysis 100(1 + 0.8) (0.18 4 52 + 100s + 10000) (0000 Ex. 10 The block diagram shown in figure represents a heat treating oven. The set point (desired temperature) is 1000°C. What is steady state temperature? R= 1000 Sol. : For the system, Gis) = H(s) = R(s) = 20000 (1+) (1+ 0.18) (1+ 0.005s) ‘20000 (1) 440.15) (440.005 cls) 1 and input is step of 1000 1000 System is Type 0 system. For step input, " Steady state error, Css = Cos, 20000 ™ Gis) H oe <9 9 0) HES) = rer) (I+ O.1s) (1+ 0.0058) 20000 A . Where A = magnitude of step input eK — 1000 _ o,04999 1+ 20000 Steady state temperature, Lim tox city W Css desired ~ es. 1000 - 0.04999 999.95° C. Scanned with CamScannerTime Response Analysis 288 ‘el sy state error with RD = OTE steady state ¢ Control Engineering 1 below, calculate the Ex. 11 In the system sho Go) His) = — For ramp input, Lim Lim 30 5S8) HG) = og where Solved Problems On Transient Response Ex. 12 A unit step input is applied to the unity feedback system for which open loop TF. Gls) = 8 ring s(s+8) i) Its closed loop T.F. ii) Natural frequency of oscillations “wy, * ui) Damping ratio &' iv) Damped frequency of oscillat * ; Sol. : Standard form of closed loop TF. MONS “og Scanned with CamScannerControl Engineering 209 Time Response Analysis. Cs) G(s) Rs) > ae Ha H(s) = 1 unity feedback 16 = _S(s+8)_ 16 14 + 85416 s(s+8) Order of denominator is two hence system is second order. Standard representation of a T.F, of a second order system RS) s? 4 2G0, stan Comparing the two expressions. @% =16 hence oy = 4 rad/sec i.e, natural frequency of oscillations then 2. Eo, a4 2 &= 1 ie. damping ratio. Hon =8 ie. system is critically damped. Now damped frequency of oscillations is given by, @g = @, y1-&? = 0 rad/sec As system is critically damped there are no oscillations and hence no damped frequency of oscillations. Wp = 4rad/sec, §=1, og =0 rad/sec 64 Ex. 13 For a system having T.F. = ~=———— * * 3? 455 +64 , for unity step input determine )@, fi) & ii) @g , iv) Time for peak overshoot Tp. Sol. : From, T.F. it is clear that system is second order hence comparing given, Tf. with standard T.F. of second order system ———°n s+ B oq stor or = 64 4. @, = 8 rad/sec Qo, = 5 BO, = 25 v & = 03125 og = Oy yl - &? = 8Y1-(0.3125)? ag = 7.599 rad/sec Time for peak over shoot, Tp = = sec.= —*_ = 0.4134 sec oa 7.599 fy = Brad/sec, &=0.3125, wg = 7.599 rad/sec, Tp = 0.134 sec Scanned with CamScannerTime Response Analysig_ 200 rind its rise time, peak time, Ex. 14 A second order system is given by 2468+ input. Also calculate it step input. Als pak overshoot and settling time if subjected (0 an sep expression for its output response. w2 Sol. : Comparing the T.F. with the standard form 5242 ston wy = 6 @, = 25 and £ = 06 wo, = 5 0 = tan = 0.9272 radians og = Oy y1-8? = 5\1-(0.6)? = 4 rad/sec 1 = 28 og x- 0.9272 0.5535 sec nok = 2.20785 ; 5 sec oa -n§ /\i-2 % My = e ME 100 = 948 % T= 1.33 sec font and cy = 1-£ sin (og t+0) . vine? -3t = e =1- lar sin(4t + 0.9272) C(t) = 1 ~ 1.5625 e7* sin (at y 0.9272) Ex. 15. For the system shown in system shown in the figure obtain the a te : " is closed i ic frequency and expression for the output response if aie win i fed to unit step input. cs) KD XY) mim Rls) Scanned with CamScanner_Time Response Analys's Control Engineering _291___ 20 1 Cs) DEH) 20 a Re), 2 3 (rer Now though TLE. is not in standard form, denominator always reflect 2 2% and tof from middle term and the last term respectively comparing s? 456424 with s? 4250 +0, o, = 24 Oq = 4.8989 rad 2b, g = 0.51031 oy = @q Yi-82 = 42129 rad/sec G 7 ) in standard form. So writing Now for C(t) we can use standard expression for s Cs) 20. 24 RS” tal 24 [st 455424 For the bracket term use standard expression, and then C(Q) can be obtained by 0 multiplying this expression by constant 5". 20 c(t) = ai sin (wg t+ o| yi-8? 1-5 foe? i radians = 1.03 radians Cit) = 3 [1-1.1028e 775! sin (4.2129 t+ 1.03)] @ = tan” 2 i Ex. 16 A system is given by differential equation, fhe + 8y = 8x, where y = output te and_x = input, Determine al time domain specifications for unit step input Sol: System siren equation is, fy, ae To find TE. i take laplace transform from above equation and neglect initial x65) F say = 8x conditions. s? y(s) + 4s y(s) + 8 y(s) = 8 x(s) Scanned with CamScannery(s) Ie a4s+8] = BNO) . ye) 8 TE) sheds 8 > er system ——— > g this with standard T.F. of second order s¥ 57 + 2 Oy SHO, 83 rad /sec 2.002 rad/sec [TE nt og = op yi-G2 = 2.83 yi - (0.7067) Tp = lime for peak overshoot z %_ _1.57sec 2 ——— = 0706/1 ~(0.706)? 100 e100 = @ 78 %0:706/ Y1=(0.706) = 4.33% T, = settling time =—4- = —__4 =a. sec Eo, 0.7067 2.83 ~bant ct) = 1-2 sin (vg t +0) ar Where 0 = tan? [ =). 45° = Gand = 07067 2.87 ¢ cw) = 1 per n(23) 4 Cit) = 1- 141 72 sin(2442) 4 ©, = 2.83 rad/sec T, = 5 | dg = 42.002 rad/sec % we 1.57 sec | T, = 2 sec p = 4.33% 5 = 0.7067 EX 17 Ope L& Open loop T.P. of tinity feedback systems jg Ge) sk ~ (Is Tse where K and T | “are constants. Determy | . ine factor by woh overshoot of unit Y ehich gain “Ke of unit step response be reduced from 25% 4 pa be multiplied so that 10 25%, . (Bangalore Univ. April 98 ~ Scanned with CamScannerSig ‘eis Control Engineering 293 Time Response Analysis Sol. : Closed loop T.F. = = 8047s) 14K ts? + s(1+Ts) Dividing by “T’ to make coefficient of s? in denominator as “1’. Till coefficient of s? is 1 in the denominator, it can not be compared with the standard form of the equation. 2 Comparing with standard form Sn 8? 42a, stor 2 Oy = Bon B= Now for 75% Mp, 5 0.75 In(0.75) = -xe/ fie? | | 0.2876 = xe/\i-2 | 0.08271 (1 - 2) = n?&? 8.38 x10 (1-&?) = &? f | a £? = 831035 «1079 & = 0.0901 | and for Mp = 25% Let & = &} | cans/ | 25 = 100¢-*8Y | i 1.9218 (1-§3)= 76? i E? = 0.1629 i E, = 0.4037 } ie. to change % M, from 75% to 25%, & must be changed from 0.0911 to | &1 = 0.4037 1) Now —, i 2VKT r¢ i | é Scanned with CamScannerResponse Analysin, Control Engineering Now “T is constant. G 4) For 5 and let K=K fate -e) 04037 = AGF Vv ) atio of equations (1) and 2 0.2256554 = 0.05092 = alZ K, = 0.05092 K ie. gain °K’ must be multiplied by 0.05092 25%. Ex. 18 A certain control system is described by ayo, 7a. 12 Y(i) = 12 X(b. Find its output response for unit step input, at Y(t) = output, X(t) = input Sol. : Taking Laplace of the equation neglecting initial conditions. s? ¥(s) +75 Y(s) +12 Y(s) = 12 X(s) Yo) X(8)_ 5? +75+12 0 change % overshoot from 75% to the differential equation, TF. is Comparing denominator with standard form on = Vi2, 2bo, =7 «6 = 1010363 As & > 1, system is overdamy As ped, hene il i oscillations. Hence standard expression for C(t) cannot be weet ‘ei not contain Now input is unit step, so X(s) = 1 , 7 = 1/s. Substituting in T. Yo) « 2 ig -F. we get, 562475412) Use partial fraction method = 12 =. Sis+ 3) (s+ 4) where, A=1, B=-4, Cx3 Scanned with CamScannerSis Control Engineorin, ee eens __Time Response Analysis. 4 Ys) = LL ay : "430 7M 2) ‘Or a cor . _ Ex.-19 a mitrol system Sown im figure, find the values of K and K, sa that the MUNPINS ratio Of system is 0.6 and s Hing time is 0.1 see, Use T, = 22. Assume unit step input, Eo, e cls) Sol.: Using Block diagram reduction rule, reduction of inner loop is, ; 7 100 | 1+0.25+ 100K; overall G(s) = K-10 LK 140.2s+100Ky 20s ~ s[1+100K, +025] H() = 1 j — 5K Go) 5(1+100K, +025) Tr = —S6)__ j 1+G@H@) 7, SK | s(1¥100 Ky + 0.25) _ _ 5K 0.2s? + s(1 +100 Ky) +5K But coefficient of s? in the denominator must be ‘unity’ to compare it with the standard form. So dividing it by 0.2. Cs) | 25K R(s)_ 5 45s(1+100K,) + 25K o2 = 25K, a, =5VK 2b, = 5(1+100K,) _ 5(1+100K,) Be Ov Scanned with CamScanner296 Time Response Analysis Control Engineering ce g 6 = 0.1 sec Now g= oe ant . Using T= fon 42 ie. ot = py = 53.33 rad/sec oy, = 5VK K = 113.78 and g = SUK) _ 96 . 10VK Ky = 0.118 5 eter e EX. 20 For a system having G(s) = ——! H(s) = 1. Determine (s+1)(s+3) ‘ i) Characteristic equation ii) @, and & iii) Time at which Ist undershoot will occur. iv) Time period of oscillations. 2) No. of cycles output will perform before settling down. 15 Ce) _6+DG+3) as R@) yy © st 45418 (s+1)(s+3) Hence characteristic equation is, s? 445418 = 0 ny = VIB = 4.2426 rad/sec 2Eo, = 4 — = 04714 Now time at which C(t) will experience maxima and minima is given by, 28 where n= n= Scanned with CamScannerTime Response Analy 297 _ This can be shown as follows. 2 ety t for Ist undershoot = 2% “ 3.74166 rad/sec t for Ist undershoot = — 4166 = 1.6792 time ; , . 2x Now time period of oscillations is related to ag by relation, mg = > where T = Time period 2 T= 22,28 oq 3.74166 = 1.6792 see/cycle For 1 cycle, output will take 1,6792 see, Now the time for 1 cycle is known and if it is known to us that what is the time required by the system to achieve steady state, we can find how many cycles output will perform before reaching steady state. T, = A= 2s0c 8 Fon So 1.6792 sec for one cycle, how many cycles output will perform in 2 sec, 2 Total no, of cycles) = —“— ° aa 1.6792 = 1.191 Output will perform 1.191 cycles before reaching the steady state. This can be shown as in figure. The frequency in Hz if required can be obtained as, = 0.5955 Hz Scanned with CamScannerTime Response An, 291 Control Engineering 234 wries RL-C cireuit, with R= 12, L rc ‘Assume input as a step of 10V. Ex. 21. Obtain the step response and output is taken across “R The circuit is as shown in following figure. Sol. : Loe a (MSC Lo vi) R Volt) To find ee ; , We can write the equations as, Vis) = sLi(s) + L1G) + RIG) sc while V,(s) = Ks) R Vv we ) in first equation vie) = Se [urd =R Substituting I(s) = Vols) p.2 Vis) = Cc its) = R [s Li +1+sRC] Vols) _ sCR Ss: we Vis) s? LC-+sRC+#1 esky lh L Lc Substituting R=L=C=1 Vols) _ Vi(s) Ss? +541 Comparing the denominator 2 o, = 1 2 @, = 1 2Eo, = 1 : “$= 05 Note : i : . The standard expression for C(t) is derived for TE. the f in ie form On and not for s? + Bon Stor he T.F. having ‘s' terms in its numerator. Hence standard expression for C : ) paral actor ne (t) cannot be used and Output must be calculated by direct Scanned with CamScannerNow s°+s41 Completing square in denominator Vos) = —— 10 stesety3 ii (a To adjust this in the form 7 (sa)? +(@)? Vols) = Taking Laplace Inverse, 1 20 -5t V3 V,(t) = Se 2 sint we TB a} Volt) = 11.54 e795" sin 0.866 t Ex. 22 For an unity feedback system, output response is observed as, Cit) = 1 + 0.504 e~ 3" —1,504e7 778", Determine its damping ratio and natural frequency of oscillations of the system. Assume unit step input. Sol. : To determine the & and @, we must have C.L.T.F. =a which can be obtained 5 by taking Laplace of C(t) and R(t) respectively. Cia) = 2+ 0504 __1504 erg) (s+3.07) (s+ 2.18) and Rt) = 13 Ris) = V/s 1, 0504 __1.504 Cs) _ s 5+307 84218 RE) 1 s 0504s 1.5045 pee s+3.07 s+2.18 Scanned with CamScannerTime Response Analysig 300" Control Engineering 0.504 s(s+ 2.18) 504 s (s+ 3.07) + 307)(5+ 218) 4.6172 6.6926 + Cs) _ i RG) $2 45.258 + 6.6926 ng denominator with standard form 2 = 6.6926 On = 2.587 rad /sec 2B on, = 5.25 & = 1.0146 As & > 1, system is overdamped. Ex. 23 A position control systent drives a Toad through a 50 : 1 gear ratio. The inertia of aft is 20 x 10-5 kg-m? and friction is 60x 10° N-m sec. The torque ved is 20 rpm. Calculate motor § constant is 0.04 N-m per degree of error, if output spee DE fe, Let T = Torque produced proportional to ‘0; ~ 09° T = K[;-9] 0, = Reference input position 6, = Output position Torque is used to drive load of M.I. ‘J’ and friction “B’ where, Taking Laplace we have, KOi(s) = KO,(s)+] 5? 0,(5) + Bo (s) ol ols) _ K * ee ae B 8i(s) a2 = () Js?+Bs+K 48,,K Jy Scanned with CamScannerControl Engineering 301 Time Response Analysis BL avyK 0.04 N-m/deg, = 0.04 57.3 N-m/radian 1/50 K at motor shaft = 0.04 x 57.3/50 Oy = f = | BOESS23 15.13 rad /see J > 50x 20% 10° K Gear ratio 2 TR x Ex. 24° For system shown in, figure, find the error using dynamic error coefficient method for input of 6 + 5t+ S tier Piont Load Re) 70 (s als) ‘5 BAGH) 15x 200(s+ 1) x 0.02 (s+ 1)(s+2)(s+5) G(s) = = — 2 _ (+2) (0+8) Now for dynamic coefficient method, exs(t) = Ko R(t)+ Ky R(Q) + Kg R'(t). where Ko = Fs) Lim dF,(s) Kr= 50 ds Lim d? F,(s) Ko 540 as? 1 where A) = apne Scanned with CamScannerTime Response Analysis. Control Engineering 1 = F,(s)=— = 0.1428 Fe =5 _ _ Lim d Ke so Was kK - lim (s? +7s +70) (28 +7)-(s? +75 +10) (25 +7) 1 so (s? +75 +70)? _ Lim 253 +215? + 189+ 490-259 — 215? - 695-70 soo (s? + 75-470)? _ Lim 120s+420 80 (5? 475470)? 2 = 20 ~ 00857 (70) ii 2 Ky = Lim d F,(s) 530 Gs? _ Lim a[22]- 4 120s+ 420 ds | (s? +7547)? 830 ds} ds . im [seems 70)? -(120)~ (1208+.430)-2 (<2 + 78+ 70)(2s+7) 70 (s? +754 705 \ . (70)? 120-4202 70%7 (70) ess(t) = KoR() + Ky RW+K, RY) Rit) = 6 45¢ 7 = 7.346 x 19°3 RW) = 5 46 Ri = 6 Scanned with CamScannerControl Engineering 303 Time Response Analysis. Css (8) = 0.142816 + 5t + 3¢?] + 0.0857 [5 + 6t]+ 7.346% 10 x6 0.8568 + 0.714t + 0.4284 t? + 0.4285 + 0.5142 t + 0.04407 Css(t) = 0.4284 t? + 1.2982 ¢4 1.3204 Advantage of this method is, i) It gives variation of error as a function of time and ii) For any input, other than standard input error can be determined. Ex. 25 A closed loop system has two complex conjugate poles at $1, sz zj. Determine the form of transfer function and values of ey , Ty, T,, 7; and My assuming standard second order system. Sol. : 5 8? + Bags +o2 while $1, 82 = -24j1 Denominator of T.F = (5 +2 + jl) (s + 2-jl) = (s+2)? -(jl = st44545 ow os Rs) s? 44505 Comparing with standard form . os oR = 5, and Ew, = og = = 0,999 rad/sec = 1 rad/sec 1 yl 2 0 = tan” rad § = 0.4636 radians Tp = — = 3.1415 sec T; = = 2sec = 0.1867% Scanned with CamScannerControl Engineering _ Ex. 26 A systent has unit ste and ramp response, Sol. : For unit step input Ris) taking laplace of C(t), Cts) rest R(s) For impulse response, 304 Time Response Analysis, O18, Determine its impulse response » respouse of C(I) = 1 — zero initial conditions. Assume § ed) Ce) = 97 Cit) = LE fC} = 0.10%" 5 For aap: tel = 1 Impulse response s Cs) = or A,B,C (s+0.1) s* 8 s+0.1 A(s + 0.1) + Bs(s + 0.1) + Cs? = B+C = 0, A+01B=0, O.1A = 01 A=1 B c= 10 Ci) = 0 ,_10 sts 8401 Ramp response cit) = LT IC@)) = t-10+ 10¢0" Scanned with CamScannercontol Engineering 925 Time Response Analysis px. 32 The open Toop transfer fiction of unity foadback System is given by G(s) 40 ‘ Determine steady state error using ri) = (34409 low error series for input given bel sol. : GG) = msg For dynamic coefficient, error series ess (= Ky R()+ K) (t+ Ky R(t) to RW = B+40t=steae R(t) = 34st Ri) = 8 Lim - Ky = F, (s oF 59 AO) x, = Him ahi s70 ds — Lim d? F, (3) 2 530 dst 1 . 1 (0.2s* +s) where AO = eqn a eT) s(1+0.2s) +5 +40) (0.45 + 1)- (0.2s* +5) (O4s +1) (0.28? + s+ 40) dF, + 04s? + 165+ 0.2s7 + 5+ 40-0.0: (0.28? +s (40 + 16s © (02s? + 5440) ds’ 2524 5440)? Uo) Scanned with CamScannerTime Response Analyais Control Engineering —— 828 - Lim d FG) s— 40+ 0) . Ki = .00 ds (02% 0404 40° = 0,025 Lim d? F, (0) 2 16- (49) 2) GOD Kr = 50) -— cao)" = 0.00875 eg (= 0 Rit)+ 0.025 (3 + Bit 0.00875 « 8 = 0.2t + 0.145 lems From Bangalore University Papers J K, and Kp 50 that non in figure, find the values of assume ranit step input (Bangalore Univ. June - 93) Sokved Pro Ex. 33 For a control sysiemt My = 25% and Tp Ris) C{s) Sol. : G(s) oKi | His)=1+Ko 8 s Ki Ki Ge) st 1+ G(s) Hi K Sask» SK, Kps+Ky TR. = :. Comparing with standard form, oy = {Ki 2 w, = ~o1 Bay =Ky-Ky, B= 2 YK Ke Now Mp is function of & alone, fo My = er NES 5100 = othe 25 = ele « 100 0.25 = In (0.25) = ~ 1.3862 = —2™5_ Scanned with CamScanner____Time Response Analysis. & = 0.4037 Now T= 2. 4 a M = Now On Ky = ws 0.7369 e-lp and § = 5 VRi-Ky = 0.4037 Kz = 0.9405 Ex. 34 The open loop T.F. of unity feedback system is Cls) = near For this: system s(1+Ts; overshoot reduces from 0.6 to 0.2 due to change in °K’ only. TK, -1 Show that 1 7 = 43:33 where Ky and Ky are values of K for 0.6 and 0.2 overshoot respectively. (Bangalore Univ. Mar.-99) Gs) kt Ch TR = » H@)=1 Sol.: Closed loop inc (s) _K_ K _ 80473) 2 KT ~ 44k ST+stK 2,1 5,K s(1+Ts) TOT oh aring, with standard form — 7 Comparing v $2 4+2%Zoq ston K = oh TF “ 1 g Bon =z Let = Now for Mp = 06 7 Scanned with CamScannerhe static error coefficients for Wie unity feedback contyop 7 Determine the static error co open loop transfer fictions are K Ystems Why i) Gls) = eds 200) (14 2s) (14 ds) ii) ce = SUeeces 2 (9? 4284 10) Find also the error for unit step and unit ramp inputs. Determine “type (Bangalore Uniy and onder’ of the systems, - Sol. : Consider the system with, i) co) = —“__, 1) s(s? + 45+ 200) Gis) Hg) =| — KL s(s? + 45+ 200) (K/ 200) comparing with, Go) Hg) = KO4T 9)... s} (14 Ty 8)... jl pe of this system is 1, K Hence ty p= Lim Gis) H(s) = Lim —_(K/200) so soo fy = 0 ¥ = Bim 8 Gis) Hs) = Lim tS) S50 ‘ 4s «x 200 Scanned with CamScanner