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THESIS This is to certify that the thesis entitled “Comparative Study of Several Methods for Determinetion of Transient Peeponse to a Unit Step Input Frequency'fiesponse Curves for a Stable System" presented by Kurt Elliott Utley has been accepted towards fulfillment of the requirements for _L._S_ degree in _E...E._ a Major professog Jl) Date 1""13L351 1954 0-169 COMPARATIVE STUDY OF SEVERAL METHODS FOR DETERMINATION OF TRANSIENT RESPONSE TO A UNIT STEP INPUT FROM FREQUENCY RESPONSE CURVES FOR A STABLE SYSTEM By Kurt Elliott Utley \ A THESIS Submitted to the School of Graduate Studies of Michigan State College of Agriculture and Applied Science in partial fulfillment of the requirements for the degree of EASTER OF SCIENCE Department of Electrical Engineering l95h THESIS PREFACE In the synthesis and design of a servomechanism the frequency response curves are of great importance, being an indication of system stability etc. Such curves are readily obtained, as a rule, either from experimental data or by evaluating the analytical expression of the system. However, from the standpoint of safety and a consideration of the system time requirements, the transient response of the system to discontinuous disturbances should also be known. Excessively high peaks are to be avoided as are semi—sustained oscillations. Unfortunately, the transient response is difficult to determine, either analytically or experimentally, hence relationships are sought between the transient response and the more easily obtained frequency response. It is the purpose of this thesis to present in an organized manner three different methods of finding the transient response to a unit step input from the frequency response for a stable system. A problem will be solved by each of the methods and comparisons made among the three with respect to the advantages and disadvantages of each. This paper can be used as a guide for applying any of the three methods to a suitable problem. 3312 91 ACKNCHLEDGFQLNTS The author wishes to express his sincere thanks to Dr. J. A. Strelzoff for suggesting the topic of this thesis and for his assistance in its preparation, and to the Michigan State College Library for its aid in procuring microfilmed material. TABLE OF CONTENTS GENERAL Introduction Three methods for finding transient response Method I Method II Method III SOLUTION OF A PROBLEM The Problem Solution by I-bthod I Solution by Method II Solution by Method III Comparison and Discussion Bibliography 10 15 21 2h 29 35 ho hS At the present time techniques for the synthesis of closed-loop servomechanism systems or improvement of existing systems are based primarily upon the frequency response of the system: the output, or some function of the output plotted against frequency in amplitude and phase as the sinusoidal input varies over a wide range of frequencies beginning at zero c.p.s. These responses may indicate maximum gain and stability, as well as other characteristics. However, it is necessary to have a knowledge of the transient response of the system as well, since all closed-loop systems are likely to be subjected to types of discontinuous inputs which cause transients. The simple closing of a switch, the action of a circuit breaker, compo- nent failure in an adjacent system or a sudden change of load may cause transient disturbances. Such disturbances are often negligible, but in a lightly damped system oscillatory transients of damaging magnitude may result, an annoying train of oscillations persisting for some time may be initiated, or both conditions may occur. Hence a knowledge of the -transient response of the system is important to the designer. If the input to a specific system is always to be of a certain specialized nature it is better to study the transient response to this typical input, of course, but often the response of the system to arbi- trary inputs is determined by comparison with response to a unit step input: a wave form of zero amplitude for t (time) S O, and of unity amplitude for t 2: 0. Thus the response of a system to a unit step can be considered a standard, and the words "transient response", as used in this paper, refer to the response to a unit step unless other— wise specified. A basic present-day procedure for finding the response (both transient and steady state) of a system to a certain input is to apply t operational calculus e.g. the Laplace transform, to the differential equation of the output, provided such an equation is available, trans- forming from a time domain to a functional domain, to obtain the charac- teristic equation and find its roots, and then by means of an inverse transformation return the problem again to the time domain in such a manner that the solution to the original differential equation is obtained.1 However, due to the fact that the desired roots of the characteristic equation are obtained only after the solution of a polynomial of possibly high order, this analytic method for finding the transient response may be exceedingly slow and laborious. Another means of determining transient response is by the uti- lization of a transient analyzer, in which inductances, capacitances, resistors and electronic amplifiers are used as analogues of masses, springs and dampers. Unfortunately, this analyzer cannot be applied until much is known about the individual components of the system, and unless the study is to run over an extended period of time, and the analyzer is flexible and well instrumented, the analytic method, though slow, may take less time.2 A third method is to find the transient response of the system from the frequency response curves, which are readily obtainable from the analytic expression of the system or from experimental observations. 1. Gordon Brown and Donald Campbell, Principles of Servomechanisms, John Wiley & Sons, Inc., New York, l9h8:’p:'66. 2. Ibid., p. 19. It is the purpose of this paper to consider three such methods and to make a comparison among them as to ease of understanding, ease of ap- plication, limiting conditions, speed, accuracy and limits of application to involved problems. The methods are: I. To determine the roots of s in the characteristic equation of the system by graphic means, from the frequency response curves, and to use either graphic or operational calculus procedures to obtain the transient solution, given the roots.1 II. To apply an approximate inverse transform to a succession of trapezoidal waves obtained from the frequency response curve of the output 690(juj) , yielding an approximate transient response.2 III. To approximate the unit step input by a half cycle of a square wave, resolve the square wave into its sinusoidal components, and to obtain the transient response by summing the system frequency response (known from the output over input frequency response curve as 91 All three of these methods presuppose a system linear in the test range, (JuJ)) to a sufficient number of these component waves.3 and stable over the frequencies involved. l. walter Evans, "Graphic Analysis of Control Systems", AIEE Trans- actions, Vol. 67, pps. ShY—SSl. 2. ‘Brown and Campbell, op. cit., pps. 332-365. 3. C. A. A. “ass and E. G. Hayman, "An Approximate method of Deriving the Transient Response of a Linear System from the Frequency Response", Royal Aircraft Establishment, C. P. 113 Technical Note no.6h} lh8, November,_l9SI. Hethod I Determine the roots of s in the characteristic equation from the frequency response curves by graphic means and use either graphic or Operational calculus procedures to obtain the transient solution, given the roots.1 The basic problem in finding the transient response is to determine the roots of the differential equations'which correspond to the ex- ponential transient terms dominating the response. The function which describes the system from output to error, €%?(s), or the inverse, €%¢(s), is a function of the complex variable 5, which has damping (6‘) as its real part and frequency (00) as its imaginary part. The imaginary axis of thee plane corresponds to '3: jet) .2. The frequency plot of %°(jw) obtained by setting 5 :30 is simply one line of a conformal map, with the roots of s lying at the value of the variable (the point in the 5 plane, d'r+JoJ| ) which makes the function €%B(s) equal to -l for a system with unity feed back, or equal to -K for a gain of K in the feedback loop. That the root lies at a value such that the function gum) is equal to -l is shown by the following illustration of a simple unity feedback system: 1. Walter R. Evans, op. cit., pps. Sh7-SSl. 2. Brown and Campbell, op cit., :- 171. error. input 91. 5 output, 9o 9 + e ' ...___K ,e 3‘ (5) = ——° (5) SCHST) 69° 6° : n + i. (5 £33 ) <9 Fig. 1. Unity Feedback System 1 £3 €90 K K 6o 511+ 3+ K The roots of s in the expression t< the expression equal to zero, or by finding values of s which will make are obtained by setting £13): -1 l, which gives the same effect. For a general case, Method I goes as follows: 1. Break the feedback loop and apply a sinusoid; make a locus ploti of error (error = input in this instance) over output as a function of frequency, either analytically, letting 3:305 in the %¢(s) expression, or by taking observational data. 2. Considering the vector locus plot to be the base line from which the complex roots of the main damped sinusoidal term can be found, construct an orthogonal lattice of - 6‘: constant and Ju)=. constant lines and determine the complex roots of s by noting where on this lattice the point —l falls. The justification for this step is easily shown: 1. The distinction between 83(3) and gone) lies in the fact that {%3(s) may lie almost anywhere in the plane, depending on the complex value of ”-0”: Jo) while goal) is a special case of 5° (3) where 5: Jo), and is a single curved line. has»... Let s=-a'+jw, from whic -—l, and 50(8) is a function awzJ' of s, which has a particular derivative with respect to s at each point in the region of interest. 4am saws = Jase §&(§é3cs))= %(%(5))'§57 .- -1 gig-0(5)) (u) The change of the function with respect to —-o- is seen to be dis- placed 90° ccw. from the change with respect toJoJ’ , which lies along the gouul) curve. The change of the function with respect to s is 90. (W. from the 5-000) curve. For equal changes in -a" and Jo) , a set of curvilinear squares will be formed with conformal map properties as shown. If the expression of go“) is known, points on this lattice near the -1 point may be plotted by solving the expression using dif- ferent values of -o" and J'co in s ="'O'+J(U’ . 333 (5-35)) I I I “ -AO'-9' 5.65)) / l A; / \ "‘ / 4410 / fi -_ - g”.-- /_ . 5- 1442/4. ¢=-,3(g--HJI \ Fig. 2. General gym) locus showing curvilinear squares. For this illustrative example the root lies at s = —O.9 4-3 2.8, and since complex roots occur in conjugate pairs, 3 is also = -O.9 -J 2.8. This procedure is valid only for systems linear in the test range, with results applicable to the range. This restriction is necessary since the justification of the curvilinear lattice depends on the deriva- tive being independent of the amplitude of the function, depending only on the nature of the function. If the 53(3) expression is not available, and the $230”) curve has been experimentally obtained, a sketch of the curvilinear squares may be made quite accurately without knowing points in the lattice, provided the baseline, or the éi§yuo locus is not so strongly curved as to cause undue distortion and overlapping near the -l point. In general, this method is much more accurate if points on the curvilinear system are carefully plotted by several evaluations of the €%b(s) expression using complex values of s which will cause :33 (s) to fall near the -1 point. Once the complex roots are known, assuming a single pair is had, theyr can be divided out of the polynomial expression for 3—;(3). The simplified expression may then be either solved by straight forward algebraic methods, or perhaps graphically from the original plot, having found the general location of the real roots by inspection of the simplified function. The author of this paper has found that, once the complex conjugate roots have been found, the real roots are most easily obtained through factoring the simplified expression, since plotting the curvilinear squares in the region of u): 0 is rather critical. 3. Having found the roots, determine the general form1 and the amplitude of the transient solution for each root by means of operational calculus. The general form of the solution due to the conjugate complex -a’+JuD)t + K16(«Y-mart (5) t For a real root, 3 .p (200.) = K3 e'" (6) roots sz-O‘ajw is 171(15): KI e‘ The amplitude of a transient is given in terms of its root by: I 3 35 9° 5=$I"O'|+JLOI . .5; - 9.: ' a. e :3 at Since 99(3) differs from 90(8) by only a constant, ”(50(9) 5-5—(5-(59 Thus, the angle of the derivative can be read directly from the plot (See angle/3 , Fig. 2) and the magnitude can be found by taking the change in £§;(s) (distance from origin) divided by the change in s (as measured on curvilinear lattice) and averaging over several measure- ments near the -1 points Latting .5: 06a (0‘: +0"... 77%: and 7% 3:6»: be’ the transient solution for ccnjugate complex roots 3 3-0: gin). and SIL 'O'I-jwl is: (‘60‘3JUI7t (-fi-Jw.)t | Flt) : ‘——'5'_Ze "' + i 54- Gueu' sC ’ 53-0. «$an Che“! bl e ‘ = '63-'30; (8) From relationships between the denominators, 0': Ch , b.3b3.’ Oh- -°u, 6.:- --,62. “hi +a‘ 4 8‘) -O"t Ut- .- de D: {I (t) = 1. Stanford Goldman, Transformation Calculus and Electrical Transients, Prentice-Hall, Inc., New York, l9h9, p. 319. ’G’t fi(t) : 5T6; (050).“: -°\¢ ‘80) (9b) For real roots, 5: ,0 , the transient solution is: e-pt file) : 03¢“. 038" (10) Since there is no j term in s, a, and ,6: will be 0 or 'n , and : e": Fl(t) 8 03b: (11) The entire transient solution is then: fit.) 2 co: (m: - are.) + 0: b3 (12) Occasionally systems will have two pairs of quadratic roots in about the same frequency range, and the g—gljw) plot will circle the -1 point with the result that the curvilinear lattice built up from one side will overlap that built up from the other side. a series of successive approximations will simplify. Start with a pair of conjugate roots sug- gested by one lattice and divide the 9:.(3) function by these. This 1 or analytically. The resulting plot will may be done graphically suggest a second pair, and i;i(s) can be divided by these to give more closely the :first pair. Continued use of this method warrants the use of specialized protractors and pivoted scales, as explained by those who developed this method, to simplify the reading of angles, the plotting of the curvi— linear lattice, and any necessary graphic division. The author of this paper has attempted to explain this method without mention of special equipment. An illustrative problem will be worked by this method a little later to more completely explain the procedure. 1. Brown and Campbell, op. cit., pp. 165-166. lO HZTHOD II Apply an approximate inverse transform to a succession of trape- zoidal waves obtained from the output frequency response curve 6005)), and obtain an approximate transient response.1 Let the overall output function of a system be H(s)= 90(8). The general problem of finding h(t), the transient response, knowing H(s) is in performing the inverse Laplace transform. It is assumed throughout the following treatment that: l. H(s) may be written as a ratio of two rational polynomials in s with real and constant coefficients. 2. Lim 151(3): 0 3-9-00 3. H(s) has no poles on the imaginary axis or in the right half- plane. In general the inverse transform is given by the integral (+10 I ha : -— J H6) at5 ds ) 21” c-Jao (13) where the path of integration is parallel to the imaginary axis and c is a value such that the path of integration is to the right of all poles of H(s). Since by assumption there are no poles on or to the right of the imaginary axis, the path of integration may be made this axis. +4» - I I Hts) e“ ds S h t - ‘--7 o <) M ”no (it) let s =:Ju5 H(s): ReH(jw)+J° Im mu) 1. 1313-, pp 332-365. 2. David Bidder, The Iaplace Transform, Princeton University Press, Princeton, New Jersey, I9hl, p. 2&1. Re and Im refer to real and imaginary parts of the function. Make these substitutions into 114. I ‘9” h(t,)= 731- .I.” [Re chw)cos cot -Im H(jw)6m wt] doc _I__ "a ' ' dw + 2." -0. [ReH(Jw)smaJt-tlmH(Jw)cos wt] (15) Since H(s) is a ratio of polynomials in s with real and constant coefficients, Re “(3(0) is an even function in w, and Im H(Jm) is an odd function in w. Cos wt, by series expansion, is an even function in a.) , and sin wtis odd. The bracketed term of the first integral is then an even function in w , while that of the second integral is odd. Since the limits of integration are from ~09 to +09 , the value of the first integral is twice that from O to on , while the value of the second integral is zero. Equation 15 then becomes: no a. nu) = -1'Tj°RcH(Jo)cos cotdw - Jfijolm How) am wt do (16) Both integrals contain functions of time in the sine and cosine terms, the first integral being an even fimction of time, the second, odd. For the integration indicated in equation 11; along a closed path which takes in the imaginary axis from-J00 to-ano, plus a, large semi-circle in the right half-plane, the value of h(t) is zero for all negative values of time, since, by assuming no poles of H(s) in the right half- plane, the sum of the residues at the poles of H(s)et5 within this contour is zero.1 This means that numerically, the two functions of time in equation 16 are equal for all values of tine. l. E. A. Ghillemin, The Mathematics of Circuit Analysis, John Wiley 8: Sons, Inc. , New York,‘l9h9, p. 301;. 12 Iii-Emcee»: «0| = 1—4.- I I. ng)smwtdwl (17) And for positive values oquime: h(t) =2~fijo Re H(jw)cos wt duo (18) This is the exact inverse transform for an H(s) satisfying the stated conditions. Few practical problems are of such a form the Re H(ja)) can be integrated as such, so a general approximate procedure is used. 1. Plot Re H(Jw) againstao and approximate the curve with a series of straight line segments. See Fig. h. 2. Yrite the straight line approximatibn as a sum of trapezoidal functions, and apply equation 18 or a simplification to each one. The sum of the resulting time functions is an approximate h(t). The general trapezoidal function used to approximate Re HQJQD) is shown in Fig. 3. Let Re H(Jw) aproximation = 11(0)) 03r+£§I-w -'- P: 4 6 ( 24. 9) an 0) (db Area oftraPo. A: = no). I'. *A. Al (O b+wa (0‘: , i 2. 00. a). wt A‘ = wb- (00. 2. i Fig. 3. General trapezoidal function. I’Mt) : Zfijomwfiaswt dm m m ”Hmajim w .- ma): Zfiojnmowtdw 4-17] r. 2A. )Coswtdw (On-AI ExFahd: (On-Al (OI-+4: __ PI QQJHCZPI4£3I) Ith)- 2%fi-J;fcusaut’+/3‘J(s+p(s+p) - I505 9. I515 91-“): -§—($)+l ___, [(5+a¢)"+5"](5+x)($+fl) o lSQs 4__J output _ _ 5 6° (541°)(5+5)_ ($163+ 51 l (B ’ ’ \ (29a) (30) 22 Note the ssh ction of K2 = -6 in equation which makes 3-38) factarable. By inversion, e. (5) = I50 5 3t- [(s+ot)1+51] (5+ K)($+P) (31a) Since, for a unit step input, F(s): ‘é‘ 9 '50 9°65) = Tcz+a)2+d'j(s+l)(s+p) (32a) The s coefficient of 150 in the numerator of the r.h. side of equation is necessary to cancel the 3 introduced in the denominator when removing 91(3 kg- form the expression, which would violate thaconditions of Method II that there be no poles at zero. In the system of Fig. 5 assign values: 0&3 1.2 K= 6 .6 = 2.5 .0= 3.3 Applying the inverse Laplace transform to ISO (5+I.z)"+ 2.5“](s+6)(s +3.3) (32b) yields the transient solution 69(5) = [ God): - 3.35 e"'”sin(2.51 + I025} L85 3'6“ 5- '6 rm (33) which is shown plotted in Figs. 7, 9, and 11. At this point, the vector locus diagram has been made for 2-3 3w) (Fig. 6, note) and from this, vector locus plots of £30m), required for Method I (Fig. 6 and note), 3:01») , required by Method III (Fig. 10) and Re eoq’m) , required by Method II. 23 Although these curves were obtained analytically for the most part, it is possible to carry through the graphic solutions of nethods II and III without knowning the analytic expression for the system. It is almost necessary, however, to know the analytic expressions in order to obtain an accurate solution by hethod I, since, in the event of strongly curved gljw) near the —1 point, the curvilinear lattice is plotted by evalu- ating «iébfis) for several complex values of s. 00 man. :0 winrle mar %:Uw3ngle Re 90003) 0 F on -00 O 90 .09 .1 10.10 -85.5 .10 85.5 .99 .2 5.10 -81 .20 81 .98 .5 2.0 —67.5 .h9 67.5 .91 .8 1.25 -53.6 .30 53.6 .81 1.0 .99 -h3.6 1.01 A3.6 .69 1.5 .60 -lh.5 1.66 1h.5 .26 2.0 .h9 11.8 2.0h -11.8 -.20 3.0 .h9 79 2.06 -79 -.67 h.0 .79 125 1.26 -l25 -.26 6.0 2.05 169 .h9 -169 -.015 8.0 h.31 192 .23 -192 .006 10.0 8.33 207 .12 -207 .005 20.0 55.50 237 .02 —237 ..8x16‘ Table Tabular data for plotting system frequency response curves. 21; METHOD I To solve: Problem given in Fig. 5. Given: the $000)) locus plot, baseline of the 4-54.30.) system. Find the conjugate roots or s by determining where the point -1+jO lies on the-(rum coordinate system. In Fig. 6 a rough sketch indicates that the point ~1+JO will lie between 0": -2 anda'z -l, andw=2 and (0:: 3. In order to plot the curvilinear squares more accurately, substitute for . O U 8 0 O s in the analytic expressmn of -é-°(s), whose expanSIOn lS: _§, (5) = 5‘4’ ”.753 +49.5$1-3|5 + I52.3 90 I505 (29b) the four values obtaining respectively . E. -JI75.1 : .. + z . s l 32 50(5) 90 e r. -1+j3 = 1.10 6'1"“ 3 -2 +j2 :. .85 3"“76 1. 12 e'5 '7‘ = -2 +33 Plot these points on Fig. 6 and draw the orthogonal lines. The complex conjugate roots are read as s 3' -1.25:!'-j2.6. The two remaining roots may be found by dividing the quadratic 2 2. portion of the output expression &(s) by (3+ 1.25) 4' 2.6 54+".7s3 +4955“ H95 + I523 2 (5+ L25? +2.62 $249.25 + ISAG = (s+e.23)(5+2..97) (314) 5,: -2.97 : 53=‘6-,23 The transient solution resulting from all the roots is e-I.15t Cm (2.- 6t-a._5Q + 8—1-97t e- “13“ DJ): Ozeabieh + case‘sboei; (35) Mt): 25 We 06nd) , , - and») .éégoufrequency response curve, low frequency portion. Q 3'30“” frequency response , curve. Note: used only ~ this 3"" to derive %’3(Jw) locus; § enlarge scale x 5 and 1 displace one unit to left. f;fis.5 1 T A1% Mum-«9' ~ ”I" LL, ‘ 0‘. . -\~4k4_‘ .- -..._-.t-. Fig. 6. Frequency sponse curves 5390’) and 35,00) . - a I a 26 .. L251: In £—' C°3(2.6t-°‘I"46I , CM): 0. 6°“ = 5. = ‘LZb +321, = 2-86 ewes b __ Age.) | - AS $35.3-|,15+32~0 b, is the average of several Ar?“ )ratios in the region of the root. A3 AgC (5) ratio Between 3 = :71; g 1 .02 .02 :33: 1 .05 .05 24,33 /2_ .2u1 .17 3:}: 1 .20 .20 I; I .111; .11 .6. can be read directly from the plot as the angle between the real axis and the (082.6 line at the point -1+JO 3 ~ 0 I Thus the first term becomes: 8"”‘cos (1.6: -II5.-8 -60) -I 25*- =- I e I .6t+ 4.2. 2.5“ .H 35 anZ 9 ) (36) smce cos(2.6t — I758) = - sin (2.61: + 94.2). e-zmt 31:“.b; e6; ’ 018°“: 51" -z.97 +JO = —2.97 eJo bi: Find absolute value of derivative algebraically. — 94(6) 55 Go ) 5:52: ~23“) _§_ (5‘4 II.75’+49.85‘+II95 + 52.3 as I505 ) b1: 5= - 2.97 b;: 0065 27 The line of constant frequency associated with the root 5 = -2.97 passing through the point ~1+jO is the a): 0 line, hence the angle of intersection of this line with the real axis is 0° or 180° . At S: -2.97, 30(8): g-l, While at S: -2097+A, 30(3)) '1. Hence, 591 180.. (As-a- decreases, thew const. line runs left to right). Then £1.97t z e-2.97£ : 5J5 e -2.9-’ t. me“: D; e“ -Z..97 x .065 en” (37) e- 6.2.31: ' For the third term, , 03 e" D; 8" 0,6“3 3 53 = ‘6.23+JO = -6.23 b3 : O." o 6,: 0 . . e-mst ant . The third term 18 : - .4 e' o -6.23a-o.u ' 6 (3 ) The entire transient solution is the sum of equations 36, 37, and 38. 'utsin(2.6t+ 911.2.) h(t) : -3.18e" + 5.18e“°”‘_1.u6 e"°‘“ (39) Compare the plot of this output waveform with the exact solution in Fig. 7. The merits and disadvantages of this method will be dis- cussed at a later point and compared with the two methods remaining to be illustrated. O [x O (11) --—<| CI. .mm>nso mmcoomm: pcoflmcmhe 0 0 1 l O ‘6’.-. o c o 9 0.. r {lolOIO HI! r o lil‘l 'J -O 9 O . 1 O f O 1 L ' f o O | l o -o 1 O O O O l 1J1 fi # fl ld 1 1 1 d u 4- l1 d 4. 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The output function, I50 £9 ' ’I- 0(5) 54 4, “.7 53 4.49.8514 ”95 + 52.3 (320) fulfills the conditions necessary for aoplication of this method, namely: that 9.(s) be written as a ratio of two rational polynomials in s, with real and constant coefficients; that lim. 9;g);o; and that Q(s) have no poles on the imaginary axis or in the right half-plane. With regard to this last condition: This method was specifically developed to determine the response of a system to a unit impulse, whose Laplace transform 9&5) :1. Because of this, the expression of gfls) is identi- cal to that of‘Eixs) alone, and the generalized h(s) in the explanation of hethod II is equal to €§f(s)=:€a‘s). In using this method to find the response to a unit step, (whose transform e‘(s)=—:!;), 90(5), not 3:38)! must satisfy the stated conditions. Since 90(5) =é-gzs) it is seen that a pole at s: 0 will exist in 99(3) unless the expression of'é;?(s) already contains a zero at the origin (an s in the numerator). In such a case the pole and zero will cancel, and €39(s) will satisfy the given condition that there be no poles on the imaginary axis or in 30 the right half-plane. The method cannot be used to find the unit s_tgp response of any system save one whose transfer function 33-33) has one or more zeros at the origin, whereas it can be used to find the unit muse response to any system whose transfer function has no poles at the origin. Since the transfer function of the problem has a zero at the origin, 99‘s 3 .505 ‘ [( 5+ ..7.)’- +2.5‘](s+33)(5+ 6) (31b) 'I'nen 90(5) - é—gtm a '5° [(54 L2.)2 +25’](S+33Ys+e) (32b) and fulfills the three conditions. The data necessary to plot Reeouw) has been entered in column 3, Table 5. Plot the function and approximate the curve with straight line segments in Fig. 8. Write the straight line approximation as a sum of trapezoidal function and apply n - 2A.: 3m (.th an: Ant hm ' §u (out Ant (,5) where A“ is the area of each trapezoid, (0-. is the mean base, A“ is half the difference between the bases. From Fig. 8, tabulate trapezoidal data. Evaluate h(t) over a sufficient range of t values (see Table 7) and plot, in Fig. 9, along with the exact solution for purposes of com- parison. .nogmsflxoumgm 3338me :3: @3360". .m .ufim 36 1.0 .N.0 0.0 N6 .Vd 0A 0.. 0. mm M. ‘ . 3 z c .1 ‘- l:3-Ipo=. 5 Sr II-Lhu- I ......=-LII.1- 2.; I l 02.- mm was" 32 T Fre “)3 (in op q. Lower Freq. w “an .. Trap. Height w“ w» ( 412—) (9%) A. I .175 .30 .86 .58 .27 .10 II .51; .86 1.50 1.17 .31. .63 III .26 1.50 1.78 1.6h .lh .h3 Iv '200 1400 600 500 100 -100 v -.50 3.0 h.o 3.5 .50 -1.75 VI .56 1.78 2.35 2.07 .29 1.16 VII .11; 2.35 2.7 2.53 .18 .36 Table 6. Trapezoidal Data. _._ _g_ 0.. gm .581. our-2'71. 0.6 am not am .361: h“) n[ O .551; an + 3 M61. .361; Q4 gm [.64t sin .I4t _ I am 5t 5m 1: + 3 L641: .l4t 5t :5 _ am 3.21: am .5t _, 5m 2.071. am .291: "75 33% .5t + I b 2.071; 2.9t gm L5§t 5m .Iat + 0’36 2.53t .IBt ' (1.0) The merits and disadvantages of this method will be discussed at a later point and compared with the other two methods, the last of which will be illustrated immediately. 33 .oom m.m I o owcwh p pm>o mvaowmgmhp 5 mo coammopmxm Apvs mo :oepm:am>m .w manme rr 00. 00.- 0H.u 0a.: 00. mm. mH.H 0:. NH. 40.: .Jw.x 0H. 00.: 00.: 0H.u as. 0:.H 0~.H 0N. 0H. 00.: sum :0. :0. H0.1 00.: No.1 H0. 0H. 0m. mm. 0m. HHp m0. 00.- H0.. mm.. 00.: mm. 05. :0.H 0H.H ~H.H He OH. mOOI JHOI. 3H0- QHO mm. ONOI NNOHI HCOHI mNoH’... P 00. H0.. H0. mo. 00. 00.- NH. 40.- 00.- 00.H- >H N0.. 00.- 00.- 00.- 00. 0N. mm. 0:. me. me. HHH “0.: m0.. 50. 0H. Hm. me. am. H0. m0. m0. HH No. 00. No. m0. m0. mo. 0H. 0H. 0H. OH. H N.m 0.0 e.w 0.0 0.H N.H 0. e. 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L L . . 1 a . 4 4 0 . q _ J _ _ 9.90.0... 4990 alooflowv+utrk+of17 00¢...To _ _ _ _ o o o 6 a + o o . o o...A o f ollo.‘.|-. .~-¢. 0.0.5 0 0 f 11...: + f 04 0-0 9 f c . .ooo.ol.§.~lIT3.+o. .+ v..- «10+.0v...f.onor¢.+o . . . _ 0669.09}. 0400? V400... 9r00+onOOOJILv¢tQ¢oOOO a _ . p 1 IILILfLIolHoliTII+|1I1|> T m 1rl+LILII+I| . _ , vuf+u§9l1fi o o 00 .0 o.+.¢lv o 4 9A «.100 +9 o.o. lfkv+Ll+ol.O 0|-TJI1 _ - 00.9 o... 4... 4o. ..q1|»_!fT,....u 00.0 o... Tw¢.ol _. _ _ . _ a... .4... coat. 0.6 rf9-o-fuf¢oo +o+4|f+¢owala+rf _ _ _ ....o .0.. *.9.0L «0. Alfho +3... «.9. -4po-vTola-_olh. . .L P P . P L h r p . H F b i Cl 35 METHOD III To solve: problem given in Fig. 5. Given: frequency response curves of the transfer function, 33"», Fig. 10. Determine a); , the fundamental frequency of the square wave, which approximates a unit step input. Set u); = 29+ 20%, where (do is frequency 5 of maximum gain. In Fig. 10, (00: 2.5. ,2.5 20% = 0.5+ 0.1 = 0.6 . Mr 7” Test to see if the mafimnn contribution of the llth harmonic of (1);: 0.6 3 2% of the contribution of 0». Contribution of U: 6.6: A. .. e: u ' H (a) Contribution of 0;: 0.6: {%}=._g% = .056) 2% Raise (q. to 0.7 and test as before. The contribution of the llth harmonic 2r 3% ofwf contribution. The 13th harmonic contribution = 1.7% that Ofwf . This is satisfactorily close, and 0);: 0.7 will be used; the 13th harmonic will be used in the solution. Indicate harmonic frequencies of (of: 0.7 on Fig. 10 and make the 90 . following tables from the age») curves. .5305 panama 639$ no «.050 .9“ .ME .i- 4 -I - 4 . 1117 -.1 .- A 1 nil..- I H 4.- -H- J. - -.-I [TIIIAfITII IQ..IIwII*IIIIOIooI IIII-IIII. IIIIIII|10«.I.IIIIIO I.0III+I. IIIO*III. «III-u L1 0 I I I O ,O-IH-I 0 I I § 7 t I — I I I 0 ”Q I I I O I I I I I I O 0 O + I I I v a I I 0 fl 6- I I I * I I I I * I 0' I I. — I I 0 In .¢ 9 I 0 I w 0 I 4 I I I I I — II.‘ I I I .0 .IIIO * I I I I * I I o . 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I I AT 0 I I .I v 0 0| 0. f- .Avl-I....0I ”III- .AT 7 a. 0 I 1* . + I . I I o I A. 4 I .4 w _I .I 4. I. A a + 4 0 v .4 «- 0-.0. .le-n- HAT 4 T 4 7 .. In ._7 4| + li'T-flj v 4 [0 Au. A. -HI I. TIArII :w A. fl 4 .7 4 .Av 1T .k 4 I0 .A.l+ .4 7 4 A1 .I .0- 0. 4. A. I- I .4 I 7 I a + I .T n.#. I. I .- I .4 9 0| til-1.41.4 h kl o w 4 .A 0 H «I4IJI 14+ . A 441T»! Ala 4.. +I +r-7 a. A1 I a-A tT-1.+.I+ 4 +- fllT W1. A. a a 4.- 4 A 0 0 + a I a 4 0- I v H 0 I + :I-IIJIIIA III-- 70 I 4 I I . T +,I Alf-Ll I A I 0 r I + 0-.. A1 4 4. fIIIT-_-.+.lI_.--.IIT+-0.I + A'- .I + A I I H A. v-0 H-“ a L w H I +- ql Ar 4. 4 I 0... 4i- I a I *1... o . k. . o . . AI 4 . II..;-..-.T4 IA o + . + +4 . 9 .W+-.- 7.; I. . . . f i + .. “Ta I...» I _o +-A I aid-p... 4.H-+* ++4 .9 1+- 0.» - f“! . #1-»- l L . r p >L|pL Lip L. F» s L 7% L _ _ ) _ TLL r > _ u . I . I . e F I . P» p I A- B _ -..I..I. \ . I I: .- Phase Shift Gain 1; 1”,. n (2n-1)wf Bu.-. Azn-I ‘iT Zn-I 1 0.7 56 0.7 .15 2 2.1 -21 .2.12 .15 3 3.5 -107 1.58 .20 h h.9 -1b9 .85 .08 S 6-3 -17b .82 .03 6 7.7 -190 .28 .01 7 9.1 —202 .15 .008 Table 9. Gain and phase shift data of g-tuw’) at odd harmonics 1-7 of 00F='O.7 . 1172 IN SECONDS n t 0 .2 .1. .8 1.2 1.6 2.0 2.1. 2.8 3.2 1 S6 6h 72 88 76 60 11 27 12 -5 2 -21 3 27 75 S7 9 -39 -87 -bS 3 3 ~73 -67 ~27 5h h6 -35 -65 15 85 3 b “31 -87 ’37 75 -7 -01 61 17 ~85 27 S -6 —78 -29 6h -81 A6 -10 -2h 60 -8h 6 10 —78 -13 16 -20 23 —26 30 -33 37 7 22 -51 -56 90 -57 21 10 -82 76 -70 Table 10. Aigle (2n—1)35°w,t+132,.- table with angles in degrees converted into quadrant I & IV equivalents. E'ultiply the sine of each term in the n: lst row of Table 19 by the corresnondinr; 11: let value of 2 am, Table 9; repeat with other n TV 212-! values to get final table, Table 11 yielding solution. Plot in Fig. 11 with exact solution. 38 .omm N.m I o mmcmg p pm>o mQCmcooEoo zoomsvmnm m>mz cheddm N 00% coflmmopQXm Apvph mo nowpwsam>m .HH mHDmB 000.- 00.- 00.- 00.- 00. 00. 00.0 0:. 00. 00.- 000 000.- 000. m00.- 000. 000.- 000. 000.- 000.- moo. 0 moo. m00.- moo. 000.- 000. 000.- 000. 000.- 00.- 000. 0 m0.- 000. 000.- m00.- 00. 00.- 000. 000.- 00.- 000.- m 00. 00.- 00. 00. 00.- 00. 00. m0.- 00.- 00.- 0 00. 00. m0. 00.- 00.- m0. 00. 00.- 00.- 00.- m 00. 0m.- m0.- 00.- 00. 00. 00. 00. 00.- 00.- 0 .400:- QO. HN. Hm. mm. 4140 m4. m3. OJ. NM. H 0.0 0.0 0.0 0.0 0.0 0.0 0. 0. 0. 00 0 0020000 20 0300 .mm>pdo mmCogmmm pamflmcmhb .HH .mflh 39 1 0 01 1 a 0 1 .1 1| . 1 1 a | 1 1 _ - 1 1 1 1 1 1 a 1 q 1 . . . . . . 4 . . . . . . H A ., _ — . _ 0'0 0 a 0 i F 0-0 4 1r 0 0 + 0 k 0 o 0 0| I0 0 0 0 .T0l.’ 0 0 _ 0 v .0- 0|. 0 0 0 0 o 0 o 0 0 0 0 0 O .0..QI. 0 6 o 0 o a 0 0 0 0 A 0 0. 0 0 0 0|. 0 01%. 0l-0 0 o — 9 O ' .011 1|. 0 O L O 0 O 0 + 0 .0! 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The foregoing solutions were arrived at without the use of such specialized equipment suggested.by the original authors. Therefore, the remarks made by the author of this paper express his opinion as to the relative merits and limitations of the three methods treated without special devices which might have affected the findings. Some general conditions apply to all three methods. The system under consideration must be both linear and stable in order that the methods apply. Understanding of procedure: Method III, the square wave approximation, is readily understood with a minimum of explanation. Its procedure is logical, simple and straight-forward. method I, next in order, pre- supposes an acquaintance with conformal mapping and complex variables, the procedure for’finding the conjugate complex roots is easily followed. Finding the magnitudes of the transient terms corresponding to the roots is more difficult. This requires a knowledge of transformation calculus, the principles of which must be accepted on the basis of a mathematical proof. Mbthod II has the least easily understood approach. Here a } mathematical development alone is used to develop the final expression of the transient output by means of the inverse Laplace transform, and bl the end expression is not obvious from an examination of the original inverse transform. Applicability: Method III applies to any'stable system, whatever the form of the curve of the system transfer function, and does not require that the analytic expression of the curve be known. Method II will not find the response to a unit step for a system whose output expression or curve indicates a pole at 3:0, thus violating the con- ditions necessary for the mathematical steps leading to the simplified expression for time response. Hence, the method.will.determine a unit step response only for a system whose transfer function has one or more zeroes at s : O as well as having no poles at s: O. The method is ap- plicable, however, to many more systems ifitis used to determine the response to a unit impulse, since only the condition rejecting systems with poles at s = 0 applies. This method also will work without knowing the analytic expression of the transfer function. Method I is not readily applicable to a system having two or more pairs of conjugate complex roots, especially if the roots lie at about the same point on the complex.plane. This is due to the fact that the errordto-output frequency reSponse curve will be strongly bent for such a system, making difficult the plotting of the complex curvilinear square system. In such a case, and indeed, in the general case, where a high degree of accuracy is required, this method should not be applied to a system whose curves cannot be expressed analytically, because such expressions are necessary for the accurate plotting of the curvilinear squares, and for the algebraic determination of the real roots. 12 Accuracy: Method II, by increasing the number of trapezoids used in approximating the curve, may be made as accurate as desired, at the expense of time and labor. Ibwever, even for a small number of trape- zoids used to obtain a rough approximation, the result closely follows the exact solution. Method III gives a solution that may agree to within a few percent of the exact response, depending upon the careful and tested selection of w; , the fundamental frequency of the square wave. Method I offers the least accurate solution of the three, es- pecially at the beginning of the time range. The accuracy may be improved by making the necessary curvilinear plot to a larger scale and by plotting more points, again at the cost of time and labor. For all but the most critical requirements, however, the solution by Method I should suffice. Speed: Method III is most quickly carried out, with the data necessary to evaluate the summation expression being read directly from the transfer function curves without delay. Method I allows the rough determination of the complex roots of 3 almost immediately if the frequency response curve is not too strongly bent. The approximate real roots also follow quickly from dividing the transfer function by the complex roots. However, the magnitudes of the transient terms corresponding to all of the roots require quite some time to find. Method II is regarded as the most time consuming, especially if a large number of trapezoids is considered, since obtaining the expressiml for the trapezoids and then evaluating the large summation requires slow, painstaking work. 1L3 Advantageous features and general evaluation: Method I yields the equation of the transient reSponse curve. This is desirable, since a change in the transient response due to an altered parameter may be found.without again going through the entire procedure for finding the response, merely by considering the effect of the parameter change on the correct term or terms in the equation. This feature is not found in the other two methods. This method also has the advantage of yielding very quickly the approximate roots of the system.from which the time response is foundl The accuracy of Method II is its chief advantage. In many cases this will offset the additional time used.in arriving at the solution. Method III, after due consideration, appears to be the best general method for obtaining a solution quickly and accurately, for, to repeat: the principles and.procedures are most clear, the limitations are much fewer in number than those for the other methods, the findings agree favorably with the exact solution and are quickly obtained. Although Methods I and II may have the desirable features of speed and accuracy respectively, these are combined in Method III along with others to make this one of the most favorable graphic means shown for determining transient reSponse from frequency re3ponse curves. Method I Method II Method III Understanding of procedure 2nd 3rd lst Applicability' 3rd 2nd lst Accuracy' 3rd potentially 2nd lst Speed potentially" 3rd 2nd lst General Gives analyt- High ac- Most rea- Advantages ic formula curacy' sonable of output possible. approach; waveform; best gen- change of eral system para- method . meters readily' handled; quickly'gives approximate response and natural fre- quency'of oscillation. Table 12. Tabular summary of Iiscussion. hS BIBLIOGRAPHY Brown, Gordon S. and Donald P. Campbell. Principles of Servomechanisms, John Wiley & Sons, Inc., New York, 19148. Chestnut, Harold, and Robert W. Mayer. Servomechanisms and Regulating System Design, John mey and Sons, Inc., New York, 1951. Evans, Walter R. "Graphic Analysis of Control Systems", AIEE Transactions, vol. 67, pp. Sin-551. - Gardner, Murray F. and John L. Barnes. Transients in Linear Systems, John Paley & Sons, Inc. , New York, 1952. Goldman, Stanford. Transformation Calculus and Electircal Transients, Prentice-Hall Inc., New York, 19119. Guillemin, E. A. The Mathematicsof Circuit Analysis, John Wiley & Sons, Inc. , New York, 1919. Sokolnikoff, Ivan S. Advanced Calculus, McGraw-Hill Book Company, Inc., New York, 1939. Wass, C. A. A. and E. G. Hayman. "An Appronmate Method of Deriving the Transient Response of a linear System From the Frequency Response" @yal Aircraft EstablishmentJ c.p. 113 Technical Note no. GU. 1L8, November, 1951. Widder, David. The Laplace Transform, Princeton University Press, Princeton, 1914.1. "‘ ".‘ Lil’s-s "‘x! 77:5 65 67;“! 3 7? ’ST